Principles of Communications, 7th Edition Solution Manual by Ace Test Banks

Principles of Communications, 7th Edition BY Ziemer, Tranter

Chapter 2

Signal and Linear System Analysis 2.1

Problem Solutions

Problem 2.1 a. For the single-sided spectra, write the signal as x1 (t) = 10 cos(4 t + =8) + 6 sin(8 t + 3 =4) = 10 cos(4 t + =8) + 6 cos(8 t + 3 =4

=2)

= 10 cos(4 t + =8) + 6 cos(8 t + =4) i h = Re 10ej(4 t+ =8) + 6ej(8 t+ =4) For the double-sided spectra, write the signal in terms of complex exponentials using Euler’s theorem: x1 (t) = 5 exp[j(4 t + =8)] + 5 exp[ j(4 t + =8)] +3 exp[j(8 t + 3 =4)] + 3 exp[ j(8 t + 3 =4)] The spectra are plotted in Fig. 2.1. b. Write the given signal as h i x2 (t) = Re 8ej(2 t+ =3) + 4ej(6 t+ =4) to plot the single-sided spectra. For the double-side spectra, write it as x2 (t) = 4ej(2 t+ =3) + 4e j(2 t+ =3) + 2ej(6 t+ =4) + 2e j(6 t+ =4) The spectra are plotted in Fig. 2.2. 1

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS c. Change the sines to cosines by subtracting =2 from their arguments to get x3 (t) = 2 cos (4 t + =8

=2) + 12 cos (10 t

= 2 cos (4 t 3 =8) + 12 cos (10 t h i = Re 2ej(4 t 3 =8) + 12ej(10 t =2)

=2)

= ej(4 t 3 =8) + e j(4 t 3 =8) + 6ej(10 t

=2)

+ 6e j(10 t

=2)

Spectral plots are given in Fig. 2.3. d. Use a trig identity to write 3 sin (18 t + =2) = 3 cos (18 t) and get x4 (t) = 2 cos (7 t + =4) + 3 cos (18 t) i h = Re 2ej(7 t+ =4) + 3ej18 t

= ej(7 t+ =4) + e j(7 t+ =4) + 1:5ej18 t + 1:5e j18 t

From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes 2 and 3 at frequencies of 3.5 and 9 Hz, respectively, and the phase spectrum consists of a line of height =4 at 3.5 Hz. The double-sided amplitude spectrum consists of lines of amplitudes 1, 1, 1.5, and 1.5 at frequencies of 3.5, -3.5, 9, and -9 Hz, respectively. The double-sided phase spectrum consists of lines of heights =4 and =4 at frequencies 3.5 Hz and 3:5 Hz, respectively. e. Use sin (2 t) = cos (2 t

=2) to write

x5 (t) = 5 cos (2 t h = Re 5ej(2 t = 2:5ej(2 t

=2) + 4 cos (5 t + =4) i =2) + 4ej(5 t+ =4)

=2)

+ 2:5e j(2 t

=2)

+ 2ej(5 t+ =4) + 2e j(5 t+ =4)

From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes 5 and 4 at frequencies of 1 and 2.5 Hz, respectively, and the phase spectrum consists of lines of heights =2 and =4 at 1 and 2.5 Hz, respectively. The double-sided amplitude spectrum consists of lines of amplitudes 2.5, 2.5, 2, and 2 at frequencies of 1, -1, 2.5, and -2.5 Hz, respectively. The double-sided phase spectrum consists of lines of heights =2, =2, =4, and =4 at frequencies of 1, -1, 2.5, and -2.5 Hz, respectively.

2.1. PROBLEM SOLUTIONS

Single sided

Double sided 6

Amplitude

6 4

2 0

2 3 f, Hz

0 -5

0 f, Hz

0.5

0.6

Phase, rad

0.8

0.4 0.2

0 -0.5

0 0

2 3 f, Hz

f. Use sin (10 t + =6) = cos (10 t + =6

-5

=2) = cos (10 t

x6 (t) = 3 cos (4 t + =8) + 4 cos (10 t =3) i h = Re 3ej(4 t+ =8) + 4ej(10 t =3)

= 1:5ej(4 t+ =8) + 1:5e j(4 t+ =8) + 2ej10 t

=3) to write

=3)

+ 2e j(10 t

=3)

From this it is seen that the singe-sided amplitude spectrum consists of lines of amplitudes 3 and 4 at frequencies of 2 and 5 Hz, respectively, and the phase spectrum consists of lines of heights =8 and =3 at 2 and 5 Hz, respectively. The double-sided amplitude spectrum consists of lines of amplitudes 1.5, 1.5, 2, and 2 at frequencies of 2, -2, 5, and -5 Hz, respectively. The double-sided phase spectrum consists of lines of heights =8, =8, =3, and =3 at frequencies of 2, -2, 5, and -5 Hz, respectively.

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Single sided

Double sided 5

8 6

Amplitude

4 2 0

2 1

2 f, Hz

0 -4

-2

0 f, Hz

-2

0 f, Hz

1 Phase, rad

0.5

0.5 0 -0.5 -1

2 f, Hz

-4

2.1. PROBLEM SOLUTIONS

Single sided

Double sided 6 Amplitude

Amplitude

4 2 0

-5

0 f, Hz

-5

0 f, Hz

f, Hz 0 Phase, rad

Phase, rad

1 -0.5 -1

0 -1

-1.5 0

f, Hz

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Problem 2.2 By noting the amplitudes and phases of the various frequency components from the plots, the result is x(t) = 4ej(8 t+ =2) + 4e j(8 t+ =2) + 2ej(4 t = 8 cos (8 t + =2) + 4 cos (4 t =

8 sin (8 t) + 4 cos (4 t

=4)

+ 2e j(4 t

=4)

Problem 2.3 a. Not periodic because f1 = 1= Hz and f2 = 3 Hz are not commensurable. b. Periodic. To …nd the period, note that 30 6 = 3 = n1 f0 and = 15 = n2 f0 2 2 Therefore

15 n2 = 3 n1

Hence, take n1 = 1, n2 = 5; and f0 = 3 Hz (we want the largest possible value for f0 with n1 and n2 integer-valued). c. Periodic. Using a similar procedure as used in (b), we …nd that n1 = 4, n2 = 21; and f0 = 0:5 Hz. d. Periodic. Using a similar procedure as used in (b), we …nd that n1 = 4, n2 = 7; n3 = 11; and f0 = 0:5 Hz. e. Periodic. We …nd that n1 = 17, n2 = 18; and f0 = 0:5 Hz. f. Periodic. We …nd that n1 = 2, n2 = 3; and f0 = 0:5 Hz. g. Periodic. We …nd that n1 = 7, n2 = 11; and f0 = 0:5 Hz. h. Not periodic. The frequencies of the separate terms are incommensurable. i. Periodic. We …nd that n1 = 19, n2 = 21; and f0 = 0:5 Hz. j. Periodic. We …nd that n1 = 6, n2 = 7; and f0 = 0:5 Hz.

2.1. PROBLEM SOLUTIONS

Problem 2.4 a. The single-sided amplitude spectrum consists of a single line of amplitude 5 at 6 Hz and the phase spectrum consists of a single line of height =6 rad at 6 Hz. The double-sided amplitude spectrum consists of lines of amplitude 2.5 at frequencies 6 Hz. The double -sided phase spectrum consists of a line of height =6 at -6 Hz and a line of height =6 at 6 Hz. b. Write the signal as x2 (t) = 3 cos(12 t

=2) + 4 cos(16 t)

From this it is seen that the single-sided amplitude spectrum consists of lines of heights 3 and 4 at frequencies 6 and 8 Hz, respectively, and the single-sided phase spectrum consists of a line of height =2 radians at frequency 6 Hz (the phase at 8 Hz is 0). The doublesided amplitude spectrum consists of lines of height 1.5 and 2 at frequencies of 6 and 8 Hz, respectively, and lines of height 1.5 and 2 at frequencies 6 and 8 Hz, respectively. The double-sided phase spectrum consists of a line of height =2 radians at frequency 6 Hz and a line of height =2 radians at frequency 6 Hz. c. Use the trig identity cos x cos y = 0:5 cos (x + y) + 0:5 cos (x

y) to write

x3 (t) = 2 cos 20 t + 2 cos 4 t From this we see that the single-sided amplitude spectrum consists of lines of height 2 at 2 and 10 Hz, and the single-sided phase spectrum is 0 at these frequencies. The double-sided amplitude spectrum consists of lines of height 1 at frequencies of 10, 2, 2, and 10 Hz. The double-sided phase spectrum is 0. d. Use trig identies to get x4 (t) = 4 sin (2 t) [1 + cos (10 t)] = 4 sin (2 t)

2 sin (8 t + ) + 2 sin (12 t)

= 4 cos (2 t h = Re 4ej(2 t

=2) + 2 cos (8 t + =2) + 2 cos (12 t i =2) + 2ej(8 t+ =2) + 2ej(12 t =2)

= 2ej(2 t

=2)

+ 2e j(2 t

=2)

+ ej(8 t+ =2) + e j(8 t+ =2) + ej(12 t

=2)

+ e j(12 t

From this we see that the single-sided amplitude spectrum consists of lines of heights 4, 2, and 2 at frequencies 1, 4, and 6 Hz, respectively and the single-sided phase spectrum is =2 radians at 1 and 6 Hz and =2 radians at 4 Hz. The double-sided amplitude spectrum

=2)

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

consists of lines of height 2 at frequencies of 1 and 1 Hz and of height 1 at frequencies of 4, -4, 6, and -6 Hz. The double-sided phase spectrum is =2 radians at -1, 4, and -6 Hz and =2 radians at 1, -4, and 6 Hz. e. Clearly, from the form of the cosine sum, the single-sided amplitude spectrum has lines of heights 1 and 7 at frequencies of 3 and 15 Hz, respectively. The single-sided phase spectrum is zero. The double-sided amplitude spectrum has lines of heights 0.5, 0.5, 3.5, and 3.5 at frequencies of 3, -3, 15, and -15 Hz, respectfully. The double-sided phase spectrum is zero. f. The single-sided amplitude spectrum has lines of heights 1 and 9 at frequencies of 2 and 10.5 Hz, respectively. The single-sided phase spectrum is =2 radians at 10.5 Hz and 0 otherwise. The double-sided amplitude spectrum has lines of heights 0.5, 0.5, 4.5, and 4.5 at frequencies of 2, -2, 10.5, and -10.5 Hz, respectfully. The double-sided phase spectrum is =2 radians at -10.5 Hz and =2 radians at 10.5 Hz and 0 otherwise. g. Convert the sine to a cosine by subtracting =2 from its argument. It then follows that the single-sided amplitude spectrum is 2, 1, and 6 at frequencies of 2, 3, and 8.5 Hz and 0 otherwise. The single-sided phase spectrum is =2 radians at 8.5 Hz and 0 otherwise. The double-sided amplitude spectrum is 1, 1, 0.5, 0.5, 3, and 3 at frequencies of 2, 2, 3, 3 8:5, and 8.5 Hz, respectively, and 0 otherwise. The double-sided phase spectrum is =2 radians at a frequency of 8:5 Hz and =2 radians at a frequency of 8.5 Hz. It is 0 otherwise.

Problem 2.5 a. This function has area Area =

sin( u) 2 du = 1 ( u)

sin( t= ) 2 dt ( t= )

where a tabulated integral has been used for sinc2 u. A sketch shows that no matter how small is, the area is still 1. With ! 0; the central lobe of the function becomes narrower and higher. Thus, in the limit, it approximates a delta function.

2.1. PROBLEM SOLUTIONS

b. The area for the function is Z1

Area =

exp( t= )u (t) dt =

exp( u)du = 1

A sketch shows that no matter how small is, the area is still 1. With ! 0; the function becomes narrower and higher. Thus, in the limit, it approximates a delta function. R 1 R1 c. Area = (1 jtj = ) dt = As ! 0, the function becomes 1 (t) dt = 1. narrower and higher, so it approximates a delta function in the limit. Problem 2.6 1 (t) to write (2t 5) = a. Make use of the formula (at) = jaj 1 5 t 2 and use the sifting property of the -function to get 2

Ia =

1 2

5 2

1 exp 2

5 2

[2 (t

5=2)] =

25 1 + exp [ 5] = 3:1284 8 2

b. Impulses at 10, 5, 0, 5, 10 are included in the integral. Use the sifting property after writing the expression as the sum of …ve integrals to get Ib = ( 10)2 + 1 + ( 5)2 + 1 + 02 + 1 + 52 + 1 + 102 + 1 = 255 c. Matching coe¢ cients of like derivatives of -functions on either side of the equation gives A = 5, B = 10, and C = 3. d. Use I = 14

1 (at) = jaj (t) to write

4 ( 3=4) + tan

(4t + 3) = 14 3 4

= 14

t + 34 .

The integral then becomes

+ tan ( 7:5 ) =

9:277

1013 .

e. Use property 5 of the unit impulse function to get d2 d Ie = ( 1)2 2 cos 5 t + e 3t t=2 = 5 sin 5 t 3e 3t t=2 dt dt h i = (5 )2 cos 5 t + 9e 3t = (5 )2 cos 10 + 9e 6 = 246:73 t=2

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Problem 2.7 (a), (c), and (e) are periodic. Their periods are 2 s (fundamental frequency of 0.5 Hz), 2 s, and 3 s, respectively. The waveform of part (c) is a periodic train of triangles, each 2 units wide, extending from -1 to 1 spaced by 2 s ((b) is similar except that it is zero for t < 1 thus making it aperiodic). Waveform (d) is aperiodic because the frequencies of its two components are incommensurable. The waveform of part (e) is a doubly-in…nite train of square pulses, each of which is one unit high and one unit wide, centered at ; 6; 3; 0; 3; 6; . Waveform (f) is identical to (e) for t 1=2 but 0 for t < 1=2 thereby making it aperiodic. Problem 2.8 a. The result is j(10 t

=2)

1 1 + e j6 t + ej6 t + ej(10 t 2 2

=2)

j6 t

x(t) = cos (6 t)+2 cos (10 t

=2) = Re e

+Re 2e

j6 t

= Re e

j(10 t

+ 2e

b. The result is x(t) = e j(10 t

=2)

c. The single-sided amplitude spectrum consists of lines of height 1 and 2 at frequencies of 3 and 5 Hz, respectively. The single-sided phase spectrum consists of a line of height =2 at frequency 5 Hz. The double-sided amplitude spectrum consists of lines of height 1, 1/2, 1/2, and 1 at frequencies of 5; 3; 3; and 5 Hz, respectively. The double-sided phase spectrum consists of lines of height =2 and =2 at frequencies of 5 and 5 Hz, respectively. Problem 2.9 a. Power. Since it is a periodic signal, we obtain Z T0 Z T0 1 1 2 P1 = 4 cos (4 t + 2 =3) dt = 2 [1 + cos (8 t + 4 =3)] dt = 2 W T0 0 T0 0 where T0 = 1=2 s is the period. The cosine in the above integral integrates to zero because the interval of integratation is two periods. b. Energy. The energy is E2 =

Z 1

2 t 2

u (t)dt =

Z 1 0

e 2 t dt =

1 J 2

=2)

2.1. PROBLEM SOLUTIONS

c. Energy. The energy is

E3 =

Z 1

2 t 2

u ( t)dt =

Z 0

e2 t dt =

1 J 2

d. Energy. The energy is

E4 =

= =

Z T

dt 1 = lim 2 2 + t2 ) T !1 ( T

lim

T !1

lim

T !1

tan 1

= lim

dt 1 + (t= )2

tan 1 (T = )

T !1

Z T

tan 1 ( T = )

e. Energy. Since it is the sum of x2 (t) and x3 (t), its energy is the sum of the energies of these two signals, or E5 = 1= J. f. Energy. The energy is

E6 = = = = = =

lim

T !1

lim

T !1

lim

T !1

lim

T !1

lim

Z T h

u (t)

(t 1)

i2 1) dt

u (t

Z T h

e 2 t u2 (t) e

2 t

Z T

2 (t 1)

dt +

e 2 t dt0 +

e 2 2

1 1 = 2

1) + e 2 (t 1) u2 (t

u (t) u (t

8 0 < e 2 t T e 2

(t 1)

Z T

T !1 :

1 2

Z T 0

1 e 2

T 1

e 2 2

e 2 (t 1) dt

1 Z T 1 t0

e 2 t dt0 9 1= ;

i 1) dt

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Problem 2.10 a. Power. Since the signal is periodic with period 2 =!, we have

! P = 2

Z 2 =! 0

! A jsin (!t + )j dt = 2 2

Z 2 =! 0

A2 f1 2

cos [2 (!t + )]g dt =

A2 W 2

b. Neither. The energy calculation gives Z T Z T (A )2 dt (A )2 dt p p p dt ! 1 E = lim dt = lim 2 + t2 T !1 T !1 + jt jt T T The power calculation gives 1 P = lim T !1 2T

Z T

(A )2 dt (A )2 p ln dt = lim 2 + t2 T !1 2T T

c. Energy: E=

Z 1 0

1 A t exp ( 2t= ) dt = A2 8 2 2

1 + T 2= 2 p 1 + 1 + T 2= 2

=0W

W (use a table of integrals)

d. Energy: This is a "top hat" pulse which is height 2 for jtj =2, height 1 for =2 < jtj , and 0 everywhere else. Making use of the even symmetry about t = 0, the energy is ! Z =2 Z E=2 22 dt + 12 dt = 5 J 0

e. Energy. The signal is a "house" two units wide and one unit up to the eves with a equilateral triangle for a roof. Because of symmetry, the energy calculation need be carried out for positive t and doubled. The calculation is Z 1 1 2 2 2 8 14 E=2 (2 t)2 dt = (2 t)3 = + = J 3 3 3 3 0 0 f. Power. Since the two terms are harmonically related, we may add their respective powers and get A2 B 2 P = + W 2 2

2.1. PROBLEM SOLUTIONS

Problem 2.11 a. Using the fact that the power contained in a sinusoid is its amplitude squared divided by 2, we get 22 P = =2W 2 b. This is a periodic train of "box cars" 3 units high, 2 units wide, and occurring every 4 units (period of 4 seconds). The power calculation is Z 1 1 2 32 2 P = 3 dt = = 4:5 W 4 1 4 c. This is a train of triangles 1 unit high, 4 s wide, and occuring every 6 s. Using the waveform period centered at 0, the power calculation is 1 P = 6

Z 2

t 2

dt =

12 63

t 2

3 2

= 0

2 W 9

d. This is a train of "houses" each of which is 2 s wide, 1 unit high to the eves, with an isoceles triangle on top for the roof. They are separated by 4 s (the period). Using the even symmetry of each house, the power calculation is 2 Pd = 4

Z 1

t) dt =

1 (2 t)3 2 3

= 0

1 2

1 3

23 3

7 W 6

Problem 2.12 a. The energy is E =

Z 1 0

= 36

( 3+j4 )t

Z 1

e 6t dt =

dt = 36

Z 1

e( 3+j4 )t e( 3 j4 )t dt

0 1

e 6t 36 =6J 6 0

The power is 0 W. b. This signal is a "top hat" pulse which is 2 for 2 t 4, 1 for 0 t < 2 and 4 < t and 0 everywhere else. It is clearly an energy signal with energy E=2

12 + 2

22 + 2

12 = 12 J

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Its power is 0 W. c. This is a power signal with power Z T Z 1 49 T 49 j6 t j6 t P = lim 49e e u (t) dt = lim dt = = 24:5 W T !1 2T T !1 2T 0 2 T Its energy is in…nite. 2

d. This is a periodic signal with power P = 22 = 2 W. Its energy is in…nite. e. This is neither an energy nor a power signal. Its energy is in…nite and its power is 1 T !1 2T

P = lim

Z T

1 t3 T !1 2T 3

t2 dt = lim

1 2T 3 !1 T !1 2T 3

= lim T

f. This is neither an energy nor a power signal. Its energy is Z 1 t 1 dt = ln (t)j1 E= 1 !1 1

and its power is 1 P = lim T !1 2T Problem 2.13

Z T

1 ln (t)j1 1 =0 T !1 2T

t 1 dt = lim

a. This is a cosine burst from t = 6 to t = 6 seconds. The energy is E1 = R6 2 0 12 + 12 cos (12 t) dt = 6 J

6 cos

b. The energy is

E2 =

Z 1h 1

jtj=3

2t=3

2=3

dt = 2

Z 1

e 2t=3 dt (by even symmetry)

=3J 0

Since the result is …nite, this is an energy signal. c. The energy is E3 =

Z 1

f2 [u (t)

u (t

8)]g dt =

Z 8 0

4dt = 32 J

2 (6

t) dt =

2.1. PROBLEM SOLUTIONS

Since the result is …nite, this is an energy signal. d. Note that r (t) ,

Z t

u( )d =

0; t < 0 t; t 0

which is called the unit ramp. Thus the given signal is a triangle between 0 and 20. The energy is Z 1 Z 10 2 2000 10 2 E4 = [r (t) 2r (t 10) + r (t 20)] dt = 2 t2 dt = t3 0 = J 3 3 1 0 where the last integral follows because the integrand is a symmetrical triangle about t = 10. Since the result is …nite, this is an energy signal. Problem 2.14 a. This is a cosine burst nonzero between 0 and 2 seconds. Its power is 0. Its energy is Z Z 2 1 2 2 [1 + cos (20 t)] dt = 1 J E1 = cos (10 t) dt = 2 0 0 b. This is a periodic sequence of triangles of period 3 s. Its energy is in…nite. Its power is Z 2 2 4 P2 = (1 t=2)2 dt = J 3 0 9 c. This is an energy signal. Its power is 0. Using evenness of the integrand, its energy is Z 1 Z 1 Z 1 2t 2 2t E3 = 2 e cos (2 t) dt = e dt + e 2t cos (4 t) dt 0

1 2 + J 2 4 + 16 2

. d. This is an energy signal. Its energy is E4 = 2

Z 1 0

t)2 dt =

2 (2 3

t)3

= 0

14 J 3

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Problem 2.15 a. Use the exponential representation of the sine to get the Fourier coe¢ cients as (note that the period = 12 f10 ). x1 (t) =

ej2 f0 t

e j2 f0 t 2j

from which we …nd that X 1 = X1 =

1 e j4 f0 t 4

2 + ej4 f0 t

1 1 ; X0 = 4 2

All other coe¢ cients are zero. b. Use the exponential representations of the sine and cosine to get 1 1 1 x2 (t) = ej2 f0 t + e j2 f0 t + ej4 f0 t 2 2 2j

1 j4 f0 t e 2j

Therefore, the Fourier coe¢ cients for this case are X 1 = X1 =

1 1 and X2 = X 2 = 2 2j

All other coe¢ cients are zero. c. Use a trig identity to write this signal as x3 (t) =

1 1 sin 8 f0 t = ej8 f0 t 2 4j

1 j8 f0 t e 4j

The fundamental frequency is 4f0 Hz. From this it follows that the Fourier coe¢ cients are X1 = X 1 =

1 4j

All other coe¢ cients are zero. d. Use trig identities and the exponential forms of cosine to write this signal as x4 (t) = =

1 3 cos 2 f0 t + cos 6 f0 t 4 4 3 j2 f0 t 3 j2 f0 t 1 j6 f0 t 1 j6 f0 t + e + e + e e 8 8 8 8

2.1. PROBLEM SOLUTIONS

The fundamental frequency is f0 Hz. It follows that the Fourier coe¢ cients for this case are 3 1 X 1 = X1 = ; X 3 = X3 = 8 8 All other Fourier coe¢ cients are zero. e. Use trig identies to write x5 (t) = =

1 1 1 sin (2 f0 t) sin (6 f0 t) + sin (10 f0 t) 2 4 4 1 j2 f0 t 1 j2 f0 t 1 j6 f0 t 1 1 e e e + e j6 f0 t + ej10 f0 t 4j 4j 8j 8j 8j

1 j10 f0 t e 8j

The fundamental frequency is f0 Hz. It follows that the Fourier coe¢ cients for this case are j j j X 1 = X1 = ; X 3 = X3 = ; X 5 = X5 = 4 8 8 All other Fourier coe¢ cients are zero. f. Use trig identities to write x6 (t) = =

1 1 1 cos (6 f0 t) cos ( f0 t) cos (11 f0 t) 2 4 4 1 j6 f0 t 1 j6 f0 t 1 j f0 t 1 j f0 t 1 j11 f0 t e + e e e e 4 4 8 8 8

1 j11 f0 t e 8

The fundamental frequency is f0 =2 Hz. It follows that the Fourier coe¢ cients for this case are 1 1 1 X 1 = X1 = ; X 12 = X12 = ; X 22 = X22 = 8 4 8 All other Fourier coe¢ cients are zero. Problem 2.16 The expansion interval is T0 = 4 so that f0 = 1=4 Hz. The Fourier coe¢ cients are Z Z 2 2 2 1 2 2 jn( =2)t Xn = 2t e dt = t (cos n t=2 j sin n t=2) dt = 4 2 4 2 Z 2 2 2 n t dt = 2t cos 4 0 2 which follows by the oddness of the second integrand and the eveness of the …rst integrand. Let u = n t=2 to obtain the form Z 2 3 n 2 16 Xn = u cos u du = ( 1)n n 6= 0 (use a table of integrals) 2 n (n ) 0

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

If n = 0, the integral for the coe¢ cients is 1 X0 = 4

Z 2

2t2 dt =

8 3

The Fourier series is therefore 8 x (t) = + 3

1 X

( 1)n

n= 1; n6=0

16 jn( =2)t e (n )2

Problem 2.17 Parts (a) through (c) were discussed in the text. For (d), break the integral for Xn up into a part for t < 0 and a part for t > 0. Then use the odd half-wave symmetry condition.

The development follows:

Xn = = =

1 T0 1 T0 1 T0

1 = T0 ( =

T0 =2

x (t) e j2 nf0 t dt +

T0 =2

x (t) e j2 nf0 t dt +

x (t) e j2 nf0 t dt

Z T0 =2

x t0 + T0 =2 e j2

nf0 (t0 +T0 =2)

T0 =2

x (t) e

j2 nf0 t

T0 =2

Z 0

Z T0 =2

x t e

j2 nf0 t0 jn

x (t) e

j2 nf0 t

( 1)

0; n even R T0 =2 2 x (t) e j2 nf0 t dt; n odd T0 0

Z T0 =2 0

x t e

dt ; t0 = t + T0 =2

dt ; f0 = 1=T0

T0 =2

j2 nf0 t0

2.1. PROBLEM SOLUTIONS

Problem 2.18 This is a matter of integration. Only the solution for part (b) will be given here. integral for the Fourier coe¢ cients is Xn = =

= = =

Z T0 =2 A sin (! 0 t) e dt = ej!0 t e j!0 t e jn!0 t dt 2jT0 0 0 # "Z Z T0 =2 T0 =2 A e j(1+n)!0 t dt ej(1 n)!0 t dt 2jT0 0 0 2 3 T0 =2 T0 =2 j(1 n)! t j(1+n)! t 0 0 A 4 e e 5 2jT0 j (1 n) ! 0 j (1 + n) ! 0 0 0 " # A ej(1 n) 1 e j(1+n) 1 + ; n 6= 1 (! 0 T0 =2 = ) 4 1 n 1+n A T0

A 4 (

Z T0 =2

The

jn! 0 t

( 1)n + 1 ( 1)n + 1 + ; n 6= 1 n 1+n 0; n odd and n 6= A ; n even (1 n2 )

For n = 1, the integral is X1 =

A 2jT0

Z T0 =2

ej!0 t

e j!0 t e j!0 t dt

Z T0 =2

e j2!0 t dt =

jA =X 1 4

This is the same result as given in Table 2.1. Problem 2.19 a. Use Parseval’s theorem to get Pjnf0 j

N X

n= N

jXn j2 =

N X

n= N

A T0

sinc2 (nf0 )

where N is an appropriately chosen limit on the sum. We are given that only frequences for which jnf0 j 1= are to be included. This is the same as requiring that jnj 1= ( f0 ) =

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

T0 = = 2: Also, for a pulse train, Ptotal = A2 =T0 and, in this case, Ptotal = A2 =2: Thus 2

Pjnf0 j 1= Ptotal

2 X A2

n= 2

A 2

sinc2 (nf0 )

1 X sinc2 (nf0 ) 2 n= 2

= = b. In this case, jnj Pjnf0 j 1= Ptotal

c. In this case, jnj Pjnf0 j 1= Ptotal

1 1 + 2 sinc2 (1=2) + sinc2 (1) 2" # 1 2 2 1+2 = 0:9053 2

5, Ptotal = A2 =5, and 5

1 X = sinc2 (n=5) 5 n= 5 h io 1n = 1 + 2 (0:9355)2 + (0:7568)2 + (0:5046)2 + (0:2339)2 5 = 0:9029 10, Ptotal = A2 =10, and

10 1 X sinc2 (n=10) 10 n= 10

1 1+2 10 = 0:9028 =

d. In this case, jnj

(0:9836)2 + (0:9355)2 + (0:8584)2 + (0:7568)2 + (0:6366)2 + (0:5046)2 + (0:3679)2 + (0:2339)2 + (0:1093)2

20, Ptotal = A2 =20, and

Pjnf0 j 1= Ptotal

20 1 X sinc2 (n=20) 20 n= 20 ( ) 20 X 1 1+2 sinc2 (n=20) 20 n=1

= 0:9028

2.1. PROBLEM SOLUTIONS

Problem 2.20 a. The integral for Yn is Z Z T0 1 1 jn! 0 t y (t) e dt = x (t Yn = T0 T0 T0 0 Let t0 = t

t0 ) e jn!0 t dt; ! 0 = 2 f0

t0 , which results in Z T0 t0 1 0 Yn = x t0 e jn!0 t dt0 e jn!0 t0 = Xn e j2 nf0 t0 T0 t0

b. Note that y (t) = A cos ! 0 t = A sin (! 0 t + =2) = A sin [! 0 (t + =2! 0 )] Thus, t0 in the theorem proved in part (a) here is can be expressed as 1 sin (! 0 t) = ej!0 t 2j

=2! 0 . By Euler’s theorem, a sine wave 1 j!0 t e 2j

1 Its Fourier coe¢ cients are therefore X1 = 2j and X 1 = proved in part (a), we multiply these by the factor

e jn!0 t0 = e jn!0 (

=2! 0 )

1 2j .

According to the theorem

= ejn =2

For n = 1, we obtain Y1 = For n =

1 j =2 1 e = 2j 2

1, we obtain

1 j =2 1 e = 2j 2 which gives the Fourier series representation of a cosine wave as Y 1=

1 1 y (t) = ej!0 t + e j!0 t = cos ! 0 t 2 2 We could have written down this Fourier representation directly by using Euler’s theorem. Problem 2.21 a. Use the Fourier series of a square wave (specialize the Fourier series of a pulse train) with A = 1 and t = 0 to obtain the series 1=

1 1 + 3 5

1 + 7

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Multiply both sides by 4 to get the series in the problem statement. Hence, the sum is 4 . b. Use the Fourier series of a triangular wave as given in Table 2.1 with A = 1 and t = 0 to obtain the series 1=

4 4 4 4 4 4 + + 2+ 2+ 2+ + 25 2 9 2 9 25 2

Multiply both sides by 8 to get the series in given in the problem. Hence, its sum is 8 . Problem 2.22 a. In the expression for the Fourier series of a pulse train (Table 2.1), let t0 = and = T0 =4 to get n nf0 A exp j Xn = sinc 4 4 4

T0 =8

The spectra are shown in Fig. 2.4. b. The amplitude spectrum is the same as for part (a) except that X0 = 3A 4 . Note that this can be viewed as having a sinc-function envelope with zeros at multiples of 3T4 0 . The phase spectrum can be obtained from that of part (a) by subtracting a phase shift of for negative frequencies and adding for postitive frequencies (or vice versa). The Fourier coe¢ cients are given by Xn =

3A sinc 4

3n 4

exp

3 nf0 4

See Fig. 2.4 for amplitude and phase plots. Problem 2.23 a. Use the rectangular pulse waveform of Table 2.1 specialized to xa (t) = 2A

T0 =4 T0 =2

A; jtj < T0 =2

and periodically extended. Hence, from Table 2.1, we have 2AT0 =2 nT0 =2 2 nT0 =4 sinc exp j T0 T0 T0 = Asinc (n=2) exp ( j n=2) ; n 6= 0 sin (n =2) = A exp ( j n=2) ; n 6= 0 n =2

XnA =

2.1. PROBLEM SOLUTIONS

Part (a)

Part (b) 0.6 Amplitude

Amplitude

0.6 0.4 0.2 0

-5

0 f, Hz

0.2 0

-5

0 f, Hz

-5

0 f, Hz

2 Phase, rad

0.4

0 -2

-5

0 f, Hz

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

where the superscript A refers to xA (t). The dc component is 0 so X0A = 0. The Fourier coe¢ cients are therefore X0A = 0 X1A =

j2A= ; X A1 = j2A=

X2A = 0 = X A2 X3A = j2A=3 ; X A3 =

j2A=3

b. Note that xA (t) =

dxB (t) dt

where A = T2B = 4B T0 obtained from matching the amplitude of xA (t) with the slope of 0 =2 xB (t) or B = T40 A. The relationship between spectral components is therefore XnA = (jn! 0 ) XnB = j2 nf0 XnB or XnB =

XnA T0 XnA = j2 nf0 j2 n

where the superscript A refers to xA (t) and B refers to xB (t). For example, X1B =

j2A= = j2 f0

2B 2

= X B1

Problem 2.24 a. This is a decaying exponential starting at t = 0 and its Fourier transform is Z 1 Z 1 X1 (f ) = A e t= e j2 f t dt = A e (1= +j2 f )t dt 0

(1= +j2 f )t

Ae 1= + j2 f

A 1 + j2 f

= 0

A 1= + j2 f

2.1. PROBLEM SOLUTIONS

b. Since x2 (t) = x1 ( t) we have, by the time reversal theorem, that X2 (f ) = X1 (f ) = X1 ( f ) A = 1 j2 f c. Since x3 (t) = x1 (t)

x2 (t) we have, after some simpli…cation, that X3 (f ) = X1 (f ) X2 (f ) A A = 1 + j2 f 1 j2 f j4A f = 1 + (2 f )2

d. Since x4 (t) = x1 (t) + x2 (t) we have, after some simpli…cation, that X4 (f ) = X1 (f ) + X2 (f ) A A = + 1 + j2 f 1 j2 f 2A = 1 + (2 f )2 This is the expected result since x4 (t) is really a double-sided decaying exponential. e. By part a and the delay theorem X5 (f ) = X1 (f ) e j10 f =

A e j10 f 1 + j2 f

f. By parts a and e and superposition h X6 (f ) = X1 (f ) 1

i A e j10 f =

1 e j10 f 1 + j2 f

Problem 2.25 a. Using a table of Fourier transforms and the time reversal theorem, the Fourier transform of the given signal is X (f ) =

1 + j2 f

1 j2 f

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Note that x (t) ! sgn(t) in the limit as transform as ! 0, we deduce that F [sgn (t)] =

! 0.

1 j2 f

Taking the limit of the above Fourier 1 1 = j2 f j f

b. Using the given relationship between the unit step and the signum function and the linearity property of the Fourier transform, we obtain F [u (t)] = =

1 1 F [sgn (t)] + F [1] 2 2 1 1 + (f ) j2 f 2

c. The same result as obtained in part (b) is obtained.

Problem 2.26 a. One di¤erentiation gives dxa (t) = dt

0:5)

2:5)

Two di¤erentiations give d2 xa (t) = (t) dt2

2) + (t + 3)

Application of the di¤erentiation theorem of Fourier transforms gives (j2 f )2 Xa (f ) = 1 =

1 e j2 f ej3 f

ej f

= 2 (cos 3 f

1 e j4 f + 1 e j6 f e j f + e j3 f e j3 f

cos f ) e j3 f

where the time delay theorem and the Fourier transform of a unit impulse have been used. Dividing both sides by (j2 f )2 , we obtain Xa (f ) =

cos f cos 3 f j3 f 2 (cos 3 f cos f ) e j3 f = e 2 2 2f 2 (j2 f )

2.1. PROBLEM SOLUTIONS

Use the trig identity sin2 x = 12

1 2 cos 2x or cos 2x = 1

2 sin2 x to rewrite this result as

2 sin2 (0:5 f ) 1 + 2 sin2 (1:5 f ) j3 f e 2 2f 2 sin2 (0:5 f ) + sin2 (1:5 f ) j3 f = e 2f 2 sin2 (1:5 f ) j3 f sin2 (0:5 f ) j3 f = e e 2f 2 2f 2 " # 2 2 sin 1:5 f sin 0:5 f = 1:52 0:52 e j3 f 1:5 f 0:5 f

Xa (f ) =

0:52 sinc2 (0:5f ) e j3 f

1:52 sinc2 (1:5f )

This is the same result as would have been obtained by writing t

xa (t) = 1:5

1:5 1:5

0:5

1:5 0:5

and using the Fourier transform of the triangular pulse along with the superposition and time delay theorems. b. Two di¤erentiations give (sketch dxb (t) =dt to see this) d2 xb (t) = (t) dt2

2 (t

1) + 2 (t

Application of the di¤erentiation theorem gives (j2 f )2 Xb (f ) = 1

2e j2 f + 2e j6 f

e j8 f

Dividing both sides by (j2 f )2 , we obtain Xb (f ) =

2e j2 f + 2e j6 f 4 2f 2

e j8 f

Further manipulation may be applied to this result to convert it to h i Xb (f ) = sinc2 (f ) e j2 f e j6 f h i = sinc2 (f ) ej2 f e j2 f e j4 f = 2j sin (2 f ) sinc2 (f ) e j4 f

which would have resulted from Fourier transforming the waveform written as xb (t) =

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS c. Two di¤erentiations give (sketch dxc (t) =dt to see this) d2 xc (t) = (t) dt2

2 (t

1) + 2 (t

2 (t

3) + (t

Application of the di¤erentiation theorem gives (j2 f )2 Xc (f ) = 1

2e j2 f + 2e j4 f

2e j6 f + e j8 f

Dividing both sides by (j2 f )2 , we obtain Xc (f ) =

2e j2 f + 2e j4 f 2e j6 f + e j8 f (j2 f )2

This result may be further arranged to give 2

Xc (f ) = sinc (f ) e

j2 f

j6 f

= 2 cos (2 f ) sinc2 (f ) e j4 f

which would have resulted from Fourier transforming the waveform written as xc (t) =

1) +

d. Two di¤erentiations give (sketch dxd (t) =dt to see this) d2 xd (t) = (t) dt2

2 (t

1) + (t

1:5) + (t

2:5)

2 (t

3) + (t

Application of the di¤erentiation theorem gives (j2 f )2 Xd (f ) = 1

2e j2 f + e j3 f + e j5 f

2e j6 f + e j8 f

Dividing both sides by (j2 f )2 , we obtain Xd (f ) =

2e j2 f + e j3 f + e j5 f (j2 f )2

2e j6 f + e j8 f

This result may be further arranged to give h i Xd (f ) = sinc2 (f ) e j2 f + 0:5e j4 f + e j6 f

which would have resulted from Fourier transforming the waveform written as xd (t) =

1) + 0:5 (t

2) +

2.1. PROBLEM SOLUTIONS

Problem 2.27 See the solutions to Problem 2.26. Problem 2.28 a. The steps in …nding the Fourier transform for (i) are as follows: (t)

! sinc (f )

(t) exp [j4 t] (t

1) exp [j4 (t

! sinc (f

1)]

! sinc (f

2) 2) exp ( j2 f )

The steps in …nding the Fourier transform for (ii) are as follows: (t)

! sinc (f )

(t) exp [j4 t]

! sinc (f

(t + 1) exp [j4 (t + 1)]

! sinc (f

2) 2) exp (j2 f )

b. The steps in …nding the Fourier transform for (i) are as follows: (t) (t

1) exp [j4 (t

1)]

! sinc (f )

! sinc (f ) exp ( j2 f )

1) exp (j4 t)

! sinc (f

2) exp [ j2 (f

2)] = sinc (f

2) exp ( j2 f )

which follows because exp( jn2 ) = 1 where n is an integer. The steps in …nding the Fourier transform for (ii) are as follows: (t) (t + 1) (t + 1) exp [j4 (t + 1)]

! sinc (f )

! sinc (f ) exp (j2 f )

(t + 1) exp (j4 t)

! sinc (f

2) exp [j2 (f

2)] = sinc (f

2) exp (j2 f )

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Problem 2.29

a. The time reversal theorem states that x ( t) so

xa (t)

! =

! X ( f ) 6= X (f ) if x (t) is complex,

1 1 1 1 X1 (f ) + X1 ( f ) = sinc (f 2) exp ( j2 f ) + sinc ( f 2 2 2 2 1 1 sinc (f 2) exp ( j2 f ) + sinc (f + 2) exp (j2 f ) 2 2

2) exp (j2 f )

Note that

xa (t) = =

1 2 1 2

1) exp [j4 (t

1)] +

1) exp (j4 t) +

1 2

( t

1) exp [j4 ( t

1)]

(t + 1) exp ( j4 t) (by the eveness of

(u) )

b. Similarly to (a), we obtain

xb (t)

1 1 1 ! X2 (f )+ X2 ( f ) = sinc (f 2 2 2

1 2) exp (j2 f )+ sinc (f + 2) exp ( j2 f ) 2

2.1. PROBLEM SOLUTIONS

Problem 2.30 a. The result is X1 (f ) = 2sinc (2f ) exp ( j2 f ) b. The result is X2 (f ) = 2

1 2

f 2

exp ( j2 f )

c. The result is X3 (f ) = 8sinc2 (8f ) exp ( j4 f ) d. The result is X4 (f ) = 4 (4f ) exp ( j6 f ) e. The result is X5 (f ) = 5 = 5

1 2 f 2

f 2

exp ( j2 f ) + 5

1 2

f 2

exp (j2 f )

cos (2 f )

f. The result is X6 (f ) = 16sinc2 (8f ) exp ( j4 f ) + 16sinc2 (8f ) exp (j4 f ) = 32sinc2 (8f ) cos (4 f )

Problem 2.31 a. This is an odd signal, so its Fourier transform is odd and purely imaginary. b. This is an even signal, so its Fourier transform is even and purely real. c. This is an odd signal, so its Fourier transform is odd and purely imaginary. d. This signal is neither even nor odd, so its Fourier transform is complex. e. This is an even signal, so its Fourier transform is even and purely real. f. This signal is even, so its Fourier transform is real and even.

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Problem 2.32 In the Poisson sum formula, we identify p (t) = (t=2) which has Fourier transform P (f ) = 2sinc(2f ). Thus, for this case, the Poisson sum formula becomes 1 X

p (t

mTs ) =

m= 1

1 X

4m 2

m= 1 1 X

m= 1

1 1 X 2sinc 4 n= 1

2n 4

ej2 (n=4)t = fs

1 X

P (nfs ) ej2 nfs t

n= 1

1 X 1 n j (n=2)t = sinc e 2 2 n= 1

4m 2

The fundamental frequency is 0.25 Hz and the Fourier coe¢ cients are X0 = 1=2; X1 = X 1 = 12 sinc 12 = 1 , X2 = X 2 = 0, X3 = X 3 = 21 sinc 32 = 31 , etc. Problem 2.33 a. The Fourier transform of this signal is X1 (f ) =

10 2 = 5 + j2 f 1 + j2 f =5

Thus, the energy spectral density is G1 (f ) =

4 1 + (2 f =5)2

b. The Fourier transform of this signal is f 2

X2 (f ) = 5 Thus, the energy spectral density is X2 (f ) = 25

f 2

= 25

c. The Fourier transform of this signal is 3 X3 (f ) = sinc 2 so the energy spectral density is G3 (f ) = 94 sinc2

f 2

2.1. PROBLEM SOLUTIONS

d. The Fourier transform of this signal is X4 (f ) =

3 sinc 4

f +5 2

+ sinc

so the energy spectral density is G4 (f ) =

9 sinc 16

5 2

+ sinc

f +5 2

1 + j2 f

Problem 2.34 a. Use the transform pair t

x1 (t) = e

u (t)

Using Rayleigh’s energy theorem, we obtain the integral relationship Z 1 Z 1 Z 1 Z 1 df 1 2 2 jX1 (f )j df = df = jx1 (t)j dt = e 2 t dt = 2 + (2 f )2 2 1 1 1 0 b. Use the transform pair 1

Z 1

x2 (t) = Rayleigh’s energy theorem gives Z 1 jX2 (f )j2 df 1

Z 1

! sinc ( f ) = X2 (f )

sinc ( f ) df = 1 2

Z 1

dt =

jx2 (t)j2 dt =2 =2

dt 2

c. Use the transform pair e

jtj

2 e jtj or 2 2 + (2 f ) 2

1 2 + (2 f )2

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

The desired integral, by Rayleigh’s energy theorem, is Z 1 2 e 2 jtj 1 df = dt 2 2 + (2 f )2 1 4 1 Z 1 1 1 e 2 t 2 1 2 t e dt = = 4 2 0 2 2 2 0 4 3

Z 1

I3 = = d. Use the transform pair

! sinc2 ( f )

The desired integral, by Rayleigh’s energy theorem, is Z 1 Z 1 2 jX4 (f )j df = sinc4 ( f ) df I4 = 1 1 Z 1 Z 1 2 2 = (t= ) dt = 2 [1 (t= )]2 dt 2 =

Z 1

u)2 du =

1 1 3

= 0

2 3

Problem 2.35 a. The convolution operation gives 8 < 0; 1 1 e y1 (t) = : 1 e (t

+1=2)

;

1=2)

t 1=2 1=2 < t + 1=2 +1=2) ; t > + 1=2

b. The convolution of these two signals gives y2 (t) =

(t) + trap (t)

where trap(t) is a trapezoidal function given by 8 0; t < 3=2 or t > 3=2 > > < 1; 1=2 t 1=2 trap (t) = 3=2 + t; 3=2 t < 1=2 > > : 3=2 t; 1=2 t < 3=2

2.1. PROBLEM SOLUTIONS

c. The convolution results in Z 1 e y3 (t) =

j j

(

t) d =

Z t+1=2

j j

t 1=2

Sketches of the integrand for various values of t give the following cases: 8 R t+1=2 > t 1=2 > < tR 1=2 e d ; R 0 t+1=2 y3 (t) = e d ; 1=2 < t 1=2 t 1=2 e d + 0 > > : R t+1=2 e d ; t > 1=2 t 1=2

Integration of these three cases gives 8 1 e (t+1=2) e (t 1=2) ; < 1 y3 (t) = e (t 1=2) e (t+1=2) ; : 1 e (t 1=2) e (t+1=2) ; d. The convolution gives

y4 (t) =

Z t

t 1=2 1=2 < t 1=2 t > 1=2

x( )d

Problem 2.36 a. The inverse FT of (f ) is sinc(t). By the time delay theorem, the inverse Fourier transform of (f ) exp ( j4 f ) is sinc(t 2). The product of this and 2 cos (2 f ) in the frequency domain has an inverse Fourier transform which is the convolution of their respective Fourier transforms. Thus x1 (t) = sinc (t = sinc (t b. The inverse Fourier transform of

2) [ (t 3) + sinc (t

1) + (t + 1)] 1)

(f =2) is 2sinc2 (2t). By the time delay theorem

x2 (t) = 2sinc2 [2 (t

2:5)]

c. The inverse Fourier transform of (f =2) is 2sinc(2t). By the modulation theorem, f +4 f 4 the inverse Fourier transform of + is sinc(2t) cos (8 t). By the time 2 2 delay theorem x3 (t) = sinc [2 (t 4)] cos [2 (t 4)]

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Problem 2.37 a. From before, the total energy is E1; total = 21 . The Fourier transform of the given signal is 1 X1 (f ) = + j2 f so that the energy spectral density is G1 (f ) = jX1 (f )j2 =

1 2 + (2

f )2

By Rayleigh’s energy theorem, the normalized inband energy is Z W

E1 (jf j W ) =2 E1; total

df 2 = tan 1 2 2 + (2 f )

2 W

b. The total energy is E2; total = . The Fourier transform of the given signal and its energy spectral density are, respectively, X2 (f ) = sinc (f ) and G2 (f ) = jX2 (f )j2 =

sinc2 (f )

By Rayleigh’s energy theorem, the normalized inband energy is E2 (jf j W ) 1 = E2; total

Z W

sinc2 (f ) df = 2

sinc2 (u) du

The integration must be carried out numerically. c. The total energy is Z 1h E3; total = e

1 2

2 +

dt =

1 = 2

Z 1h

e 2 t

2e ( + )t + e 2 t

( + ) 2

4 + ( + ) ( )2 = ( + ) 2 ( + )

The Fourier transform of the given signal and its energy spectral density are, respectively, X3 (f ) =

1 + j2 f

2.1. PROBLEM SOLUTIONS

and 1 + j2 f

G3 (f ) = jX3 (f )j = 1 2 + (2 f )2 1 2 + (2 f )2

= =

1 + (2 f )2 j2 f + j2 f 1 2 Re + 2 2 + (2 f )2 2 + (2 f )2 + (2 f )2 2 3 ( ) j2 f + (2 f )2 5 1 2 Re 4 + 2 2 2 2 2 + (2 f ) + (2 f )2 + (2 f ) 2 Re

2 + (2

f )2

1 2 + (2 f )2

2 1 + j2 f 1 1 + + j2 f j2 f

+ (2 f )2

2 + (2

+ (2 f )

1 + (2 f )2

The normalized inband energy is E3 (jf j W ) 1 = E3; total E3; total

Z W

G3 (f ) df

The …rst and third terms may be integrated easily as inverse tangents. The second term may be integrated after partial fraction expansion: + (2 f )2

2 2 + (2

+ (2 f )

where

A 2 + (2

2 +

Therefore = = = = =

1 E3; total

Z W

1 E3; total 1 E3; total 1 E3; total 2 ( )

2 and B = 2

1 2 + (2

A 2

tan

2 W

tan 1

2 W

B + (2 f )2

A) tan 1 1

+ tan 1

2 + (2

A=2

E3 (jf j W ) E3; total

tan 1

2 W

tan

f) 1

2 W

+ 1 tan 1 2 W

2 W

B + + (2 f )2 #

B) tan 1

2 W

tan 1

2 W

tan 1

2 W

1 df + (2 f )2

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Pulse

Fraction of total energy 1 E1/E1,tot

x 1(t)

1 0.5 0 -2

0.5 0

αt E2/E2,tot

x 2(t)

0.5

τW 1

β = 2α

E3/E3,tot

x 3(t)

0.5

t/τ

0.5 0 -2

0 -2

4 W/ α

αt

0.5 0

2 W/ α

Plots of the signals and inband energy for all three cases are shown in Fig. 2.5

2.1. PROBLEM SOLUTIONS

Problem 2.38 a. By the modulation theorem X (f ) = =

AT0 4 AT0 4

sinc (f sinc

1 2

f0 ) f f0

T0 T0 + sinc (f + f0 ) 2 2 1 f 1 + sinc +1 2 f0

b. Use the superposition and modulation theorems to get X (f ) =

AT0 4

sinc

f 2f0

1 1 sinc 2 2

f f0

+ sinc

1 2

f +2 f0

c. In this case, p(t) = x(t) and P (f ) = X(f ) of part (a) and Ts = T0 : From part (a), we have AT0 n 1 n+1 P (nf0 ) = sinc + sinc 4 2 2 Using this in (2.149), we have the Fourier transform of the half-wave recti…ed cosine waveform as 1 X n 1 n+1 A X(f ) = sinc + sinc (f nf0 ) 4 2 2 n= 1 Note that sinc(x) = 0 for integer values of its argument and it is 1 for its argument 0. Also, use sinc(1=2) = 2= ; sinc(3=2) = 2=3 , etc. to get X (f ) =

(f ) +

A [ (f 4

A [ (f 15

f0 ) + (f + f0 )] +

A [ (f 3

2f0 ) + (f + 2f0 )]

4f0 ) + (f + 4f0 )] +

Problem 2.39 Signals x1 (t), x2 (t), and x6 (t) are real and even. Therefore their Fourier transforms are real and even. Signals x3 (t), x4 (t), and x5 (t) are real and odd. Therefore their Fourier transforms are imaginary and odd. Using the Fourier transforms for a square pulse and a triangle along with superposition, time delay, and scaling, the Fourier transforms of these signals are the following. a. X1 (f ) = 2sinc2 (2f ) + 2sinc(2f ) b. X2 (f ) = 2sinc(2f ) sinc2 (f )

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS c. X3 (f ) = sinc(f ) ej f sinc(f ) e j f = 2j sin ( f )sinc(f ) d. X4 (f ) = sinc2 (f ) e j2 f sinc2 (f ) ej2 f =

2j sin (2 f )sinc2 (f )

e. By duality sgn(t) ! j= ( f ) so, by the convolution theorem of Fourier transforms, X5 (f ) =sinc2 (f ) j= ( f ). The convolution cannot be carried out in closed form, but it is clear that the result is imaginary and odd. f. By the modulation theorem X6 (f ) = 12 sinc2 (f even.

1) + 21 sinc2 (f + 1) which is real and

Problem 2.40 a. Write 6 cos (20 t) + 3 sin (20 t) = R cos (20 t

)

= R cos cos (20 t) + R sin sin (20 t) Thus R cos

= 6 cos (20 t)

R sin

= 3 sin (20 t)

Square both equations, add, and take the square root to obtain p p R = 62 + 32 = 45 = 6:7082

Divide the second equation by the …rst to obtain tan

= 0:5 = 0:4636 rad

So x (t) = 3 +

45 cos (20 t

0:4636)

Following Example 2.19, we obtain Rx ( ) = 32 +

45 cos (20 2

)

b. Taking the Fourier transform of Rx ( ) we obtain Sx (f ) = 9 (f ) +

45 [ (f 4

10) + (f + 10)]

2.1. PROBLEM SOLUTIONS

Problem 2.41 Use the facts that the power spectral density integrates to give total power, it must be even, and contains no phase information. a. The total power of this signal is 22 =2 = 2 watts which is distributed equally at the frequencies 10 hertz. Therefore, by inspection we write S1 (f ) = (f

10) + (f + 10) W/Hz

b. The total power of this signal is 32 =2 = 4:5 watts which is distributed equally at the frequencies 15 hertz. Therefore, by inspection we write S2 (f ) = 2:25 (f

15) + 2:25 (f + 15) W/Hz

c. The total power of this signal is 52 =2 = 12:5 watts which is distributed equally at the frequencies 5 hertz. Therefore, by inspection we write S3 (f ) = 6:25 (f

5) + 6:25 (f + 5) W/Hz

d. The power of the …rst component of this signal is 32 =2 = 4:5 watts which is distributed equally at the frequencies 15 hertz. The power of the second component of this signal is 52 =2 = 12:5 watts which is distributed equally at the frequencies 5 hertz. Therefore, by inspection we write S4 (f ) = 2:25 (f

15) + 2:25 (f + 15) + 6:25 (f

5) + 6:25 (f + 5) W/Hz

Problem 2.42 Since the autocorrelation function and power spectral density of a signal are Fourier transform pairs, we may write down the answers by inspection using the Fourier transform pair A cos (2 f0 t) ! A2 (f f0 ) + A2 (f + f0 ). The answers are the following. a. R1 ( ) = 8 cos (30

); Average power = R1 (0) = 8 W.

b. R2 ( ) = 18 cos (40

); Average power = R2 (0) = 18 W.

c. R3 ( ) = 32 cos (10

); Average power = R3 (0) = 32 W.

d. R4 ( ) = 18 cos (40

) + 32 cos (10

); Average power = R4 (0) = 50 W.

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Problem 2.43 The autocorrelation function must be (1) even, (2) have an absolute maximum at and (3) have a Fourier transform that is real and nonnegative.

= 0,

a. Acceptable - all properties satis…ed; b. Acceptable - all properties satis…ed; c. Not acceptable - none of the properties satis…ed; d. Acceptable - all properties satis…ed; e. Not acceptable - property (3) not satis…ed; f. Not acceptable - none of the properties satis…ed.

Problem 2.44 2 Given that the autocorrelation function of x (t) = A cos (2 f0 t + ) is Rx ( ) = A2 cos (2 f0 ) (special case of Ex. 2.19), the results are as follows. a. R1 ( ) = 2 cos (10

);

b. R2 ( ) = 2 cos (10

);

c. R3 ( ) = 5 cos (10

) (write the signal as x3 (t) = Re 5 exp j tan 1 (4=3) exp (j10 t) );

d. R4 ( ) = 2 +2 cos (10 2

) = 4 cos (10

Problem 2.45 This is a matter of applying (2.151) by making the appropriate identi…cations with the parameters given in Example 2.20 Problem 2.46 Fourier transform both sides of the di¤erential equation using the di¤erentiation theorem of Fourier transforms to get [j2 f + a] Y (f ) = [j2 bf + c] X (f ) Therefore, the frequency response function is H (f ) =

Y (f ) c + j2 bf = X (f ) a + j2 f

2.1. PROBLEM SOLUTIONS

a = 1; b = 2; c = 0

a = 1; b = 0; c = 3

mag(H)

1.5 1

0.5

0 -5

angle(H), rad

0 -5

0 -1 -2 -5

0 f, Hz

0 -1 -2 -5

The amplitude response function is

and the phase response is

q c2 + (2 bf )2 jH (f )j = q a2 + (2 f )2

arg [H (f )] = tan 1

2 bf c

tan 1

2 f a

Amplitude and phase responses for various values of the constants are plotted in Figure 2.6. Problem 2.47 a. Use the transform pair Ae

u (t)

to …nd the unit impulse response as h1 (t) = e 7t u (t)

A + j2 f

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS b. Long division gives 7 7 + j2 f

H2 (f ) = 1 Use the transform pair (t) (a) to get

! 1 along with superposition and the transform pair in part h2 (t) = (t)

7e 7t u (t)

c. Use the time delay theorem along with the result of part (a) to get h3 (t) = e 7(t 3) u (t

d. Use superposition and the results of parts (a) and (c) to get h4 (t) = e 7t u (t)

e 7(t 3) u (t

Problem 2.48 Use the transform pair for a sinc function to …nd that Y (f ) =

f 2B

f 2W

a. If W < B, it follows that Y (f ) = because

f 2B

= 1 throughout the region where

f 2W f 2W

is nonzero.

b. If W > B, it follows that Y (f ) = because

f 2W

= 1 throughout the region where

f 2B f 2B

is nonzero.

c. Part (b) gives distortion because the output spectrum di¤ers from the input spectrum. In fact, the output is y (t) = 2Bsinc (2Bt) which clearly di¤ers from the input for W > B.

2.1. PROBLEM SOLUTIONS

Problem 2.49 a. Replace the capacitors with 1=j!C which is their ac-equivalent impedance. Call the junction of the input resistor, feedback resistor, and capacitors 1. Call the junction at the positive input of the operational ampli…er 2. Call the junction at the negative input of the operational ampli…er 3. Write down the KCL equations at these three junctions. Use the constraint equation for the operational ampli…er, which is V2 = V3 , and the de…nitions for ! 0 , Q, and K to get the given transfer function as H (j!) = Vo (j!) =Vi (j!). The node equations are V1

Vi R

+ j!CV1 +

Vo R

+ j!C(V1

j!C(V2

V2 ) = 0

V1 ) +

V2 R Vo

= 0

V3 V3 + = 0 Rb Ra (constraint on op amp input) V2 = V3 b. See plot given in Figure 2.7. c. In terms of f , the transfer function magnitude is K jH(f )j = p s 2

(f =f0 ) 1

f f0

2 2

+ Q12

f f0

It can be shown that p the maximim of jH(f )j is at f = f0 . By substitution, this maximum is jH(f0 )j = KQ= p2. To …nd the 3-dB bandwidth, we must …nd the frequencies for which jH(f )j = jH(f0 )j= 2. This results in Q p =s 2

(f3 =f0 ) 1

f3 f0

2 2

+ Q12

f3 f0

which reduces to the quadratic equation f3 f0

1 Q2

f3 f0

+1=0

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Using the quadratic formula, the solutions to this equation are f 3 f0

f 3 f0

1 2

1 2+ 2 Q

1 1 2+ 2 = 2 Q r 1 1 1 Q

1 Q

2+ r

1 Q2

1 4Q2

1 Q

f0 =

f0 Q

1 ; Q >> 1 2Q

Therefore, the 3-dB bandwidth is, for large Q, approximately B = f3

f 3

1 2Q

d. Combinations of components giving RC = 2:2508

10 4 seconds

and Ra = 2:5757 Rb will work.

Problem 2.42 a. By voltage division, with the inductor replaced by j2 f L, the frequency response function is R2 + j2 f L R2 =L + j2 f H1 (f ) = = R1 + R2 + j2 f L (R1 + R2 ) =L + j2 f By long division H1 (f ) = 1

R1 =L R1 +R2 + j2 L

Using the transforms of a delta function and a one-sided exponential, we obtain h1 (t) = (t)

R1 exp L

R1 + R2 t u (t) L

2.1. PROBLEM SOLUTIONS

|H(f)|, dB

10 0 -10 f0 = 999.9959 Hz; B 3 dB = 300.0242 Hz -20 2 10

angle(H(f)), radians

2 1 0 -1 -2 2 10

10 f, Hz

b. Substituting the ac-equivalent impedance for the inductor and using voltage division, the frequency response function is H2 (f ) = =

R2 jj (j2 f L) j2 f LR2 where R2 jj (j2 f L) = R1 + R2 jj (j2 f L) R2 + j2 f L j2 f L (R1 k R2 ) =L R2 R2 1 = R2 R1 + R2 RR1+R R + R (R + j2 f L 1 2 1 k R2 ) =L + j2 f 1 2

R2 where R1 k R2 = RR11+R . Therefore, the impulse response is 2

h2 (t) =

R2 R1 + R2

(t)

R 1 R2 exp (R1 + R2 ) L

R1 R2 t u (t) (R1 + R2 ) L

2 Both have a high pass amplitude response, with the dc gain of the …rst circuit being R1R+R 2 and the second being 0; the high frequency gain of the …rst is 1 and that of the second is R2 R1 +R2 .

Problem 2.51 Application of the Payley-Wiener criterion gives the integral Z 1 f2 I= df 2 1 1+f

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

which does not converge. Hence, the given function is not suitable as the frequency response function of a causal LTI system. Problem 2.52 a. The condition for stability is Z 1 Z 1 jexp ( jh1 (t)j dt = 1 Z11 exp ( = 2 0

which follows because jcos (2 f0 t)j

jtj) cos (2 f0 t)j dt t) jcos (2 f0 t)j dt < 2

Z 1

exp (

t) dt =

1. Hence this system is BIBO stable.

b. The condition for stability is Z 1 Z 1 jh2 (t)j dt = jcos (2 f0 t) u (t)j dt 1 1 Z 1 = jcos (2 f0 t)j dt ! 1 0

which follows by integrating one period of jcos (2 f0 t)j and noting that the total integral is the limit of one period of area times N as N ! 1: This system is not BIBO stable. c. The condition for stability is Z 1 Z 1 1 jh3 (t)j dt = u (t 1) dt 1 1 jtj Z 1 dt = ln (t)j1 = 1 !1 t 1 This system is not BIBO stable. d. The condition for stability is Z 1 Z 1 jh4 (t)j dt = je t u (t) e ((t 1)) u (t 1) jdt 1 1 Z 1 e t e (t 1) dt = 1 + e < 1 0

This system is BIBO stable.

2.1. PROBLEM SOLUTIONS

e. The condition for stability is Z 1

jh5 (t)j dt =

Z 1 1

t 2 jdt = 1 < 1

This system is BIBO stable. f. The condition for stability is Z 1

jh6 (t)j dt =

Z 1

sinc (2t) dt =

1 <1 2

This system is BIBO stable. Problem 2.53 The energy spectral density of the output is Gy (f ) = jH (f )j2 jX (f )j2 where jH (f )j2 = X (f ) =

25 16 + (2 f )2 1 1 ; Gx (f ) = jX (f ) j2 = 3 + j2 f 9 + (2 f )2

Hence Gy (f ) = h

25 ih i 9 + (2 f )2 16 + (2 f )2

Plots of the input and output energy spectral densities are left to the student. Problem 2.54 Using the Fourier coe¢ cients of a half-recti…ed sine wave from Table 2.1 and noting that those of a half-recti…ed cosine wave are related by Xcn = Xsn e jn =2

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

The fundamental frequency is 10 Hz. The ideal rectangular …lter passes all frequencies less than 31 Hz and rejects all frequencies greater than 31 Hz. Therefore

y (t) = = = =

A j jA j =2 j20 t A jA j =2 j20 t A j j40 t e e j40 t + e e + + e e e e 3 4 3 ! 4 ! A A ej(20 t =2) + e j(20 t =2) 2A ej(40 t ) + e j(40 t ) 2 2j 3 2 A 2A sin (20 t =2) cos (40 t 2 3 A A 2A + cos (20 t) + cos (40 t) 2 3 A

)

2.1. PROBLEM SOLUTIONS

Problem 2.55 a. The energy spectral density is given by 2 1 = + j2 f

G1 (f ) = jX1 (f ) j2 =

1 2 + (2

f )2

Since Etotal = 21 , the 90% energy containment bandwidth is given by Z B90 Z B90 1 2 1 2 f 0:9 df = 2 = 2 2 df; u = 2 + (2 f )2 2 0 0 1+ 2 f Z 2 B90 = du 1 1 = tan 1 (2 B90 = ) = 2 1 + u 0 or B90 =

tan (0:45 ) = 1:0049

b. For this case, using X2 (f ) =

(f =2W ) and Etotal = 2W; we obtain Z B90 0:9 (2W ) = 2 df = 2B90 0

or B90 = 0:9W

c. For this case, using X3 (f ) = sinc(f ) and E3 = , we obtain Z B90 2 0:9 = 2 sinc2 (f ) df 0 Z B90 or 0:45 = sinc2 (u) du 0

Numerical integration gives B90 = 0:9= d. For this case, using X4 (f ) = obtain

R sinc2 (f ) and E4 = 2 0 (1 Z B90

2 1:8 =3 = 2 sinc4 (f ) df 0 Z B90 or 0:3 = sinc4 (u) du 0

Numerical integration gives B90 = 0:35=

t= )2 dt = 2 =3, we

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS e. For this case, using X5 (f ) =

0:9=

2 2 +(2

f )2

= 2

R1 and E5 = 2 0 exp ( 2 t) dt = 1= , we obtain

Z B90 0

or 0:225

Z 2 B90 = 0

2 2 df 2 + (2 f )2

du (1 + u2 )2

; u = 2 f=

Numerical integration gives 2 B90 = = 1:18 or B90 = = 0:188 or B90 = 0:188 : Problem 2.56 The outputs are the inputs phase shifted by =2 radians for frequencies greater than 0 and =2 for frequencies less than 0 Hz. a. y1 (t) = exp [j100 t

j =2] =

j exp (j100 t);

b. y2 (t) = 21 exp [j100 t

j =2] + 12 exp [ j100 t + j =2] = sin (100 t);

1 exp [j100 t c. y3 (t) = 2j

j =2]

1 2j exp [

j100 t + j =2] =

cos (100 t);

d. The spectrum of the input is X4 (f ) = 2sinc(2f ). The spectrum of the output is Y4 (f ) =

2sinc (2f ) exp ( j =2) ; f > 0 2sinc (2f ) exp (j =2) ; f < 0

The time domain output signal is the inverse Fourier transform of this, which is y4 (t) =

Z 1 0

2sinc (2f ) exp ( j =2) exp (j2 f t) df + Z 1 0

sinc (2f ) exp (j2 f t) df + 2j

Z 0

2sinc (2f ) exp (j =2) exp (j2 f t) df

sinc ( 2u) exp ( j2 ut) ( du) ; u =

Z11

Z 1 2j sinc (2f ) exp (j2 f t) df + 2j sinc (2u) exp ( j2 ut) du; sinc2u is even 0 0 Z 1 = 2 sinc (2f ) [ j exp (j2 f t) + j exp ( j2 f t)] df ; rewrite 2nd int in terms of f Z0 1 = 4 sinc (2f ) sin (2 f t) df =

This requires numerical integration. A plot is given in Fig. 2.8.

2.1. PROBLEM SOLUTIONS

2 1.5 1

x 4(t), y 4(t)

0.5 0 -0.5 -1 -1.5 -2 -3

-2

-1

0 t, s

Problem 2.57 a. Amplitude distortion; no phase distortion. The output for (a) is ya (t) = 4 cos 48 t = 4 cos 48

+ 10 cos 126 t 150 1 + 10 cos 126 t 300

150 1 300

b. No amplitude distortion; phase distortion. The output for (b) is yb (t) = 2 cos 126 t = 2 cos 126

+ cos (170 t) 150 1 + cos (170 t) 300

c. No amplitude distortion; no phase distortion.

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

The output for (c) is yc (t) = 2 cos 126

1 300

+ 6 cos 144

1 300

d. No amplitude distortion; no phase distortion. The output for (d) is yd (t) = 4 cos 10

1 300

+ 16 cos 50

Problem 2.58 a. The frequency response function corresponding to this impulse response is H1 (f ) = The group delay is

3 =q 5 + j2 f

exp

25 + (2 f )2

Tg1 (f ) = =

1 d tan 1 2 df 1 1 2 2 5 2 1 + 2 5f

j tan 1

2 f 5

5 25 + (2 f )2

The phase delay is Tp1 (f ) =

1 (f )

2 f

tan 1 =

2 f 5

2 f

b. The frequency response function corresponding to this impulse response is 5 2 H2 (f ) = 3 + j2 f 5 + j2 f 5 (5 + j2 f ) 2 (3 + j2 f ) = (3 + j2 f ) (5 + j2 f ) 19 + j6 f = 15 (2 f )2 + j16 f q h i 361 + (6 f )2 exp j tan 1 619f = rh h i i2 15 (2 f )2 + (16 f )2 exp j tan 1 15 16(2 ff )2

2.1. PROBLEM SOLUTIONS

Therefore 2 (f ) = tan

6 f 19

16 f 15 (2 f )2

tan 1

The group delay is Tg2 (f ) =

1 d tan 1 2 df 8 > 1 < 2 > : 8 > 1 <

6 f 19

16 f 15 (2 f )2

tan 1 1 6 f 19

6 19

1+( ) 16 (15 (2 f )2 ) 16 f [ 2(2 f )(2 )]

1 2

16 f 15 (2 f )2

[15 (2 f )2 ] 16

9 > =

15 + (2 f ) i2 > 15 (2 f )2 + (16 f )2 ; h i 8 15 + (2 f )2

19 (6 ) 2 2 > : 361 + (6 f )

57 + 361 + (6 f )2 225 + 34 (2 f )2 + (2 f )4

The phase delay is Tp2 (f ) =

2 (f )

2 f

tan 1 =

16 f 15 (2 f )2

tan 1

6 f 19

2 f

The group and phase delays for (a) and (b) are shown in Fig. 2.9. c. The frequency response is H (f ) =

f 2B

exp ( j2 t0 f )

The group delay is 1 d [ 2 t0 f ] ; B f B 2 df = t0 ; B f B and 0 otherwise

Tg (f ) =

The phase delay is

Tp (f ) =

( 2 t0 f ) = t0 ; 2 f

B and 0 otherwise

9 > = > ;

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Tg1(f), Tp1(f), s

0.2 Group delay Phase delay

0.15 0.1 0.05 0 -2 10

-1

10 f, Hz

Tg2(f), Tp2(f), s

0.5

-0.5

-1 -2 10

-1

10 f, Hz

d. The frequency response function is H (f ) = =

5 2 exp ( j2 t0 f ) 3 + j2 f 3 + j2 f 2 [2:5 exp ( j2 t0 f )] 3 + j2 f

The phase shift function is (f ) =

sin 2 t0 f 2:5 cos 2 t0 f

tan 1

2 f 3

The phase delay is Tp (f ) =

2 f sin 2 t0 f 1 tan 1 = 2 f 3 2:5 cos 2 t0 f

The group delay is Tg (f ) = = Problem 2.59

1 d 2 f sin 2 t0 f tan 1 = 2 df 3 2:5 cos 2 t0 f 3 1 2:5 cos 2 t0 f 2 + 9 + (2 f ) (2:5 cos 2 t0 f )2

2.1. PROBLEM SOLUTIONS

a. The amplitude response is 2 jf j q 64 + (2 f )2 9 + (2 f )2

jH (f ) j = r b. The phase response is (f ) =

f 4

tan 1

+ tan 1

2 f 3

tan 1

sgn (f )

2 f 3

c. The phase delay is Tp (f ) =

1 tan 1 2 f

f 4

sgn (f )

d. The group delay is Tg (f ) =

1=8 1=3 2 + 1 + ( f =4) 1 + ( f =3)2

1 (f ) 4

Problem 2.60 In terms of the input spectrum, the output spectrum is Y (f ) = X (f ) + 0:1X (f ) X (f ) f 10 f + 10 = 2 + 4 4 f + 10 f 10 + +0:4 4 4 f 10 f + 10 = 2 + 4 4 f 20 f +0:4 4 +8 +4 4 4

10 4

f + 10 4

f + 20 4

where (f ) is an isoceles triangle of unit height going from -1 to 1. The student should sketch the output spectrum given the above analytical result. Problem 2.61 a. The amplitude response is 2 jf j

jH(f )j = q

4 2 f 2 )2 + (0:3 f )2

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

|H(f)|

6 4 2 0 -3

-2

-1

-2

-1

0 f, Hz

θ(f), radians

2 1 0 -1 -2 -3

b. The phase response is (f ) =

sgn (f )

tan 1

0:3 f 9 4 2f 2

These are shown in Fig. 2.10. c. The phase delay is Tp (f ) =

1 2

0:3 f 9 4 2f 2

sgn (f )

tan 1

0:3 f 9 4 2f 2

d. The group delay is Tg (f ) =

1 d 2 df 2 1 6 4 2 81

sgn (f )

0:3

(f ) 1+

0:3 f 9 4 2f 2

1:35 + 0:6 2 f 2 71:91 2 f 2 + 16 4 f 4

1 (f ) 2

2f 2

+ (0:3 f ) 8 4

2 f 2 )2

3 7 5

2.1. PROBLEM SOLUTIONS

Problem 2.62 Let u = 2 t. We then have y (t) = [cos (u) + cos (3u)]3 = cos3 (u) + 3 cos2 (u) cos (3u) + 3 cos (u) cos2 (3u) + cos3 (3u) Use the trig identities cos2 (z) = cos3 (z) = cos (w) cos (z) =

1 [1 + cos (2z)] 2 3 1 cos (z) + cos (3z) 4 4 1 1 cos (w z) + cos (w + z) 2 2

to get 1 3 3 cos (u) + cos (3u) + [1 + cos (2u)] cos (3u) 4 4 2 3 3 1 + cos (u) [1 + cos (6u)] + cos (3u) + cos (9u) 2 4 4 5 3 3 1 = 3 cos (u) + cos (3u) + cos (5u) + cos (7u) + cos (9u) 2 2 4 4 3 3 1 5 = 3 cos (2 t) + cos (6 t) + cos (10 t) + cos (14 t) + cos (18 t) 2 2 4 4

y (t) =

Problem 2.63 Write the transfer function as H (f ) = H0 e j2 f t0

f 2B

e j2 f t0

Use the inverse Fourier transform of a constant, the delay theorem, and the inverse Fourier transform of a rectangular pulse function to get h (t) = H0 (t

t0 )

2BH0 sinc [2B (t

Problem 2.64 a. The Fourier transform of this signal is p X (f ) = A 2 b2 exp

2 2 2f 2

t0 )]

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

By de…nition, using a table of integrals, Z 1 p 1 T = jx (t)j dt = 2 x (0) 1 Similarly, 1 W = 2X (0) Therefore,

Z 1

1 jX (f )j df = p 2 2 1 p

2 2W T = p 2 2

b. The Fourier transform of this signal is X (f ) = The pulse duration is 1 T = x (0) The bandwidth is W = Thus,

Z 1

1 2X (0)

2A= 1 + (2 f = )2

Z 1

2W T = 2

jx (t)j dt =

jX (f )j df = 2

Problem 2.65 a. The poles for a second order Butterworth …lter are given by !3 s1 = s2 = p (1 j) 2 where ! 3 is the 3-dB cuto¤ frequency of the Butterworth …lter. function is H (s) = h

!3 s+ p (1 2

Its s-domain transfer

!2 !2 i h3 i= p 3 !3 s2 + 2! 3 s + ! 23 j) s + p (1 + j) 2

Letting ! 3 = 2 f3 and s = j! = j2 f , we obtain H (j2 f ) =

4 2f 2 +

4 2 f32 = 2 (2 f3 ) (j2 f ) + 4 2 f32

f2 p3 f 2 + j 2f3 f + f32

2.1. PROBLEM SOLUTIONS

b. If the phase response function of the …lter is (f ), the group delay is Tg (f ) =

1 d [ (f )] 2 df

For the second-order Butterworth …lter considered here, ! p 2f f 3 (f ) = tan 1 f32 f 2 Therefore, the group delay is Tg (f ) = =

" 1 d tan 1 2 df

2f3 f 2 f3 f 2

f f2 + f2 1 1 + (f =f3 )2 p 3 34 p = 2 f3 + f 4 2 f3 1 + (f =f3 )4

This is plotted in Fig. 2.11. c. Use partial fraction expansion of H (s) =s and then inverse Laplace transform it to get the given step response. The expansion is !2 A Bs + C p 3 p = + 2 2 2 s s s + 2! 3 s + ! 3 s + 2! 3 s + ! 23 p where A = 1; B = 1; C = 2! 3 H (s) s

This allows H (s) =s to be written as H (s) s

1 s

p s + 2! 3 p s2 + 2! 3 s + ! 23 p s + 2! 3 p s2 + 2! 3 s + ! 23 =2 ! 23 =2 + ! 23 p s + 2! 3 p 2 s + ! 3 = 2 + ! 23 =2 p p p s + ! 3 = 2 ! 3 = 2 + 2! 3 p 2 s + ! 3 = 2 + ! 23 =2 p p s + !3= 2 !3= 2 p 2 p 2 s + ! 3 = 2 + ! 23 =2 s + ! 3 = 2 + ! 23 =2

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Step response for a 2nd-order BW filter 1.4

0.3

1.2

0.25

0.2

0.8 y s (t)

f3Tg(f)

Group delay for a 2nd-order BW filter 0.35

0.15

0.6

0.1

0.4

0.05

0.2

0 -1 10

10 f/f3

0.5 f3t

Using the s-shift theorem of Laplace transforms, this inverse transforms to the given expression for the step response (with ! 3 = 2 f3 ). Plot it and estimate the 10% and 90% times from the plot. From the MATLAB plot of Fig. 2.10, f3 t10% 0:08 and f3 t90% 0:42 so that the 10-90 % rise time is about 0:34=f3 seconds. Problem 2.66 a. Slightly less than 0.5 seconds; b & c. Use sketches to show. Problem 2.67 a. The leading edges of the ‡at-top samples follow the waveform at the sampling instants. b. The spectrum is Y (f ) = X (f ) H (f ) where X (f ) = fs

1 X

n= 1

X (f

nfs )

2.1. PROBLEM SOLUTIONS

and H (f ) = sinc (f ) exp ( j f ) The latter represents the frequency response of a …lter whose impulse response is a square pulse of width and implements ‡at top sampling. If W is the bandwidth of X (f ), very little distortion will result if 1 >> W . Problem 2.68 a. The sampling frequency should be large compared with the bandwidth of the signal. b. The output spectrum of the zero-order hold circuit is

Y (f ) = sinc (Ts f )

1 X

X (f

nfs ) exp ( j f Ts )

n= 1

where fs = Ts 1 . For small distortion, we want Ts << W

Problem 2.69 Use trig identities to rewrite the signal as a sum of sinusoids: x (t) = 10 cos2 (600 t) cos (2400 t) = 5 [1 + cos (600 t)] cos (2400 t) = 5 cos (2400 t) + 2:5 cos (1800 t) + 2:5 cos (3000 t) The lowpass recovery …lter can cut o¤ in the range 1.5+ kHz to 3 kHz where the superscript + means just above and the superscript means just below. The lower of these is the highest frequency of x (t) and the larger is equal to the sampling frequency minus the highest frequency of x (t). The minimum allowable sampling frequency is just above 3 kHz. Problem 2.70 For bandpass sampling and recovery, all but (b) and (e) will work theoretically, although an ideal …lter with bandwidth exactly equal to the unsampled signal bandwidth is necessary. For lowpass sampling and recovery, only (f) will work.

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Problem 2.71 The Fourier transform is 1 X (f 2

1 f0 ) + X (f + f0 ) 2 1 1 + [ jsgn (f ) X (f )] (f f0 ) e j =2 + (f + f0 ) ej =2 2 2 1 1 = X (f f0 ) [1 sgn (f f0 )] + X (f + f0 ) [1 + sgn (f + f0 )] 2 2

Y (f ) =

Noting that

and

1 [1 sgn (f f0 )] = u (f0 f ) 2 1 [1 + sgn (f + f0 )] = u (f + f0 ) 2

this may be rewritten as Y (f ) = X (f

f0 ) u (f0

f ) + X (f + f0 ) u (f + f0 )

f Thus, if X (f ) = 2 (a unit-height rectangle 2 units wide centered at f = 0) and f0 = 10 Hz, Y (f ) would consist of unit-height rectangles going from 10 to 9 Hz and from 9 to 10 Hz.

Problem 2.72 a. x ba (t) = cos (! 0 t

1 T !1 2T lim

=2) = sin (! 0 t), so Z T

x (t) x b (t) dt = = =

1 T !1 2T lim

Z T

sin (! 0 t) cos (! 0 t) dt

T Z T

1 sin (2! 0 t) dt T 2

1 cos (2! 0 t) T =0 T !1 2T 4! 0 T lim

b. Use trigonometric identities to express x (t) in terms of sines and cosines. Then …nd the Hilbert transform of x (t) by phase shifting by =2. Multiply x (t) and x b (t) together term by term, use trigonometric identities for the product of sines and cosines, then integrate. The integrand will be a sum of terms similar to that of part (a). The limit as T ! 1 will be zero term-by-term.

2.1. PROBLEM SOLUTIONS

c. For this case x b (t) = A exp (j! 0 t j =2) = jA exp (j! 0 t), take the product, integrate over time to get Z T Z T 1 1 x (t) x b (t) dt = lim [A exp (j! 0 t)] [jA exp (j! 0 t)] dt lim T !1 2T T !1 2T T T Z jA2 T = lim exp (j2! 0 t) dt = 0 T !1 2T T by periodicity of the integrand Problem 2.73 a. Note that F [jb x(t)] = j [ jsgn (f )] X (f ). Hence x1 (t) = = =

2 1 2 1 x (t) + jb x (t) ! X1 (f ) = X (f ) + j [ jsgn (f )] X (f ) 3 3 3 3 2 1 + sgn (f ) X (f ) 3 3 1 3 X (f ) ; f < 0 X (f ) ; f > 0

b. It follows that x2 (t)

3 3 x (t) + jb x (t) exp (j2 f0 t) 4 4 3 ) X2 (f ) = [1 + sgn (f f0 )] X (f 4 0; f < f0 = 3 f0 ) ; f > f0 2 X (f =

f0 )

c. This case has the same spectrum as part (a), except that it is shifted right by W Hz. That is, 2 1 x (t) + jb x (t) exp (j2 W t) 3 3 2 1 ! X3 (f ) = + sgn (f W ) X (f 3 3

x3 (t)

d. For this signal x4 (t)

1 jb x (t) exp (j W t) 3 2 1 ! X4 (f ) = sgn (f W=2) X (f 3 3 =

2 x (t) 3

W=2)

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Problem 2.74 The Hilbert transform of the given signal is x b(t) = 2 sin (52 t)

The signal xp (t) is

xp (t) = x (t) + jb x(t) = 2 cos (52 t) + j2 sin (52 t) = 2ej52 t a. We have, for f0 = 25 Hz, x ~ (t) = xp (t) e j2 f0 t = 2ej52 t e j50 t = 2ej2 t = 2 cos (2 t) + j2 sin (2 t)

xR (t) = 2 cos (2 t) xI (t) = 2 sin (2 t) b. For f0 = 27 Hz, x ~ (t) = 2ej52 t e j54 t = 2e j2 t = 2 cos (2 t)

j2 sin (2 t)

xR (t) = 2 cos (2 t) xI (t) =

2 sin (2 t)

c. For f0 = 10 Hz, x ~ (t) = 2ej52 t e j20 t = 2ej32 t = 2 cos (32 t) + j2 sin (32 t)

xR (t) = 2 cos (32 t) xI (t) = 2 sin (32 t) d. For f0 = 15 Hz, x ~ (t) = 2ej52 t e j30 t = 2ej22 t = 2 cos (22 t) + j2 sin (22 t)

2.1. PROBLEM SOLUTIONS

xR (t) = 2 cos (22 t) xI (t) = 2 sin (22 t) e. For f0 = 30 Hz, x ~ (t) = 2ej52 t e j60 t = 2e j8 t = 2 cos (8 t)

j2 sin (8 t)

xR (t) = 2 cos (8 t) xI (t) =

2 sin (8 t)

f. For f0 = 20 Hz, x ~ (t) = 2ej52 t e j40 t = 2ej12 t = 2 cos (12 t) + j2 sin (12 t)

xR (t) = 2 cos (12 t) xI (t) = 2 sin (12 t)

Problem 2.75 For t < =2, the output is zero. For jtj y (t) =

=2, the result is

f )2

2 + (2

n cos [2 (f0 +

f) t

]

(t+ =2)

cos [2 (f0 +

f) t

]

tan 1

cos [2 (f0 +

For t > =2, the result is y (t) =

( =2) e

2 + (2

f )2

n e

In the above equations,

is given by =

cos [2 (f0 +

o f) t + ]

2.2

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

Computer Exercises

Computer Exercise 2.1 % ce2_1.m: Amplitude spectra and Fourier series synthesized % for various periodic waveforms % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clear all; clf N = input(’Number of harmonics in Fourier sum => ’); T = input(’Period of periodic waveform => ’); A = input(’Amplitude of waveform => ’); n = -N:1:N; I_type = input(’1 = rect pulse train; 2 = HR sinewave; 3 = FR sinewave; 4 = triang pulse train; 5 = triangle wave; 6 = sawtooth wave => ’); X = zeros(size(n)); if I_type == 1 tau = input(’Width of rectangular pulse => ’); t0 = input(’Delay of rectangular pulse center => ’); d = tau/T; X = A*d*sinc(n*d).*exp(-j*2*pi*n*t0/T); elseif I_type == 2 I = …nd(n == 1 j n == -1); II = …nd(rem(n, 2) == 0); III = …nd(rem(abs(n), 2) == 1 & (n ~= 1 & n ~= -1)); X(I) = -0.25*j*n(I)*A; X(II) = A./(pi*(1-n(II).*n(II))); X(III) = 0; elseif I_type == 3 X = 2*A./(pi*(1-4*n.*n)); elseif I_type == 4 tau = input(’Half width of triangle pulse => ’); t0 = input(’Delay of triangle pulse => ’); d = tau/T; X = A*d*sinc(n*d).*sinc(n*d).*exp(-j*2*pi*n*t0/T); elseif I_type == 5 I = …nd(rem(abs(n), 2) == 1); X(I) = 4*A./(pi^2*n(I).^2); elseif I_type == 6

2.2. COMPUTER EXERCISES

I = …nd(n~=0); X(I) = 2*A*(-exp(-j*2*pi*n(I))+(1 - exp(-j*2*pi*n(I)))./(j*2*pi*n(I)))./(j*2*pi*n(I)); end subplot(2,1,1), stem(n, abs(X)),xlabel(’n’), ylabel(’jX_nj’), ... if I_type == 1 title([’Rectangular pulse train; period = ’, num2str(T), ’; delay = ’, num2str(t0), ’; ampli = ’, num2str(A)]) elseif I_type == 2 title([’Half-recti…ed sinewave; period = ’, num2str(T),’; ampli = ’, num2str(A)]) elseif I_type == 3 title([’Full-recti…ed sinewave; period = ’, num2str(T),’; ampli = ’, num2str(A)]) elseif I_type == 4 title([’Triangle pulse train; ’, num2str(2*N+1), ’terms. A = ’, num2str(A), ’, ntau = ’, num2str(tau), ’, T = ’, num2str(T) ’s; d = ’, num2str(d), ’; t_0 = ’, num2str(t0), ’s’]) elseif I_type == 5 title([’Triangle waveform; period = ’, num2str(T),’; ampli = ’, num2str(A)]) elseif I_type == 6 title([’Sawtooth waveform; period = ’, num2str(T),’; ampli = ’, num2str(A)]) end fn = n./T; t = -T:T/500:T; x = real(X*exp(j*2*pi*fn’*t)); subplot(2,1,2), plot(t, x), xlabel(’t’), ylabel(’x(t)’), ... >> ce2_1 Number of harmonics in Fourier sum => 25 Period of periodic waveform => 2 Amplitude of waveform => 1 1 = rect pulse train; 2 = HR sinewave; 3 = FR sinewave; 4 = triang pulse train; 5 = triangle wave; 6 = sawtooth wave => 6 >> Computer Exercise 2.2 % ce2_2.m: Plot of line spectra for half-recti…ed, % full-recti…ed sinewave, square wave, and triangle wave % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clear all; clf

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Sawtooth waveform; period = 2 ; ampli = 1 0.4

|Xn|

0.3 0.2 0.1 0 -25

-20

-15

-10

-5

0 n

-0.5

0 t

0.5

1.5

x(t)

1 0 -1 -2 -2

-1.5

-1

waveform = input(’Enter type of waveform: 1 = HR sine; 2 = FR sine; 3 = square; 4 = triangle: ’); A = 1; n_max = 13; % maximum harmonic plotted; odd n = -n_max:1:n_max; if waveform == 1 X = A./(pi*(1+eps - n.^2)); % O¤set 1 slightly to avoid divide by zero for m = 1:2:2*n_max+1 X(m) = 0; % Set odd harmonic lines to zero X(n_max+2) = -j*A/4; % Compute lines for n = 1 and n = -1 X(n_max) = j*A/4; end elseif waveform == 2 X = 2*A./(pi*(1+eps - 4*n.^2)); elseif waveform == 3 X = abs(4*A./(pi*n+eps)); for m = 2:2:2*n_max+1 X(m) = 0; end elseif waveform == 4 X = 4*A./(pi*n+eps).^2; for m = 2:2:2*n_max+1

2.2. COMPUTER EXERCISES

X(m) = 0; end end [arg_X, mag_X] = cart2pol(real(X),imag(X)); % Convert to magnitude and phase if waveform == 1 for m = n_max+3:2:2*n_max+1 arg_X(m) = arg_X(m) - 2*pi; % Force phase to be odd end elseif waveform == 2 m = …nd(n > 0); arg_X(m) = arg_X(m) - 2*pi; % Force phase to be odd elseif waveform == 4 arg_X = mod(arg_X, 2*pi); end subplot(2,1,1),stem(n, mag_X),ylabel(’jX_nj’) if waveform == 1 title(’Half-recti…ed sine wave spectra’) elseif waveform == 2 title(’Full-recti…ed sine wave spectra’) elseif waveform == 3 title(’Spectra for square wave with even symmetry ’) elseif waveform == 4 title(’Spectra for triangle wave with even symmetry’) end subplot(2,1,2),stem(n, arg_X),xlabel(’nf_0’),ylabel(’angle(X_n)’) >> ce2_2 Enter type of waveform: 1 = HR sine; 2 = FR sine; 3 = square; 4 = triangle: 1 >> Computer Exercise 2.3 % ce2_3.m: FFT plotting of line spectra for half-recti…ed, full-recti…ed sinewave, square wave, % and triangular waveforms % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % % User de…ned functions used: pls_fn( ); trgl_fn( ) % clf

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Half-rectified sine wave spectra 0.4

|Xn|

0.3 0.2 0.1 0 -15

-10

-5

-10

-5

0 nf0

angle(Xn)

2 0 -2 -4 -15

I_wave = input(‘Enter type of waveform: 1 = positive squarewave; 2 = 0-dc level triangular; 3 = half-rect. sine; 4 = full-wave sine: ’); T = 2; del_t = 0.001; t = 0:del_t:T; L = length(t); fs = (L-1)/T; del_f = 1/T; n = 0:9; if I_wave == 1 x = pls_fn(2*(t-T/4)/T); X_th = abs(0.5*sinc(n/2)); disp(‘’) disp(‘0 - 1 level squarewave’) elseif I_wave == 2 x = 2*trgl_fn(2*(t-T/2)/T)-1; X_th = 4./(pi^2*n.^2); X_th(1) = 0; % Set n = even coe¢ cients to zero (odd indexed because of MATLAB) X_th(3) = 0; X_th(5) = 0; X_th(7) = 0; X_th(9) = 0; disp(‘’)

2.2. COMPUTER EXERCISES

disp(‘0-dc level triangular wave’) elseif I_wave == 3 x = sin(2*pi*t/T).*pls_fn(2*(t-T/4)/T); X_th = abs(1./(pi*(1-n.^2))); X_th(2) = 0.25; X_th(4) = 0; % Set n = odd coe¢ cients to zero (even indexed because of MATLAB) X_th(6) = 0; X_th(8) = 0; X_th(10) = 0; disp(‘’) disp(‘Half-recti…ed sinewave’) elseif I_wave == 4 x = abs(sin(pi*t/T)); % Period of full-recti…ed sinewave is T/2 X_th = abs(2./(pi*(1-4*n.^2))); disp(‘’) disp(‘Full-recti…ed sinewave’) end X = 0.5*¤t(x)*del_t; % Multiply by 0.5 because of 1/T_0 with T_0 = 2 f = 0:del_f:fs; Y = abs(X(1:10)); Z = [n’Y’X_th’]; disp(‘Magnitude of the Fourier coe¢ cients’); disp(‘’) disp(‘n FFT Theory’); disp(‘__________________________’) disp(‘’) disp(Z); subplot(2,1,1),plot(t, x), xlabel(‘t’), ylabel(‘x(t)’) subplot(2,1,2),plot(f, abs(X),’o’),axis([0 10 0 1]),... xlabel(‘n’), ylabel(‘jX_nj’) % Unit-width pulse function % function y = pls_fn(t) y = stp_fn(t+0.5)-stp_fn(t-0.5); function y = stp_fn(t) % Function for generating the unit step % y = zeros(size(t));

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS I = …nd(t >= 0); y(I) = ones(size(I)); function y = trgl_fn(t) % This function generates a unit-high triangle centered % at zero and extending from -1 to 1 % y = (1 - abs(t)).*pls_fn(t/2); % % End of script …le

A typical run follows with a plot given in Fig. 2.13. >> ce2_3 Enter type of waveform: 1 = positive squarewave; 2 = 0-dc level triangular; 3 = half-rect. sine; 4 = full-wave sine: 3 Half-recti…ed sinewave Magnitude of the Fourier coe¢ cients n FFT Theory __________________________ 0 0.3183 0.3183 1.0000 0.2501 0.2500 2.0000 0.1062 0.1061 3.0000 0.0001 0 4.0000 0.0212 0.0212 5.0000 0.0001 0 6.0000 0.0091 0.0091 7.0000 0.0000 0 8.0000 0.0051 0.0051 9.0000 0.0000 0 Computer Exercise 2.4 Make the time window long compared with the pulse width. Computer Exercise 2.5 % ce2_5.m: Finding the energy ratio in a preset bandwidth % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % % User de…ned functions used: pls_fn( ); trgl_fn( )

2.2. COMPUTER EXERCISES

x(t)

0.5

0.2

0.4

0.6

0.8

1 t

1.2

1.4

1.6

1.8

5 n

|Xn|

0.5

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS

% I_wave = input(’Enter type of waveform: 1 = rectangular; 2 = triangular; 3 = half-rect. sine; 4 = raised cosine: ’); tau = input(’Enter pulse width: ’); per_cent = input(’Enter percent total desired energy: ’); clf T = 20; f = []; G = []; del_t = 0.001; t = 0:del_t:T; L = length(t); fs = (L-1)/T; del_f = 1/T; % n = [0 1 2 3 4 5 6 7 8 9]; if I_wave == 1 x = pls_fn((t-tau/2)/tau); disp(’’) disp(’Rectangular pulse’) elseif I_wave == 2 x = trgl_fn(2*(t-tau/2)/tau); disp(’’) disp(’Triangular pulse’) elseif I_wave == 3 x = sin(pi*t/tau).*pls_fn((t-tau/2)/tau); disp(’’) disp(’Half sinewave’) elseif I_wave == 4 x = abs(sin(pi*t/tau).^2.*pls_fn((t-tau/2)/tau)); disp(’’) disp(’Raised sinewave’) end X = ¤t(x)*del_t; f1 = 0:del_f*tau:fs*tau; G1 = X.*conj(X); NN = ‡oor(length(G1)/2); G = G1(1:NN); ¤ = f1(1:NN); f = f1(1:NN+1); E_tot = sum(G);

2.2. COMPUTER EXERCISES

E_f = cumsum(G); E_W = [0 E_f]/E_tot; test = E_W - per_cent/100; L_test = length(test); k = 1; while test(k) <= 0 k = k+1; end B = k*del_f; if I_wave == 2 tau1 = tau/2; else tau1 = tau; end subplot(3,1,1),plot(t/tau, x), xlabel(’t/ntau’), ylabel(’x(t)’), axis([0 2 0 1.2]) if I_wave == 1 title([’Energy containment bandwidth for rectangular pulse of width ’, num2str(tau),’ seconds’]) elseif I_wave == 2 title([’Energy containment bandwidth for triangular pulse of width ’, num2str(tau),’ seconds’]) elseif I_wave == 3 title([’Energy containment bandwidth for half-sine pulse of width ’, num2str(tau),’ seconds’]) elseif I_wave == 4 title([’Energy containment bandwidth for raised cosine pulse of width ’, num2str(tau),’ seconds’]) end subplot(3,1,2),semilogy(¤*tau1, abs(G./max(G))), xlabel(’fntau’), ylabel(’G(f)’), axis([0 10 1e-5 1]) subplot(3,1,3),plot(f*tau1, E_W,’–’), xlabel(’fntau’), ylabel(’E_W’), axis([0 4 0 1.2]) legend([num2str(per_cent), ’% bandwidth X (pulse width) = ’, num2str(B*tau)],4) % This function generates a unit-high triangle centered % at zero and extending from -1 to 1 % function y = trgl_fn(t) y = (1 - abs(t)).*pls_fn(t/2); % Unit-width pulse function

CHAPTER 2. SIGNAL AND LINEAR SYSTEM ANALYSIS Energy containment bandwidth for half-sine pulse of width 2 seconds

x(t)

1 0.5 0

0.2

0.4

0.6

0.8

1 t/τ

1.2

1.4

1.6

1.8

5 fτ

G(f)

1 0.5 95% bandwidth X (pulse width) = 1.1 0

0.5

1.5

2 fτ

2.5

3.5

% function y = pls_fn(t) y = stp_fn(t+0.5)-stp_fn(t-0.5); % % End of script …le A typical run follows with a plot given in Fig. 2.14. >> ce2_5 Enter type of waveform: 1 = positive squarewave; 2 = triangular; 3 = half-rect. sine; 4 = raised cosine: 3 Enter pulse width: 2 Enter percent total desired energy: 95 Computer Exercise 2.6 The program for this exercise is similar to that for Computer Exercise 2.5, except that the waveform is used in the energy calculation. Computer Exercise 2.7 Use Computer Example 2.2 as a pattern for the solution .

Chapter 3

Basic Modulation Techniques 3.1

Problems

Problem 3.1 The demodulated output, in general, is  () = Lp{ () 2 cos[   +  ()]} where Lp {•} denotes the lowpass portion of the argument. With  () =   () cos [   + 0 ] the demodulated output becomes  () = Lp {2  () cos [   + 0 ] cos [   +  ()]} Performing the indicated multiplication and taking the lowpass portion yields  () =   () cos [ () − 0 ] If () = 0 (a constant), the demodulated output becomes  () =   () cos [0 − 0 ] Letting  = 1 gives the error  () =  () [1 − cos (0 − 0 )] The mean-square error is E 2 ® D 2  () =  () [1 − cos (0 − 0 )]2 1

CHAPTER 3. BASIC MODULATION TECHNIQUES

where h·i denotes the time-average value. Since the term [1 − cos (0 − 0 )] is a constant, we have 2 ® 2 ®  () =  () [1 − cos (0 − 0 )]2

Note that for 0 = 0 , the demodulation carrier is phase coherent with the original modulation carrier, and the error is zero. For  () =  0  we have the demodulated output  () =   () cos ( 0  − 0 ) Letting  = 1, for convenience, gives the error  () =  () [1 − cos (0  − 0 )] giving the mean-square error E 2 ® D 2  () =  () [1 − cos ( 0  − 0 )]2

In many cases, the average of a product is the product of the averages. (We will say more about this in Chapters 4 and 5). For this case E 2 ® 2 ®D  () =  () [1 − cos (0  − 0 )]2

Note that 1 − cos (0  − 0 ) is periodic. Taking the average over an integer number of periods yields D E ® [1 − cos ( 0  − 0 )]2 = 1 − 2 cos ( 0  − 0 ) + cos2 (0  − 0 ) = 1+

Thus

3 1 = 2 2

2 ® 3 2 ®  () =  () 2

Problem 3.2 We are given that () =

5 X 10 =1



sin (2 ) =

³ ´ cos 2  −  2

5 X 10 =1

With a carrier of () = 100 cos (200)

3.1. PROBLEMS

we have  () =

5 h X 500 n h i  io cos (200 + 2 )  − + cos (200 − 2 )  +  2 2 =1

The complex exponential Fourier series is  () =

5 h h X 250 n i  io exp (200 + 2 )  − + exp (200 − 2 )  +  2 2 =1

5 h h X 250 n i  io exp (−200 − 2 )  + + exp (−200 + 2 )  − +  2 2 =1

The top row represents the positive frequency terms and the second row represents the negative frequency terms. The transmitted power is ¶ ¶ µ µ 5 X 250 2 1 1 1 1  = + = 182960 kW 2 = 125000 1 + + +  4 9 16 25 =1

Problem 3.3 A full-wave rectifier takes the form shown in Figure 3.1. The waveforms are shown in Figure 3.2, with the half-wave rectifier on top and the full-wave rectifier on the bottom. The message signal is the envelopes. Decreasing exponentials can be drawn from the peaks of the waveform as depicted in Figure 3.3(b) in the text. It is clear that the full-wave rectified  () defines the message better than the half-wave rectified  () since the carrier frequency is eﬀectively doubled. Problem 3.4 Using

yields the table below. Part a b c

® 2 2 ()   = 1 + 2 h2 ()i

2 ®  ()  = 02  = 03  = 04 =7 =1 13   = 132%   = 291%   = 506%   = 1404%   = 25% 13   = 132%   = 291%   = 506%   = 1404%   = 25% 1   = 385%   = 826%   = 1379%   = 3289%   = 50%

CHAPTER 3. BASIC MODULATION TECHNIQUES

xC  t  yD  t 

Figure 3.1: Full wave rectifier.

2 1.5 1 0.5 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

2 1.5 1 0.5 0

Figure 3.2: Output waveforms for a half-wave rectifier (top) and a full-wave rectifier (bottom).

3.1. PROBLEMS

mn  t  1 t

0 T -1

Figure 3.3: Normalized message signal for Problem 3.5. Problem 3.5 By inspection, the normalized message signal is as shown in Figure 3.3. Thus  2 0≤≤  () =   2 and µ ¶ µ ¶ µ ¶ Z 2 ® 2  2 2 2 2 2 21  3 1   = =  () =  0    3 2 3

Also

 [1 + ] = 40  [1 − ] = 10 This yields

40 1+ = =4 1− 10 1 +  = 4 − 4 5 = 3

Thus  = 06 Since the index is 06, we can write  [1 + 06] = 40

CHAPTER 3. BASIC MODULATION TECHNIQUES

This gives  =

40 = 25 16

This carrier power is 1 1  = 2 = (25)2 = 3125 Watts 2 2 The eﬃciency is   = Thus

(06)2

¡1¢

036 3 2 ¡ 1 ¢ = 336 = 0107 = 107% 1 + (06) 3  = 0107  + 

where  represents the power in the sidebands and  represents the power in the carrier. The above expression can be written  = 0107 + 0107 This gives  =

0107  = 9748 Watts 10 − 0107

Problem 3.6 ® Using (3.14) with  = 100,  = 07 and 2 () = 1 gives 2 ®  () = 05(100)2 + 05(100)2 (07)2 The power in the sidebands is

 = 245 kW The total power is 745 kW. Thus the eﬃciency is   =

245 = 03289 = 3289% 745

This may be checked by using ® 2 2 () (07)2   = = 03289 = 1 + 2 h2 ()i 1 + (07)2 The modulation trapezoid is shown below.

3.1. PROBLEMS

70 30

‐30 ‐70

Figure 3.4: Modulation trapezoid for Problem 3.6. Problem 3.7 For the first signal − + 5( −  ) = 0 so that = Also

5  6

 () =  ()

∙ ¸ 2® 30 1 2 5 2 (−1) + (5) = =5  =  6 6 6

For  = 07 the eﬃciency is

® 2 2 () (07)2 (5) = = 07101 = 7101%   = 1 + 2 h2 ()i 1 + (07)2 (5) and for  = 1 the eﬃciency is ® 2 2 () (1)2 (5)   = = = 08333 = 8333% 1 + 2 h2 ()i 1 + (1)2 (5) For the second signal −5 + ( −  ) = 0

CHAPTER 3. BASIC MODULATION TECHNIQUES

so that = Also 2® 1  = 

  6

1  () =  () 5

µ ¶2 ∙ ¸ 1 5 2 2 5 + (−5) = = 02 (1) 5 6 6 25

For  = 07 the eﬃciency is

® 2 2 () (07)2 (02) = = 00893 = 893%   = 1 + 2 h2 ()i 1 + (07)2 (02) and for  = 1 the eﬃciency is ® 2 2 () (1)2 (02) = = 01667 = 1667%   = 1 + 2 h2 ()i 1 + (1)2 (02)

Problem 3.8 (a) With () = 9 cos(20) − 8 cos(60) we can use MATLAB and either plot the waveform or use a root finding algorithm to find the most negative value of (). Plotting the waveform shows that it varies between −129 and +129. Therefore  () = 1 129 [9 cos(20) − 8 cos(60)] i ® ¡ 1 ¢2 ¡ 1 ¢ h 2 2 = 04357 + (8) (b) 2 () = 129 (9) 2 £ ¤ £ ¤ (c)  = (08)2 (04357)  1 + (08)2 (04357) = 02180 = 2180% (d) The expression for  () is ¸ ∙ 1 (9 cos(20) − 8 cos(60)) cos 200  () = 110 1 + 129 = −341085 cos(140) + 383721 cos(180) +110 cos(200)

+383721 cos(220) − 341085 cos(260) The spectrum is illustrated below. The positive and negative terms dedined in the table below so that the magnitude spectrum is even an the phase spectrum is odd.

3.1. PROBLEMS

f 0

C DE

Figure 3.5: Spectrum of transmitted signal using AM.

Component Frequency (Hz) Magnitude Phase A 70 341085 − B 90 383721 0 C 100 110 0 D 110 383721 0 E 130 341085  -A −70 341085  -B −90 383721 0 -C −100 110 0 -D −110 383721 0 -E −130 341085 −

Problem 3.9 (a) With () = 9 cos(20) + 8 cos(60) the most negative value of () = −17 and falls 1 [9 cos(20) − 8 cos(60)] at  = 005. Therefore  () = 17 i 2 ® ¡ 1 ¢2 ¡ 1 ¢ h 2 2 (b)  () = 17 = 02509 + (8) (9) 2 £ ¤ £ ¤ (c)  = (08)2 (02509)  1 + (08)2 (02509) = 01384 = 1384%

CHAPTER 3. BASIC MODULATION TECHNIQUES

(d) The expression for  () is ¸ ∙ 1  () = 110 1 + (9 cos(20) + 8 cos(60)) cos 200 17 = 258824 cos(140) + 291176 cos(180) +110 cos(200) +291176 cos(220) + 258824 cos(260) The spectrum is as in the previous problem. Components are identified below. Note that the change in sign makes phase spectrum everywhere zero. Component Frequency (Hz) Magnitude Phase A 70 258824 0 B 90 291176 0 C 100 110 0 D 110 291176 0 E 130 258824 0 -A −70 258824 0 -B −90 291176 0 -C −100 110 0 -D −110 291176 0 -E −130 258824 0 Problem 3.10 The modulator output  () = 40 cos 2 (200)  + 5 cos 2 (180)  + 5 cos 2 (220)  can be written  () = [40 + 10 cos 2 (20) ] cos 2 (200)  or

¸ ∙ 10 cos 2 (20)  cos 2 (200)   () = 40 1 + 40

By inspection, the modulation index is

=

10 = 025 40

Since the component at 200 Hertz represents the carrier, the carrier power is  =

1 (40)2 = 800 Watts 2

3.1. PROBLEMS

The components at 180 and 220 Hertz are sideband terms. Thus the sideband power is 1 1 (5)2 + (5)2 = 25 Watts 2 2

 = Thus, the eﬃciency is   =

 25 = 03030 = 330% =  +  800 + 25

Problem 3.11 The sideband power is  =

2 2 + = 2 2 2

The eﬃciency is   =

 2 2 2 = =  +  (2 2) +  2 2 + 2 2

The carrier power is  = Thus =

2 = 200 2

√ 400 = 20

This gives   =

 2 = = 03  +  200 +  2

Therefore 60 =  2 − 03 2 = 07 2 and =

p 6007 = 92582

We determine the modulation index by writing ¸ ∙ 2 cos(2(20) cos(2(200))  () =  1 +  The modulation index is = Summarizing

2(92582) 2 = = 09285  20

CHAPTER 3. BASIC MODULATION TECHNIQUES

 = 20

 = 92582

 = 09258

Problem 3.12 The modulator output  () = 25 cos 2 (150)  + 5 cos 2 (160)  + 5 cos 2 (140)  is

¸ ∙ 10 cos 2 (10)  cos 2 (150)   () = 25 1 + 25 Thus, the modulation index, , is 10 = = 04 25 The carrier power is 1  = (25)2 = 3125 Watts 2 and the sideband power is 1 1  = (5)2 + (5)2 = 25 Watts 2 2 Thus, the eﬃciency is 25   = = 00741 3125 + 25

Problem 3.13 (a) By plotting  () or by using a root-finding algorithm we see that the minimum value of  () is  = −3432. Thus  () = 05828 cos (2 ) + 02914 cos (4 ) + 05828 cos (10 ) The AM signal is  () =  [1 + 08 ()] cos 2  = 02331 cos 2 ( − 5 ) 

+01166 cos 2 ( − 2 ) 

+02331 cos 2 ( −  )  + cos 2 

+02331 cos 2 ( +  )  +01166 cos 2 ( + 2 )  +02331 cos 2 ( + 5 ) 

3.1. PROBLEMS

The spectrum is drawn from the expression for  (). It contains 14 discrete components as shown. Comp 1 2 3 4 5 6 7 (b) Since

the eﬃciency is

Freq − − 5 − − 2 − −  − − +  − + 2 − + 5

Amp 01166 00583 01166 05 01166 00583 01166

Comp 8 9 10 11 12 13 14

Freq  − 5  − 2  −    +   + 2  + 5

Amp 01166 00583 01166 05 01166 00583 01166

i 2 ® 1h (05838)2 + (02914)2 + (05838)2 = 03833  () = 2 ® 2 2 () (08)2 (03833) = = 01970   = 1 + 2 h2 ()i 1 + (08)2 (03833)

Problem 3.14 (a) From Figure 3.35  () =  () + cos    With the given relationship between  () and  () we can write  () = 4 { () + cos   } + 2 { () + cos   }2 which can be written  () = 4 () + 4 cos    + 22 () + 4 () cos    + 1 + 1 cos 2   The equation for  () is more conveniently expressed  () = 1 + 4 () + 22 () + 4 [1 +  ()] cos    + cos 2  (b) The spectrum illustrating the terms involved in  () is shown in Figure 3.6. The center frequency of the filter is  and the bandwidth must be greater than or equal to 2 . In addition,  −   2 or   3 , and  +   2 . The last inequality states that    , which is redundant since we know that   3 .

CHAPTER 3. BASIC MODULATION TECHNIQUES

m  t 



Filter Characteristic



 m2  t 

fc  W

fc  W

2 fc

Figure 3.6: Generation of an AM signal using a nonlinear operation. (c) From the definition of  () we have  () =   () so that  () = 4 [1 +   ()] cos    It follows that  = 01 =  Thus  = 01 (d) This method of forming a DSB signal avoids the need for an analog multiplier. Problem 3.15 With () = 4 cos (2 ) + cos (4 ) we have

Therefore

() b = 4 sin (2 ) + sin (4 )

 () = 5 [4 cos (2 ) + cos (4 )] cos (2 )±5 [4 sin (2 ) + sin (4 )] sin (2 )

3.1. PROBLEMS

f ‐fc

Figure 3.7: Generation of upper-sideband SSB using sideband filtering. This gives  () = 10 cos [2 ( +  ) ] + 25 cos [2 ( + 2 ) ] +10 cos [2 ( −  ) ] + 25 cos [2 ( − 2 ) ]

∓10 cos [2 ( +  ) ] + 25 cos [2 ( + 2 ) ] ±10 cos [2 ( −  ) ] + 25 cos [2 ( − 2 ) ] Note that by using the top, +, sign we have  () = 20 cos [2 ( −  ) ] + 5 cos [2 ( − 2 ) ] The upper-sideband (frquency components above  ) cancel and we are left with only lowersideband terms (frquency components below  ). Note that by using the bottom, +, sign we have  () = 20 cos [2 ( +  ) ] + 5 cos [2 ( + 2 ) ] The lower-sideband (frquency components below  ) cancel) and we are left with only uppersideband terms (frquency components above  ). The spectrum in each case is obvious. Problem 3.16 Consider the filter transfer function shown below From the above figure, and comparing it with Figure 3.10 in the text, it is clear that the uppersideband filter is given by

CHAPTER 3. BASIC MODULATION TECHNIQUES

1 1 [1 + sgn ( −  )] + [1 − sgn ( +  )] 2 2 Going through the same steps as in the text (pp 125-127) gives  ( ) =

Problem 3.17 We have

1 1  () =   () cos    −   b () sin   2 2 2 () =

which is

∙

¸2 1 1   () cos    ±   b () sin   2 2

1 1 1 b cos(  ) sin( )+ 2  b 2 () [1 − cos(2  ] 2 () = 2 2 () [1 + cos(2 )]± 2 ()() 4 2 4 or ¤ 1 £ b 2 () + 2 () cos(2  ) ± ()() b sin(  ) −  b 2 () cos(2  ) 2 () = 2 2 () +  4 Therefore, yes, components are created at twice the carrier frequency. Problem 3.18 We assume that the VSB waveform is given by  () =

1  cos (  −  1 )  2 1 +  (1 − ) cos (  + 1 )  2 1 +  cos (  + 2 )  2

We let  () be  () plus a carrier. Thus  () =  () +  cos   It can be shown that  () can be written  () = 1 () cos   () + 2 () sin    where   cos  1  + cos  2  +  2 ¶ 2 µ   sin  1  − sin  2  2 () =  + 2 2

1 () =

3.1. PROBLEMS

‐A

f ‐W

Figure 3.8: Coherent demodulation of VSB. Also  () =  () cos (   + ) where  () is the envelope and is therefore the output of an envelope detector. It follows that q  () = 12 () + 22 ()

For  large, 1 () À 2 () for all  so that 2 () can be neglected. Therefore,  () = |1 ()|, which is 12  () + , where  is a dc bias. Thus if the detector is ac coupled,  is removed and the output  () is  () scaled by 12 .

Problem 3.19 Coherent demodulation translates the positive-frequency portion of the spectrum to  = 0 and to  = 2 . The negative-frequency portion of the spectrum is translated to  = 0 and to  = −2 . The following figure illustrates the portion of the spectrum about  = 0. The portion of the spectrum is the portion between −A and A. Using the required symmetry properties for a VSB filter, the two portions of the translated spectra are conjugate symmetric. Thus they add resulting in the spectrum shown in the bottom pane. Problem 3.20 The result is shown in the following figure. Problem 3.21

CHAPTER 3. BASIC MODULATION TECHNIQUES

Desired Signal



1 Local Oscillator



1   2 Signal at Mixer Output

 2   IF

21   2



Image Signal

 21   2 Image Signal at Mixer Output

 3 2  21

2

Figure 3.9: Result for Problem 3.20.

3.1. PROBLEMS

Since high-side tuning is used, the local oscillator frequency is  =  +  where  , the carrier frequency of the input signal, varies between 10 and 30 MHz. The ratio is  + 30 =  + 10 where  is the  frequency expressed in  . We make the following table  MHz 04 06 08 10 50 100

 29231 28868 28519 28182 23333 20000

A plot of  as a function of  is the required plot. However, we see that  is nearly constant and decreases as  increases. Therefore, in this case, a large IF frequency should be used consistant with hardware requirements and availability. Problem 3.22 With  = 455 kHz we have for high-side tuning  =  +  = 1100 + 455 = 1555 kHz  =  + 2 = 1100 + 910 = 2010 kHz and for low-side tuning we have  =  −  = 1100 − 455 = 665 kHz

 =  − 2 = 1100 − 910 = 210 kHz With  = 2500 we have for high-side tuning  =  +  = 1100 + 2500 = 3600 kHz  =  + 2 = 1100 + 5000 = 6100 kHz and for low-side tuning we have  =  −  = 1100 − 2500 = −1400 kHz

 = 1400 kHz

 =  − 2 = 1100 − 5000 = −3900 kHz  = 3900 kHz

CHAPTER 3. BASIC MODULATION TECHNIQUES

In the preceding development for low-side tuning, recall that the spectra are symmetrical about  = 0. Problem 3.23 We assume single-tone interference as in the body of the text. Therefore we assume  () = () cos (2 ) +  cos (2 ) Therefore 2 () =

2 2 2  () [1 + cos (4 )] + 2 cos (2 ) cos (2 ) +  [1 + cos (4 )] 2 2

Ignoring the dc terms, we have 2 () =

2 2 2  () cos (4 )+ [cos (2( +  ) ]+ [cos (2( −  ) ]+  cos (4 ) 2 2

We therefore see that the squaring operation generates a component at twice the carrier frequency and a component at twice the interference frequency. A pair of sidebands are also generated about the carrier frequency, above and below the carrier frequency, and separated from the carrier frequency by the interference frequency. Problem 3.24 From (3.96) the transfer function of the holding operation is ( ) = 

sin (2 ) exp (− ) 2

Sampling reproduces this about  = 0 and all harmonics of the sampling frequency. We consider the terms about  = 0 and  =  . Ignoring the phase, which is linear and therefore only induces a delay, we have 1 ( ) = 

sin (2  ) ∗  [( ) +  ( −  )] 2 

Finally we have 1 ( ) =  

sin (2 ) sin (2( −  ) ) +   2  2( −  )

The equalizer will have the transfer function  ( ) =

2  1   sin(2 )

3.1. PROBLEMS

1 where In order for the frequency response to be bounded we must have 2   1 or   2  is the highest frequency in the message signal. We must also have 2 − 2   −1. To plot this requires that  be set as a parameter.

Problem 3.25 Let  be the peak-to-peak value of the data signal. The peak error is 05% (±025%) and the peak-to-peak error is 001 . The required number of quantizating levels is  = 100 ≤ 2 =  001 so we choose  = 128 and  = 7. The bandwidth is  = 2  log2  = 2 (7) The value of  is estimated by assuming that the speech is sampled at the Nyquist rate. Then the sampling frequency is  = 2 = 8 kHz. Each sample is encoded into  = 7 pulses. Let each pulse be  with corresponding bandwidth 1 . For our case = Thus the bandwidth is

1 1 =  2 

1 = 2  = 2 log2  

and so  = 1. For  = 1  = 2 (8 000) (7) = 112 kHz

Problem 3.26 The message signal is () = 3 sin[2(10)] + 4 sin[2(20)] The derivative of the message signal is  () = 60 cos[2 (10) ] + 160 cos[2 (20)]  The maximum value of  ()  is obviously 280 and the maximum occurs at  = 0. Thus 0 ≥ 220  or 220 220  ≥ = 4400 = 0 005

CHAPTER 3. BASIC MODULATION TECHNIQUES

x1 (BW  W )

x2 (BW  W ) x3 (BW  2W )

x4 (BW  4W )

x5 (BW  4W )

Output

Figure 3.10: Commutator configuration for Problem 3.27. Thus, the minimum sampling frequency is 4400 Hz. Problem 3.27 One possible commutator configuration is illustrated in Figure 3.10. The signal at the point labeled “output” is the baseband signal. The minimum commutator speed is 2 revolutions per second. For simplicity the commutator is laid out in a straight line. Thus, the illustration should be viewed as it would appear wrapped around a cylinder. After taking the sample at point 12 the commutator moves to point 1. On each revolution, the commutator collects 4 samples of 4 and 5 , 2 samples of 3 , and one sample of 1 and 2 . The minimum transmission bandwidth is X  =  +  + 2 + 4 + 4  = 

= 12

Problem 3.28 If the speed of the comutator is doubled from 2W to 4W, the sampling frequency is obviously doubled. The sampling frequencies are 4W, 4W, 8W, 16W, and 16W, respectively. The required transmission freqency is doubled. However, the guard frequencies are doubled.

3.1. PROBLEMS

Figure 3.11: Single-sided spectrum for Problem 3.30. Problem 3.29 The minimum baseband bandwidth is  =  +  + 2 + 5 + 7 = 16 A suitable scheme for this takes a bit of thought. However, the problem is not much diﬀerent than when we consider the requirement that a sum of sinusoids be periodic.To be periodic, the frequencies must be commensurable (measurable by the same standard). In this case all bandwidths must be factors of a common bandwidth, which is 70W. Thus,the communator must have 70 distinct psitions.There are many possibilities one is shown in the Table below. Single Bandwidth Communator Positions 7 10, 20, 30, 40, 50, 60, 70 5 1, 15, 29, 43, 57 2 2, 37  3  4

Note Spacing = 70 7 = 10 70 Spacing = 5 = 14 Spacing = 70 2 = 35 Spacing = 70 1 = 70 70 Spacing = 1 = 70

The positions are arbitrary but the spacings are not. The communator rotates at a minimum rate of 2W. Note that there are many unused positions. These unused positions can be used for other signals as long as the commensurable conditions hold.klmmmmmmmmmmmmmmmmmmmmmm Problem 3.30 The single-sided spectrum for  () is shown in Figure 3.11. From the definition of  () we have  () = 1  ( ) + 2  ( ) ∗  ( )

CHAPTER 3. BASIC MODULATION TECHNIQUES

Yf a1

2a2W

a2W f

f 2  f1

2 f2

f1  f 2

2 f2

Figure 3.12: Output spectrum for Problem 3.54. The spectrum for  ( ) is given in Figure 3.12. Demodulation can be a problem since it may be difficult to filter the desired signals from the harmonic and intermodulation distortion caused by the nonlinearity. As more signals are included in  (), the problem becomes more difficult. The difficulty with harmonically related carriers is that portions of the spectrum of  ( ) are sure to overlap. For example, assume that 2 = 21 . For this case, the harmonic distortion arising from the spectrum centered about 1 falls exactly on top of the spectrum centered about 2 .

3.2

Computer Exercises

Computer Exercise 3.1 The MATLAB program is % File: ce3_1.m t = 0:0.001:1; fm = 1; fc =10; m = 4*cos(2*pi*fm*t-pi/9) + 2*sin(4*pi*fm*t); [minmessage,index] = min(m); mncoefs = [4 2]/abs(minmessage) mn = m/abs(minmessage);

3.2. COMPUTER EXERCISES

plot(fm*t,mn,’k’), grid, xlabel(’Normalized Time’), ylabel(’Amplitude’) mintime = 0.001*(index-1); dispa = [’The minimum of m(t) is ’, num2str(minmessage,’%15.5f’), ... ’and falls at ’,num2str(mintime,’%15.5f’), ’ s.’]; disp(dispa) % % Now we compute the efficiency. % mnsqave = mean(mn.*mn); % mean-square value a = 0.5; % modulation index asq = a^2; efficiency = asq*mnsqave/(1+asq*mnsqave); dispb = [’The efficiency is ’, num2str(efficency,’%15.5f’),’ .’]; disp(dispb) % % Now compute the carrier amplitude. % cpower = 50; % carrier power camplitude = sqrt(2*cpower); % carrier amplitude % % Compute the magnitude and phase spectra. % xct = camplitude*(1+a*mn).*cos(2*pi*fc*t); Npts = 1000; fftxct = fft(xct,Npts)/Npts; s1 = [conj(fliplr(fftxct(2:25))) fftxct(1:25)]; mags1 = abs(s1); angles1 = angle(s1); for k=1:length(s1) if mags1(k) = 0.01 angles1(k)=0; end end % % Plot the magnitude and phase spectra. % figure subplot(2,1,1), stem(-24:24,mags1,’.’) xlabel(’frequency’); ylabel(’magnitude’) subplot(2,1,2), stem(-24:24,angles1,’.’) xlabel(’frequency’); ylabel(’phase’)

CHAPTER 3. BASIC MODULATION TECHNIQUES

1.5

Amplitude

0.5

-0.5

-1

0.1

0.2

0.3

0.4 0.5 0.6 Normalized Time

0.7

0.8

0.9

Figure 3.13: Message signal. % End of script file. Executing the program gives: » ce3_1 mncoefs = 0.9165 0.4583 The minimum of m(t) is -4.36424 and falls at 0.43500 s. The efficiency is 0.11607 . The program also generates Figures 3.13 and ??. Computer Exercise 3.2 The MATLAB code written for Computer Exercise 3.2 follows. % File: ce3_2.m t = 0:0.001:1; fm = 1;

3.2. COMPUTER EXERCISES

magnitude

0 -25

-20

-15

-10

-5

0 frequency

-20

-15

-10

-5

0 frequency

phase

1 0 -1 -2 -25

Figure 3.14: Spectra of AM signal.

CHAPTER 3. BASIC MODULATION TECHNIQUES fc =10; m = 2*cos(2*pi*fm*t) + cos(4*pi*fm*t); mhil = 2*sin(2*pi*fm*t) + sin(4*pi*fm*t); xctusb = 0.5*m.*cos(2*pi*fc*t)-0.5*mhil.*sin(2*pi*fc*t); xctlsb = 0.5*m.*cos(2*pi*fc*t)+0.5*mhil.*sin(2*pi*fc*t); % % Plot time-domain signals. % subplot(2,1,1), plot(t,xctusb) xlabel(’time’); ylabel(’USB signal’) subplot(2,1,2), plot(t,xctlsb) xlabel(’time’); ylabel(’LSB signal’) % % Determine and plot spectra. % Npts = 1000; fftxctu = fft(xctusb,Npts)/Npts; fftxctl = fft(xctlsb,Npts)/Npts; sxctusb = [conj(fliplr(fftxctu(2:25))) fftxctu(1:25)]; magsusb = abs(sxctusb); sxctlsb = [conj(fliplr(fftxctl(2:25))) fftxctl(1:25)]; magslsb = abs(sxctlsb); figure subplot(2,1,1), stem(-24:24,magsusb) xlabel(’frequency’); ylabel(’USB magnitude’) subplot(2,1,2), stem(-24:24,magslsb) xlabel(’frequency’); ylabel(’LSB magnitude’) % End of script file.

Executing the code gives Figures 3.15 and 3.16. Computer Exercise 3.3 In this computer example we investigate the demodulation of SSB using carrier resertion for several values of the constant .for an assumed message signal. % File: ce3_3.m t = 0:0.001:1; fm = 1; m = 2*cos(2*pi*fm*t) + cos(4*pi*fm*t); mhil = 2*sin(2*pi*fm*t) + sin(4*pi*fm*t);

3.2. COMPUTER EXERCISES

USB signal

2 1 0 -1 -2 0

0.1

0.2

0.3

0.4

0.5 time

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5 time

0.6

0.7

0.8

0.9

LSB signal

2 1 0 -1 -2

Figure 3.15: Time-domain waveforms.

CHAPTER 3. BASIC MODULATION TECHNIQUES

USB magnitude

0.8 0.6 0.4 0.2 0 -25

-20

-15

-10

-5

0 frequency

-20

-15

-10

-5

0 frequency

LSB magnitude

0.8 0.6 0.4 0.2 0 -25

Figure 3.16: Amplitude spectra.

3.2. COMPUTER EXERCISES

k = 1; ydt = sqrt((k+m).*(k+m)+mhil.*mhil); ydt1 = ydt-k; k = 10; ydt = sqrt((k+m).*(k+m)+mhil.*mhil); ydt2 = ydt-k; k = 100; ydt = sqrt((k+m).*(k+m)+mhil.*mhil); ydt3 = ydt-k; subplot(3,1,1), plot(t,m,’k’,t,ydt1,’k--’) ylabel(’k=1’) subplot(3,1,2), plot(t,m,’k’,t,ydt2,’k--’) ylabel(’k=10’) subplot(3,1,3), plot(t,m,’k’,t,ydt3,’k--’) ylabel(’k=100’) % End of script file. Executing the preceding MATLAB program gives the results illustrated in Figure 3.17. In the three plots the message signal is the solid line and the dashed line represents the output resulting from carrier reinsertion. It can be seen that the demodulated output improves for increasing  and that the distortion is negligible for  = 100.

Computer Exercise 3.4 Two MATLAB programs are generated for this Computer Exercise, ce3_4a ancd ce3_4b. The first of these two programs illustrates the message signal, the VSB signal and the VSB signal after carrier reinsertion. The second program illustrates the spectrum of a VSB signal The MATLAB code for the first program is % File: ce3_4a.m clear all t = 0:0.001:1; m = cos(2*pi*t)-cos(4*pi*t)+cos(6*pi*t); e1 = 0.64; e2 = 0.78; e3 = 0.92; fc = 25; A = 100; ct = A*cos(2*pi*fc*t); xct = e1*cos(2*pi*(fc+1)*t)+(1-e1)*cos(2*pi*(fc-1)*t)-... e2*cos(2*pi*(fc+2)*t)-(1-e2)*cos(2*pi*(fc-2)*t)+... e3*cos(2*pi*(fc+3)*t)+(1-e3)*cos(2*pi*(fc-3)*t)+... cos(2*pi*(fc+4)*t)+cos(2*pi*(fc+5)*t)+cos(2*pi*(fc+6)*t);

CHAPTER 3. BASIC MODULATION TECHNIQUES

k=1

4 2 0 -2

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

k=10

4 2 0 -2

k=100

4 2 0 -2

Figure 3.17: Illustration of demodulation resertion for  = 1, 10, and 100.

3.2. COMPUTER EXERCISES

m(t)

2 0 -2 -4

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5 Time

0.6

0.7

0.8

0.9

xc(t)

xc(t)+c(t)

-5

102 100 98

Figure 3.18: Illustration of VSB (message signal, VSB signal, and VSB signal plus carrier reinsertion). xdem = abs(xct+ct); subplot(3,1,1), plot(t,m), ylabel(’m(t)’) subplot(3,1,2), plot(t,xct), ylabel(’xc(t)’) subplot(3,1,3), plot(t,xdem), ylabel(’xc(t)+c(t)’) xlabel(’Time’), axis([0 1 97 103]) % End of script file. The result of demodulation using carrier reinsertion if shown in the bottom pane. The message signal can be recovered using a dc-coupled envelope detection. We now consider the spectrum of a VSB signal. We change the message signal to include a carrier component and to include six components above the carrier. In addition, all components in the message signal are set equal in order to illustrate the action of the VSB filter. The MATLAB program is % File: ce3_4b.m clear all t = 0:0.001:1; e1 = 0.64; e2 = 0.78; e3 = 0.92; fc = 25; ct = 0.5*cos(2*pi*fc*t);

CHAPTER 3. BASIC MODULATION TECHNIQUES

1.4

VSB magnitude spectrum

1.2

0.8

0.6

0.4

0.2

25 30 frequency

Figure 3.19: VSB spectrum. xct = ct+e1*cos(2*pi*(fc+1)*t)+(1-e1)*cos(2*pi*(fc-1)*t)-... e2*cos(2*pi*(fc+2)*t)-(1-e2)*cos(2*pi*(fc-2)*t)+... e3*cos(2*pi*(fc+3)*t)+(1-e3)*cos(2*pi*(fc-3)*t)+... cos(2*pi*(fc+4)*t)+cos(2*pi*(fc+5)*t)+cos(2*pi*(fc+6)*t); Npts = 1000; fftxctu = fft(xct,Npts)/Npts; magvsb=2*abs(fftxctu(1:51)); freq=0:50; stem(freq, magvsb) xlabel(’frequency’); ylabel(’VSB magnitude spectrum’) % End of script file. The resulting spectrum follows. Note that there is a carrier component at  = 25. There are three attenuated components above the carrier and three non-attenuated components. Below the carrier there are the complements of the three attenuated components above the carrier.

Computer Exercise 3.5 The MATLAB program used appears below. Given the annotations, the code should be understandable. (Thanks to Dr. John Tranter of the University of Minnesota for his help on this problem.)

3.2. COMPUTER EXERCISES

% File CE3.5.m clear all; close all; f1 = 70; f2 = 300; % sine frequencies fmin = min(f1,f2); % use lowest freq. for 1 cycle on time axis A1 = 3; A2 = 1.5; % sine amplitudes samp = [15 60]; % sample period (in MATLAB increments) N = 1500; % discretize signal with N points hl = 1; % staircase amplitude t = 1/N/fmin:1/N/fmin:1/fmin; % x-axis in seconds x = A1*sin(2*pi*f1*t)+A2*sin(2*pi*f2*t); % build our signal for kk = 1:2 d(1:samp(kk))=0; for ii=samp(kk)+1:samp(kk):N dd(ii) = sign(x(ii) - d(ii-samp(kk))); % this is our delta d(ii:ii+samp(kk)-1) = d(ii-samp(kk)) + hl*dd(ii); end d = d(1:N); % trim/pad extra samples (if sample dd = [dd zeros(1,N-numel(dd))]; % period does not exactly divide N) subplot(3,2,[kk kk+2]),plot(t,x,’k’),... axis([t(1) t(N) min([x d]) max([x d])]) hold on,stairs(t,d,’r’),xlabel(’Time (s)’) if (kk == 1) str = ’no slope overload)’; else str = ’slope overload)’; end title(strcat(’Stairstep Approximation (’,str)); subplot(3,2,kk+4),stem(t(1:samp:end),dd(1:samp:end),’marker’,’none’) axis([t(1) t(N) -1.2 1.2]),xlabel(’Time (s)’) title(strcat(’Delta Modulation (’,str)); clear d; clear dd; % diff. sample rate = diff. dimensions... end % This is a neat trick that automatically maximizes your figure screen_size = get(0, ’ScreenSize’); set(figure(1), ’Position’, [0 0 screen_size(3) screen_size(4) ] ); % End of script file. Executing the program yields the following output.

CHAPTER 3. BASIC MODULATION TECHNIQUES Stairstep Approximation (no slope overload)

Stairstep Approximation (slope overload)

5 4 4 3 3 2 2 1 1 0

-1

-2

-3

-3 -4

Time (s) Delta Modulation (no slope overload)

-4

-3

x 10

0.5

-0.5

-1

6 8 10 Time (s) Delta Modulation (slope overload)

14 -3

x 10

-1 2

8 Time (s)

14 -3

x 10

8 Time (s)

14 -3

x 10

Figure 3.20: Delta modulation without slope overload (left two panes) and with slope overload (right two panes). Computer Exercise 3.6 The MATLAB code follows. % File ce3_6.m clear all t=0:0.001:(2-0.001); m=zeros(1,2000); x=zeros(1,2000); m=cos(2*pi*t)-cos(4*pi*t)+cos(6*pi*t); tau=5; for k=1:2000 if rem(k,10)==0 x(k)=1; end end xsamp=m.*x; h=zeros(1,10); for k=1:tau h(k)=1; end xsh=conv(xsamp,h);

3.2. COMPUTER EXERCISES

Samp & Hold Output

-2

-4 0.5

0.55

0.6

0.65

0.7

0.75 Time

0.8

0.85

0.9

0.95

Approximation

-1

-2 0.4

Message Reconstruction 0.6

0.8

1 Time

1.2

1.4

1.6

Figure 3.21: PAM and reconstruction. (Top pane - Sample and hold output. Bottom pane - original message signal and reconstructured signal.) [B,A]=butter(3,0.05); y=filter(B,A,xsh); subplot(2,1,1), plot(t(500:1000),xsh(500:1000)) axis([0.5 1 -4 2]) xlabel(’Time’); ylabel(’Samp & Hold Output’) subplot(2,1,2), plot(t(500:1500),y(500:1500),’k’,... t(500:1500),0.5*m(500:1500),’k--’) xlabel(’Time’); ylabel(’Approximation’) legend(’Message’,’Reconstruction’,’Location’,’South’) % End of script file. Note that for this example, the simulation sampling frequency is 1000 Hz. The message signal is sampled every 10 simulation steps. Thus  = 100 Hz. The width of the PAM pulse is 5 simulation steps so  = 0005 s. Thus   = 05. The 3 dB frequency of the Butterworth filter is 0.5(1000/2)=250 Hz. Other values for   are easily tried.

Chapter 4

Angle Modulation and Multiplexing 4.1

Problems

Problem 4.1 By definition  () =  cos [   +   ()] =  cos [  +  ( − 0 )] The waveforms are shown below. The the top pane represents the unmodulated signal. The middle pane is for is for  = , and the bottom-right pane is for  = −38. The plots illustrate 2 s of data with a frequency 1 Hz. The parameter 0 = 0 for all plots.

Problem 4.2 Once again, by definition  () =  cos [   +   ()] =  cos [  +  ( − 0 )] The waveforms are shown below. The the top pane represents the unmodulated signal. The middle pane is for is for  = −2, and the bottom-right pane is for  = 38. The plots illustrate 2 s of data with a frequency 1 Hz. The parameter 0 = 0 for all plots. Problem 4.3 With

³ ´ () =  sin 2  + 6 1

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

Amplitude

1 0 -1 -1

-0.8

-0.6

-0.4

-0.2

0 Time

0.2

0.4

0.6

0.8

-0.8

-0.6

-0.4

-0.2

0 Time

0.2

0.4

0.6

0.8

-0.8

-0.6

-0.4

-0.2

0 Time

0.2

0.4

0.6

0.8

Amplitude

1 0 -1 -1

Amplitude

1 0 -1 -1

Figure 4.1: Time-domain plots for Problem 4.1.

Amplitude

1 0 -1 -1

-0.8

-0.6

-0.4

-0.2

0 Time

0.2

0.4

0.6

0.8

-0.8

-0.6

-0.4

-0.2

0 Time

0.2

0.4

0.6

0.8

-0.8

-0.6

-0.4

-0.2

0 Time

0.2

0.4

0.6

0.8

Amplitude

1 0 -1 -1

Amplitude

1 0 -1 -1

Figure 4.2: Time-domain plots for Problem 4.2.

4.1. PROBLEMS

we have for AM h ³  ´i cos (2 )  () =  cos 1 +  sin 2  + 6

where  is the modulation index. Thus  () can be written ½µ ¾ ¶ ³ ³  h ´  ´i 1+  () =  Re exp 2  + − exp −2  − exp (2 ) 2 6 6 or

 () =  Re This gives

³ ³ n³ o h  ´   ´i´ exp 2  + − − exp −2  − + exp (2 ) 1+ 2 6 2 6 2

 () =  Re

³ ³ n³ o h ´  ´i´ exp 2  − − exp −2  + exp (2 ) 1+ 2 3 3

The phasor diagram is the sama as in Figure 4.4 except that the vectors rpresenting the modulation have inital angles ± 3  For FM we have ³ ´ cos (2 )  () =  cos (2 ) −  sin 2  + 6

³ ³  ´  ´ cos 2( − cos 2(  () =  cos (2 ) +  −  ) −  +  ) + 2 6 2 6

The phasor diagram is obvious.

Problem 4.4 Let  () =  cos    where   = 2 . This gives o n 2 () =  Re    cos   Expanding a Fourier series gives

 cos   =

∞ X

  

=−∞

where   = 2

Z 

−

 cos   −  

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

With  =   , the Fourier coeﬃcients become Z  1  =  cos  −  2 − ¡ ¢ Since cos  = sin  + 2 Z   1  [ sin(+ 2 )−]   = 2 −

With  =  + 2 , the preceding becomes 1 2

Z 32

  2

 = This gives  = 



 [ sin −+ 2 ] 

−2

1 2

Z 

[ sin −]





−

where the limits have been adjusted by recognizing that the integrand is periodic with period 2. Thus   =  2  () and

(

  

2 () =  Re  Taking the real part yields 2 () = 

∞ X

 ( +  2 )

 () 

=−∞

)

h  i  () cos (  +   )  + 2 =−∞ ∞ X

The amplitude spectrum is therefore the same as in the preceding problem. The phase spectrum is the same as in the preceding problem except that  2 is added to each term. Problem 4.5 Since sin() = cos( − 2 ) we can write

which is Since

³ ´  3 () =  sin(  +  sin   ) =  cos    − +  sin   2 n o 3 () =  Re ( −2)  sin    sin   =

∞ X

=−∞

 () 

4.1. PROBLEMS

we have

(

( −2)

3 () =  Re  Thus

∞ X

3 () = 

 

 ()

=−∞

Thus

∞ X

)

o n  () Re ( + −2)

h i  () cos ( +   )  − 2 =−∞ ∞ X

Note that the amplitude spectrum of 3 () is identical to the amplitude spectrum for both 1 () and 2 (). The phase spectrum of 3 () is formed from the spectrum of 1 () by adding −2 to each term. For 4 () we write ³ ´  4 () =  sin(  +  cos   ) =  cos    − +  cos    2 Using the result of the preceding problem we write ( ) ∞ X  (  +  ( −2) 2 )  ()  4 () =  Re  =−∞

This gives 4 () = 

∞ X

=−∞

Thus 4 () = 

o n    () Re ( + − 2 + 2 )

i h   () cos ( +   )  + ( − 1) 2 =−∞ ∞ X

Compared to 1 (), 2 (), 3 () and 4 (), we see that the only diﬀerence is in the phase spectrum. Problem 4.6 From the problem statement  () =  cos [2 (40)  + 10 sin (2 (5) )] Since  = 40 and  = 5 (it is important to note that  is an integer multiple of  ) there is considerable overlap of the portion of the spectrum centered about  = −50 with

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

1 0.8 0.6

Magnitude

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1

0.02

0.04

0.06

0.08 0.1 0.12 Frequency

0.14

0.16

0.18

0.2

Figure 4.3: Time-domain signal for Problem 4.6. the portion of the spectrum centered about  = 50 for large  (such as 10). This can also be seen in the time domain. We can see from the time domain plot (top pane) of Figure ??. Thus, this is not a true FM signal. Problem 4.7 We are given 0 (5) = −0178 and 1 (5) = −0328. Using +1 () =

2  () − −1 () 

with  = 5 we have 2 +1 (5) =  (5) + −1 (5) 5 With  = 1, 2 (5) = =

2 (1)1 (5) + 0 (5) 5 2 (−0328) + 0178 = 00468 5

With  = 2, 3 (5) = =

2 (2)2 (5) − 1 (5) 5 2 (2)00468 + 0328 = 03648 5

4.1. PROBLEMS

0.3

Magnitude

0.25

0.2

0.15

0.1

0.05

0 1000

1200

1400

1600

1800 2000 2200 Frequency

2400

2600

2800

3000

Figure 4.4: Finally with  = 3 we have 4 (5) = =

2 (3)3 (5) − 2 (5) 5 2 (3) (03648) − 00468 = 0 5

Problem 4.8 The amplitude and phase spectra follow directly from the table of Fourier-Bessel coeﬃcients, or one can write a simple MATLAB program as was done here.. The single-sided magnitude spectrum is shown in Figure ??. The magnitude spectrum is plotted assuming  = 1. The student should verify that this spectrum is correct and the student should plot the phase spectrum using both a table of Bessel functions and the ﬀt MATLAB function. The table of Bessel functions approach is trivial. The other approach takes a bit of thought to get as clean plot. Problem 4.9 (a) Since the carrier frequency is 1000 Hertz, the general form of  () is £ ¤  () =  cos 2 (1000)  + 40 sin(52 ) =  cos [2 (1000)  +  ()]

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

The phase deviation,  (), is therefore given by  () = 40 sin(52 )

rad

The frequency deviation is  = 400 cos(52 )  or

rad/sec

1  200 =  cos(52 ) 2  

(b) Since the carrier frequency is 1000 Hertz, the general form of  () is  () =  cos [2 (1000)  − 2(400)] =  cos [2 (1000)  +  ()] The phase deviation is  () = −2 (400) 

rad

(Note that we substracted the phase of the unmodulated carrier from the instantaneous carrier.) The frequency deviation is  = −2 (400)  or

rad/sec

1  = −400 2 

Problem 4.10 (a) Since the carrier frequency is 1000 Hertz, the general form of  () is £ ¤  () =  cos 2 (1000)  − 2(1000) + 2(1200)2 =  cos [2 (1000)  +  ()] The phase deviation is

 () = −2 (1000)  + 2(1200)2

rad

and the frequency deviation is  = −2000 + 2(2400)  or

1  = −1000 + 2400 2 

rad/sec

4.1. PROBLEMS

(b) Since the carrier frequency is 1000 Hertz, the general form of  () is h √i  () =  cos 2 (1000)  − 2(100) + 10  =  cos [2 (1000)  +  ()]

The phase deviation is

√ √  () = −2(100) + 10  − 2(1000) = −2(100) + 10  and the frequency deviation is 1 1  5 = −2(100) + (10) − 2 = −2(100) + √  2 

1  5 = −100 + √ 2  2 

Problem 4.11 (a) The phase deviation is () = 2(20)

Z 

which is, for 0 ≤  ≤ 8, () = 40

¸ 1 4Π ( − 4)  8

Z 

∙

4 = 160

Thus () = 0,

0

() = 160 () = 160(8)

0≤≤8 8

(b) The frequency deviation is, by definition ∙ ¸ 1  1 = 80Π ( − 4) = 80 2  8 for 0 ≤  ≤ 8 and is zero for   0 and   8. (c) The peak frequency deviation is 80 Hz. (d) The peak phase deviation is 160(8) = 1280 radians. (e) The transmitted power is  =

(100)2 = 5000 2

rad/sec

rad

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

Problem 4.12 (a) The phase deviation is () = 0, () = = () = = = = () =

3 µ ¶ Z  160 2 9 4 ( − 3) = − − 3 + 9 40 3 2 2 3 3 ¡ ¢ 80 2  − 6 + 9  3≤≤6 3 ¶ ¶ Z µ Z µ 4 4 4 − ( − 6)  = 240 + 40 12 −   240 + 40 3 3 6 µ6 ¶ 41 2 240 + 40 12( − 6) − ( − 36) 32 µ ¶ 2 2 240 + 40 −  + 12 − 48 3 ¢ 80 ¡ 2 6≤≤9 240 + − + 18 − 72  3 480 9

(b) The frequency deviation in Hz is ¸ 1  () = 80Λ ( − 6) 3 ∙

(c) The peak frequency deviation is 4 (20) = 80 Hz. (d) The peak phase deviation is, since the area of Λ[(−6)3] = 3 is 2 (3) (4) = 24 (20) = 480 (e) The modulator output power is 1 1  = 2 = (100)2 = 5000 Watts 2 2

Problem 4.13 For all three message signals illustrated in Figure 4.37, the frequency deviation in Hz is simply the given message signal with the ordinate values multiplied by 10. For the first message signal (the top signal given in Figure 4.37) the phase deviation in radians is given Z Z 

 () = 2



 ()  = 20

 () 

4.1. PROBLEMS

For 0 ≤  ≤ 1, we have  () = 20

Z 

2 = 202

For 1 ≤  ≤ 2  () = (1) + 20

Z  1

¡ ¢ (5 − )  = 20 + 100 ( − 1) − 10 2 − 1

= 20 + 100 − 100 − 102 + 10 = −70 + 100 − 102 For 2 ≤  ≤ 3  () = (2) + 20

Z 

3 = 90 + 60 ( − 2) = 60 − 30

Z 

2 = 150 + 40( − 3) = 30 + 40

For 3 ≤  ≤ 4  () = (3) + 20

Finally, for   4 we recognize that  () =  (4) = 190. The required figure results by plotting these curves. The phase deviation for the second message signal (bottom right) is given again by  () = 2

Z 

 ()  = 20

For 0 ≤  ≤ 1, we have  () = 20

Z 

 () 

 = 102

For 1 ≤  ≤ 2

Z  ¡ ¢  () = (1) + 20 ( − 2)  = 10 + 10 2 − 1 − 40( − 1) 1 ¢ ¡ = 10 2 − 4 + 4 = 10 ( − 2)2

For 2 ≤  ≤ 4

 () = (2) + 20

Z  2

¡ ¢ (6 − 2) = 0 + 20 (6) ( − 2) − 20 2 − 4

= −20(2 − 6 + 8)

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

Finally, for   4 we recognize that  () = (4) = 0 (Note that the area under the curve from 0 to 4 is obviously zero). The required figure follows by plotting these expressions. Finally, the phase deviation in radians for the third message signal is again given by Z  Z   ()  = 20  ()   () = 2 For 0 ≤  ≤ 1, we have  () = 20

Z  0

For 1 ≤  ≤ 2  () = (1) + 20

Z  1

For 2 ≤  ≤ 25  () = (2) + 20

Z  2

(−2) = −202

2 = −20 + 40 ( − 1) = −60 + 40

¡ ¢ (10 − 4) = 20 + 20 (10) ( − 2) − 20(2) 2 − 4

= 20(−22 + 10 + 7) For 25 ≤  ≤ 3

 () = (25) − 20

Z 

25

2 = 15 − 40( − 25)

= (−250 + 500 + 140) − 40( + 25) = 390 − 40 − 100 = 290 − 40 For 3 ≤  ≤ 4  () = (3) + 20

Z  3

(2 − 8)  = 170 + 20(2 − 9) − 20(8)( − 3)

= 20(2 − 8 + 235) Finally, for   4 we recognize that  () = (4) = 20(16 − 32 + 235) = 150. The required figure follows by plotting these expressions. Problem 4.14 (a) The peak deviation is (20)(5) = 100 and  = 10. Thus, the modulation index is 100 10 = 10. (b) The magnitude spectrum, scaled to  , is the Fourier-Bessel spectrum of Figure 4.5 with  = 10

4.1. PROBLEMS

(c) Since  is not ¿ 1, this is not narrowband FM. The bandwidth exceeds 2 . (d) For phase modulation,  (5) = 10 or  = 2. Problem 4.15 The results are given in the following table. The deviation ratio is found by definition and the bandwidth follows from Carson’s rule. Part a b c d

 20 200 2000 20000

 = 10  =  1500 0.0133 0.1333 1.3333 13.3333

Problem 4.16 From  () = 

∞ X

 = 2 ( + 1)  30.39 kHz 34 kHz 70 kHz 430 kHz

 () cos[(  +  ) ]

=−∞

we obtain

∞ 2 ® 1 2 X  () =   2 () 2 =−∞ 

Also

2 ® 2 ®  () =  cos2 [   +  ()]

which, assuming that   À 1 so that  () has no dc component, is 2 ® 1 2  () =  2

This gives

∞ 1 2 1 2 X 2  =   () 2  2 =−∞ 

from which

∞ X

2 () = 1

=−∞

Problem 4.17 Since

1  () = 2

Z 

−

−(− sin )



1  = 2

Z 

−

( sin −) 

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

we can write 1  () = 2

Z 

1 cos ( sin  − )  +  2 −

Z 

−

sin ( sin  − ) 

The imaginary part of  () is zero, since the integrand is an odd function of  and the limits (− ) are even. Thus Z  1 cos ( sin  − )   () = 2 − Since the integrand is even 1 

 () =

Z  0

cos ( sin  − ) 

which is the first required result. With the change of variables  =  − , we have Z 1  cos [ sin ( − ) −  ( − )] (−1)   () =  0 Z 1  = cos [ sin ( − ) −  + ]   0 Since sin ( − ) = sin , we can write Z 1  cos [ sin  +  − ]   () =  0 Using the identity cos ( − ) = cos  cos  + sin  sin  with  =  sin  +  and  =  yields  () =

1  +

Z 

cos [ sin  + ] cos () 

1 

Z 

sin [ sin  + ] sin () 

Since sin () = 0 for all , the second integral in the preceding expression is zero. Also cos () = (−1)

4.1. PROBLEMS

Thus  () = (−1) However

1 − () = 

1 

Z 

cos [ sin  + ] 

Thus  () = (−1) − () or equivalently − () = (−1)  ()

Problem 4.18 (a) Peak frequency deviation = 10(8) = 80 Hz. (b) Since Z () = 2(8)

10 cos(20) =

160 sin(20) = 8 sin(20) 20

the peak phase deviation is 8 rad. (c)  = 8 (d) The power at the filter input is

 =

(10)2 2 = = 50 W 2 2

The power at the filter output is determined by the number of spectral components passed by the filter. The bandwith of the filter is 70 Hz, which is 35 Hz each sidfe of the carrier. Since  = 10 the filter passes 3 terms each side of the carrier. Thus the power at the filter output is # " 3 3 X 2 X 2 2 2 2  =  (8) =  (8) 0 (8) + 2 2 =−3  2 =1 £ ¤ = 50 (0172) + 2(0235)2 + 2(0113)2 + 2(0291)2 = 1674 W

(e) The input and output spectra are determined from the corresponding Fourier-Bessel spectra. Problem 4.19

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

We wish to find  such that  = 02 (8) + 2

 X

=1

02 (8) ≥ 080

This gives  = 7, yielding a power ratio of  = 08593. The bandwidth is therefore  = 2 = 2 (7) (250) = 3500 Hz For  ≥ 09, we have  = 8 for a power ratio of 09592. This gives  = 2 = 2 (8) (250) = 4000 Hz

Problem 4.20 From the given data, we have 1 = 110

kHz

1 = 005

2 =  (005) = 20

This gives =

20 = 400 005

and 1 =  (100)

kHz = 44 MHz

The two permissible local oscillator frequencies are 01 = 100 − 44 = 56

02 = 100 + 44 = 144

MHz MHz

The center frequency of the bandpass filter must be  = 100 MHz and the bandwidth is ¡ ¢  = 2 ( + 1)  = 2 (20 + 1) (10) 103 or

 = 420 kHz

Problem 4.21 For the circuit shown  ( ) =

  ( ) = 1  ( )  + 2  + 2 

4.1. PROBLEMS

or  ( ) = where

1 ³ ´ 1 +  2  − 21 

10−3  = = 10−6  3  10 ¡ ¢¡ ¢ =  = 103 10−9 = 10−6

 = 

A plot of the amplitude response shows that the linear region extends from approximately 54 kHz to118 kHz. Thus an appropriate carrier frequency is  =

118 + 54 = 86 2

kHz

The slope of the operating characteristic at the operating point is measured from the amplitude response. The result is ¢ ¡  ∼ = 8 10−6 Problem 4.22 We can solve this problem by determining the peak of the amplitude response characteristic. This peak falls at 1  = √ 2  ¡ ¢ It is clear that   100 MHz. Let  = 150 MHz and let  = 0001 10−12 . This gives =

(2)2 2 

¡ ¢ = 1126 10−3

We find the value of  by trial and error using plots of the amplitude response. An appropriate value for  is found to be 1  Ω. With these values, the discriminator constant is approximately ¢ ¡  ≈ 85 10−9 Problem 4.23 From (4.117)

¡ ¢  0 2 + 2∆  + 2   = lim 3 →0  +  2 +   +  

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

1. For first-order PLL ( = 0,  = 0) this is ¡ ¢  0 2 + 2∆  + 2 0 2 + 2∆  + 2   = lim = lim →0 →0 3 +  2 2 +   Thus 0 2 + 2∆  + 2 0 + (2∆ ) + (22 ) = lim →0 →0 2 +   1 + ( ) 0 =0 6= 0, ∆ = 0, and  = 0:   = lim →0 1 + ( ) 0 + (2∆ ) 2∆ = 6 = 0, ∆ 6= 0, and  = 0:   = lim →0 1 + ( )  0 + (2∆ ) + (22 ) 2 = lim =∞ 6 = 0, ∆ 6= 0, and  6= 0:   = lim →0 →0 1 + ( )  

  = lim For 0 For 0 For 0

2. For a second-order PLL ( 6= 0,  = 0) this is ¡ ¢  0 2 + 2∆  + 2 0 2 + 2∆  + 2 = lim   = lim →0 →0 3 +  2 +   2 +   +   Thus

2   which is zero for  = 0 and nonzero for  6= 0 for any finite values of 0 and ∆ . 3. For a third-order PLL ( 6= 0,  6= 0) this is ¡ ¢  0 2 + 2∆  + 2 0 = =0   = lim 3 2 →0  +   +   +       =

for any finite values 0 , ∆ , and . Problem 4.24 The Costas PLL is shown in Figure 4.26. The output of the top multiplier is  () cos    [2 cos (   + )] =  () cos  +  () cos (2  + ) which, after lowpass filtering, is  () cos . The quadrature multiplier output is  () cos    [2 sin (   + )] =  () sin  +  () sin (2   + ) which, after lowpass filtering, is  () sin . The multiplication of the lowpass filter outputs is  () cos  () sin  = 2 () sin 2

4.1. PROBLEMS

d / dt

 A

Figure 4.5: Phase-plane plot for a Costas PLL. as indicated. Note that with the assumed input  () cos   and VCO output 2 cos (  + ), the phase error is . Thus the VCO is defined by   = = −  ()   This is shown in Figure 4.5. Since the   intersection is on a portion of the curve with negative slope, the point A at the origin is a stable operating point. Thus the loop locks with zero phase error and zero frequency error. Problem 4.25 ¡ ¢ With  () =  cos(20 ), we desire 0 () =  cos 2 73 0 . Assume that the VCO output is a pulse train with frequency 13  . The pulse should be narrow so that the seventh harmonic is relatively large. The spectrum of the VCO output consists of components separated by 1 3 0 with an envelope of sinc(  ), where  is the pulse width. The center frequency of the bandpass filter is 73 0 and the bandwidth is on the order of 13 0 as shown in Figure.4.6. Problem 4.26 The phase plane is defined by  = ∆ −  sin  () at  = 0,  =   , the steady-state phase error. Thus −1

  = sin

∆ 

−1

= sin

∆ 2 (100)

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING This component (the fundamental) tracks the input signal

Bandpass filter passband

1 f0 3

7 f0 3

Figure 4.6: Filtering scheme for Problem 3.44. For ∆ = 2 (30)

30 100

50 100

80 100

−1

 = sin

For ∆ = 2 (50) −1

  = sin For ∆ = 2 (80)

−1

  = sin For ∆ = −2 (80)

−1

  = sin

−80 100

= 1746 degrees

= 30 degrees

= 5313 defrees

= −5313 degrees

For ∆ = 2 (120), there is no stable operating point and the frequency error and the phase error oscillate (PLL slips cycles continually). Problem 4.27 From the definition  () =  −  ()

4.1. PROBLEMS

it is clear that in the limit as  → ∞,  (0) = ∞, and  () = 0 for  6= 0. Also Z ∞ Z ∞ ¯∞  () =  −   = −−  −−  ¯0 = 1 −∞

Therefore the given  () satisfies all properties of an impulse function in the limit as  → ∞. Problem 4.28 From the definition of the transfer function 

+ +

Θ ()   () ³ ´ = = Φ ()  +   ()  +  + 

which is

+

 ( + )  ( + ) Θ () = = 2 Φ ()  ( + ) +  ( + )  + ( + )  +  

Therefore 2 + 2   +  2 = 2 + ( + )  +   This gives  = and =

Problem 4.29 Since

p  

 +  √ 2  

Θ ()  ( + ) = Φ ()  ( + ) +  ( + )

we first determine () = so that

 ( + )  ( + ) Ψ() =1− = Φ ()  ( + ) +  ( + )  ( + ) +  ( + ) ∙

0 2∆ 2 + 2 + 3 Ψ() =    ∙

¸∙

0 2 + 2∆  + 2 Ψ() = 2

 ( + )  ( + ) +  ( + )

¸∙

( + )  ( + ) +  ( + )

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

Multiplying by  and taking the limit gives ¸∙ ¸ ∙ ( + ) 2   = lim 0 + 2∆ + →0   ( + ) +  ( + ) or

¸∙ ¸ ∙  2   = lim 2∆ + →0  

Thus For 0 6= 0, ∆ = 0, and  = 0:

  = 0

For 0 6= 0, ∆ 6= 0, and  = 0:

  = lim

For 0 6= 0, ∆ 6= 0, and  6= 0:

2∆  2    = lim =∞ →0   →0

Problem 4.30 Since the phase error is assumed small, so that cos [()] ≈ 1, the input from the top branch (see Figure 4.26 in the text.) to the multiplier preceding the loop filter is simply the phase error, () cos[()] ≈ (). The other input to the multiplier, for suﬃciently small phase error, is () sin[()] ≈ ()() Thus, the input to the loop filter preceding the VCO is 2 ()(). Since () is a step function 2 () = () = ( − 0 ) so that the input to the loop filter is ( − 0 )(), which is exactly the loop filter input to a PLL. We therefore see that this problem is identical to Example 4.8 in the text and the phase error as a function of time is that given by (4.140) with  replaced by  − 0 . The parameter  in the loop filter transfer function is  =   as defined by (4.136). This problem illustrates the equivalence of a Costas PLL to a PLL in cases where the phase error is small. Problem 4.31 With the assumed VCO output represented by −() we have the output of the complex phase detector defined as  ()[()−()] =  () cos[() − ()] +  () sin[() − ()] This is Figure 4.26 with the top branch representing the real component  () cos[()] and the bottom branch representing the quadrature, or imaginary, component  () sin[()]. Costas PLLs are often simulated using a complex multiplication in this manner.

4.1. PROBLEMS

Problem 4.32 For  =  we can write, from (4.146),  () =  [cos    + cos (  +  ) ] which is  () =  [(1 + cos   ) cos    − sin    sin  ] This yields  () =  () cos [   +  ()] where ¸ sin     () = tan 1 + cos    ∙ ¸   −1 = tan tan = 2 2 −1

This gives 1   () = 2 

∙

2  2

1 =  2

For  = − , we get 1  () = −  2 Finally, for  À  we see from the phasor diagram that  () ≈  () =   and  () =

  (2 ) =  2 

Problem 4.33 Let  be the peak-to-peak value of the data signal. The peak error is 020% and the peak-to-peak error is 0004 . The required number of quantizating levels is  = 250 ≤ 2 =  0004 so we choose  = 256 and  = 8. The bandwidth is  = 2  log2  = 2 (8)

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

Figure 4.7: Single-sided spectrum for Problem 3.54. The value of  is estimated by assuming that the speech is sampled at the Nyquist rate. Then the sampling frequency is  = 2 = 8 kHz. Each sample is encoded into  = 8 pulses. Let each pulse be  with corresponding bandwidth 1 . For our case =

1 1 =  2 

Thus the bandwidth is 1 = 2  = 2 log2  = 2 = 2  and so  = 1. For  = 1  = 2 (8 000) (8) = 112 kHz

Problem 4.34 The single-sided spectrum for  () is shown in Figure 4.7. From the definition of  () we have  () = 1  ( ) + 2  ( ) ∗  ( ) The spectrum for  ( ) is given in Figure 4.8. Demodulation can be a problem since it may be difficult to filter the desired signals from the harmonic and intermodulation distortion caused by the nonlinearity. As more signals are included in  (), the problem becomes more difficult. The difficulty with harmonically related carriers is that portions of the spectrum of  ( ) are sure to overlap. For example, assume that 2 = 21 . For this case, the harmonic distortion arising from the spectrum centered about 1 falls exactly on top of the spectrum centered about 2 .

4.2. COMPUTER EXERCISES

Yf a1

2a2W

a2W f

f 2  f1

2 f2

f1  f 2

2 f2

Figure 4.8: Output spectrum for Problem 3.54.

4.2

Computer Exercises

Computer Exercise 4.1 This Computer Exercise is a simple modification of Computer Exercise 4.1. Understanding the relationship between Figure 4.7 and the result of this exercise is, however, important. The MATLAB code follows. % File: c4ce1.m fs = 500; delt = 1/fs; t = 0:delt:1-delt; npts = length(t); fm = 20; fd = [10 20 100]; for k=1:3 beta = fd(k)/fm; cxce = exp(i*beta*sin(2*pi*fm*t)); as = (1/npts)*abs(fft(cxce)); evenf = [as(fs/2:fs) as(1:fs/2-1)]; fn = -fs/2:fs/2-1; subplot(3,1,k); stem(fn,2*evenf,’.’) ylabel(’Amplitude’) end

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

Amplitude

0 -250

-200

-150

-100

-50

100

150

200

250

-200

-150

-100

-50

100

150

200

250

-200

-150

-100

-50

100

150

200

250

Amplitude

0 -250 1

0.5

0 -250

Figure 4.9: FM spectra with  held constant and modulation indices of 0.5, 1 and 5. % End of script file. Executing the program gives the following spectral plots.

Computer Exercise 4.2 For a square-wave message signal, the phase deviation will be a triangular-wave. The triangular-wave message signal is generated using a 6-term Fourier series. The MATLAB code follows. % File: c4ce2.m fs = 1000; % sampling frequency delt = 1/fs; % sampling increment t = 0:delt:1-delt; % time vector npts = length(t); % number of points fn = (0:(npts/2))*(fs/npts); % frequency vector for plot % Generate a triangular-wave signal as the phase deviation corresponding % to a square-wave message signal using Fourier series and plot.

4.2. COMPUTER EXERCISES

fm = 5; b = [1 -1/9 1/25 -1/49 1/81 -1/121]; c = 2*pi*fm*[1 3 5 7 9 11]; m = zeros(1,length(t)); for n=1:length(c) m=m+b(n)*sin(c(n)*t); end m = m/max(m); plot(t,m), xlabel(’Time’) % Determine spectra. fd1 = 0.10; xct1 = sin(2*pi*250*t+fd1*m); % modulated carrier asxc1 = (2/npts)*abs(fft(xct1)); % amplitude spectrum ampspec1 = asxc1(1:((npts/2)+1)); % positive frequency portion fd2 = 1.0; xct2 = sin(2*pi*250*t+fd2*m); % modulated carrier asxc2 = (2/npts)*abs(fft(xct2)); % amplitude spectrum ampspec2 = asxc2(1:((npts/2)+1)); % positive frequency portion fd3 = 10.0; xct3 = sin(2*pi*250*t+fd3*m); % modulated carrier asxc3 = (2/npts)*abs(fft(xct3)); % amplitude spectrum ampspec3 = asxc3(1:((npts/2)+1)); % positive frequency portion % Plot spectra. figure % new figure subplot(3,1,1) stem(fn,ampspec1,’.k’); ylabel(’Magnitude - fd=0.1’) subplot(3,1,2) stem(fn,ampspec2,’.k’); ylabel(’Magnitude - fd=1’) subplot(3,1,3) stem(fn,ampspec3,’.k’); xlabel(’Frequency’), ylabel(’Magnitude - fd=10’) % End of script file. Executing the program yields the results illustrated in Figures 4.10 and 4.11 , respectively. Figure 4.10 shows the triangular-wave message signal. Figure 4.11 shows the spectrum. As described in the text, all sum and diﬀerence frequencies are present. This non-linear eﬀect is most easily seen for  = 10 since higher deviations spread the spectrum more than small deviation constants.

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

1 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1

0.1

0.2

0.3

0.4

0.5 Time

0.6

0.7

0.8

0.9

Figure 4.10: Triangular wave representing the phase deviation. Computer Exercise 4.3 The MATLAB program for generating Figures 4.24 and 4.25 is given in the text in Computer Example 4,4 (See page 192). The purpose of this Computer Exercise is to gain experience in executing these simulations. Since the program is given, it will be executed for only one case:  = 10 Hz, ∆ = 40 Hz and  = 0707. The results follow.

The other three cases are easily developed by simply entering the appropriate step sizes. Computer Exercise 4.4 The labeling is straightforward. The appropriate model is illustrated in Figure 4.19. The output of the phase detector is s1. The output of the sinusoidal nonlinearity is s2. The next several lines of code implement the loop filter. The input to the VCO is vco_in and the VCO output of the VCO output is vco_out. The main sources of error are as follows: The output of the phase detector is not the diﬀerence between the current input phase and current VCO output but the diﬀerence between the current input phase and the previous VCO output. Thus, there is a one sample delay induced in the simulation loop. This reduces the phase margin of the PLL. Another error source is the approximation of integration by a discrete-time integrator. Sampling,as we have seen, can induce errors. Computer Exercise 4.5

Magnitude - fd=10

Magnitude - fd=1

Magnitude - fd=0.1

4.2. COMPUTER EXERCISES

0.5

0 0

100

150

200

250

300

350

400

450

500

100

150

200

250

300

350

400

450

500

100

150

200

250 300 Frequency

350

400

450

500

0.5

0 1

0.5

Figure 4.11: Spectra for FM signals with a square-wave message signal.

Input Frequency and VCO Freqeuncy 50

Frequency - Hertz

-10

0.1

0.2

0.3

0.4 0.5 0.6 Time - Seconds

0.7

0.8

0.9

Figure 4.12: Input frequency and VCO frequency.

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING Phase Plane 50

Frequency Error - Hz

-10

0.5

1.5 2 Phase Error / pi

2.5

3.5

Figure 4.13: Phase plane for Computer Exercise 4.3. The MATLAB code for the PLL simulation with a line added for entering the sampling frequency follows. % File: c4ce5.m % beginning of preprocessor clear all % be safe fdel = input(’Enter frequency step size in Hz  ’); fn = input(’Enter the loop natural frequency in Hz  ’); zeta = input(’Enter zeta (loop damping factor)  ’); fs = 2000; % default sampling frequency fs = input(’Enter the sampling frequency (default = 2000)  ’); npts = fs ; % default number of simulation point T = 1/fs; t = (0:(npts-1))/fs; % time vector nsettle = fix(npts/10); % set nsettle time as 0.1*npts Kt = 4*pi*zeta*fn; % loop gain a = pi*fn/zeta; % loop filter parameter filt_in_last = 0; filt_out_last=0; vco_in_last = 0; vco_out = 0; vco_out_last=0; % end of preprocesso - beginning of simulation loop for i=1:npts if insettle

4.2. COMPUTER EXERCISES fin(i) = 0; phin = 0; else fin(i) = fdel; phin = 2*pi*fdel*T*(i-nsettle); end s1 = phin - vco_out; s2 = sin(s1); % sinusoidal phase detector s3 = Kt*s2; filt_in = a*s3; filt_out = filt_out_last + (T/2)*(filt_in + filt_in_last); filt_in_last = filt_in; filt_out_last = filt_out; vco_in = s3 + filt_out; vco_out = vco_out_last + (T/2)*(vco_in + vco_in_last); vco_in_last = vco_in; vco_out_last = vco_out; phierror(i)=s1; fvco(i)=vco_in/(2*pi); freqerror(i) = fin(i)-fvco(i); end % end of simulation loop - beginning of postprocessor kk = 0; while kk == 0 k = menu(’Phase Lock Loop Postprocessor’,... ’Input Frequency and VCO Frequency’,... ’Phase Plane Plot’,... ’Exit Program’); if k == 1 plot(t,fin,t,fvco), grid title(’Input Frequency and VCO Freqeuncy’) xlabel(’Time - Seconds’) ylabel(’Frequency - Hertz’) pause elseif k == 2 plot(phierror/2/pi,freqerror), grid title(’Phase Plane’) xlabel(’Phase Error / 2*pi’) ylabel(’Frequency Error - Hz’) pause

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING Phase Plane 50

Frequency Error - Hz

-10

0.5

1.5

2 2.5 Phase Error / 2*pi

3.5

4.5

Figure 4.14: Phase-plane plot with  = 400̇. elseif k == 3 kk = 1; end end % end of postprocessor % End of script file. The preceding program was executed twice; once for  = 400 and once for  = 1000. The siumulation for  = 400 is shown in Figure 4.14. We note that the curves, especially for large frequency error, are not smooth, This is an indication that the samplingt frequency is too low. We therefore increase the sampling frequency from 400 to 1000. The results are illustrated in Figure 4.15. We now see that the curve defining the phase plane is much smoother. In addition, for  = 400 we see that the PLL skips 4 cycles while acquiring lock. For  = 1000 we see that the PLL skips 3 cycles while acquiring lock. By executing the simulation a number of times

4.2. COMPUTER EXERCISES

Phase Plane 50

Frequency Error - Hz

-10

0.5

1.5 2 Phase Error / 2*pi

2.5

Figure 4.15: Phase-plane plot with  = 1000

3.5

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

we see that additional increases in the sampling frequency have no eﬀect on the results. An eﬀective way to determine an appropriate sampling frequency is to set the sampling frequency at a high value and then reduce the sampling frequency until the results change and the waveforms are distorted. At this point the sampling frequency is too low. Computer Exercise 4.6 A rectangulator integrator is defined by y[n] = y[n − 1] + Tx[n − 1] so that, in the simulation program, the line filt_out = filt_out_last + (T2) ∗ (filt_in + filt_in_last) is replaced by filt_out = filt_out_last + T ∗ filt_in_last The VCO output is modified in exactly the same way. The revised MATLAB code follows. % File: c4ce6.m % beginning of preprocessor clear all % be safe fdel = input(’Enter frequency step size in Hz  ’); fn = input(’Enter the loop natural frequency in Hz  ’); zeta = input(’Enter zeta (loop damping factor)  ’); npts = 2000; % default number of simulation points fs = 2000; % default sampling frequency T = 1/fs; t = (0:(npts-1))/fs; % time vector nsettle = fix(npts/10); % set nsettle time as 0.1*npts Kt = 4*pi*zeta*fn; % loop gain a = pi*fn/zeta; % loop filter parameter filt_in_last = 0; filt_out_last=0; vco_in_last = 0; vco_out = 0; vco_out_last=0; % end of preprocesso - beginning of simulation loop for i=1:npts if insettle fin(i) = 0; phin = 0; else

4.2. COMPUTER EXERCISES fin(i) = fdel; phin = 2*pi*fdel*T*(i-nsettle); end s1 = phin - vco_out; s2 = sin(s1); % sinusoidal phase detector s3 = Kt*s2; filt_in = a*s3; filt_out = filt_out_last + T*filt_in_last; filt_in_last = filt_in; filt_out_last = filt_out; vco_in = s3 + filt_out; vco_out = vco_out_last + T*vco_in_last; vco_in_last = vco_in; vco_out_last = vco_out; phierror(i)=s1; fvco(i)=vco_in/(2*pi); freqerror(i) = fin(i)-fvco(i); end % end of simulation loop - beginning of postprocessor kk = 0; while kk == 0 k = menu(’Phase Lock Loop Postprocessor’,... ’Input Frequency and VCO Frequency’,... ’Phase Plane Plot’,... ’Exit Program’); if k == 1 plot(t,fin,t,fvco), grid title(’Input Frequency and VCO Freqeuncy’) xlabel(’Time - Seconds’) ylabel(’Frequency - Hertz’) pause elseif k == 2 plot(phierror/2/pi,freqerror), grid title(’Phase Plane’) xlabel(’Phase Error / pi’) ylabel(’Frequency Error - Hz’) pause elseif k == 3 kk = 1; end

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING Input Frequency and VCO Freqeuncy 50

Frequency - Hertz

-10

0.1

0.2

0.3

0.4 0.5 0.6 Time - Seconds

0.7

0.8

0.9

Figure 4.16: PLL behavior with a sampling frequency of 2000 Hz. end % end of postprocessor % End of script file. We compare the results of rectangular and trapezoidal integration are compared using the result of Computer Exercise 4.3, which uses trapezoidal integration. The simulation parameters used in Computer Exercise are:  = 10 Hz, ∆ = 40 Hz and  = 0707. The result is as follows: Reducing the sampling frequency to 500 Hz results in the following: We see that with a reduced sampling frequency the PLL skips four cycles rather than three and takes slightly longer to lock.. How would the PLL respond with a sampling frequency of 500 Hz and trapezoidal integration? This is left as an exercise for the student. In general the integration algorithm should be as accurate as possible (trapezoidal integration is better than rectanglar integration). Approximation accuracy becomes more important as the sampling frequency is reduced. Computer Exercise 4.7 The MATLAB program for the PLL with the limiting phase detector follows. We note that the phase detector has been modified and a line has been added for entering the phase detector constant .

4.2. COMPUTER EXERCISES

37 Input Frequency and VCO Freqeuncy

Frequency - Hertz

-10

0.1

0.2

0.3

0.4 0.5 0.6 Time - Seconds

0.7

0.8

0.9

Figure 4.17: PLL behavior with a sampling frequency of 500 Hz. % File: c4ce7.m % beginning of preprocessor clear all % be safe fdel = input(’Enter frequency step size in Hz  ’); fn = input(’Enter the loop natural frequency in Hz  ’); zeta = input(’Enter zeta (loop damping factor)  ’); AA = input(’Enter phase detector constant A  ’); npts = 2000; % default number of simulation points fs = 2000; % default sampling frequency T = 1/fs; t = (0:(npts-1))/fs; % time vector nsettle = fix(npts/10); % set nsettle time as 0.1*npts Kt = 4*pi*zeta*fn; % loop gain a = pi*fn/zeta; % loop filter parameter filt_in_last = 0; filt_out_last=0; vco_in_last = 0; vco_out = 0; vco_out_last=0; % end of preprocessor - beginning of simulation loop for i=1:npts if insettle fin(i) = 0; phin = 0; else

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING fin(i) = fdel; phin = 2*pi*fdel*T*(i-nsettle); end s1 = phin - vco_out; s2 = sin(s1); % sinusoidal phase detector if s2-AA s2=-AA; elseif s2AA s2=AA; end s3 = Kt*s2; filt_in = a*s3; filt_out = filt_out_last + (T/2)*(filt_in + filt_in_last); filt_in_last = filt_in; filt_out_last = filt_out; vco_in = s3 + filt_out; vco_out = vco_out_last + (T/2)*(vco_in + vco_in_last); vco_in_last = vco_in; vco_out_last = vco_out; phierror(i)=s1; fvco(i)=vco_in/(2*pi); freqerror(i) = fin(i)-fvco(i); end % end of simulation loop - beginning of postprocessor kk = 0; while kk == 0 k = menu(’Phase Lock Loop Postprocessor’,... ’Input Frequency and VCO Frequency’,... ’Phase Plane Plot’,... ’Exit Program’); if k == 1 plot(t,fin,t,fvco), grid title(’Input Frequency and VCO Freqeuncy’) xlabel(’Time - Seconds’) ylabel(’Frequency - Hertz’) pause elseif k == 2 plot(phierror/2/pi,freqerror), grid title(’Phase Plane’) xlabel(’Phase Error / 2*pi’)

4.2. COMPUTER EXERCISES

Phase Plane 50

Frequency Error - Hz

-10

-20

6 Phase Error / 2*pi

Figure 4.18: PLL acquisition with  = 05. ylabel(’Frequency Error - Hz’) pause elseif k == 3 kk = 1; end end % end of postprocessor % End of script file.

We see that with  = 05, the loop slips 5 cycles, whereas with ̇  1 (limiter absent), the PLL slips only three cycles with the parameters given. The acquisition time is therefore increased. Computer Exercise 4.8 The imperfect PLL is defined in Problem 4.28 as a PLL in which the loop filter is defined as +  () =  + 

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

A non-zero value of  moves the away from the origin so that the integration is no longer "perfect". In this example we use a more eﬃcient and more general simulation model. The simulation model is developed in detail in Chapter 6 of the textbook: W. H. Tranter, K. S. Shanmugan, T. S. Rappaport, and K. L. Kosbar, Principles of Communication Systems Simulation with Wireless Applications, Printice Hall PTR, 2004. The MATALB code follows. (Note: The integration routine used here is the transposed direct form II realization of a trapezoidal integrator.) % File: c4ce8.m clear all % be safe disp(’ ’) % insert blank line fdel = input(’Enter the size of the frequency step in Hertz  ’); fn = input(’Enter the loop natural frequency in Hertz  ’); lambda = input(’Enter lambda, the relative pole offset  ’); disp(’ ’) disp(’Accept default values:’) disp(’ zeta = 1/sqrt(2) = 0.707,’) disp(’ fs = 200*fn, and’) disp(’ tstop = 1’) dtype = input(’Enter y for yes or n for no  ’,’s’); if dtype == ’y’ zeta = 1/sqrt(2); fs = 200*fn; tstop = 1; else zeta = input(’Enter zeta, the loop damping factor  ’); fs = input(’Enter the sampling frequency in Hertz  ’); tstop = input(’Enter tstop, the simulation runtime  ’); end npts = fs*tstop+1; % number of simulation points t = (0:(npts-1))/fs; % default time vector nsettle = fix(npts/10); % set nsettle time as 0.1*npts tsettle = nsettle/fs; % set tsettle % % The next two lines establish the loop input frequency and phase % deviations. % fin = [zeros(1,nsettle),fdel*ones(1,npts-nsettle)]; phin = [zeros(1,nsettle),2*pi*fdel*t(1:(npts-nsettle))];

4.2. COMPUTER EXERCISES

disp(’ ’) % insert blank line % w2b=0; w2c=0; s5=0; phivco=0; %initialize twopi = 2*pi; % define 2*pi twofs = 2*fs; % define 2*fs G = 2*pi*fn*(zeta+sqrt(zeta*zeta-lambda)); % set loop gain a = 2*pi*fn/(zeta+sqrt(zeta*zeta-lambda)); % set filter parameter a1 = a*(1-lambda); a2 = a*lambda; % define constants phierror = zeros(1,npts); % initialize vector fvco = zeros(1,npts); % initialize vector % beginning of simulation loop for i = 1:npts s1 = phin(i) - phivco; % phase error s2 = sin(s1); % sinusoidal phase detector s3 = G*s2; s4 = a1*s3; s4a = s4-a2*s5; % loop filter integrator input w1b = s4a+w2b; % filter integrator (step 1) w2b = s4a+w1b; % filter integrator (step 2) s5 = w1b/twofs; % generate fiter output s6 = s3+s5; % VCO integrator input w1c = s6+w2c; % VCO integrator (step 1) w2c = s6+w1c; % VCO integrator (step 2) phivco = w1c/twofs; % generate VCO output phierror(i) = s1; % build phase error vector fvco(i) = s6/twopi; % build VCO input vector end % end of simulation loop freqerror=fin-fvco; % build frequency error vector kk = 0; while kk == 0 k = menu(’Phase Lock Loop Postprocessor’,... ’Input Frequency and VCO Frequency’,... ’Input Phase and VCO Phase’,... ’Frequency Error’,’Phase Error’,’Phase Plane Plot’,... ’Phase Plane and Time Domain Plots’,’Exit Program’); if k == 1 plot(t,fin,’k’,t,fvco,’k’) title(’Input Frequency and VCO Freqeuncy’) xlabel(’Time - Seconds’);ylabel(’Frequency - Hertz’);

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING pause elseif k ==2 pvco=phin-phierror;plot(t,phin,t,pvco) title(’Input Phase and VCO Phase’) xlabel(’Time - Seconds’);ylabel(’Phase - Radians’); pause elseif k == 3 plot(t,freqerror);title(’Frequency Error’) xlabel(’Time - Seconds’);ylabel(’Frequency Error - Hertz’); pause elseif k == 4 plot(t,phierror);title(’Phase Error’) xlabel(’Time - Seconds’);ylabel(’Phase Error - Radians’); pause elseif k == 5 ppplot elseif k == 6 subplot(211);phierrn = phierror/pi; plot(phierrn,freqerror,’k’);grid; title(’Phase Plane Plot’);xlabel(’Phase Error /Pi’); ylabel(’Frequency Error - Hertz’);subplot(212) plot(t,fin,’k’,t,fvco,’k’);grid title(’Input Frequency and VCO Freqeuncy’) xlabel(’Time - Seconds’);ylabel(’Frequency - Hertz’); subplot(111) elseif k == 7 kk = 1; end end % End of script file. The routine is executed with the following dialog:  ce4_8 Enter the size of the frequency step in Hertz  40 Enter the loop natural frequency in Hertz  10 Enter lambda, the relative pole offset  0.1 Accept default values:

4.2. COMPUTER EXERCISES

43 Input Frequency and VCO Freqeuncy

Frequency - Hertz

-10

0.1

0.2

0.3

0.4 0.5 0.6 Time - Seconds

0.7

0.8

0.9

Figure 4.19: PLL acquisition with  = 10 Hz, ∆ = 40 Hz,  = 0707 and  = 01 zeta = 1/sqrt(2) = 0.707, fs = 200*fn, and tstop = 1 Enter y for yes or n for no  y The result is shown in the following figure. Comparing with the result of Computer Example 4.4 we see that the see that the number of cycles slipped increases from three to five. The acquisition time is increased accordingly. Computer Exercise 4.9 The loop filter for a third-order PLL is defined by (See Table 4.3)  () = (2 +  + )2 This gives the transfer function () =

 2 +   +    (2 +  + )2 =  +  (2 +  + )2 3 +  2 +   +  

The characteristic equation is () = 3 +  2 +   +  

CHAPTER 4. ANGLE MODULATION AND MULTIPLEXING

We see that for small loop gain  the system exhibits a third-order pole at the origin. As  increases, a pair of these poles become conjugates and move into the right-hand plane, which means that the system is unstable. As  continues to increase the complexconjugate pair move back into the left-half plane and the system becomes stable. Therefore a third-order PLL is unstable for small values of the loop gain but stable for larger values of loop gain. The MATLAB program follows. % File c4ce9 clear all clf a = .70; b = .70; Kt0 = 0.01; %D = s^3 +Kt*(s^2 + a*s^2 + b); for k = 1:4000 Kt = k*Kt0; dd = [1 Kt Kt*a Kt*b]; q = roots(dd); if k == 1 dd q elseif k == 4000 dd q end plot(q,’.’) if k == 1 hold on axis([-3 1 -2 2]) axis (’square’) grid xlabel(’real’), ylabel(’imaginary’) end end The pole migration is illustrated in the following root-locus plot.

Computer Exercise 4.10 As mentioned in the text, there are many ways to work this problem. Rather than develop the simulation we oﬀer hints and point out pitfalls. Problem 4.25 illustrated one technique.

4.2. COMPUTER EXERCISES

2 1.5 1

imaginary

0.5 0 -0.5 -1 -1.5 -2 -3

-2.5

-2

-1.5

-1 real

-0.5

0.5

Figure 4.20: Root-locus plot for Computer Exercise 4.9. ¡ ¢ With  () =  cos(20 ), we desire 0 () =  cos 2 75 0 . Assume that the VCO output is a pulse train with frequency 15  . The pulse should be narrow so that the seventh harmonic is relatively large. The spectrum of the VCO output consists of components separated by 1 5 0 with an envelope of sinc(  ), where  is the pulse width. There are two problems with this technique. The first issue is that the output of the VCO, since it is a pulse, is rich in harmonic content and getting the PLL to lock may be diﬃcult. This issue can be solved by having the VCO output not be a pulse but a sinusoid having a quiscent frequency of 15 0 . We then have a second PLL output that is a pulse train having frequency 15 0  We desire to track the 7th harmonic. Designing the filter can be diﬃcult since adjacent harmonics must be rejected. This problem can be addressed by incorporating a second PLL designed to track the desired harmonic.

Chapter 5

Principles of Baseband Digital Data Transmission 5.1

Problem Solutions

Problem 5.1 a. Split phase or bipolar RZ; b. NRZ change or NRZ mark; c. Split phase; d. NRZ mark; e. Polar RZ; f. Unipolar RZ. Problem 5.2 These are a matter of …lling in the details given at the ends of Examples 5.1 through 5.5. Problem 5.3 This is a matter of adapting Figures 5.2 top and bottom to the data sequence given in the problem statement. Problem 5.4 This is a matter of adapting Figure 5.2, second plot, to the data sequence given in the problem statement. 1

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION

Problem 5.5 This is a matter of adapting Figure 5.2, third - …fth plots, to the data sequence given in the problem statement. Problem 5.6 a. 4 kbps; b. 2 kbps; c. 2 kbps; d. 4 kbps for …rst null and 2 kbps for second null.

Problem 5.7 The step response is given in Problem 2.65c as ys (t) = 1

exp

2 f3 p 2

2 f3 2 f3 cos p + sin p 2 2

u (t)

The speci…ed input is x (t) = u (t)

2u (t

T ) + u (t

2T )

Using the linear, time invariant properties of the system, we can immidiately write down the output as y (t) = ys (t)

2ys (t

T ) + ys (t

2T )

Plots are given in Figure 5.1 for the two cases speci…ed in the problem statement. : Problem 5.8 The references should be to Equations (5:30) and (5:29), respectively. The result for the response follows by noting that the response to Au (t) is A [1 exp ( t=RC)] u (t) and the responsse to Au (t T ) is A [1 exp ( (t T )=RC)] u (t T ). When superposition is applied, the result is that given by (5:30) :

5.1. PROBLEM SOLUTIONS

3 f3T = 2

y step(t)

1 0 -1 -2

0.5

1.5 t/T f3T = 20

2.5

0.5

1.5 t/T

2.5

y step(t)

1 0 -1 -2

Problem 5.9

The reference should be to (5:37). Clearly, with 1 and 0 for jf j > 1+ 2T = 2T :

= 0, the spectrum is T for jf j

1 2T

1 = 2T

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION

Problem 5.10 Because PRC (f ) is even, it follows that Z 1 1 pRC (t) = F [PRC (f )] = 2 PRC (f ) cos (2 f t) df 0

Z (1

= 2T

Z (1+ )=2T

)=2T

cos (2 f t) df + T

sin = 2T +T

2 t Z (1+ )=2T (1

sin = T +

T 2

sin +T

1+ T

cos

sin T

2 t 1

sin +T

2 t T

cos sin

2 t h T f sin T

1+ T

+ 2 ft

+2 t

sin = T +

2h t

T sin 2

T sin

sin +T

1+ 2T

1+ T

T= + 2 t 1 2T

2 ft

T sin 2

1 2T

f T

2 ft

2 t

i (1+ )=2T

)=2T

2 t 1 2T

+ 2 f t + cos

i (1+ )=2T (1

2 t

1 2T

)=2T

cos (2 f t) df

1+ T

)=2T

2 t Z (1+ )=2T

sin = T

1 + cos

1+ T

i i

T sin 2 T sin

h h

1 2T

T t

T= + 2 t 1 2T

2 T= 2 t 2 T= 2 t sin ( t=T ) cos ( t=T ) 1 1 1 t = + cos t=T 2 t=T T =2 t 1 T =2 t + 1 T 1 sin ( t=T ) = 1 t=T ) 2 cos ( t=T 1 (T =2 t) " # (T =2 t)2 sin ( t=T ) cos ( t=T ) = t=T ) = sinc (t=T ) 2 cos ( t=T 1 (T =2 t) 1 (2 t=T )2

i i sin

t T

)=2T

5.1. PROBLEM SOLUTIONS

Problem 5.11 Nyquist’s pulse shaping criterion is a su¢ cient condition. Each of these spectra, being built of square and triangular shapes, will have regularly spaced zero crossings in the time domain. a. This spectrum has the zero-ISI property. The corresponding pulse function is p1 (t) = 2W sinc (2W t) + W sinc (W t) k The …rst sinc-function has zero crossings at tk = 2W seconds (k an integer) and the second k one has zero crossings at tk = W seconds (k an integer). The appropriate sample interval 1 is ts = W seconds for 0 ISI (both terms must contribute zero ISI).

b. This spectrum has the zero-ISI property. The corresponding pulse function is p2 (t) = 2W sinc2 (2W t) + W sinc (W t) k seconds (k an integer) and the The …rst sinc–squared function has zero crossings at tk = 2W k second one has zero crossings at tk = W seconds (k an integer). The appropriate sample 1 interval is ts = W seconds for 0 ISI (both terms must contribute zero ISI).

c. This spectrum has the zero-ISI property. The corresponding pulse function is p3 (t) = 4W sinc (4W t)

W sinc2 (W t)

k The …rst sinc–squared function has zero crossings at tk = 4W seconds (k an integer) and the k second one has zero crossings at tk = W seconds (k an integer). The appropriate sample 1 interval is ts = W seconds for 0 ISI (both terms must contribute zero ISI).

d. This spectrum has the zero-ISI property. By the modulation theorem, the corresponding pulse function is p4 (t) = 2W sinc (W t) cos (2 W t) 1 The zero crossings occur where the sinc-function is zero or at integer multiples of tk = W seconds.

e. This spectrum has the zero-ISI property. The corresponding pulse function is p5 (t) = 2W sinc2 (2W t)

W sinc2 (W t)

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION

k The …rst sinc–squared function has zero crossings at tk = 2W seconds (k an integer) and the k second one has zero crossings at tk = W seconds (k an integer). The appropriate sample 1 interval is ts = W seconds

Problem 5.12 The MATLAB program is given below. The plot of the desired frequency response functions follows the program. % Problem 5_12 % Computation and plotting of transmitter and receiver frequency responses % for raised cosine signaling through lowpass Butterworth …ltered channel % clear all A = char(’-’,’:’,’–’,’-.’); clf f3dB = input(’Enter channel …lter 3-dB frequency, Hz => ’); n_poles = input(’Enter number of poles for Butterworth …lter repr channel => ’); Rb = input(’Enter data rate, bits/s => ’); beta0 = input(’Enter vector of betas (max no = 3) => ’); Lbeta = length(beta0); for n = 1:Lbeta beta = beta0(n); f1 = 0:.1:(1-beta)*Rb/2; HR1 = sqrt(1+(f1/f3dB).^(2*n_poles)); f2 = (1-beta)*Rb/2:.1:(1+beta)*Rb/2; HR2 = sqrt(1+(f2/f3dB).^(2*n_poles)).*cos(pi*(f2-(1-beta)*Rb/2)./(2*beta*Rb)); f3 = (1+beta)*Rb/2:.1:Rb; HR3 = zeros(size(f3)); f = [f1 f2 f3]; HR = [HR1 HR2 HR3]; plot(f,HR,A(n,:), ’LineWidth’, 1.5),xlabel(’{nitf}, Hz’),... ylabel(’j{nitH_R}({nitf})j or j{nitH_T}({nitf})j’) if n == 1 hold on end end title([’Bit rate = ’,num2str(Rb),’bps; channel …lter 3-dB frequency = ’,num2str(f3dB), ’Hz; no. of poles = ’,num2str(n_poles)]),... if Lbeta == 1 legend([’nbeta = ’,num2str(beta0(1))],1) elseif Lbeta == 2

5.1. PROBLEM SOLUTIONS

Bit rate = 5000 bps; channel filter 3-dB frequency = 5000 Hz; no. of poles = 1 1.4

β=1 β = 0.5

1.2

|HR(f )| or |HT(f )|

0.8

0.6

0.4

0.2

500

1000

1500

2000

2500 f, Hz

3000

3500

4000

4500

5000

legend([’nbeta = ’,num2str(beta0(1))],[’nbeta = ’,num2str(beta0(2))],1) elseif Lbeta == 3

legend([’nbeta = ’,num2str(beta0(1))],[’nbeta = ’,num2str(beta0(2))],... [’nbeta = ’,num2str(beta0(3))],1) end

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION

Problem 5.13 Use (5:37) and set the maximum frequency equal to 7 kHz: 1+ 2T

= 7 kHz

But T1 = 9 kbps, so 1+

9 14 5 1 = = 0:556 9 9

Problem 5.14 a. Consider P (f ) = 2 (f =2W ) (f =W ) which is a trapezoid 1 unit high going from 2W to 2W on its base and.going from W to W on its top. Clearly, superimposing it with its translates by 1=T = 3W=2; 2=T = 3W; ::: produces a constant height of 1 so Nyquist’s pulse shaping criterion is satis…ed. The zero crossings of the time domain pulse are spaced by 2=3W seconds. b. Use the transform pair (t) $ sinc2 (f ) with duality to produce the transform pair sinc2 (t) $ (f ). Then use scale change to get sinc2 (at) $ a 1 (f =a). Apply this to P (f ) to get p (t) = 4W sinc2 (2W t) W sinc2 (W t)

Problem 5.15 a. The channel response matrix is 2

[Pc ] = 4

1:0 0:1 0:2 1:0 0:02 0:2

3 0:01 0:1 5 1:0

5.1. PROBLEM SOLUTIONS Its inverse is

9 2

1:0217 1 4 0:0216 [Pc ] = 0:0626

3 0:0209 0:0163 5 1:0217

0:0163 1:0423 0:0216

The equalizer coe¢ cients are the middle column: [

1] = [

0:01631

1:0423

n pc [(m

n) T ]

0:0216]

b. The output samples are given by peq (mT ) =

1 X

n= 1

[peq ] = [ 0:0121 0:0000 1:0000 0:0000

0:0214]

Problem 5.16 a. The sample values pc ( 4T ) = 0:001 and pc (4T ) = The channel response matrix is 2

1:0 0:2 0:02 0:005 0:003

6 6 [Pc ] = 6 6 4

0:1 1:0 0:2 0:02 0:005

0:01 0:1 1:0 0:2 0:02

0:003 should have been given.

0:001 0:01 0:1 1:0 0:2

3 0:001 0:001 7 7 0:01 7 7 0:1 5 1:0

0:0046 0:0227 0:1111 1:0438 0:2111

3 0:0018 0:0046 7 7 0:0218 7 7 0:1067 5 1:0218

Its inverse is 2

6 6 [Pc ] = 6 6 4

1:0218 0:2111 0:0651 0:0234 0:0101

0:1067 1:0438 0:2179 0:0673 0:0234

0:0218 0:1111 1:0451 0:2179 0:0651

The equalizer coe¢ cients are the middle column: [

2 ] = [0:0218

0:1111 1:0451

0:2179 0:0651]

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION b. The output samples are given by 2 X

peq (mT ) =

n pc [(m

n) T ]

n= 2

[peq ] = [0:0000 0:0000 1:0000 0:0000 0:0000]

Problem 5.17 a. The output in terms of the input is y (t) = x (t) + x (t

Using the superposition and time delay theorems, its Fourier transform is Y (f ) = X (f ) [1 +

exp ( j2

m f )]

The frequency response function is Y (f ) = [1 + exp ( j2 m f )] X (f ) q jH (f )j = [1 + cos (2 m f )]2 + sin2 (2 q = 1 + 2 cos (2 m f ) + 2 H (f ) =

mf )

b. From the block diagram

z (t) =

N X

i y [t

]

i=1

Using the superposition and time delay theorems, the Fourier transform of z (t) is Z (f ) = Y (f )

N X

i exp [

j2 (i

]

i=1

The frequency response function is N

Z (f ) X Heq (f ) = = Y (f ) i=1

i exp [

j2 (i

]

5.1. PROBLEM SOLUTIONS

c. From part a, 1 H (f )

1 1 + exp ( j2 = 1 exp ( j2 =

mf ) mf ) +

Equate the above to Heq (f ) with

m =

exp ( j4

: Clearly

mf )

i =(

exp ( j6

)i 1 ; i = 1; 2; :::; N:

Problem 5.18 a. The channel response matrix is 2

1:0 4 0:1 [Pc ] = 0:07 Its inverse is

0:9898 [Pc ] 1 = 4 0:1031 0:0590

0:07 1:0 0:1

0:0639 0:9864 0:1031

The equalizer coe¢ cients are the middle column: [

1] = [

3 0:05 0:07 5 1:0 3 0:0540 0:0539 5 0:9898

0:0639

0:9864

n pc [(m

n) T ]

0:1031]

b. The output samples are given by peq (mT ) =

1 X

n= 1

[peq ] = [ 0:0428 0:0000 1:0000 0:0000 0:0619] Problem 5.19 a. The channel response matrix is 2 1:0 6 0:1 6 [Pc ] = 6 6 0:07 4 0:05 0:03

0:07 1:0 0:1 0:07 0:05

0:05 0:07 1:0 0:1 0:07

mf ) +

0:02 0:05 0:07 1:0 0:1

3 0:01 0:02 7 7 0:05 7 7 0:7 5 1:0

:::

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION

Its inverse is

[Pc ]

0:9883 0:1054 0::0621 0:0307 0:0163

6 6 =6 6 4

0:0617 0:9823 0:1088 0:0635 0:0370

0:0502 0:0569 0:9801 0:1088 0:0621

0:0248 0:0485 0:0569 0:9823 0:1054

The equalizer coe¢ cients are the middle column: [

2 ] = [0:0502

0:0569 0:9801 0:1088

3 0:0154 0:0248 7 7 0:0502 7 7 0:0617 5 0:9886 0:0621]

b. The output samples are given by peq (mT ) =

2 X

n pc [(m

n) T ]

n= 2

[peq ] = [0:0000 0:0000 1:0000 0:0000 0:0000] Problem 5.20 a. For a timing jitter of zero PE0 = or z0 =

1 exp ( z0 ) = 10 6 2 ln 2 + 6 ln 10

= 13:12 = 11:18 dB b. For a timing jitter of 5% PE5% = =

1 1 exp ( z) + exp [ z (1 2 0:05)] 4 4 1 1 exp ( z) + exp ( 0:9z) = 10 6 4 4

Trial and error provides the value z5% = 14:054 = 11:48 dB Degradation = 11:48

11:18 = 0:3 dB

c. PE0 = or z0 =

1 exp ( z0 ) = 10 4 2 ln 2 + 4 ln 10

= 8:52 = 9:3 dB

5.1. PROBLEM SOLUTIONS

z5% = 9:07 = 9:6 dB Degradation = 9:6

9:3 = 0:3 dB

Problem 5.21 a. The step response of a lowpass RC …lter is ys; RC (t) = [1 = exp ( t=RC)] u(t) Using superposition and time invariance properties of the …lter, the output is y (t) = ys; RC (t)

2ys; RC (t

T ) + 2ys; RC (t

2T )

b. The output is the negative of the one found in part (a). c. The output is the step response. d. The output is the negatuve if the step response. A plot is given in Fig. 5.3. Problem 5.22 1+ 2T

1+ R = fmax 2

so 1+ 10 = 5 2 1+ = 1 = 0 Problem 5.23 We have, from the raised cosine spectrum expression, that 1+ 2T

1+ R = fmax 2

2fmax R

ys; RC (t

3T )

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION T/RC = 10 1 0.8 0.6

step response

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -1

We must have

-0.5

0.5

1 t, seconds

1.5

2.5

0 so 2fmax R

2fmax

1 results in R

fmax :

or Note that requiring that

Problem 5.24 The modi…ed program follows with the spectrum shown in Fig. 5.2. The spectral line at the bit rate does not appear to be much di¤erent than that of Fig. 5.16 of the text. % Problem 5.24 % nsym = 1000; nsamp = 50; lambda = 0.7; [b,a] = butter(3,2*lambda/nsamp); l = nsym*nsamp; % Total sequence length y = zeros(1,l-nsamp+1); % Initalize output vector x =2*round(rand(1,nsym))-1; % Components of x = +1 or -1 for i = 1:nsym % Loop to generate info symbols k = (i-1)*nsamp+1; y(k) = x(i);

Amplitude

5.1. PROBLEM SOLUTIONS

1 0 -1 0

100

200

300

400

500

600

100

200

300

400

500

600

Amplitude

1.5 1 0.5 0

Spectrum

0.5

200

400

600

800

1000 1200 FFT Bin

1400

1600

1800

2000

end datavector1 = conv(y,ones(1,nsamp)); % Each symbol is nsamp long subplot(3,1,1), plot(datavector1(1,200:799),’k’, ’LineWidth’, 1.5) axis([0 600 -1.4 1.4]), ylabel(’Amplitude’) …ltout = …lter(b,a,datavector1); % datavector2 = …ltout.*…ltout; datavector2 = abs(…ltout); subplot(3,1,2), plot(datavector2(1,200:799),’k’, ’LineWidth’, 1.5), ylabel(’Amplitude’) y = ¤t(datavector2); yy = abs(y)/(nsym*nsamp); subplot(3,1,3), stem(yy(1,1:2*nsym),’k.’) xlabel(’FFT Bin’), ylabel(’Spectrum’) % End of script …le. They are about the same. Problem 5.25 Given that NF F T = 5; 000. Since this corresponds to fs = 50 Hz, the spectral line at the bit rate corresponds to 50 1000=5000 = 1 Hz: Problem 5.26

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION

Expand the signal as h i xPSK (t) = Ac cos 2 fc t + d (t) h i 2 h i = Ac cos d (t) cos (2 fc t) sin d (t) sin (2 fc t) 2i h h 2 i cos (2 fc t) sin sin (2 fc t) = Ac cos h i2 h i 2 = Ac cos cos (2 fc t) sin sin (2 fc t) 2h i h 2i = Ac cos cos (2 fc t) d (t) sin sin (2 fc t) 2 2

The …rst term is clearly carrier and the second modulation. From the …rst line, the total power is PT = 12 A2c From the last line the power in the carrier component is PC = 1 2 2 2 Ac cos 2 . The fraction of total power in the carrier component is h i PC = cos2 PT 2 We want cos2

= 0:1 or cos

0:1 or

Problem 5.27 Equation (5:69) is xFSK (t) = Ac cos 2 fc t + kf where d (t) =

Z t

cos 1

0:1 = 0:7952

d( )d

1 in T = 1=1000 = 0:001 second intervals. The peak frequency deviation is f kf

5.2

1 kf dmax hertz = 10000 hertz 2 = 20000 radians/volt (or unit of d (t) )

Computer Exercises

Computer Exercise 5.1

5.2. COMPUTER EXERCISES

% ce5_1: Simulation of a BB pulse transmission system …ltered % by Butterworth …lter % N_order = input(’Enter order of Butterworth …lter => ’); BWT_sym = input(’Enter bandwidth X sym dur product => ’); T_sym = 1; BW = BWT_sym/T_sym; sam_sym = 100; [num den] = butter(N_order, 2*BW/sam_sym); N_sym = 20; ts = T_sym/sam_sym; n_samp = N_sym*sam_sym; t = 0:ts:(n_samp-1)*ts; s = 3279; rand(’state’, s); ss = sign(rand(1, N_sym) - 0.5); % Bipolar data = 0.5*(ss+1); % Unipolar x1 = zeros(size(t)); for n = 1:N_sym x1 = x1 + ss(n)*unit_pulse((t-(n-1)*T_sym-T_sym/2)/T_sym); %NRZ change end y1 = …lter(num, den, x1); x2 = zeros(size(t)); sgn = 1; for n = 1:N_sym if n == 1 x2 = x2 + ss(1)*unit_pulse((t-(n-1)*T_sym-T_sym/2)/T_sym); %NRZ mark else x2 = x2 + sgn*ss(1)*unit_pulse((t-(n-1)*T_sym-T_sym/2)/T_sym); end if n < N_sym if data(n+1) == 1 sgn = -sgn; elseif data(n+1) == 0 sgn = sgn; end end end y2 = …lter(num, den, x2); x3 = zeros(size(t));

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION for n = 1:N_sym x3 = x3 + data(n)*unit_pulse(2*(t-(n-1)*T_sym-T_sym/4)/T_sym); %Unip RZ end y3 = …lter(num, den, x3); x4 = zeros(size(t)); for n = 1:N_sym x4 = x4 + ss(n)*unit_pulse(2*(t-(n-1)*T_sym-T_sym/4)/T_sym); %Polar RZ end y4 = …lter(num, den, x4); x5 = zeros(size(t)); sgn = 1; for n = 1:N_sym x5 = x5 + sgn*data(n)*unit_pulse(2*(t-(n-1)*T_sym-T_sym/4)/T_sym); %Bip

RZ if data(n) > 0 sgn = -sgn; end end y5 = …lter(num, den, x5); x6 = zeros(size(t)); %Split phase for n = 1:N_sym x6 = x6 + ss(n)*(unit_pulse(2*(t-(n-1)*T_sym-T_sym/4)/T_sym) ... -unit_pulse(2*(t-(n-1)*T_sym-3*T_sym/4)/T_sym)); end y6 = …lter(num, den, x6); …gure(1) subplot(6,1,1), plot(t, x1, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’NRZ change’), title(’Baseband binary data formats; un…ltered’) subplot(6,1,2), plot(t, x2, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’NRZ mark’) subplot(6,1,3), plot(t, x3, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’Unipolar RZ’) subplot(6,1,4), plot(t, x4, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’Polar RZ’) subplot(6,1,5), plot(t, x5, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’Bipolar RZ’) subplot(6,1,6), plot(t, x6, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’Split phase’), xlabel(’Time, seconds’) …gure(2) subplot(6,1,1), plot(t, y1, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ...

5.2. COMPUTER EXERCISES ylabel(’NRZ change’), ... title([’Butterworth …lter; order = ’, num2str(N_order), ’; ... BW = ’, num2str(BWT_sym), ’/T_b_i_t’]) subplot(6,1,2), plot(t, y2, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’NRZ mark’) subplot(6,1,3), plot(t, y3, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’Unipolar RZ’) subplot(6,1,4), plot(t, y4, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’Polar RZ’) subplot(6,1,5), plot(t, y5, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’Bipolar RZ’) subplot(6,1,6), plot(t, y6, ’LineWidth’, 1.5), axis([min(t) max(t) -1.2 1.2]), ... ylabel(’Split phase’), xlabel(’Time, seconds’) % End of script …le function y = unit_step(t) % Function for generating the unit step % y = zeros(size(t)); I = …nd(t >= 0); y(I) = ones(size(I)); function y = unit_pulse(t) % Unit rectangular pulse function % y = unit_step(t+0.5) - unit_step(t-0.5); A typical run follows. Plots appesr in Figures 5.5 and 5.6. >> ce5_1 Enter order of Butterworth …lter => 2 Enter bandwidth X sym dur product => 1

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION

Split phaseBipolar RZ Polar RZUnipolar RZNRZ markNRZ change

Baseband binary data formats; unfiltered 1 0 -1 0

8 10 12 Time, seconds

1 0 -1 1 0 -1 1 0 -1 1 0 -1 1 0 -1

Split phaseBipolar RZ Polar RZUnipolar RZNRZ markNRZ change

5.2. COMPUTER EXERCISES

Butterworth filter; order = 2; BW = 1/Tbit 1 0 -1 0

8 10 12 Time, seconds

1 0 -1 1 0 -1 1 0 -1 1 0 -1 1 0 -1

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION

Computer Exercise 5.2 % ce5_2: Computation and plotting of transmitter and receiver frequency responses % for raised cosine signaling through lowpass Butterworth …ltered channel % clf clear all A = char(’-’,’:’,’–’,’-.’,’-o’,’-x’); clf f3dB = input(’Enter channel …lter 3-dB frequency, Hz => ’); n_poles = input(’Enter number of poles for BW …lter representing channel => ’); Rb = input(’Enter data rate, bits/s => ’); beta0 = input(’Enter vector of betas => ’); delf = 100; Lbeta = length(beta0); for n = 1:Lbeta beta = beta0(n); f1 = 0:delf:(1-beta)*Rb/2; HR1 = sqrt(1+(f1/f3dB).^(2*n_poles)); f2 = (1-beta)*Rb/2:delf:(1+beta)*Rb/2; HR2 = sqrt(1+(f2/f3dB).^(2*n_poles)).*cos(pi*(f2-(1-beta)*Rb/2)./(2*beta*Rb)); f3 = (1+beta)*Rb/2:delf:Rb; HR3 = zeros(size(f3)); f = [f1 f2 f3]; HR = [HR1 HR2 HR3]; plot(f,HR,A(n,:), ’LineWidth’, 1.5) if n == 1 hold on xlabel(’{nitf}, Hz’),ylabel(’j{nitH_R}({nitf})j or j{nitH_T}({nitf})j’) title([’Bit rate = ’,num2str(Rb),’bps; ch …lter 3-dB freq = ’,num2str(f3dB),’Hz; no. of poles = ’,... num2str(n_poles) ’; nbeta = ’, num2str(beta0)]) end end

A typical run follows with a plot given in Fig. 5.7. >> ce5_2 Enter channel …lter 3-dB frequency, Hz => 2000 Enter number of poles for Butterworth …lter representing channel => 2

5.2. COMPUTER EXERCISES

Bit rate = 5000 bps; ch filter 3-dB freq = 2000 Hz; no. of poles = 2; β = 0 2

0.5

1.8 1.6

|HR(f )| or |HT(f )|

1.4 1.2 1 0.8 0.6 0.4 0.2 0

500

1000

1500

2000

2500 f, Hz

3000

3500

4000

4500

Enter data rate, bits/s => 5000 Enter vector of betas => [0 .5 1]

Computer Exercise 5.3 % ce5.3: Design of an N-tap zero-forcing equalizer; input and % output plotted % pc = [-0.03 0.05 -0.1 0.02 -0.05 0.2 1 0.3 -0.07 0.03 -0.2 0.05 -0.1]; n_tap = input(’Number of taps => ’); nn = ‡oor(n_tap/2)+1; n_samp = length(pc); n_center = ‡oor(n_samp/2)+1; if n_tap > n_center disp(’Not enough samples for speci…ed equalizer length’) end Pc = []; for k = 1:n_tap

5000

CHAPTER 5. PRINCIPLES OF BASEBAND DIGITAL DATA TRANSMISSION Pc(:, k) = pc(n_center-k+1:n_center+n_tap-k)’;

end Pcinv = inv(Pc); A = Pcinv(:,nn); disp(’The equalizer tap vector is:’) disp(A) nT = [-(n_center-1):n_center-1]; peq = []; n_eq = n_samp-n_tap+1; nTc = [-‡oor(n_eq/2):‡oor(n_eq/2)]; for n = 1:n_eq Nd = n-1; pcd = pc(Nd+1:n_tap+Nd)’; peq(n) = A’*‡ipud(pcd); end subplot(2,1,1), stem(nT, pc), axis([min(nT) max(nT) -0.5 1.5]), ... xlabel(’{nitn}’), ylabel(’{nitp_c}({nitn})’) title([’Input sequence and zero-forced equalizer output for ’, num2str(n_tap), ’-tap equalizer’]) subplot(2,1,2), stem(nTc, peq), axis([min(nT) max(nT) -0.5 1.5]), ... xlabel(’{nitn}’), ylabel(’{nitp}_e_q({nitn})’) % End of script …le A typical run follows with a plot given in Fig. 5.8. >> ce5_3 Number of taps => 5 The equalizer tap vector is: 0.1581 -0.3160 1.2068 -0.4405 0.2577

Computer Exercise 5.4: See Problem 5.24 solution

5.2. COMPUTER EXERCISES

Input sequence and zero-forced equalizer output for 5-tap equalizer 1.5

pc (n)

1 0.5 0 -0.5 -6

-4

-2

0 n

-4

-2

0 n

1.5

peq(n)

1 0.5 0 -0.5 -6

Chapter 6

Overview of Probability and Random Variables 6.1

Problem Solutions

Problem 6.1 S = sample space is the collection of the 21 parts. Let Ai = event that the pointer stops on the ith part, i = 1; 2; :::; 21. These events are exhaustive and mutually exclusive. Thus P (Ai ) = 1=21 a. P (even number) = P (2 or 4 or

or 20) = 10=21;

b. P (A21 ) = 1=21; c. P (4 or 5 or 9) = 3=21 = 1=7; d. P (number > 10) = P (11; 12;

; or 21) = 11=21:

Problem 6.2 a. Use a tree diagram similar to Fig. 6.2 to show that P (3K; 2A) =

4 52

4 3 3 2 3 4 3 2 3 4 3 2 4 3 3 2 3 + 3 + + 3 51 50 49 48 51 50 49 48 51 50 49 48 51 50 49 48

= 9:2345

10 6 1

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES b. Use the tree diagram of Fig. 6.2 except that we now look at outcomes that give 4 of a kind. Since the tree diagram is for a particular denomination, we multiply by 13 and get P (4 of a kind)

4 3 52 51 = 0:0002401

= 13

2 1 48 50 49 48

48 3 2 1 48 4 3 2 1 + 51 50 49 48 52 51 50 49 48

c. The …rst card can be anything. Given a particular suit on the …rst card, the probability of suit match on the second card is 12=51; on the third card it is 11=50; on the fourth card it is 10=49; on the …fth and last card it is 9=48. Therefore, P (all same suit) = 1

12 11 10 9 = 0:001981 51 50 49 48

d. For a royal ‡ush P (A; K; Q; J; 10 of same suit) =

4 1 1 1 1 = 1:283 52 51 50 49 48

10 8

e. The desired probability is P (QjA; K; J; 10 not all of same suit) =

4 = 0:0833 48

Problem 6.3

P (A; B) = P (A) P (B) P (A; C) = P (A) P (C) P (B; C) = P (B) P (C) P (A; B; C) = P (A) P (B) P (C)

Problem 6.4 a. According to (6.11) P (A \ B) = P (A) P (B) for statistical independence. For the given probabilities, 0:2 0:5 = 0:1 6= 0:4 so they are not statistically independent.

6.1. PROBLEM SOLUTIONS

b. According to (6.6) P (A [ B) = P (A) + P (B)

P (A \ B) = 0:2 + 0:5

0:4 = 0:3:

c. A and B being mutually exclusive implies that P (AjB) = P (BjA) = 0. A and B being statistically independent implies that P (AjB) = P (A) and P (BjA) = P (B). The only way that both of these conditions can be true is for P (A) = P (B) = 0. Problem 6.5 a. The result is P (AB) = 1

q 2 (1

P (1 or more links broken) = 1

b. If link 4 is removed, the result is P (ABj link 4 removed) = 1

c. If link 2 is removed, the result is P (ABj link 2 removed) = 1

d. Removal of link 4 is more severe than removal of link 2. Problem 6.6 Using Bayes’rule P (AjB) =

P (BjA) P (A) P (B)

where, by total probability P (B) = P (BjA) P (A) + P BjA P A = P (BjA) P (A) + 1

P BjA

P A

= (0:95) (0:45) + (0:35) (0:55) = 0:62 Therefore P (AjB) =

(0:95) (0:45) = 0:6895 0:62

Similarly, P AjB =

P BjA P (A) P B

with P B = P BjA P (A) + P BjA P A = (0:05) (0:45) + (0:65) (0:55) = 0:38

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

(Note also that P B = 1

P (B).) Thus P AjB =

(0:05) (0:45) = 0:0592 0:38

Problem 6.7 a. P (A2 ) = 0:3; P (A2 ; B1 ) = 0:05; P (A1 ; B2 ) = 0:05; P (A3 ; B3 ) = 0:05; P (B1 ) = 0:15; P (B2 ) = 0:25; P (B3 ) = 0:6: b. P(A3 jB3 ) = 0:083; P (B2 jA1 ) = 0:091; P (B3 jA2 ) = 0:333: Problem 6.8 See the tables below for the results. Outcomes (1; 1) (1; 2) ; (2; 1) (1; 3) ; (3; 1) ; (2; 2) (1; 4) ; (4; 1) ; (2; 3) ; (3; 2) (1; 5) ; (5; 1) ; (2; 4) ; (4; 2) ; (3; 3) (a) (1; 6) ; (6; 1) ; (2; 5) ; (5; 2) ; (3; 4) ; (4; 3) (2; 6) ; (6; 2) ; (3; 5) ; (5; 3) ; (4; 4) (3; 6) ; (6; 3) ; (4; 5) ; (5; 4) (4; 6) ; (6; 4) ; (5; 5) (5; 6) ; (6; 5) ((6; 6))

(b)

Outcomes (1; 1) ; (1; 3) ; (3; 1) ; (2; 2) ; (1; 5) ; (5; 1) ; (2; 4) ; (4; 2) ; (3; 3) ; (2; 6) ; (6; 2) ; (3; 5) ; (5; 3) ; (4; 4) ; (4; 6) ; (6; 4) ; (5; 5) ; (6; 6) (1; 2) ; (2; 1) ; (1; 4) ; (4; 1) ; (2; 3) ; (3; 2) ; (1; 6) ; (6; 1) ; (2; 5) ; (5; 2) ; (3; 4) ; (4; 3) ; (3; 6) ; (6; 3) ; (4; 5) ; (5; 4) ; (5; 6) ; (6; 5)

X1 = 2 3 4 5 6 7 8 9 10 11 12

P spots up

P (X2 = xi )

18=36 = 1=2

P (X1 = xi ) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

6.1. PROBLEM SOLUTIONS

Problem 6.9 First, make a table showing outcomes giving the two values of the random variable: Outcomes (H = head, T = tail) X P (X = xi ) TTH, THT, HTT, HHH (odd # of heads) 0 4=8 = 1=2 HHT, HTH, THH, TTT (even # of heads) 1 4=8 = 1=2 The cdf is zero for x < 0, jumps by 1/2 at x = 0, jumps another 1/2 at x = 1, and remains at 1 out to 1. The pdf consists of an impulse of weight 1/2 at x = 0 and an impulse of weight 1/2 at x = 1. It is zero elsewhere. Problem 6.10 a. As x ! 1, the cdf approaches 1. Therefore B = 1. Assuming continuity of the cdf, Fx (12) = 1 also, which says A 124 = 1 or A = 4:8225 10 5 . b. The pdf is given by dFX (x) = 4 4:8225 10 5 x3 u (x) u (12 dx = 1:929 10 4 x3 u (x) u (12 x)

fX (x) =

The graph of this pdf is 0 for t < 0, the cubic 1:929 for t > 12.

10 4 x3 for 0

12, and 0

c. The desired probability is P (X > 5) = 1

FX (5) = 1

4:8225

10 5

54 = 1

0:0303 = 0:9699

d. The desired probability is

P (4

X < 6) = FX (6)

FX (4)

= 4:8225

10 5

= 4:8225

10 5 64

4:8225

10 5

= 0:0502 Problem 6.11 a. A =

where the condition

b. B =

where the condition

R1 0

1 Be

x dx = 1 must be satis…ed; x dx = 1 must be satis…ed;

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES c. C = e where the condition d. D = 1= where the condition

R1 1

x dx = 1 must be satis…ed;

0 Ddx = 1 must be satis…ed.

Problem 6.12 a. Factor the joint pdf as fXY (x; y) =

p A exp [ jxj] A exp [ jyj]

With proper choice of A, the two separate factors are marginal pdfs. Thus X and Y are statistically independent. b. Note that the pdf is the volume between a plane which intersects the x and y coordinate axes one unit out and the fXY coordinate axis at C. First …nd C from the integral Z 1Z 1 y Z 1Z 1 fXY (x; y) dxdy = 1 or C (1 x y) dxdy = 1 1

This gives C = 6. Next …nd fX (x) and fY (y) as ( Z 1 x 3 (1 fX (x) = 6 (1 x y) dy = 0

and fY (y) =

Z 1 y

6 (1

y) dx =

(

3 (1

x)2 ; 0 x 0; otherwise

y)2 ; 0 y 0; otherwise

Since the joint pdf is not equal to the product of the two marginal pdfs, X and Y are not statistically independent. Problem 6.13 RR a. fXY (x; y) dxdy = 1 gives C = 1=32; b. fXY (1; 1:5) = 1+1:5 32 = 0:0781

c. fXY (x; 3) = 0 because y = 3 is in the domain where fXY is zero. d. By integration, fY (y) = (1+2y) ; 0 8 fXjY (xjy) =

2, so the desired conditional pdf is

1 (1 + xy) fXY (x; y) (1 + xy) = 32 = ; 0 1 fY (y) 4 (1 + 2y) 8 (1 + 2y)

4; 0

6.1. PROBLEM SOLUTIONS

Substitute y = 1 to get the asked for result: fXjY (xj3) =

1+x 1+x = ; 0 4 (1 + 2) 12

Problem 6.14 a. To …nd A, evaluate the double integral Z 1Z 1 Z 1Z 1 Axye (x+y) dxdy fXY (x; y) dxdy = 1 = 0 1 1 Z 10 Z 1 x = A xe dx ye y dy 0

= A

Thus A = 1. b. Clearly, fX (x) = xe x u (x) and fY (y) = ye y u (y) c. Yes, because the joint pdf factors into the product of the marginal pdfs.

Problem 6.15 a. Use normalization of the pdf to unity: Z 1 Z 1 2 x u (x ) dx = x 2 dx =

x 1

Hence, this is a pdf for any value of . b. The cdf is FX (x) =

Z x

z 2 u (z

) dz =

0; x < =x) ; x >

c. The desired probability is P (X

10) = 1

P (X < 10) = 1

FX (10) =

1; =10;

10 < 10

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

Problem 6.16 The result is 8 2 < expp( y=2 ) ; y i 1 2 2y fY (y) = fX (x)jx= py + fX (x)jx=py p = 2 y : 0; y < 0 h

Problem 6.17 a. First note that P (Y = 0) = P (X

0) = 1=2

For y > 0, transformation of variables gives dg 1 (y) dy x=g 1 (y)

fY (y) = fX (x)

Since y = g (x) = ax, we have g 1 (y) = y=a. Therefore exp fY (y) = p

y2 2a2 2

2 a2 2

; y>0

For y = 0, we need to add 0:5 (y) to re‡ect the fact that Y takes on the value 0 with probability 0.5. Hence, for all y, the result is y2

exp 1 2a2 2 u (y) fY (y) = (y) + p 2 2 a2 2 where u (y) is the unit step function. b. Note that exp fY (y) = p

y2 2 2

exp p +

( y)2 2 2

exp =2 p

y2 2 2

; y

and fY (y) = 0 otherwise, because Y is nonnegative. Also note that and y 0.

dy dx

= 1 for y

6.1. PROBLEM SOLUTIONS

Problem 6.18 a. The normalization of the pdf to unity provides the relationship Z 1

fX (x) dx = A

Z 1

e bx dx = Ae 2b =b = 1

Thus A = be2b . b. The mean is 2b

E [X] = be

Z 1

xe bx dx

1 b

= be2b = 2+

1 b

e 2b

1 b

where integration by parts was used. c. The second moment of X is E X

= be

Z 1

x2 e bx dx

2 1 1 2+ 2+ b b b 1 1 = 2 2+ 2+ b b 2b

= be

e 2b

where integration by parts was used twice. d. The variance of X is 2 X

= E X2 = 2 2+ =

1 b2

1 b

fE [X]g2 2+

1 b

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

Problem 6.19 h i2 h i2 R2 R 2 dx 4 4 1 x3 1 x2 2 a. E X 2 = 0 x2 dx = = ; E [X] = x = 2 2 3 0 3 2 2 2 0 = 1; 3 > 1 so it is 0 true in this special case. h i4 h i4 R 4 dx R4 1 x3 16 1 x2 16 2 = = ; E [X] = x = b. E X 2 = 0 x2 dx 4 4 3 0 3 4 4 2 0 = 2; 3 > 2 so it is 0 also true in this special case. n o c. Use the fact that E [X E (X)]2 0 (0 only if X = 0 with probability one). Expanding, we have E X 2

fE [X]g2 > 0 if X 6= 0.

Problem 6.20 Problem statement should refer to Table 6.4. a. The mean and variance of the Rayleigh random variable can be obtained by using tabulated de…nite integrals. The mean is r Z 1 2 2 p r r2 1 1 2 = E [R] = exp dr = 2 2 2 22 (1=2 2 ) 2 2 2 0 by using a de…nite integral for square is E R

Z 1 0

= 2

exp 2

Z 1

R1 0

r2 2 2

x2n exp

dr; let u =

= 2 2

r2 r ; du = 2 dr 2 2

u exp ( u) du

= 2 2

ax2 dx given in Appendix F. The mean

u exp ( u)j1 0 +

Z 1

exp ( u) du

Thus, the variance is var [R] = E R2 = 2 2 which is the same as given in Table 6.4.

E 2 [R] 2

(integration by parts)

6.1. PROBLEM SOLUTIONS

b. The mean and second moment of the single-sided exponential pdf are easily found in terms of tabulated integrals or by integration by parts. The mean is Z 1 E [X] = x exp ( x) dx 0 Z 1 1 u exp ( u) du = 0

= The second moment is E [X] =

Z 1 0

= = =

x2 exp (

Z 1

x) dx

u2 exp ( u) du

u exp (

1 u) 0 + 2

Z 1

u exp ( u) du

and the variance is var [X] = E X 2

E 2 [X] =

1 2

c. The mean of the hyperbolic pdf is zero by virtue of the evenness of the pdf, . For the variance of the hyperbolic pdf, consider Z 1 2 Z 1 2 x (m 1) hm 1 dx x (m 1) hm 1 dx E X2 = dx = dx m 2 (jxj + h) (x + h)m 1 0 where the second integral follows by evenness of the integrand in the …rst integral, and the absolute value on x is unnecessary because the integration is for positive x. Integrate by parts twice to obtain (" #1 Z ) 1 2x (x + h) m+1 x2 (h + h) m 1 2 m 1 E X = (m 1) h + dx m+1 m 1 0 0

The …rst term is zero if m E X 2 = 2hm 1

3. Therefore Z 1

x (x + h) m+1 dx =

The variance is the same since its mean is zero.

2h2 2) (m

; m

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES d. The mean of a Poisson random variable is E [K] =

1 X

k=0

exp (

)

exp (

)

1 X

k=1 1 X

n=0

k 1

1)!

; let n = k

exp (

) exp ( )

Its mean-square value is E K2

1 X

k=0

exp (

exp ( ) ) ) )

1 X

k=1 1 X

) k 1

(n + 1)

n=0 "1 X

Hence, var[K] = E K 2

exp (

n! 1 X

n=1 " 1 X

n=1

1)!

n=0

n 1

1)!

1 X

n=0

E 2 [K] = :

E [K] =

1 X

kpq k 1

k=1

k=0 q

k =

1 1 q ; jqj < 1 to get the sum

1 X k=1

) [ exp ( ) + exp ( )] =

e. For the geometric distribution, the mean is

Di¤erentiate the sum

kq k 1 =

1 (1

q)2

6.1. PROBLEM SOLUTIONS

Apply this to E [K] to get E [K] =

2 = p

k 2 pq k 1 =

A similar procedure results in 1 X

E K2 =

b. The cdf is 0 for x < 0 and 1 for x > B. For 0

k=1

2 + p 2 = p2 + p2

Problem 6.21 a. A = b 1

e bB

;

c. The mean is E [X] =

1 1 b

B, the result is (A=b) 1

e bx ;

e bB 1 e bB

d. The mean-square is E X2 =

e bB (1 + bB) b2

2A 1 b b2

AB 2 bB e b

where A must be substituted from part (a.). e. For the variance, subtract the result of part (c) squared from the result of part (d).

Problem 6.22 a. The integral evaluates as follows: E X 2n =

Z 1

e x2=2 x2n p dx = 2 2 2 1

where y = x= 21=2

Z 1 0

Z 2 p 2n e y 2n+1 2n 1 2n y p dy = p 2 2 y y e dy 0

. Using a table of de…nite integrals, the given result is obtained.

b. The integral is zero by virtue of the oddness of the integrand.

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

Problem 6.23

E [X] =

Z 1 1 2

E X2

= =

Z 1

Z 1 1 2

1 [u (x 4) u (x 8)] dx 8 Z 1 8 1 1 82 42 11 5) dx + xdx = (5) + = 8 4 2 8 2 2

1 (x 2

5) +

x (x

Z 1

1 (x 2

1 [u (x 4) u (x 8)] dx 8 Z 1 8 2 1 1 83 43 187 5) dx + x dx = (25) + = 8 4 2 8 3 6 5) +

x2 (x

2 X =E

E 2 [X] =

187 6

121 11 = 4 12

Problem 6.24 Regardless of the correlation coe¢ cient, the mean of Z is E fZg = E f3X

4Y g = 3E fXg

4E fY g = 3 (1)

4 (3) =

The variance of Z is n o var (Z) = E [Z E (Z)]2 n o = E [3X 4Y 3E(X) + 4E(Y )]2 o n 2 = E 3 X X 4(Y Y ) = E 9 X

= 9 2X + 16 2Y

24 X

X (Y

Y ) + 16(Y

Y )2

Putting in numbers, the results for the variance of Z are: (a) 148; (b) 122.6; (c) 59.1; (d) 21.0.

6.1. PROBLEM SOLUTIONS

Problem 6.25 The conditional pdf fXjY (xjy) is given by

fXjY (xjy) =

fXY (x; y) 2 = fY (y)

2 2 2 2 2 2

1 p

exp

exp 2

x2 2 xy+y 2 2 2 (1 2 )

exp( y 2 =(2 2 )) p 2 2 2 x 2 xy + y 2

2 2 (1

y2 2 2

exp

x2 2 xy y2 + 2) 2 2 (1 2 2

y2 2 2 (1

exp

x2 2 xy y2 + 2) 2 2 (1 2 2

exp

x2 2 xy y2 + 2) 2 2 (1 2 2

1 2

2 xy + 2 y 2 2) 2 2 2 (1 " # (x y)2 exp 2) 2 2 2 (1

1 p

exp

1 p

Note that the conditional pdf of X given Y is Gaussian with conditional mean E [XjY ] = Y 2 2. and conditional variance var(XjY ) = 1 Problem 6.26 Divide the joint Gaussian pdf of two random variables by the Gaussian pdf of Y , collect all exponential terms in a common exponent, complete the square of this exponent which will be quadratic in x and y, and the desired result is a Gaussian pdf with E fXjY g = mX +

X Y

mY ) and var (XjY ) =

2 X

The derivation follows closely that of Problem 6.25 except for the presence of nonzero means mX and mY : Problem 6.27 Convolve the two component pdfs. A sketch of the two component pdfs making up the integrand shows that there is no overlap until z > 1, and the overlap ends when z > 7. The

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

result, either obtained graphically or analytically, is given by (a trapezoid) 8 0; z < 1 > > > > < (z 1) =8; 1 z < 3 1=4; 3 z < 5 fZ (z) = > > (z 7) =8; 5 z 7 > > : 0; z > 7

Problem 6.28

a. From Table 6.4, E [X] = 0; E X 2 = 1=32 = var [X] b. By transformation of random variables, the pdf of Y is fY (y) = =

fX (x)jx=(y 4)=5

dx dy

4 8jy 4j=5 e 5

c. Using the transformation from X to Y , it follows that E [Y ] = E [4 + 5X] = 4 + 5E [X] = 4; h i E Y 2 = E (4 + 5X)2

= E 16 + 40X + 25X 2 = 16 + 40E [X] + 25E X 2 816 51 = 16 + 25=32 = = = 25:5; 32 2 = E Y2 E 2 [Y ] = 25:5 42 = 9:5

Problem 6.29 a. The characteristic function is MX (jv) =

a a

6.1. PROBLEM SOLUTIONS

b. The mean and mean-square values are E [X] =

@MX (jv) = 1=a @v v=0

and E X 2 = ( j)2

2 @ 2 MX (jv) = 2 @v 2 a v=0

respectively. d. var[X] = E X 2

E 2 [X] = 1=a2 .

Problem 6.30 The required distributions are P (k) =

n k p (1 k

p)n k (Binomial)

P (k) = P (k) =

e (k np) =[2np(1 p)] p (Laplace) 2 np (1 p) (np)k np e (Poisson) k!

Comparison tables are given below: n; p 3,1/5

k 0 1 2 3

Binomial 0.512 0.384 0.096 0.008

Laplace 0.396 0.487 0.075 0.0014

Poisson 0.549 0.329 0.099 0.020

n; p 3,1/10

k 0 1 2 3

Binomial 0.729 0.243 0.027 0.001

Laplace 0.650 0.310 0.004 1 10 6

Poisson 0.741 0.222 0.033 0.003

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES n; p 10,1/5

k 0 1 2 3 4 5 6 7 8 9

Binomial 0.107 0.268 0.302 0.201 0.088 0.026 0.005 7.9 10 4 7.4 10 5 4.1 10 6

Laplace 0.090 0.231 0.315 0.231 0.090 0.019 0.002 1.3 10 4 4.1 10 6 7.1 10 8

n; p 10,1/10

k 0 1 2 3 4 5 6 7 8 9

Binomial 0.348 0.387 0.194 0.057 0.011 1.5 10 3 1.4 10 4 3.6 10 7 9.0 10 9 1 10 10

Laplace 0.241 0.420 0.241 0.046 0.003 6 10 5 3.9 10 7 6.3 10 13 1.5 10 16 1.2 10 20

Poisson 0.135 0.271 0.271 0.180 0.090 0.036 0.012 3.4 10 3 8.6 10 4 1.9 10 4 Poisson 0.368 0.368 0.183 0.061 0.015 3.1 10 3 5.1 10 4 3.4 10 3 1 10 6 1 10 7

Problem 6.31 a. P (5 or 6 heads in 10 trials) =

k=5;6 [10!=k! (10

10! 1 k)!] 2 10 = 5!5! 1024 = 0:2461

b. P (…rst head at trial 15) = (1=2)(1=2)4 = 1=32 P 100! 100 = 0:1193; P (…rst head at c. P (50 to 60 heads in 100 trials) = 60 k=50 k!(100 k)! 2 trial 50) = (1=2)(1=2)49 = 8:89 10 16 . Problem 6.32 a. Np = 26 (25) (24) (23) = 358; 800; b. Np = (26)4 = 456; 976;

6.1. PROBLEM SOLUTIONS

c. The answers are the reciprocals of the numbers found in (a) and (b), or 2:79 and 2:19 10 6 ; respectively.

10 6

Problem 6.33 a. The desired probability is P (fewer than 3 heads in 20 coins tossed)

2 X k=0

20 k

1 2

= (1 + 20 + 190) 2 20 = 2:0123

10 4

b. var(N ) = npq = 20 (1=2) (1=2) = 5; E [N ] = np = 10. The Laplace-approximated probability is P (< 3 heads in 20 coins tossed)

2 2 X e (k 10) =10 p = 10 k=0

1 10 = 3:587 =

e 10 + e 81=10 + e 64=10 10 4

c. The percent error is % error =

3:587

10 4 2:0123 2:0123 10 4

10 4

100 = 78:25%

Problem 6.34 a. The probability of exactly one error in 105 digits, by the binomial distribution, is P 1 error in 105

105 1

= 105

1 0:3679

10 5

99;999

10 5

10 5 = 0:3679

By the Poisson approximation, it is P 1 error in 105 = exp ( 1) = 0:3679

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES b. The Poisson approximation gives P 2 errors in 105 =

105

10 5 exp 2!

105

10 5 = 0:18395

c. Use the Poisson approximation: P (more than 5 errors in 105 ) = 1

P (0; 1; 2; 3; 4; or 5 errors) 5 X (np)k exp ( np) k!

= 1

k=0

= 1

exp ( 1) 1 + 1 +

= 1

0:9994 = 0:0006

1)2 4

1 1 1 1 + + + 2 6 24 120

Problem 6.35 a. The marginal pdf for X is fX (x) = p

1 2 (4)

exp

(x 2

1 = p exp 8

1)2

(x 8

The marginal pdf for Y is of the same form except for Y and y in place of X and x, respectively. b. The joint pdf is

fXY (x; y) =

1 p 2 (2) (2) 1

1 p exp 8 0:75

0:52 (

exp

8 > < > :

x 1 2 2

2 (0:5)

2 (1

1)2

(x 1) (y 1) + (y 4 2 (1 0:52 )

xy + y 2 6

y+1

y 1 2 2 0:5 )

x 1 2

1)2

)

y 1 2

9 > = > ;

6.1. PROBLEM SOLUTIONS

The conditional pdf of X given Y = y is fXjY (xj y) = =

fXY (x; y) fY (y) p1 exp 8 0:75

1 p exp 6

( ( (

(x 1)2 (x 1)(y 1)+(y 1)2 6

p1 8

exp

1)2

(y 1)2 8

1) (y 6

1) + (y 1)

1)2

(x 6

1) (y

0:5 (y 6

1)]2

)

1)2

)

1)2 24 (

1 = p exp 6

1)2

)

0:5y 6

0:5]2

)

c. See Problem 6.26 to deduce that E [Xj Y ] = 1 + 0:5 (Y

and var [Xj Y ] = 4 1

0:5

Problem 6.36 a. K = 1= ; b. E [X] is not de…ned, but one could argue that it is zero from the oddness of the integrand for this expectation. If one writes down the integral for the second moment, it clearly does not converge (the integrand approaches a constant as jxj ! 1). c. The answer is given; d. Compare the form of the characteristic function with the answer given in (c) and do a suitable rede…nition of variables.

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

Problem 6.37 a. The characteristic function is found from MY (jv) = E ejvY

n Pn o 2 = E ejv i=1 Xi = E

( n Y i=1

jvXi2

)

But Z 1 n o x2 =2 2 2e 2 dx ejvx p E ejvXi = 2 2 1 Z 1 2 2 e (1=2 jv)x p = dx 2 2 1 =

j2v 2

1=2

which follows by rede…ning the variance of the integrand to get a Gaussian function. Thus N=2 MY (jv) = 1 j2v 2 b. The given pdf follows easily by letting

= 2 2 in the given Fourier transform pair.

c. The mean of Y is E [Y ] = N

by taking the sum of the expectations of the separate terms of the sum de…ning Y . The mean-square value of Xi2 is h i 1=2 2 2 h i d 1 j2v 2 E Xi2 = ( j)2 =3 4 dv 2 v=0 Thus, var

Xi2

=3 4

=2 4

Since the terms in the sum de…ning Y are independent, var[Y ] = 2N 4 . The mean and variance of Y can be put into the de…nition of a Gaussian pdf to get the desired approximation. d. The cases N = 2; 4; 8; and N = 16 are shown in Figure 6.1. The approximations appear to get better as N increases. Note the extreme di¤erence between exact and approximation due to the central limit theorem not being valid for N = 2; 4.

6.1. PROBLEM SOLUTIONS

0.5

0.2

chi-sq; σ 2 = 1; N = 2 Gauss approx

0.4

chi-sq; σ 2 = 1; N = 4 Gauss approx

0.15 fY (y)

fY (y)

0.3 0.2

0.1

0.05

0.1 0

y chi-sq; σ 2 = 1; N = 8 Gauss approx

0.12

y chi-sq; σ 2 = 1; N = 16 Gauss approx

0.08

0.1 0.06 fY (y)

fY (y)

0.08 0.06

0.04

0.04 0.02 0.02 0

e. Evident by direct substitution.

Problem 6.38 R 1 exp( t2 =2) exp( x2 =2) p p This is a matter of plotting Q (x) = x dx and Q (x) = on the same a 2 2 x set of axes. Problem 6.39 By de…nition, the cdf of a Gaussian random variable is h i h i (u m)2 (u m)2 Z x exp Z 1 exp 2 2 2 2 p p du = 1 du FX (x) = 2 2 2 2 1 x Change variables to

exp p

u2 2

v= with the result that

FX (x) = 1

Z 1

(x m)=

du = 1

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

A plot may easily be obtained byR a MATLAB program. It is suggested that you use the 1 MATLAB function erfc(x) = p2 x exp t2 dt. Problem 6.40 Consider the product of integrals Z 1 Z 1 exp v 2 =2 exp u2 =2 p p du dv I (x) = 2 2 x 0 Z 1 1 exp v 2 =2 1 p = dv = Q (x) 2 x 2 2

where the 1/2 comes from the fact that the …rst integral is the integral of the right half of a Gaussian pdf. Now write the product of the two integrals as an iterated integral to get Z 1Z 1 u2 + v 2 1 exp dudv I (x) = 2 0 2 x Transform the integral to polar coordinates via u = r cos ; v = r sin ; and dudv = rdrd to get Z =2 Z 1 1 1 I (x) = Q (x) = exp r2 =2 rdrd 2 2 0 x= sin Z =2 x2 1 exp d = 2 0 2 sin2 Multiplying through by 2 gives the answer given in the problem statement. Problem 6.41 The following relationships involving the Q-function will prove useful: Z b exp u2 =2 p du = Q (a) Q (b) ; b > a; 2 a Q (x) = 1 Q (jxj) ; x < 0; Q (0) = 0:5 a. The probability is P (jXj

Z 15

e (x 10) =50 p dx 50 15 Z 1 exp u2 =2 x 10 p = du; u = p 2 25 5 = Q ( 5) Q (1) = 1 Q (5) Q (1)

15) =

= 0:8413

6.1. PROBLEM SOLUTIONS

b. This probability is Z 20

e (x 10) =50 p dx 20) = 50 10 Z 2 exp u2 =2 x 10 p = du; u = p 2 25 0 = Q (0) Q (2) = 0:5 Q (2)

P (10 < X

= 0:4772 c. The desired probability is Z 25

e (x 10) =50 p dx 50 5 Z 3 exp u2 =2 x 10 p = du; u = p 2 25 1 = Q ( 1) Q (3) = 1 Q (1) Q (3)

P (5 < X

25) =

= 0:8400 d. The result is P (20 < X

Z 30

e (x 10) =50 p dx 30) = 50 20 Z 4 exp u2 =2 x 10 p = du; u = p 2 25 2 = Q (2) Q (4) = 0:0227

Problem 6.42 a. Observe that 2

Z 1

m)2 fX (x) dx (x

m)2 fX (x) dx

jx mj>k

Z = k

k 2 2 fX (x) dx

jx mj>k 2 2

P fjX

mj > k g = k 2 2 f1

P [jX

mj > k ]g

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

or P [jX

mj < k ]

b. Note that for this random variable mX = 0 and probability is P [jX

1 k2 2 X

= 22 =12 = 1=3: The actual

h p i mX j < k X ] = P jXj < k= 3 ( R p p p k= 3 dx p = k= 3; k < 3 k= 3p2 = 3 1; k

A comparison is given in the table below: k Actual prob. Ch. bound 0 0 p 1 1= 3 = 0:577 0 2 1 0.75 3 1 0.889 Note that in all cases the bound exceeds the actual probability. Problem 6.43 The desired probability is P [jXj > k ] = P [X < k ] + P [X > k ] ; k = 1; 2; 3 Z k Z 1 exp x2 =2 2 exp x2 =2 2 p p = dx + dx 2 2 2 2 1 k = 2Q (k) a. P [jXj > ] = 0:3173; b. P [jXj > 2 ] = 0:0455; c. P [jXj > 3 ] = 0:0027:

Problem 6.44

6.1. PROBLEM SOLUTIONS

a. The mean is 0 by even symmetry of the pdf. Thus, the variance is 2

= var [X] = E X 2 (because the mean is 0) Z 1 a exp ( ajxj) dx = x2 2 1 Z 1 a = 2 x2 exp ( ax) dx 2 0 Z 1 y 2 dy = 2a exp ( y) a a 0 1 1 = 1=a2 = 2a a3 2

Thus, in terms of the variance, the pdf is fX (x) =

1 exp ( jxj = ) 2

b. The desired probability is P [jXj > k ] = P [X < k ] + P [X > k ] ; k = 1; 2; 3 Z k Z 1 1 1 = exp (x= ) dx + exp ( x= ) dx 2 1 2 k Z 1 = exp ( u) du (by eveness of the integrand and u = x= ) k

= exp ( k)

Thus, the required numerical values are P [jXj > ] = 0:3679; P [jXj > 2 ] = 0:1353; P [jXj > 3 ] = 0:0498

Problem 6.45 Since both X and Y are Gaussian and the transformation is linear, Z is Gaussian. Thus, all we need are the mean and variance of Z. Since the means of X and Y are 0, so is the

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

mean of Z. Therefore, its variance is h i 2 2 2 = E Z = E (X + 2Y ) Z = E X 2 + 4XY + 4Y 2

= E X 2 + 4E [XY ] + 4E Y 2 =

2 X +4 X Y

= 3 + (4) = 19

2 XY + 4 Y

(because the means are 0)

3 4 ( 0:4) + (4) (4) p 3:2 3 = 13:457

The pdf of Z is fZ (z) =

exp

z 2 =26:915 26:915

Problem 6.46 Since both X and Y are Gaussian and the transformation is linear, Z is Gaussian. Thus, all we need are the mean and variance of Z. The mean of Z is mZ

= E [3X + Y ] = 3mX + mY = (3) (1) + 2 = 5

Its variance is 2 Z

h i h = E (Z mZ )2 = E (3X + Y 3mX h i = E (3X 3mX + Y mY )2 h i = 9E (X mX )2 + 6E [(X mX ) (Y = 9 2X + 6 X

XY +

2 Y

mY )2

h mY )] + E (Y

= 9 (3) + (6) 3 2 (0:2) + 2 p = 29 + 1:2 6 = 31:939 The pdf of Z is exp fZ (z) =

Problem 6.47

(z 5)2 =63:879 p 63:879

mY )2

6.1. PROBLEM SOLUTIONS a. fX (x) =

exp[ (x 5)2 =2] p ; 2

b. fXY (x; y) =

fY (y) =

exp[ (y 3)2 =4] p 4

exp[ (x 5)2 =2] exp[ (y 3)2 =4] p p 2 4

c. E (Z1 ) = mX + mY = 5 + 3 = 8; E (Z2 ) = mX

mY = 5

3=2

d. The variances are

h i h = E (Zi mZi )2 = E (X Y mX h i = E ((X mX ) (Y mY ))2 h i = E (X mX )2 + 2E (X mX ) E (Y

2 Zi

2 X +0+

mY )2

h mY ) + E (Y

2 Y

mY )2

= 1+2=3 e. The desired pdf is exp fZ1 (z1 ) =

(z1 8)2 =6 p 6

i (z2 2)2 =6 p 6

f. The desired pdf is exp fZ2 (z2 ) =

Problem 6.48 a. fX (x) =

exp[ (x 4)2 =6] p ; 6

b. fXY (x; y) =

fY (y) =

exp[ (y 2)2 =10] p 10

exp[ (x 4)2 =6] exp[ (y 2)2 =10] p p 6 10

c. E (Z1 ) = 3mX + mY = (3) (4) + 2 = 14; E (Z2 ) = 3mX d. The variances are h i h 2 2 = E (Z m ) = E (3X Y mX i Zi Zi h i = E (3 (X mX ) (Y mY ))2 h i = 9E (X mX )2 + 6E (X mX ) E (Y = 9 2X + 0 +

2 Y

= (9) (3) + 5 = 32

mY = (3) (4)

mY )2

2 = 10

h mY ) + E (Y

mY )2

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES e. The pdf of Z1 is exp fZ1 (z1 ) = f. The pdf of Z2 is exp fZ2 (z2 ) =

(z1 p

(z2 p

14)2 =64 64

10)2 =64 64

Problem 6.49 a. The probability of mean exceedance for a uniformly distributed random variable is Z b 1 dx P X > (a + b) = 2 b a (a+b)=2 b

= (a+b)=2

1 2

b. The probability of mean exceedance for a Rayleigh distributed random variable is r Z 1 r r2 r2 P X> = p exp dr; v = 2 2 2 2 2 2 =2 Z 1 = e v dv = e =4 = 0:4559 =4

c. The probability of mean exceedance for a one-sided exponential distributed random variable is Z 1 1 P X> exp ( x) dx = = e

6.2

1= 1

= 0:3679

Computer Exercises

Computer Exercise 6.1 % ce6_1.m: Generate an exponentially distributed random variable % with parameter a from a uniform random variable in [0, 1) % clf

6.2. COMPUTER EXERCISES

a = input(’Enter the parameter of the exponential pdf: f(x) = a*exp(-a*x)*u(x) => ’); N = input(’Enter number of exponential random variables to be generated => ’); U = rand(1,N); V = -(1/a)*log(U); [M, X] = hist(V); disp(’’) disp(’No. of random variables generated’) disp(N) disp(’’) disp(’Bin centers’) disp(X) disp(’’) disp(’No. of random variable counts in each bin’) disp(M) disp(’’) norm_hist = M/(N*(X(2)-X(1))); plot(X, norm_hist, ’o’) hold plot(X, a*exp(-a*X), ’–’), xlabel(’x’), ylabel(’f_X(x)’),... title([’Theoretical pdf and histogram for ’,num2str(N),... ’computer generated exponential random variables’]) legend([’Histogram points’],[’Exponential pdf; a = ’, num2str(a)]) A typical run follows with a plot given in Fig. 6.2. >> ce6_1 Enter the parameter of the exponential pdf: f(x) = a*exp(-a*x)*u(x) => 2 Enter number of exponential random variables to be generated => 5000 No. of random variables generated 5000 Bin centers Columns 1 through 6 0.2094 0.6280 1.0467 1.4653 1.8839 2.3026 Columns 7 through 10 2.7212 3.1398 3.5585 3.9771 No. of random variable counts in each bin Columns 1 through 5 2880 1200 543 215 90 Columns 6 through 10 38 16 9 5 4

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES Theoretical pdf and histogram for 5000 computer generated exponential random variables 1.4 Histogram points Exponential pdf; a = 2 1.2

fX(x)

0.8

0.6

0.4

0.2

0.5

1.5

2 x

2.5

3.5

Current plot held Computer Exercise 6.2 % ce6_2.m: Generate pairs of Gaussian random variables from Rayleigh and uniform RVs % clf m = input(’Enter the mean of the Gaussian random variables => ’); sigma = input(’Enter the standard deviation of the Gaussian random variables => ’); N = input(’Enter number of Gaussian random variable pairs to be generated => ’); U = rand(1,N); V = rand(1,N); R = sqrt(-2*log(V)); X = sigma*R.*cos(2*pi*U)+m; Y = sigma*R.*sin(2*pi*U)+m; disp(’’) disp(’Covarance matrix of X and Y vectors:’)

6.2. COMPUTER EXERCISES

disp(cov(X,Y)) disp(’’) [MX, X_bin] = hist(X, 20); norm_MX = MX/(N*(X_bin(2)-X_bin(1))); [MY, Y_bin] = hist(Y, 20); norm_MY = MY/(N*(Y_bin(2)-Y_bin(1))); gauss_pdf_X = exp(-(X_bin - m).^2/(2*sigma^2))/sqrt(2*pi*sigma^2); gauss_pdf_Y = exp(-(Y_bin - m).^2/(2*sigma^2))/sqrt(2*pi*sigma^2); subplot(2,1,1), plot(X_bin, norm_MX, ’o’) hold subplot(2,1,1), plot(X_bin, gauss_pdf_X, ’–’), xlabel(’x’), ylabel(’f_X(x)’),... title([’Theoretical pdfs and histograms for ’,num2str(N),’... computer generated independent Gaussian RV pairs’]) legend([’Histogram points’],[’Gauss pdf; nsigma, nmu = ’, num2str(sigma),’, ’, num2str(m)])

subplot(2,1,2), plot(Y_bin, norm_MY, ’o’) hold

subplot(2,1,2), plot(Y_bin, gauss_pdf_Y, ’–’), xlabel(’y’), ylabel(’f_Y(y)’)

A typical run follows with a plot given in Fig. 6.3.

>> ce6_2 Enter the mean of the Gaussian random variables => 2 Enter the standard deviation of the Gaussian random variables => 3 Enter number of Gaussian random variable pairs to be generated => 5000 Covarance matrix of X and Y vectors: 9.0928

0.2543

9.0777

Current plot held

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

fX(x)

Theoretical pdfs and histograms for 5000 computer generated independent Gaussian RV pairs 0.2 Histogram points 0.15 Gauss pdf; σ , µ = 3, 2 0.1 0.05 0 -10

-5

x 0.2 0.15 fY (y)

0.1 0.05 0 -10

-5

0 y

6.2. COMPUTER EXERCISES

Computer Exercise 6.3 % ce6_3.m: Generate a sequence of Gaussian random variables with speci…ed correlation % between adjacent RVs % clf m = input(’Enter the mean of the Gaussian random variables => ’); sigma = input(’Enter the standard deviation of the Gaussian random variables => ’); rho = input(’Enter correlation coe¢ cient between adjacent Gaussian random variables => ’); N = input(’Enter number of Gaussian random variables to be generated => ’); var12 = sigma^2*(1 - rho^2); X = []; X(1) = sigma*randn(1)+m; for k = 2:N m12 = m + rho*(X(k-1) - m); X(k) = sqrt(var12)*randn(1) + m12; end [MX, X_bin] = hist(X, 20); norm_MX = MX/(N*(X_bin(2)-X_bin(1))); gauss_pdf_X = exp(-(X_bin - m).^2/(2*sigma^2))/sqrt(2*pi*sigma^2); subplot(2,1,1), plot(X_bin, norm_MX, ’o’) hold subplot(2,1,1), plot(X_bin, gauss_pdf_X, ’–’), xlabel(’x’), ylabel(’f_X(x)’),... title([’Theoretical pdf and histogram for ’,num2str(N),’computer generated Gaussian RVs’]) legend([’Histogram points’],[’Gauss pdf; nsigma, nmu = ’, num2str(sigma),’, ’, num2str(m)],2) Z = X(1:50); subplot(2,1,2), plot(Z, ’x’), xlabel(’k’), ylabel(’X(k)’) title([’Gaussian sample values with correlation ’, num2str(rho), ’between samples’]) A typical run follows with a plot given in Fig. 6.4. >> ce6_3 Enter the mean of the Gaussian random variables => 0 Enter the standard deviation of the Gaussian random variables => 2 Enter correlation coe¢ cient between adjacent Gaussian random variables => 0.8 Enter number of Gaussian random variables to be generated => 250 Current plot held

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

Theoretical pdf and histogram for 250 computer generated Gaussian RVs 0.4 Histogram points fX(x)

0.3

Gauss pdf; σ , µ = 2, 0

0.2 0.1 0 -5

-4

-3

-2

-1

0 1 2 3 4 x Gaussian sample values with correlation 0.8 between samples

X(k)

-5

25 k

6.2. COMPUTER EXERCISES

Computer Exercise 6.4 % ce6_4.m: Testing the validity of the Central Limit Theorem with sums of uniform % random numbers % clf N = input(’Enter number of Gaussian random variables to be generated => ’); M = input(’Enter number of uniform random variables to be summed to generate one GRV => ’); % Mean of uniform = 0; variance = 1/12 Y = rand(M,N)-0.5; X_gauss = sum(Y)/(sqrt(M/12)); disp(’’) disp(’Estimated variance of the Gaussian RVs:’) disp(cov(X_gauss)) disp(’’) [MX, X_bin] = hist(X_gauss, 20); norm_MX = MX/(N*(X_bin(2)-X_bin(1))); gauss_pdf_X = exp(-X_bin.^2/2)/sqrt(2*pi); plot(X_bin, norm_MX, ’o’) hold plot(X_bin, gauss_pdf_X, ’–’), xlabel(’x’), ylabel(’f_X(x)’),... title([’Theoretical pdf & histogram for ’,num2str(N),’... Gaussian RVs each generated as the sum of ’, num2str(M), ’uniform RVs’]) legend([’Histogram points’],[’Gauss pdf; nsigma = 1, nmu = 0’],2) A typical run follows with a plot given in Fig. 6.5. >> ce6_4 Enter number of Gaussian random variables to be generated => 1000 Enter number of uniform random variables to be summed to generate one GRV => 10 Estimated variance of the Gaussian RVs: 0.9091 Current plot held

CHAPTER 6. OVERVIEW OF PROBABILITY AND RANDOM VARIABLES

Theoretical pdf & histogram for 1000 Gaussian RVs each generated as the sum of 10 uniform RVs 0.45 Histogram points Gauss pdf; σ = 1, µ = 0

0.4 0.35 0.3 0.25 fX(x)

0.2 0.15 0.1 0.05 0 -3

-2

-1

1 x

Chapter 7

Random Signals and Noise 7.1

Problem Solutions

Problem 7.1 The various sample functions are as follows. Sample functions for case (a) are horizontal lines at levels 2A; 0; 2A, each case of which occurs equally often (with probability 1/3). Sample functions for case (b) are horizontal lines at levels 3A; 2A; A; A; 2A; 3A, each case of which occurs equally often (with probability 1/6). Sample functions for case (c) are horizontal lines at levels 4A; 2A; 2A; 4A; or oblique straight lines of slope A or A, each case of which occurs equally often (with probability 1/6). Problem 7.2 a. For case (a) of problem 7.1, since the sample functions are constant with time and are less than or equal 2A, the probability is one. For case (b) of problem 7.1, since the sample functions are constant with time and 5 out of 6 are less than or equal to 2A, the probability is 5/6. For case (c) of problem 7.1, the probability is 5/6 because 5 out of 6 of the sample functions will be less than 2A at t = 2. b. The probabilities are now 2/3, 1/2, and 1/2, respectively. c. The probabilities are 1, 5/6, and 5/6, respectively.

Problem 7.3 a. The sketches would consist of squarewaves of random delay with respect to t = 0. 1

CHAPTER 7. RANDOM SIGNALS AND NOISE b. X (t) takes on only two values A and fX (x) =

A, and these are equally likely. Thus

1 (x 2

A) +

1 (x + A) 2

Problem 7.4 a. The sample functions at the integrator output are of the form Z t A (i) A cos (! 0 ) d = Y (t) = sin ! 0 t !0 where A is Gaussian. b. Y (i) (t0 ) is Gaussian with mean zero and standard deviation Y =

jsin (! 0 t0 )j !0

c. Not stationary (the second moment, for example, depends on time) and not ergodic.

Problem 7.5 a. By inspection of the sample functions, the mean is zero. Considering the time average correlation function, de…ned as 1 R ( ) = lim T !1 2T

Z T

1 x (t) x (t + ) dt = T0 T

x (t) x (t + ) dt

where the last expression follows because of the periodicity of x (t), it follows from a sketch of the integrand that R( ) =

A2 (1 4 =T0 ) , 0 A2 (1 + 4 =T0 ) , T0 =2

T0 =2 0

Since R ( ) is periodic, this de…nes the autocorrelation function for all . b. Yes, it is wide sense stationary. The random delay, , being over a full period, means that no untypical sample functions occur.

7.1. PROBLEM SOLUTIONS

Problem 7.6

a. The time average mean is zero. The statistical average mean is

E [X (t)] = = =

A cos (2 f0 t + ) =2

[sin (2 f0 t + )

d 2A = sin (2 f0 t + ) =2

sin (2 f0 t + =2)]

[ sin (2 f0 t) cos (2 f0 t)] p 2 2A [sin (2 f0 t + =4)]

The time average variance is A2 =2. The statistical average second moment is

E X (t)

= =

d A2 cos2 (2 f0 t + ) =2 =2 "Z Z

d +

= =

cos (4 f0 t + 2 ) d

1 + sin (4 f0 t + 2 ) 2 2

+ [sin (4 f0 t + 2 ) sin (4 f0 t + )] 2 2 A2 A2 A2 A2 + [sin (4 f0 t) + sin (4 f0 t)] = + sin (4 f0 t) 2 2 2

The statistical average variance is this expression minus E 2 [X (t)].

b. The time average autocorrelation function is

hx (t) x (t + )i =

A2 cos (2 f0 ) 2

CHAPTER 7. RANDOM SIGNALS AND NOISE The statistical average autocorrelation function is Z d R( ) = A2 cos (2 f0 t + ) cos (2 f0 (t + ) + ) =2 =2 Z A2 = [cos (2 f0 ) + cos (2 f0 (2t + ) + 2 )] d =2

= = = =

A2 2

cos (2 f0 ) +

A2 sin (2 f0 (2t + ) + 2 ) 2

cos (2 f0 ) + [sin (2 f0 (2t + ) + 2 ) sin (2 f0 (2t + ) + )] 2 2 A2 A2 cos (2 f0 ) + [sin 2 f0 (2t + ) + sin 2 f0 (2t + )] 2 A2 2A2 cos (2 f0 ) + sin 2 f0 (2t + ) 2

c. No. The random phase, not being uniform over (0; 2 ), means that for a given time, t, certain phases will be favored over others. Problem 7.7 a. The time-average mean is zero. The time-average autocorrelation function is Z 1 A 2 hY (t) Y (t + )i = sin (! 0 t) sin [! 0 (t + )] dt T0 T0 ! 0 Z A 2 1 = fcos (! 0 ) cos [! 0 (2t + )]g dt 2T0 ! 0 T0 =

1 2

A !0

cos (! 0 )

b. Again, the mean is zero because the mean of A is zero. The ensemble-average autocorrelation function is R ( ) = E [Y (t; A) Y (t + ; A)] Z 1 exp a2 =2 2a a 2 p sin (! 0 t) sin [! 0 (t + )] = da 2 2a 1 !0 =

2 a sin (! 0 t) sin [! 0 (t + )] ! 20 2 a fcos (! 0 ) cos [! 0 (2t + 2! 20

)]g

7.1. PROBLEM SOLUTIONS

c. No, it is not wide sense stationary because the ensemble-average autocorrelation function is time dependent (not a function only of time di¤erence).

Problem 7.8 From the measurements given, assuming ergodicity, we can infer that the mean is 6 V and the second moment is 62 + 49 = 85 V2 (we assume that the true-rms meter is AC coupled). Thus the variance is 49 V2 . Hence, the pdf of the noise voltage is 2

e (x 6) =2 49 e (x 6) =98 p fX (x) = p = 2 49 98

Problem 7.9 a. Suitable; b. Suitable; c. Not suitable because the Fourier transform is not everywhere nonnegative; d. Not suitable because autocorrelation functions must be even; e. Suitable; f. Suitable (note the A missing on the RHS in the …rst printing).

Problem 7.10 Use the Fourier transform pair N0 W sinc (2W ) $ with W = 103 Hz and N20 = 2

N0 2

(f =2W )

10 5 V2 /Hz to get the autocorrelation function as

RX ( ) = N0 W sinc (2W ) = 4 = 0:04 sinc 2

103

10 5

103 sinc 2

103

CHAPTER 7. RANDOM SIGNALS AND NOISE

Problem 7.11 a. The r ( ) function, (7.55), is Z 1 1 p (t) p (t + ) dt r( ) = T 1 Z 1 T =2 (t + ) = cos ( t=T ) cos dt, 0 T T T =2 1 = (1 =T ) cos , 0 T =2 2 T Now r ( ) = r (

T =2

) and r ( ) = 0 for j j > T =2. Therefore it can be written as r( ) =

1 ( =T ) cos 2

Hence, by (7.54) Ra ( ) =

A2 2

( =T ) cos

1 2T

The power spectral density is Sa (f ) =

A2 T 4

sinc2 T

+ sinc2 T

1 2T

b. The autocorrelation function now becomes Rb ( ) = A2 [2r ( ) + r ( + T ) + r (

T )]

The corresponding power spectrum is Sb (f ) = A2 T sinc2 (T f

0:5) + sin c2 (T f + 0:5) cos2 ( f T )

c. The plots are left for the student. Problem 7.12 a. The autocorrelation functions are RX ( ) = E [X (t) X (t + )] = Rn ( ) + E A2 cos (! 0 t + = Rn ( ) +

)

A2 A2 cos (! 0 ) = B ( = 0 ) + cos (! 0 ) 2 2

cos [! 0 (t + ) +

]

7.1. PROBLEM SOLUTIONS

and RY ( ) = E [Y (t) Y (t + )] = Rn ( ) + E A2 sin (! 0 t + = Rn ( ) +

)

sin [! 0 (t + ) +

]

A2 A2 cos (! 0 ) = B ( = 0 ) + cos (! 0 ) 2 2

b. The cross-correlation function is RXY ( ) = E [X (t) Y (t + )] = Rn ( ) + E A2 cos (! 0 t + = B ( = 0) +

A2 2

)

sin [! 0 (t + ) +

]

sin (! 0 )

Problem 7.13 a. By de…nition RZ ( ) = E [Z (t) Z (t + )] = E [X (t) X (t + ) Y (t) Y (t + )] = E [X (t) X (t + )]E[Y (t) Y (t + )] = RX ( ) RY ( ) b. Since a product in the time domain is convolution in the frequency domain, it follows that SZ (f ) = SX (f ) SY (f ) c. Use the transform pairs 2W sinc (2W ) and cos (2 f0 )

1 (f 2

(f =2W )

f0 ) +

1 (f + f0 ) 2

Also, RY ( ) = E f4 cos (50 t + ) cos [50 (t + ) + ]g = 2 cos (50 Using the …rst transform pair, we have RY ( ) = 500 sinc (100 )

)

CHAPTER 7. RANDOM SIGNALS AND NOISE This gives RZ ( ) = [500 sinc (100 )] [2 cos (50 = 1000 sinc (100 ) cos (50 and SZ (f ) = 5

103

25 200

)] )

f + 25 200

Problem 7.14 a. E X 2 (t) = R (0) = 12 W; E 2 [X (t)] = lim !1 R ( ) = 9 W; 2 =E X

X 2 (t)

E 2 [X (t)] = 12

9 = 3 W.

b. DC power = E 2 [X (t)] = 9 W. c. Total power = E X 2 (t) = 12 W. d. S (f ) = 9 (f ) + 15 sinc2 (5f ).

Problem 7.15 a. The autocorrelation function is RX ( ) = E [Y (t) Y (t + )] = E f[X (t) + X (t

T )][X (t + ) + X (t +

T )]g

= E [X (t) X (t + )] + E[X (t) X (t + + )] +E [X (t

T ) X (t + )] + E[X (t

= 2RX ( ) + RX (

T ) X (t +

T )]

T ) + RX ( + T )

b. Application of the time delay theorem of Fourier transforms gives SY (f ) = 2SX (f ) + SX (f ) [exp ( j2 f T ) + exp (j2 f T )] = 2SX (f ) + 2SX (f ) cos (2 f T ) = 4SX (f ) cos2 ( f T )

7.1. PROBLEM SOLUTIONS

c. Use the transform pair RX ( ) = 6 ( )

! SX (f ) = 6 sinc2 (f )

and the result of (b) to get SY (f ) = 24 sinc2 (f ) cos2 ( f =4)

Problem 7.16 a. The student should carry out the sketch. R 0+ b. DC power = 0 S (f ) df = 10 W. R1 c. Total power = 1 S (f ) df = 10 + 25=5 + 5 + 5 = 25 W

d. Power for jf j 0:2 Hz = 10 + 0:9 (25=5) = 14:5 W. Fraction of total power = 14:5=25 = 0:58 = 58%.

Problem 7.17 a. This is left for the student. b. Using the Fourier transform of a two-sided decaying exponential and the modulation theorem, the Fourier transform of the …rst is SX1 (f ) =

4 2 + [2

2 +

2 + [2

1)]

(f + 1)]2

The Fourier transform of the second is SX2 (f ) =

4 + 2 (f 2 + (2 f )2

b) + 2 (f + b)

Using the Fourier transform of the Gaussian pulse, the Fourier transform of the third is SX3 (f ) =

5p exp 2

All Fourier transforms are non-negative functions of frequency.

CHAPTER 7. RANDOM SIGNALS AND NOISE c. None of the Fourier transforms above has a unit impulse at f = 0 so no DC power. d. Evaluate the correlation function at and 5 W, respectively, for (a)-(c).

= 0 to get total power. The results are 4, 6,

e. Only the second one has a periodic component at b Hz. Problem 7.18 a. The output power spectral density is Sout (f ) = 10 6

105

f =5

Note that the rectangular pulse function does not need to be squared because its amplitude squared is unity. b. Use the Fourier transform pair 2W sinc(2W ) $ function of the output is Rout ( ) = 5

105

10 6 sinc 5

105

(f =2W ).

The autocorrelation

= 0:5 sinc 5

105

c. The output power is 0:5 W, which can be found from Rout (0) or the integral of Sout (f ). Problem 7.19 a. By assuming a unit impulse at the input, the impulse response (i.e., the response to a unit impulse) is 1 h (t) = [u (t) u (t T )] T b. Use the time delay theorem of Fourier transforms and the Fourier transform of a rectangular pulse to get H (f ) = sinc (f T ) e j f T c. The output power spectral density is S0 (f ) = d. Use F [ ( = 0 )] =

0 sinc

N0 sinc2 (f T ) 2

0 f ) to get

R0 ( ) =

N0 2T

( =T )

7.1. PROBLEM SOLUTIONS

e. By de…nition of the equivalent noise bandwidth 2 E Y 2 = Hmax BN N0

The output power is also R0 (0) = N0 =2T . Equating the two results and noting that Hmax = 1, gives BN = 1= (2T ) Hz. f. Evaluate the integral of the power spectral density over all frequency: Z 1 Z N0 N0 N0 1 2 2 sinc2 (u) du = E Y = sinc (f T ) df = = R0 (0) 2T 2T 1 2 1

Problem 7.20 a. The output power spectral density is S0 (f ) =

N0 =2 1 + (f =f3 )4

b. The autocorrelation function of the output, by inverse Fourier transforming the output power spectral density, is (a table of de…nite integrals is necessary) Z 1 N0 =2 j2 f R0 ( ) = F 1 [S0 (f )] = df 4e 1 + (f =f ) 1 3 Z 1 Z 1 cos (2 f ) cos (2 f3 x) = N0 dx 4 df = N0 f3 1 + x4 1 + (f =f3 ) 0 0 p f3 N0 p2 f3 e cos 2 f3 =4 = 2 c. Yes. R0 (0) =

f 3 N0 = N0 BN ; 2

BN = f3 =2

Problem 7.21 We have jH (f )j2 =

(2 f )2 (2 f )4 + 5; 000

Thus jH (f )j = q

j2 f j

(2 f )4 + 5; 000

This could be realized with a second-order Butterworth …lter in cascade with a di¤erentiator.

CHAPTER 7. RANDOM SIGNALS AND NOISE

Problem 7.22 a. The power spectral density is N0 2

SY (f ) =

(f =2B)

The autocorrelation function is RY ( ) = N0 B sinc (2B ) b. The transfer function is H (f ) =

A + j2 f

The power spectral density and autocorrelation function are, respectively, A2 B 2 + (2 f )2 1 + (2 f )2 1 A2 = 2 B A2 B= 2 = 2 2 2 2 1 1 + (2 f = ) 1 + (2 f ) 1 + (2 f = )2

SY (f ) = jH (f )j2 SX (f ) = =

Use the transform pair exp ( j j= 0 ) autocorrelation function is RY (t) =

A2 B=

2 1

(

! 1+(22 f0

h i e

j j

e j j=

6= 1=

a. E [Y (t)] = 0 because the mean of the input is zero. b. The frequency response function of the …lter is 1 10 + j2 f

The output power spectrum is SY (f ) = Sx (f ) jH (f )j2 1 = 1 100 + (2 f )2 0:01 = = 0:05 1 + (2 f =10)2 which is obtained applying (7.89).

1 + (2

f )2

to …nd that the corresponding

Problem 7.23

H (f ) =

2 2

2=10 1 + (2 f =10)2

7.1. PROBLEM SOLUTIONS

c. Use the transform pair exp ( j j= 0 )

! 1+(22 f0

to …nd the power spectrum as

RY ( ) = 0:05e 10j j d. Since the input is Gaussian, so is the output. Also, E [Y ] = 0 and var[Y ] = RY (0) = 0:05, so exp y 2 =2 2Y q fY (y) = 2 2Y where

2 = 0:05: Y

e. Use (7.189) with x = y1 = Y (t1 ) , y = y2 = Y (t2 ) , mx = my = 0, and Also ( )= Set

2 x =

2 y = 0:05

RY ( ) = e 10j j RY (0)

= 0:03 to get (0:03) = 0:741. Put these values into (7.189).

Problem 7.24 Use the integrals I1 = and I2 =

Z 1

b0 ds j b0 = a0 a1 1 (a0 s + a1 ) ( a0 s + a1 )

Z 1

b0 s2 + b1 ds =j 2 2 a1 s + a2 ) 1 (a0 s + a1 s + a2 ) (a0 s

b0 + a0 b1 =a2 a0 a1

to get the following results for BN . Filter Type

First Order

Chebyshev

fc =2

Butterworth

2 fc

Second Order p

f = 2 qcp 1+1= 2 1+1= 2 1 p f 2 2 c

Problem 7.25 The transfer function is H (s) = =

! 23 p p s + ! 3 = 2 + j! 3 = 2 s + ! 3 = 2 j! 3 = 2 A A p p + p p s + ! 3 = 2 + j! 3 = 2 s + ! 3 = 2 j! 3 = 2 p

CHAPTER 7. RANDOM SIGNALS AND NOISE

p where A = j! 3 = 2 with ! 3 the 3-dB frequency in rad/s. This is inverse Fourier transformed, with s = j!, to yield p p !3t h (t) = 2! 3 exp ! 3 t= 2 sin p u (t) 2 for the impulse response. Now Z 1 1 f3 !3 jh (t)j2 dt = p = p 2 1 4 2 2 2 R1 after some e¤ort. Also note that 1 h (t) dt = H (0) = 1: Therefore, the noise equivalent bandwidth, from (7.108), is f3 250 BN = p Hz = p Hz 2 2 2 Problem 7.26 a. Note that Ha; max = 2, Ha (f ) = 2 for 1 f 1; Ha (f ) = 1 for 2 f 1 f 2 and is 0 otherwise. Thus Z 1 Z 1 Z 2 1 1 2 2 BN = jHa (f )j df = 2 df + 12 df 2 Ha; 4 0 1 max 0 1 (4 + 1) = 1:25 Hz = 4

1 and

b. For this case Hb; max = 2 and the frequency response function is a 2-unit high isoceles triangle centered at 0 and 100 Hz wide so that Z 1 Z 1 1 50 2 BN = jH (f )j df = [2 (1 f =50)]2 df ; v = f =50 b 4 0 Hb;2 max 0 Z 1 1 50 = 50 (1 v)2 dv = (1 v)3 = 16:67 Hz 3 0 0 c. For this case Hc; max = 1 and for the given frequency response function Z Z 1 1 1 100 1 2 BN = df jH (f )j df = c Hc;2 max 0 1 0 100 + (2 f )2 Z 1 1 = df ; v = 2 f =10 1 + (2 f =10)2 0 Z 1 10 1 1 5 1 = dv = tan v 2 0 1 + v2 0 5 = = 2:5 Hz 2

7.1. PROBLEM SOLUTIONS

d. This frequency response function can be written in terms of piecewise functions as 8 0; f < 5 > > < 2 + f =5; 5 f < 0 Hd (f ) = 2 f =5; 0 f 5 > > : 0; f > 5

That is, it is a unit-high isoceles triangle of total width 10 centered at 0 on top of a unit-high rectangle centered at 0 of total width 10. Thus Hd; max = 2 and for the given frequency response function Z 1 Z 1 5 1 2 jH (f )j df = BN = (2 f =5)2 df; v = f =5 d 2 2 2 Hd; 0 max 0 Z 1 1 51 5 5 3 (2 v)2 dv = (2 v)3 = 1 23 = 4 0 43 12 0 = 35=12 = 2:92 Hz

Problem 7.27 By de…nition BN

1 2 Hmax Z 500

Z 1 0

jH (f )j2 df =

1 4

Z 1

500 100

Z 600 500 2 f 500 2 f 500 df + 1 df ; u = 100 100 100 400 500 Z 0 Z 1 = 100 (1 + u)2 du + 100 (1 u)2 du =

0 100 (1 + u)3 = 3 1 = 66:67 Hz

100 (1 3

u)3

= 0

100 100 + 3 3

Problem 7.28 a. Use the impulse response method. First, use partial fraction expansion to get the impulse response: Ha (f ) =

10 23

1 j2 f + 2

1 j2 f + 25

, ha (t) =

10 e 2t 23

e 25t u (t)

CHAPTER 7. RANDOM SIGNALS AND NOISE Thus BN

2 1 jh (t)j dt

= 2 = =

R1 hR

i2 = 1 2 h (t) dt 1

R 10 2 1 e 2t 23 0 R 1 10 2t 23 0 (e

e 25t

e 25t ) dt

R1 1 0 e 4t 2e 27t + e 50t dt 1 2 1 2 e 2t + 1 e 25t 1 2

2 25 2 1 4t + 27 e 27t 4e 2

1 50 2 23 = 0:463 Hz =

0 1 50t 1 50 e 0 2

(1=2

1=25)

1 4

2 1 + 27 50

1 2

50 23

529 27 100

b. Again use the impulse response method. Use the transform pair t exp ( at) u (t) $ 1 to get (a+j2 f )2 hb (t) = 100t exp ( 10t) u (t) Thus, the equivalent noise bandwidth is R1 R1 2 2 1 jh (t)j dt 0 [100t exp ( 10t)] dt BN = = hR i2 hR i2 1 1 2 h (t) dt 2 100t exp ( 10t) dt 1 1 R1 2 R 3 1 u2 exp ( u) du t exp ( 20t) dt 20 0 0 = i2 ; u = 20t; v = 10t hR i2 = h R1 1 2 v exp ( v) dv 2 t exp ( 10t) dt 2 10 1 1 R1 2 Z 1 4 10 0 u exp ( u) du = xn e x dx = n! for n integer hR i2 ; use the integral 1 0 2 8 103 1 v exp ( v) dv =

10 2! 20 = 1:25 Hz = 2 16 (1!) 16

Problem 7.29 a. Choosing f0 = f1 moves f1 right to f = 0 and f1 left to f = 0. The lowpass …ltered version of the superposition of these two spectra is the power spectrum of nc (t)h or ns (t). Itiis a v-shaped spectrum centered at f = 0 given by SLP (f ) = f 1 2 N0 1 f2 f1

7.1. PROBLEM SOLUTIONS

b. Choosing f0 = f2 moves f2 right to f = 0 and f2 left to f = 0. Thus, the baseband f spectrum can be written as SLP (f ) = 12 N0 f2 f1 : c. For this case both triangles (left and right) are centered around the origin and they f add to give SLP (f ) = 12 N0 f2 f1 . d. They are not uncorrelated for any case for an arbitrary delay. However, all cases give quadrature components that are uncorrelated at the same instant.

Problem 7.30 a. By inverse Fourier transformation, the autocorrelation function is Rn ( ) =

j j

where K = =2. b. Use the result from part (a) and the modulation theorem to get Rnp ( ) =

j j

cos (2 f0 )

c. The result is

Snc (f ) = Sns (f ) =

2 + (2

f )2

and Snc ns (f ) = 0

Problem 7.31 a. The equivalent lowpass power spectral density is given by Snc (f ) = Sns (f ) = N0 The cross-spectral density is Snc ns (f ) = 0: b. For this case Snc (f ) = Sns (f ) = N20 Snc ns (f ) =

f 2(f2 f1 ) N0 2 ; N0 2 ;

(f2 f

and the cross-spectral density is f1 ) (f2

f 0 f1 )

f f2 f1

CHAPTER 7. RANDOM SIGNALS AND NOISE c. For this case, the lowpass equivalent power spectral densities are the same as for part (b) and the cross-spectral density is the negative of that found in (b). d. For part (a) Rnc ns ( ) = 0 For part (b) Rnc ns ( ) = j = j = = = =

"Z N0 2

N0 j2 f e df + 2

(f2 f1 )

e j2 f j2

+ (f2 f1 )

Z f1 f 2 0

e j2 f j2

N0 j2 f e df 2 # (f f ) 2

N0 1 + ej2 (f2 f1 ) + e j2 (f2 f1 ) 4 N0 f2 2 cos [2 (f2 f1 ) ]g 4 N0 sin2 [ (f2 f1 ) ] h i N0 (f2 f1 )2 sinc2 [(f2 f1 ) ]

For part (c), the cross-correlation function is the negative of the above. Problem 7.32 The result is

1 1 Sn2 (f ) = Sn (f f0 ) + Sn (f + f0 ) 2 2 so take the given spectrum, translate it to the right by f0 and to the left by f0 , add the two translated spectra, and divide the result by 2. Problem 7.33 De…ne N1 = A + Nc . Then R= Note that the joint pdf of N1 and Ns is fN1 Ns (n1 , ns ) =

1 2

N12 + Ns2

exp 2

Thus, the cdf of R is FR (r) = Pr (R

r) =

Z Z

p 2

1 h

2 2

N1 +Ns2 r

(n1

A)2 + n2s

fN1 Ns (n1 , ns ) dn1 dns

7.1. PROBLEM SOLUTIONS

Change variables in the integrand from rectangular to polar with n1 =

cos

and n2 =

sin

Then

FR (r) =

Z rZ 2 0

Z r 0

exp 2

exp

2 2

1 2

1 h

2 2

A)2 + ( sin )2

( cos

+ A2

Di¤erentiating with respect to r, we obtain the pdf of R:

fR (r) =

r 2

exp

1 2

r 2 + A2

rA 2

Problem 7.34 The suggested approach is to apply

Sn (f ) = lim

T !1

n o E j= [x2T (t)]j2 2T

where x2T (t) is a truncated version of x (t). Thus, let

x2T (t) =

N X

nk (t

kTs )

k= N

The Fourier transform of this truncated waveform is

= [x2T (t)] =

N X

k= N

nk e j2 kTs

, r

d d

CHAPTER 7. RANDOM SIGNALS AND NOISE

Thus, 8 9 2 N < X = n o E j= [x2T (t)]j2 = E nk e j2 kTs : ; k= N ( N ) N X X j2 kTs j2 lTs = E nk e nl e k= N

N X

l= N

E [nk nl ] e j2 (k l)Ts

k= N l= N

N N X X

Rn (0) kl e j2 (k l)Ts

k= N l= N

N X

Rn (0) = (2N + 1) Rn (0)

k= N

But 2T = 2N Ts so that Sn (f ) = =

lim

T !1

n o E j= [x2T (t)]j2 2T

Rn (0) Ts

= lim

N !1

(2N + 1) Rn (0) 2N Ts

Problem 7.35 a. The output is 1 y (t) = T

Z t+T

x( )x(

Therefore Z 1 t+T E [y (t)] = E x( )x( )d T t Z 1 t+T = E [x ( ) x ( )] d T t Z 1 t+T = Rx ( ) d = Rx ( ) T t

7.1. PROBLEM SOLUTIONS

b. Obtain the result by writing y 2 (t) as an iterated integral: Z t+T Z t+T 1 2 x( )x( x( )x( )d y (t) = T2 t t Z t+T Z t+T 1 = x( )x( )x( )x( T2 t t

)d )d d

Take the expectation: Z t+T Z t+T 1 2 E y (t) = E x( )x( )x( )x( )d d T2 t t Z t+T Z t+T 1 E [x ( ) x ( ) x ( )x( )] d d ; identify x1 = x ( ) etc. = T2 t t Z t+T Z t+T fE [x ( ) x ( )] E [x ( )x( )] 1 +E [x ( ) x ( )] E [x ( ) x ( )] d d = T2 t t +E [x ( ) x ( )] E [x ( ) x ( )]g Z t+T Z t+T 1 = Rx2 ( ) + Rx2 ( ) + Rx ( ) Rx ( ) d d T2 t t Problem 7.36 a. In the expectation for the cross correlation function, write the derivative as a limit: dy (t + ) dt y (t + + ) = E y (t) lim

Ryy_dotp ( ) = E y (t)

= lim

y (t + )

fE [y (t) y (t + + )]

Ry ( + )

E [y (t) y (t + )]g

Ry ( )

dRy ( ) d

b. The variance of Y = y (t) is 2Y = N0 B. The variance for Z = dy dt is found from its power spectral density. Using the fact that the transfer function of a di¤erentiator is j2 f we get N0 SZ (f ) = (2 f )2 (f =2B) 2

CHAPTER 7. RANDOM SIGNALS AND NOISE so that

2 Z

= 2 2 N0

Z B

f 2 df = 2 2 N0

4 3

f3 3

B B

N0 B 3

dR ( )

The two processes are uncorrelated because Ryy_dotp (0) = dy = 0 because the derivative of the autocorrelation function of y (t) exists at = 0 and therefore must be 0. Thus, the random process and its derivative are independent. Hence, their joint pdf is

fY Z ( ;

exp

2 =2N B 0

2 N0 B

exp

=2:67 2 N0 B 3

2:67 3 N0 B 3

c. No, because the derivative of the autocorrelation function at not exist (it is a double-sided exponential).

= 0 of such noise does

7.2. COMPUTER EXERCISES

7.2

Computer Exercises

Computer Exercise 7.1 % ce7_1.m: Computes approximations to mean and mean-square ensemble and time % averages of cosine waves with phase uniformly distributed between 0 and 2*pi % clf f0 = 1; A = 1; theta = 2*pi*rand(1,100); t = 0:.1:1; X = []; for n = 1:100 X = [X; A*cos(2*pi*f0*t+theta(n))]; end EX = mean(X); AVEX = mean(X’); EX2 = mean(X.*X); AVEX2 = mean((X.*X)’); disp(‘’) disp(‘Sample means (across 100 sample functions)’) disp(EX) disp(‘’) disp(‘Typical time-average means (sample functions 1, 20, 40, 60, 80, & 100)’) disp([AVEX(1) AVEX(20) AVEX(40) AVEX(60) AVEX(80) AVEX(100)]) disp(‘’) disp(‘Sample mean squares’) disp(EX2) disp(‘’) disp(‘Typical time-average mean squares’) disp([AVEX2(1) AVEX2(20) AVEX2(40) AVEX2(60) AVEX2(80) AVEX2(100)]) disp(‘’) for n = 1:5 Y = X(n,:); subplot(5,1,n),plot(t,Y),ylabel(‘X(t,ntheta)’) if n == 1 title(‘Plot of …rst …ve sample functions’) end if n == 5

CHAPTER 7. RANDOM SIGNALS AND NOISE xlabel(‘t’) end end A typical run follows with a plot given in Fig. 7.1. >> ce7_1 Sample means (across 100 sample functions) Columns 1 through 6 -0.0121 0.0226 0.0486 0.0560 0.0421 0.0121 Columns 7 through 11 -0.0226 -0.0486 -0.0560 -0.0421 -0.0121 Typical time-average means (sample functions 1, 20, 40, 60, 80, & 100) -0.0232 -0.0654 0.0255 -0.0344 0.0055 0.0245 Sample mean squares Columns 1 through 6 0.4613 0.4538 0.5101 0.5525 0.5223 0.4613 Columns 7 through 11 0.4538 0.5101 0.5525 0.5223 0.4613 Typical time-average mean squares 0.4604 0.5015 0.4617 0.4676 0.4549 0.4611

Computer Exercise 7.2 This is a matter of changing the statement theta = 2*pi*rand(1,100); in the program of Computer Exercise 7.1 to theta = (pi/2)*(rand(1,100) - 0.5); The time-average means will be the same as in Computer Exercise 7.1, but the ensembleaverage means will vary depending on the time at which they are computed. Computer Exercise 7.3 This was done in Computer exercise 6.2 by printing the covariance matrix. The diagonal terms give the variances of the X and Y vectors and the o¤-diagonal terms give the correlation coe¢ cients. Ideally, they should be zero. % ce7_3: %

computes covariance matrix of two vectors of Gaussian random variables that are independent

7.2. COMPUTER EXERCISES

Plot of first five sample functions X(t,θ)

1 0 -1

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5 t

0.6

0.7

0.8

0.9

X(t,θ)

1 0 -1 X(t,θ)

1 0 -1

CHAPTER 7. RANDOM SIGNALS AND NOISE

4 3 2 1

0 -1 -2 -3 -4 -4

-3

-2

-1

0 X

% clf X = randn(1,1000); Y = randn(1,1000); disp(’Covariance matrix:’) disp(cov(X, Y)) plot(X, Y, ’+’), axis square, xlabel(’X’), ylabel(’Y’) A typical run follows with a plot given in Fig. 7.2. >> ce7_3 Covariance matrix: 1.0553 0.0235 0.0235 1.0341

7.2. COMPUTER EXERCISES

Computer Exercise 7.4 % ce7_4.m: plot of Ricean pdf for several values of K % clf A = char(‘-’,‘–’,‘-.’,‘:’,‘–.’,‘-..’); sigma = input(‘Enter desired value of sigma => ’); r = 0:.1:15*sigma; n = 1; KdB = []; for KK = -10:5:15; KdB(n) = KK; K = 10.^(KK/10); fR = (r/sigma^2).*exp(-(r.^2/(2*sigma^2)+K)).*besseli(0, sqrt(2*K)*(r/sigma)); plot(r, fR, A(n,:)) if n == 1 hold on xlabel(’r’), ylabel(’f_R(r)’) grid on end n = n+1; end legend([‘K = ’, num2str(KdB(1)),‘dB’],[‘K = ’, num2str(KdB(2)),‘dB’], [‘K = ’, num2str(KdB(3)),‘dB’], [‘K = ’, num2str(KdB(4)),‘dB’],[‘K = ’, num2str(KdB(5)),‘dB’], [‘K = ’, num2str(KdB(6)),‘dB’]) title([‘Ricean pdf for nsigma = ’, num2str(sigma)]) A typical run follows with a plot given in Fig. 7.3. >> ce7_4 Enter desired value of sigma => 2

CHAPTER 7. RANDOM SIGNALS AND NOISE

Ricean pdf for σ = 2 0.35 K = -10 dB K = -5 dB K = 0 dB K = 5 dB K = 10 dB K = 15 dB

0.3

0.25

fR(r)

0.2

0.15

0.1

0.05

15 r

Chapter 8

Noise in Modulation Systems 8.1

Problems

Problem 8.1 Degrees kelvin (degrees absolute) are converted to to the Celsius scale by substracting 273 (actually 273.15). Degrees Celsius are converted to degrees Fahrenheit by the familiar expression 9 degrees F = degrees C × + 32 5 Thus 9 degrees F = (kelvin − 273) + 32 5 Thus, 3 degrees on the kelvin scale is (5 − 273)

9 + 32 = −4504 degrees F 5

and 290 degrees on the kelvin scale is (290 − 273)

9 + 32 = 626 degrees F 5

Problem 8.2 The signal power at the output of the lowpasss filter is  . The noise power is 0  , where  is the noise-equivalent bandwidth of the filter. From (6.116), we know that the noise-equivalent bandwidth of an  order Butterworth filter with 3 dB bandwidth  is  () =

2 sin (2) 1

CHAPTER 8. NOISE IN MODULATION SYSTEMS

Thus, the signal-to-noise ratio at the filter output is   =

 sin (2)   =  () = 0  2 0  0 

so that  () is given by  () =

sin (2) 2

Thus we have the values given in the following table.  1 3 5 10

 () 06366 09549 09836 09959

SNR 06366 0 09549 0 09836 0 09959 0

Note that for large , sin(2)(2) ≈ 1, so that the SNR→  0  as  → ∞. As the order of the Butterworth filter increases, the amplitude response approaches that of an ideal filter. Thus the output SNR approaches that of the input SNR as  increases. Problem 8.3 The signal power is = The noise power is =

(5)2 = 125 2

Z 8

1 0  = 80 −8 2

The SNR is  =

125 80

125 −3010 125 (10 = 00015625 W/Hz )= 8 8000 Any larger value of 0 will result in an SNR30db. 0 =

Problem 8.4 We express  () as ¸ ¸ ∙ ∙ 1 1  () =  () cos 2  ± (2 )  +  +  () sin 2  ± (2 )  +  2 2

8.1. PROBLEMS

Snc  f  , Sns  f  N0

1  W 2

1 W 2

Figure 8.1: Plots of  ( ) and  ( ) for Problem 8.4. where we use the plus sign for the   and the minus sign for the . The received signal is b () sin (2  + )]  () =  [ () cos (2  + ) ∓ 

Multiplying  () +  () by 2 cos (2  + ) and lowpass filtering yields  () =   () +  () cos ( ) ±  () sin ( )

From this expression, we can show that the postdetection signal and noise powers are given by  = 2 2

 = 0 

=  2

 = 0 



This gives a detection gain of one. The power spectral densities of  () and  () are illustrated in Figure 8.1. Problem 8.5 Expanding the noise about  ± 2 gives  () =  () cos [2  ±   + ] +  () sin [2  ±   + ] where we use the plus sign for the   and the minus sign for the . The received signal is b () sin (2  + )]  () =  [ () cos (2  + ) ∓ 

Multiplying  () +  () by 2 cos (2  + ) and lowpass filtering yields  () =   () +  () cos ( ) ±  () sin ( )

CHAPTER 8. NOISE IN MODULATION SYSTEMS

We see that the postdetection signal and noise powers are, as given in the text,  = 2 2

 = 0 

 2

 = 0 



This gives a detection gain of one. The power spectral densities of  () and  () are easily plotted. Problem 8.6 The message signal is  () = 10 cos (8) Thus,  () = cos (8) so that 2 =

1 2

The eﬃciency is therefore given by   =

(05) (05)2 = 01111 = 1111% 1 + (05) (05)2

From (8.36), the detection gain is (SNR) = 2  = 02222 = −65256 dB (SNR) Relative to baseband, the output SNR is (SNR) = −65256 dB  0  If the modulation index in increased to 08, the eﬃciency becomes   =

(05) (08)2 = 02424 = 2424% 1 + (05) (08)2

This gives a detection gain of (SNR) = 2  = 04848 = −31443 dB (SNR) which, relative to baseband, is (SNR) = −3882 dB  0 

8.1. PROBLEMS

This represents an improvement of 276 dB. Problem 8.7 The first step is to determine the value of  . Since  {−  − } +  {   } = 001 First we compute Z ∞

 {   } =



µ ¶  1 − 2 22 √  = 0005  =   2

This gives  = 2576  Thus,  = 2576 and  () =

 () 2576

Thus, since 2 =  2 2 =

2 2 = = 01507 68358 2 (2576)2

Since  = 08, we have   =

01507 (08)2 = 0088 = 802% 1 + 01507 (08)2

and the detection gain is (SNR) = 2 = 01759 (SNR) Problem 8.8 The output of the predetection filter is  () =  [1 +  ()] cos [   + ] +  () cos [   +  +  ()] The noise function  () has a Rayleigh pdf. Thus  ( ) =

 −2 2  

CHAPTER 8. NOISE IN MODULATION SYSTEMS

where  is the predetection noise power. This gives 1 1  = 2 + 2 = 0  2 2 From the definition of threshold 099 =

Z  0

 −2 2   

which gives

099 = 1 − − 2 Thus, −

2 = ln (001) 2

which gives 2 = 921  The predetection signal power is i 1 1 h  =  1 + 2 2 ≈ 2 [1 + 1] = 2 2 2

which gives

2  =  = 921 ≈ 964 dB   Problem 8.9 Since  () is a sinusoid, 2 = 12 , and the eﬃciency is   =

1 2 2 2 1 2 = 2 + 2 1 + 2

and the output SNR is (SNR) =   = In dB we have (SNR) = 10 log10

 2 2 2 +  0 

2 2 + 2

+ 10 log10

For  = 03, (SNR) = 10 log10

 − 13659 0 

 0 

8.1. PROBLEMS

For  = 05, (SNR) = 10 log10

 − 9542 0 

(SNR) = 10 log10

 − 8166 0 

(SNR) = 10 log10

 − 6154 0 

For  = 06,

For  = 08,

The plot of (SNR) as a function of  0  in dB is linear having a slope of one. The modulation index only biases (SNR) down by an amount given by the last term in the preceding four expressions. Problem 8.10 Let the predetection filter bandwidth be  and let the postdetection filter bandwidth be  . The received signal (with noise) at the predetection filter output is represented  () =  [1 +  ()] cos    +  () cos    −  () sin    which can be represented  () = { [1 +  ()] +  ()} cos   −  () sin   The output of the square-law device is  () = { [1 +  ()] +  ()}2 cos2   

− { [1 +  ()] +  ()}  () cos  () +2 () sin2 cos   ()

Assuming that the postdetection filter removes the double frequency terms,  () can be written 1 1  () = { [1 +  ()] +  ()}2 + 2 () 2 2 Except for a factor of 2, this is (8.53). We now express the preceding expression as 1 1 1  () = 2 [1 + 2 () + 2 2 ()] +  [1 +  ()]  () + 2 () + 2 () 2 2 2 Let () = cos( ) + cos(2  ), which clearly has a maximum value of 2. Thus  () =

1 1 cos( ) + cos(2  ) 2 2

CHAPTER 8. NOISE IN MODULATION SYSTEMS

and 2 () =

1 1 1 1 1 + cos(2 ) + + cos(4  ) + cos( ) cos(2  ) 8 8 8 8 2

1 1 1 1 1 + cos( ) + cos(2  ) + cos(3 ) + cos(4  ) 4 4 8 4 8 Putting this together gives ¶ µ 2 1 2  1+ +  () = 2  4 ¶ ¶ ∙µ µ ¸  1  1 2   + cos(  ) + + cos(2  ) + 2 8 2 16 ¸ ∙ 1 1 2 2 cos(3  ) + cos(4  ) + 8 16   () + ∙ ¸ 1 1   cos(  ) + cos(2  )  () + 2 2 1 1 2  () + 2 () 2 2 2 () =

The PSD therefore consists of 6 basic terms as follows: Term 1: A dc term ( = 0) which is an impulse. Term 2: Impulse functions at frequencies  = ± and  = ±2 . The weights are given on line 2 of the preceding equation when squared and divided by two. Note that the terms do not have the same weight due to intermodulation distortion. Thus the demodulated output is distorted. Term 3: Impulse functions at frequencies  = ±3 and  = ±4 resulting from harmonic distortion. Term 4: The PSD of  () (flat) about  = 0 and weighted by  . Term 5: The PSD of  () (flat) translated to frequencies  = ± and  = ±2 and weighted by  2. Term 6: The PSD of 2 () (triangular) about  = 0 The PSD of 2 () is identical to the PSD of 2 () Problem 8.11 Equation (8.59) is, suppressing ,  22 +2 = {[2 + 2 ]2 } −  2 {2 + 2 } The first term is {[2 + 2 ]2 } = {4 } + {4 } + 2{2 }{2 }

8.1. PROBLEMS

where the last term follows from the independence of  and  . Since (See Problem 6.22) {4 } = {4 } = 3 4 and {2 } = {2 } =  2 we have {[2 + 2 ]2 } = 3 4 + 3 4 + 2 2  2 = 8 4 Also Thus

£ ¤2 £ ¤2 £ ¤2  2 {2 + 2 } = {2 + 2 } = {2 + 2 } = 2 2 = 4 4 £ ¤2  22 +2 =  2 {2 + 2 } − {2 + 2 } = 8 4 − 4 4 = 4 4

Problem 8.12 By definition  () = and 2 = so that

()  

2  2 1 = = 2 2 2 2    (  )

¡ ¢2 2    (SNR) = 2  = ¡  ¢2 0  0  1+ 

(SNR) =

 2 ()2 + 1 0 

The plot is easily generated. We conclude that for  À  (SNR) ≈ 2 and for  ¿ 

Problem 8.13

³  ´2    0 

(SNR) ≈ 2

 0 

CHAPTER 8. NOISE IN MODULATION SYSTEMS

Assume sinusoidal modulation since sinusoidal modulation was assumed in the development of the square-law detector. For  = 1 and  () = cos (2 ) so that 2 = 05, we have, for linear envelope detection (since  0  À 1, we use the coherent result), (SNR) =  

1    2 2 1  = = 2 1 = 0  3 0  1 + 2 0  1 + 2 2 0 

For square-law detection we have, from (6.61) with  = 1 (SNR) = 2

 2 + 2

¶2

 2  = 0  9 0 

Taking the ratio (SNR) (SNR)



= 91 0  =  3 0 

2 = −176 dB 3

This is approximately −18 dB. Problem 8.14 The transfer function of the RC highpass filter is  ( ) =

2   = 1 1 + 2   + 2

so that |( )|2 =

(2)2 1 + (2 )2

Thus, The PSD at the output of the ideal lowpass filter is ( 2 0 (2 ) | |   2, 2 1+(2 )  ( ) = | |   0, The noise power at the output of the ideal lowpass filter is =

Z 

 ( )  = 0

−

Z 

(2 )2  1 + (2)2

with  = 2 , the preceding expression becomes 0 = 2

Z 2 0

2  1 + 2

8.1. PROBLEMS

Since

2 1 =1− 1 + 2 1 + 2

we can write 0 = 2 or = which is

½Z 2 0

 −

Z 2 0

 1 + 2

¢ 0 ¡ 2 − tan−1 (2 ) 2

¸ ∙ 2 − tan−1 (2 ) 0 tan−1 (2 ) = 0  = 0  − 2 2 The output signal power is 1 2 (2 )2  = 2 | ( )|2 = 2 2 1 + (2 )2 Thus, the signal-to-noise ratio is  (2 )2 2 2 =  20 1 + (2 )2 2 − tan−1 (2 )  → 0. Note that as  → ∞ 

Problem 8.15 We can write ¶ µ ¶ µ ¶ ∙ µ ¶¸ µ  7 5 5 7 = cos 20 + = cos 20  + sin 20 +  = cos 20 − + 4 2 4 4 80 so that the receiver input is ∙ µ ¶¸ 5 () = 5 cos 20  + + 02 cos (60) + () 80 The signal power is 52 = 125 2 and the power in the components orthogonal to the signal (noise) is  =

 =

022 + 012 = 002 + 001 = 005 2

CHAPTER 8. NOISE IN MODULATION SYSTEMS

The SNR is

125 = 250 = 2398 dB 005 The delay is negative as shown. The delay cannot be negative and so we add two periods, which is 220 s. Therefore the dlay is SNR =

Delay =

5 3 8 − = s 80 80 80

Problem 8.16 From (8.101) and (8.107), we have  2  SSB  2 3 ¡ 2 ¢2  2 + 2  DSB  4  

006 =  2 + 006 =

Thus we have the following signal-to-noise ratios  2  2

 2  2

1  006 −  2 1

006 − 34

SSB

³ ´2   2

DSB

The SSB characteristic has an asymptote defined by  2 = 006 and the DSB result has an asymptote defined by

3 ¡ 2 ¢2 = 006  4   2 = 02828

The curves are shown below. The appropriate operating regions are above and to the left of the curves. It is clear that DSB has the larger operating region. Problem 8.17 This problem is identical to the previus problem except the upper bound on the meansquare-error is 0.1. Therefore

8.1. PROBLEMS

1200 USB SSB

Signal-to-Noise Ratio

1000

800

600

400

200

0.05

0.1

0.15 0.2 phase error variance

0.25

0.3

0.35

Figure 8.2: Operating regions for Problem 8.16.

 2  2

The result is shown below.

1  01 −  2 1

SSB

³ ´2  01 − 34  2

DSB

Problem 8.18 The phasor diagram for SNR À 1 is illustrated in Figure 8.4. The phasor diagram for SNR¿1 is shown in Figure 8.5. We see that the angle (), which determines the demodulated output, is approximately  (), whereas for SNR À 1 the angle () is approximately (). For () ≈ () the error in the demodulated output is small. Problem 8.19 The single-sided spectrum of a stereophonic FM signal and the noise spectrum is shown in Figure 8.6. The two-sided noise spectrum is given by  ( ) =

2  0  , 2

−∞ ∞

CHAPTER 8. NOISE IN MODULATION SYSTEMS

25 USB SSB

Signal-to-Noise Ratio

0.05

0.1

0.15 0.2 phase error variance

0.25

0.3

0.35

Figure 8.3: Operating regions for Problem 8.17.

R(t )e j (t )

rn (t )e jn ( t )

rn (t )sin[ (t )   (t )]  rn (t )sin[e (t )]

rn (t ) cos[ (t )   (t )]  rn (t ) cos[e (t )] Ac e j ( t )

Figure 8.4: Phasor diagram for SNR À 1.

8.1. PROBLEMS

rn (t )sin[ (t )   (t )]  rn (t )sin[e (t )] rn (t ) cos[ (t )   (t )]  rn (t ) cos[e (t )] rn (t )e jn ( t )

R(t )e j (t )

Ac e j (t )

Figure 8.5: Phasor diagram for SNR ¿ 1 The predetection noise powers are easily computed. For the  +  channel Z 15000 2 2 ¡ 12 ¢   2 + = 2    = 225 10 0 0 2 2 0

For the  −  channel

− = 2

Z 53000 23000

2 2 ¡ ¢   0  2  = 9114 1012 0 2 2  

Thus, the noise power on the  −  channel is over 40 times the noise power in the  +  channel. After demodulation, the diﬀerence will be a factor of 20 because of 3 dB detection gain of coherent demodulation. Thus, the main source of noise in a stereophonic system is the  −  channel. Therefore, in high noise environments, monophonic broadcasting is preferred over stereophonic broadcasting. Problem 8.20 The received FDM spectrum is shown in Figure 8.7. The k th channel signal is given by  () =   () cos 21  Since the guardbands have spectral width 3 , 11 =  + 3 +  = 5 , 21 = 1 +  + 3 +  = 10 and so on. Clearly 1 = 5 and the k th channel occupies the frequency band (5 − 1)  ≤  ≤ (5 + 1) 

CHAPTER 8. NOISE IN MODULATION SYSTEMS

Pilot

L f   R f 

Noise Spectrum

SnF  f 

L f   R f 

f  kHz 

Figure 8.6: Sterophonic FM baseband signal and noise. Since the noise spectrum is given by  =

2  0  2 , 2 

| |  

The noise power in the k th channel is Z (5+1) i  3 h (5 + 1)3 − (5 − 1)3  =   2  = 3 (5−1) where  is a constant. This gives  =

¢  3 ¡ 150 2 + 2 ∼ = 50 3  2 3

The signal power is proportional to 2 . Thus, the signal-to-noise ratio can be expressed as µ ¶2 2  (SNR) = 2 =    where  is a constant of proportionality and is a function of  , , ,0 . If all channels are to have equal (SNR) ,   must be a constant, which means that  is a linear function of . Thus, if 1 is known,  =  1   = 1 2  7. 1 is fixed by setting the SNR of Channel 1 equal to the SNR of the unmodulated channel. The noise power of the unmodulated channel is Z    2  =  3 0 =  3 0

8.1. PROBLEMS

Noise PSD

f 0

Figure 8.7: FDM signal and noise for Problem 7.23. yielding the signal-to-noise ratio 0 (SNR) =  3 3 where 0 is the signal power. For 3 = 21 and  = 15, the value of  is ¶ µ ¶2 µ 21 15 −1 15 − tan = 0047 =3 15 21 21 Expressed in dB this is  = 10 log10 (0047) = −133 dB The improvement resulting from the use of preemphasis and deemphasis is therefore 133 dB. Neglecting the tan−1 (3 ) gives an improvement of 21 − 875 = 1225 dB. The diﬀerence is approximately 1 dB. Problem 8.21 With de-emphasis we have 2   = 

∙ µ ¶¸  0  2  3 1 − tan−1  3 3

Without de-emphasis we have 1 2 2 0    =  3 

CHAPTER 8. NOISE IN MODULATION SYSTEMS

Since 2 2 2  =   

For  we have =3

3 

¶2 ∙ µ ¶¸ 3 −1  tan 1−  3

With 3 = 21 kHz and  = 15 kHz we have =3 or

21 15

¶2 ∙ µ ¶¸ 21 15 −1 tan 1− 15 21

£ ¤  = 00588 1 − 0140 tan−1 (71429)

This gives

 = 00446 The result with pre-emphasis from Example 8.3 is (SNR)  = or (SNR)  = 12755

 3

¶2

2

1  1 − 3 tan−1

 3

1 = 12755 ∗ 1251 1 − 0140 tan−1 (71429)

The first term in the above epression, 127.55, is the result from Example 8.3. The second term represents the result of not making the approximation that µ ¶ 3 −1  tan ¿1  3 Not making the approximation increases (SNR)  by approximately 25%. The plot of  as a function of 3 follows. Problem 8.22 From the plot of the signal spectrum it is clear that = Thus the signal power is =2

Z  0

 2

 2 2   =  2  3

8.1. PROBLEMS

1 0.9 0.8 0.7

0.6 0.5 0.4 0.3 0.2 0.1 0

10 W/f3

Figure 8.8: Ratio  as a function of 3 . The noise power is 0 . This yields the signal-to-noise ratio (SNR)1 =

2  3 0 

If  is reduced to  , the SNR becomes (SNR)2 =

2  3 0

This increases the signal-to-noise ratio by a factor of  . Problem 8.23 Since the signal is integrated twice and then diﬀerentiated twice, the output signal is equal to the input signal. The output signal is  () =  cos 2  and the output signal power is 1  = 2  2

| |  

For | |   , the output signal power and the SNR are zero. The noise power spectral density at  () is determined as in the previous problem. Therefore  ( ) =

0 (2 )4 2

CHAPTER 8. NOISE IN MODULATION SYSTEMS

so that the noise power is 0 (2)4  = 2

Z 

0 (2)4 (2)   = 2 − 4

Z 

 4  =

16 0  4  5 5

Thus, the signal-to-noise ratio at  () is (SNR) =

 5 2 5  = 4 5 4 4 2 160   16  0 

| |  

Problem 8.24 The signal power is  = 2

Z  0

 = 23 1 + ( 3 )2

Z 3 0

 = 23 tan−1 1 + 2

 3

Since  = 0  the signal-to-noise ratio at  () is (SNR) =

2 0

3 

tan−1

 3

The normalized (SNR) , defined by (SNR) (20 ) is illustrated in Figure 8.9. Problem 8.25 The instantaneous frequency deviation in Hz is  =   () =  () where  () is a zero-mean Gaussian process with variance  2 = 2  2 . Thus, Z ∞ || 2 2 √ | | = || = − 2  2 −∞ Recognizing that the integrand is even gives r r Z ∞ 2 1 2 −2 22  | | =   =   0  Therefore,

8.1. PROBLEMS

1 0.9 0.8

Norm alized S NR

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

5 W/f3

Figure 8.9: (SNR)D for Problem 8.24.

|| =

2    

´2

This gives (SNR) =

√ 1 + 2 3  

 

2 0 ¸ q h i 2 −2 2    + 6 exp 0    20 

3 ∙q

The preceding expression can be placed in terms of the deviation ratio, , by letting =

 

and  = 2 ( + 1) 

Problem 8.26 The input to the PLL is  () =  cos(2  + ) +  () cos(2  + ) −  () sin((2  + )

CHAPTER 8. NOISE IN MODULATION SYSTEMS

which can be written () cos[2  +  + ()] = { +  ()} cos(2  + ) −  () sin(2  + ) The phase deviation is −1

() = tan

∙

 ()  +  ()

Since the signal-to-noise ratio is suﬃciency large to justify the linear model, the phase deviation is written  () () =  We know that the PSD of the quadrature noise is 0 for | |  2, where  is the predetection bandwidth. Thus, the PSD of the input signal to the baseband (lowpass) model of the PLL is 0 | |  2  ( ) = 2   Using the PLL transfer function (The linear model is assumed valid since the input SNR is assumed large, which yields small phase errors.) gives  ( ) =  ( ) |( )|2 =

0 |( )|2 2

where ( ) is the transfer function relating the phase error Ψ( ) to the input phase Φ( ) (See (3.221)). The variance of the phase error is Z Z 0 ∞ 20 ∞ |( )|2  = 2 |( )|2   2 = 2  −∞  0 The integral represents the equivalent noise bandwidth  of the PLL based on the closedloop transfer function ( ). We have obviously assumed  ¿ , which is typically the case. Thus 20  0  =  2 = 2   where  = 2 2 is the input signal power. The input noise is assumed Gaussian. The phase error is therefore Gaussian since it is a linear transformation, through the transfer function ( ), of the input noise. The preceding expression shows that the phase error variance is the reciprocal of the signal-to-noise ratio, where the noise power is measured in the bandwidth of the transfer function ( ). Problem 8.27

8.1. PROBLEMS

PPM was defined in Chapter 3. Assume that the pulse position is represented by  () =   +  1  () where   is the nominal pulse position for () = 0 and  1 establishes the peak oﬀset from the nominal pulse position and is adjusted so that the pulse does not deviate outside of its assigned slot. We assume a zero-mean signal so that () = 0 for which  =   . The pulse is detected when the received pulse crosses a threshold. When noise is added to the pulse we can approximate the error in detecting the threshold crossing as (1 ) (1 ) =   where  is the pulse risetime, 1 is the point at which the pulse would cross the threshold in the absence of noise, and  represents the pulse amplitude. Thus, in general µ ¶2 µ ¶2   2 2  =  =     As we saw in Chapter 2, the relationship between risetime and bandwidth gives  = 12 , where  is the transmission bandwidth. We let ¶ µ 1 1 =  0 = 2 = 2 2  This gives 2 =

¶2

  =

 =

 2  

1 2

 42 

The average transmitted signal power is

where  is the sampling frequency or 12 assuming Nyquist rate sampling. Using this in the preceding equation by substituting for 2 gives 2 =

   4  

Since  1  () represents the signal, the received signal power is  =  21 2

CHAPTER 8. NOISE IN MODULATION SYSTEMS

we have (SNR)D =

 21 2 2

=  21 2

or (SNR)D =  21 2

4     

=  21 2

¶µ

= 2 21 2 2

2 

  

1 2

¶µ

4    

  

where we have used   = 1 . The peak-to-peak pulse displacement is 2 1 =  = 12 so that  1 = 14 . Substituting this into the preceding equation yields 1 (SNR)D = 2 8

 

¶2 µ

  

Therefore with all the assumptions made  = 18. Note: We have made many assumptions in this problem including Nyquist rate sampling. It is useful, however, for illustrating the thought process for applying the concepts of this chapter to modulation processes other than those considered in this chapter. Problem 8.28 The peak value of the signal is 75 (15 peak-to-peak) so that the signal power is =

1 (75)2 = 5625 2

If the A/D converter has a wordlength , there are 2 quantizing levels. Since these span the peak-to-peak signal range, the width of each quantization level is = This gives 2 =

¡ ¢ 25 = 25 2−  2

¡ ¢ 1 2 1  = (25)2 2−2 = (5208)2−2 12 12

This results in the signal-to-noise ratio   =

 2

¡ ¢ 5625 ¡ 2 ¢ = 108 22 2 5208

8.2. COMPUTER EXERCISES

8.2

Computer Exercises

Computer Exercise 8.1 From the expression for the normalized error 2 =  2 +

 2  2

the following MATLAB program results: % File: ce8_1a.m snrdb = [0 1 2]; snr = 10.^(snrdb./10); varphas = 0:0.1:1; hold on for k=1:3 nerrqpsk = varphas + (1/snr(k)); plot(varphas,nerrqpsk’) end hold off grid, xlabel(’Phase Error Variance’), ylabel(’Normalized MSE (QDSB)’) % End of script file. Executing the program gives the results illustrated in Figure 8.10. We see from the defining equation that the mse is linear in the phase error variance for a given SNR. The SNR simply provides an oﬀset (SNR = 1 (0 dB) curve is on top).

For the DSB system we have the MATLAB program % File: ce8_1b.m snrdb = [0 1 2]; snr = 10.^(snrdb./10); varphas = 0:0.1:1; hold on for k=1:3 nerrqpsk = 0.75*varphas.*varphas + (1/snr(k)); plot(varphas,nerrqpsk’) end hold off

CHAPTER 8. NOISE IN MODULATION SYSTEMS

Normalized MSE (QDSB)

1.8

1.6

1.4

1.2

0.8

0.6

0.1

0.2

0.3

0.4 0.5 0.6 Phase Error Variance

0.7

0.8

0.9

Figure 8.10: Normalized mse for QDSB and SSB with the SNR as a parameter.

8.2. COMPUTER EXERCISES

1.8

Normalized MSE (DSB)

1.6

1.4

1.2

0.8

0.6

0.1

0.2

0.3

0.4 0.5 0.6 Phase Error Variance

0.7

0.8

0.9

Figure 8.11: Normalized mse for DSB with the SNR as a parameter. grid, xlabel(’Phase Error Variance’), ylabel(’Normalized MSE (DSB)’) % End of script file. This yields the result shown in Figure 8.11. The mse is now quadradic in the phase error variance for a given SNR. The SNR again simply provides an oﬀset (SNR = 1 (0 dB) curve is on top).

Computer Exercise 8.2 The MATLAB program for solving this computer exercise follows. % File: ce8_2.m zdB = 0:50; z = 10.^(zdB/10); beta = [1 5 10 20]; hold on for j=1:length(beta) bta = beta(j);

% predetection SNR in dB % predetection SNR % modulation index vector % hold for plots % current index

CHAPTER 8. NOISE IN MODULATION SYSTEMS

80 70 20

Postdetection SNR in dB

60 50 5 40 1 30 20 10 0 -10

20 25 30 35 Predetection SNR in dB

Figure 8.12: Plot for Computer Exercise 8.2. a1 = exp(-(0.5/(bta+1)*z)); % temporary constant a2 = q(sqrt(z/(bta+1))); % temporary constant num = (1.5*bta*bta)*z; den = 1+(4*sqrt(3)*(bta+1))*(z.*a2)+(12/pi)*bta*(z.*a1); result = num./den; resultdB = 10*log10(result); plot(zdB,resultdB,’k’) end hold off xlabel(’Predetection SNR in dB’) ylabel(’Postdetection SNR in dB’) % End of script file. Executing the program gives the results shown in Figure 8.12. We see that the threshold eﬀect is less pronounced. A smaller value of  yields a smaller bandwidth and decreased detection gain. Computer Exercise 8.3

8.2. COMPUTER EXERCISES

The MATLAB program which solves this computer exercise is a modification of the program used in Computer Example 8.5 and follows: % File: ce8_3.m zdB = 0:0.1:40; % predetection SNR in dB z = 10.^(zdB/10); % predetection SNR beta = [2 4 6 8 10 15 20]; % modulation index vector hold on % hold for plots for j=1:length(beta) a2 = exp(-(0.5/(beta(j)+1)*z)); a1 = q(sqrt((1/(beta(j)+1))*z)); r1 = 1+(4*sqrt(3)*(beta(j)+1)*a2); r2 = r1+((12/pi)*beta(j)*a2); semilogy(zdB,r2,’k’) axis([10 30 0 4]) grid end hold off % release xlabel(’Predetection SNR in dB’) ylabel(’1+D_2+D_3’) % End of script file. This program gives the results shown in Figure 8.13 for  = 2, 4, 6, 8 10, 15, and 20. We see that denominator has a value of two for  = 2 at approximately 13 dB. This increases to approximately 23.5 dB at an index of 20. These values can be taken oﬀ Figure 8.13 and plotted if desired. It is clear that the predetection SNR for which the denominator of (7.148) is equal to 2 increases as the modulation index  increases.

Computer Exercise 8.4 The MATLAB program for solving this computer exercise follows. The strategy used is to plot two curves for each modulation index; one curve with the modulation not included and one curve with the modulation included. In order to keep the results from being too cluttered only two modulation indexes are used;  = 5 and  = 20. % File: ce8_4.m zdB = 0:50; z = 10.^(zdB/10); beta = [5 20]; hold on

% predetection SNR in dB % predetection SNR % modulation index vector % hold for plots

CHAPTER 8. NOISE IN MODULATION SYSTEMS

4 3.5 3

1+D2+D3

2.5 4

2 8

1.5 20 1 0.5 0 10

18 20 22 24 Predetection SNR in dB

Figure 8.13: Values of  0  for which 1 + 2 + 3 = 2.

8.2. COMPUTER EXERCISES

for j=1:length(beta) bta = beta(j); % current index a1 = exp(-(0.5/(bta+1)*z)); % temporary constant a2 = q(sqrt(z/(bta+1))); % temporary constant num = (1.5*bta*bta)*z; den1 = 1+(4*sqrt(3)*(bta+1))*(z.*a2); den2 = 1+(4*sqrt(3)*(bta+1))*(z.*a2)+(12/pi)*bta*(z.*a1); result1dB = 10*log10(num./den1); % w/o modulation result2dB = 10*log10(num./den2); % with modulation plot(zdB,result1dB,’k’,zdB,result2dB,’k-.’) end hold off legend(’w/o modulation’,’with modulation’,4) xlabel(’Predetection SNR in dB’) ylabel(’Postdetection SNR in dB’) % End of script file. Executing the program gives the results shown in Figure 8.14. We see, as expected, that above threshold the results are identical. This results since the above-threshold performance is only a function of the predetection SNR and the modulation index. Below threshold, a region in which we typically have little interest, the inclusion of modulation reduces the postdetection SNR, since the denominator of the postdetection SNR expression is increased.

Computer Exercise 8.5 (Note: Since Computer Exercise 8.4 was changed for the new edition, the first sentence for Computer Exercise 8.5 should be deleted.) The input to the PLL is assumed to be a sinusoid plus bandlimited white noise. Thus the PLL input is modeled as  () =  cos(2  + ) +  () cos(2  + ) −  () sin(2  + ) or  () = [ +  ()] cos(2  + ) −  () sin(2  + ) This can be represented as  () = () cos[2  +  + ()] where −1

() = tan

∙

¸ ∙ ¸  ()  () −1  () ≈ tan =  +  ()  

CHAPTER 8. NOISE IN MODULATION SYSTEMS

80 70

Postdetection SNR in dB

60 50 40 30 20 10 0

w/o modulation with modulation 0

20 25 30 35 Predetection SNR in dB

Figure 8.14: FM performance results with and without modulation. The approximation results since the input SNR is assumed large. We therefore use () as the input to the baseband PLL model. The SNR of the input signal is =

2 2 2 =  2 2 2

where  2 = 2 . The variance of () is  2 =

 2 1 = 2 2

In the following program we set the phase error variance. Note that this does not determine the SNR. One can set 2 at some value (such as 1) and then adjust  to give the required SNR. % File: ce8_5.m % beginning of preprocessor clear all % be safe fn = 10; zeta = 0.707; varn = input(’Enter noise variance  ’); npts = 50000; % number of simulation points fs = 2000; % sampling frequency T = 1/fs;

8.2. COMPUTER EXERCISES

t = (0:(npts-1))/fs; % time vector Kt = 4*pi*zeta*fn; % loop gain a = pi*fn/zeta; % loop filter parameter filt_in_last = 0; filt_out_last=0; vco_in_last = 0; vco_out = 0; vco_out_last=0; % end of preprocessor - beginning of simulation loop for i=1:npts phin = sqrt(varn)*randn(1); s1 = phin - vco_out; aa(1,i) = s1; s2 = sin(s1); % sinusoidal phase detector s3 = Kt*s2; filt_in = a*s3; filt_out = filt_out_last + (T/2)*(filt_in + filt_in_last); filt_in_last = filt_in; filt_out_last = filt_out; vco_in = s3 + filt_out; vco_out = vco_out_last + (T/2)*(vco_in + vco_in_last); vco_in_last = vco_in; vco_out_last = vco_out; phierror(i)=s1; end % end of simulation loop - beginning of postprocessor [N_samp,x] = hist(aa,40); % get histogram parameters bar(x,N_samp,1) xlabel(’Histogram Bin’), ylabel(’Number of Samples’) % end of postprocessor % End of script file. Executing the program gives the histogram illustrated in Figure 8.15. The histogram suggests that the pdf of the phase error is closely gaussian.

Computer Exercise 8.6 The MATLAB program for this computer exercise is a simple modification of the program used for Computer Example 8.6. The only change is for the bit error probabilities for FSK (frequency-shift keying) and BPSK (binary phase-shift keying). In order to keep the results from being cluttered results are computed for  = 8. The MATLAB program follows. % File:

ce8_6.m

CHAPTER 8. NOISE IN MODULATION SYSTEMS

4500 4000

Number of Samples

3500 3000 2500 2000 1500 1000 500 0 -2

-1.5

-1

-0.5

0 0.5 Histogram Bin

1.5

Figure 8.15: Histogram of phase error. n = 8; % wordlength snrtdB = 0:0.1:30; % predetection snr in dB snrt = 10.^(snrtdB/10); % predetection snr Pb1 = q(sqrt(snrt)); % bit error probability for FSK Pb2 = q(sqrt(2*snrt)); % bit error for BPSK Pw1 = 1-(1-Pb1).^n; % Pw for FSK Pw2 = 1-(1-Pb2).^n; % Pw for BPSK a = 2^(-2*n); % temporary constant snrd1 = 1./(a+Pw1*(1-a)); snrddB1 = 10*log10(snrd1); % postdetection snr (FSK) snrd2 = 1./(a+Pw2*(1-a)); snrddB2 = 10*log10(snrd2); % postdetection snr (BPSK) plot(snrtdB,snrddB1,’k’,snrtdB,snrddB2,’k-.’) legend(’FSK’,’BPSK’) xlabel(’Predetection SNR in dB’) ylabel(’Postdetection SNR in dB’) % End of script file. The results are shown in Figure 8.16. We see that above threshold, the postdetection SNR is independent of the modulation scheme. This, of course, was expected since above threshold

8.2. COMPUTER EXERCISES

50 FSK BPSK

Postdetection SNR in dB

40 35 30 25 20 15 10 5 0

10 15 20 Predetection SNR in dB

Figure 8.16: PCM performance for FSK and BPSK with  = 8 the postdetection SNR is a function only of wordlength. We also note that the use of BPSK extends the threshold over that of FSK due to the 3-dB improvement in bit error probability.

Computer Exercise 8.7 A little thought shows that since the SNR estimate is very sensitive to the peak of the crosscorrelation function, if the peak is not sampled, a large error can occur. As a simple test let () = 5 cos(2) and let the reference signal be h i () = 4 cos 2 − 2

Consider the following MATLAB program: % File: ce8_7a fs=1/32; t=1:6400; x=5*cos(2*pi*fs*t); y=4*cos(2*pi*fs*t-(pi/2));

CHAPTER 8. NOISE IN MODULATION SYSTEMS [gain,delay,px,py,rxy,rho,snrdb] = snrmse(x,y); a=[’The gain estimate is ’,num2str(gain),’.’]; disp(a) b=[’The delay estimate is ’,num2str(delay),’.’]; disp(b) c=[’The correlation coeﬃcient estimate is ’,num2str(rho),’.’]; disp(c) d=[’The signal-to-noise ratio estimate is ’,num2str(snrdb),’ dB.’]; disp(d) % End of script file.

Executing the program gives: The gain estimate is 0.8. The delay estimate is 8. The correlation coeﬃcient estimate is 1. The signal-to-noise ratio estimate is 150.0228+13.64376i dB. These values are easily understood. From the definition of the signals () and () the gain is 45 = 08. Note that the period is 1. The quantity   is one when () = 32 since the sampling frequency is 1/32. The period of each waveform is 32 samples. The delay is 2, which is 1/4 of a period or 8 sample periods. Therefore the cross-correlation function is sampled precisely at it’s peak. The correlation coeﬃcient is =

 (  )  

This gives, for (8.78) SNR =

 1−

Therefore  = 1 gives an SNR of ∞, which is correct since () is simply an amplitude scaled and delayed version of (). A complex number for the signal-to-noise ratio in dB appears strange. However, this is easily explained. If  is displayed to a suﬃcient number of decimal places, it will be found that , due to roundoﬀ and quantization of the calculations, is very slightly negative. When the logarithm of a negative number is taken to express SNR in dB, the result is a complex number results. Thus, all computed values are correct. In order to illustrate the result when the cross-correlation function is not sampled at the peak, the delay of () is set so that that the peak value of the crosscorrelation falls equidistant between samples. consider the following MATLAB program. % File: ce8_7b.m fs=1/32; t=1:6400; x=5*cos(2*pi*fs*t); y=4*cos(2*pi*fs*t-(pi/2)-(pi/32));

8.2. COMPUTER EXERCISES

[gain,delay,px,py,rxy,rho,snrdb] = snrmse(x,y); a=[’The gain estimate is ’,num2str(gain),’.’]; disp(a) b=[’The delay estimate is ’,num2str(delay),’.’]; disp(b) c=[’The correlation coefficient estimate is ’,num2str(rho),’.’]; disp(c) d=[’The signal-to-noise ratio estimate is ’,num2str(snrdb),’ dB.’]; disp(d) % End of script file. Executing this program gives: The gain estimate is 0.79615. The delay estimate is 9. The correlation coefficient estimate is 0.99518. The signal-to-noise ratio estimate is 20.132 dB. Note the very large drop in the SNR estimate. This issue is mitigated by using a much higher sampling frequency. This significantly increases executation time, especially for complicated systems. However in problems of this type (simulation) there is typically a trade-off between accuracy and execution time. Computer Exercise 8.8 Since the signal is a sinusoid, we can use an inverse sin function as a compressor characteristic. Note that this actually expands the signal but we will use the term ‘compressor’ since this is the first operation performed. The compressor output is a triangular wave so that the compressor output has a uniform pdf ranging from −2 to 2. Thus, with 3 bit words we have 8 quantization levels given by 1 = 0 ≤  ≤ 8

2 = 8 ≤  ≤ 4

3 = 4 ≤  ≤ 38

4 = 38 ≤  ≤ 2 5 = −8 ≤  ≤ 0

6 = −4 ≤  ≤ −8

7 = −38 ≤  ≤ −4

8 = −2 ≤  ≤ −38

The expander characteristic has a sinusoidal characteristic to restore the original sinusiodal signal. The compressor and expander characteristics are given in Figure 8.17 and the original signal, the compressed signal, and the expanded signal are shown in Figure 8.18.

CHAPTER 8. NOISE IN MODULATION SYSTEMS

1 0.8

1.5

0.6 Expander Characteristic

Compressor Characteristic

1 0.5 0 -0.5

0.4 0.2 0 -0.2 -0.4

-1 -0.6 -1.5

-0.8

-2 -1

-0.5

0 Time

0.5

-1 -2

-1

0 Time

Figure 8.17: Compressor and expander characteristics.

Orig. sig.

Compressed sig.

-1 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.1

0.2

0.3

0.4

0.5 Time

0.6

0.7

0.8

0.9

-2

Expanded sig.

-1

Figure 8.18: Original signal, compressed signal, and expanded signal.

8.2. COMPUTER EXERCISES

Computer Exercise 8.9 The computer exercise, as stated in the text, does not make sense since the compressor characteristic, as given, has no frequeny dependence. Please substitute the following. A compressor characteristic is often modeled as out () =  tanh[in ()] The purpose of this Computer Exercise is to show that this simple expression can be used to model a vast number of compressor characteristics. Consider three basic models. (a. Model 1) A compressor characteristic is often modeled as out () = (1) tanh[in ()] Using three separate values of , namely  = 03,  = 06,  = 10, and  = 15, explore the characteristics of this compressor by plotting out as a function of in over a sufficiently wide range to fully define the compressor characteristic. (b. Model 2) A compressor characteristic is often modeled as out () = (1) tanh[(1) in ()] Once again, using three separate values of , namely  = 03,  = 06,  = 10, and  = 15, explore the characteristics of this compressor by plotting out as a function of in over a sufficiently wide range to fully define the compressor characteristic. (c. Model 3) A compressor characteristic is often modeled as out () =  tanh[in ()] This model is a little different. We wish the asympotic values of out to be ±1. Select the value of  so that this condition is satisfied. Then using the same three separate values of , namely  = 03,  = 06,  = 10, and  = 15, explore the characteristics of this compressor by plotting out as a function of in over a sufficiently wide range to fully define the compressor characteristic. Solution The characteristic for (a. Model 1) is developed using the following MATLAB code: %File ce8_9a.m a=[0.3 0.6 1.0 1.5];

CHAPTER 8. NOISE IN MODULATION SYSTEMS

4 y1 y2 y3 y4

Output signal level

2 1 0 -1 -2 -3 -4 -10

-8

-6

-4

-2 0 2 Input signal level

Figure 8.19: Compressor characteristic for Model 1. x=-10:0.001:10; for j=1:4 for i=1:length(x) y1(i)=(1/a(1))*tanh(a(1)*x(i)); y2(i)=(1/a(2))*tanh(a(2)*x(i)); y3(i)=(1/a(3))*tanh(a(3)*x(i)); y4(i)=(1/a(4))*tanh(a(4)*x(i)); end end plot(x,y1,’k-’,x,y2,’k—’,x,y3,’k-.’,x,y4,’k:’) grid xlabel(’Input signal level’) ylabel(’Output signal level’) legend(’y1’,’y2’,’y3’,’y4’) % End of script file. The result is shown below. The characteristic for (b. Model 2) is developed using the following MATLAB code: %File ce8_9b.m a=[0.3 0.6 1.0 1.5]; x=-10:0.001:10;

8.2. COMPUTER EXERCISES

4 y1 y2 y3 y4

Output signal level

2 1 0 -1 -2 -3 -4 -10

-8

-6

-4

-2 0 2 Input signal level

Figure 8.20: Compressor characteristic for Model 2. for j=1:6 for i=1:length(x) y1(i)=(1/a(1))*tanh((1/a(1))*x(i)); y2(i)=(1/a(2))*tanh((1/a(2))*x(i)); y3(i)=(1/a(3))*tanh((1/a(3))*x(i)); y4(i)=(1/a(4))*tanh((1/a(4))*x(i)); end end plot(x,y1,’k-’,x,y2,’k—’,x,y3,’k-.’,x,y4,’k:’) grid xlabel(’Input signal level’) ylabel(’Output signal level’) legend(’y1’,’y2’,’y3’,’y4’) % End of script file. The result for Model 2 is shown below. The characteristic for (c. Model 3) is developed using the following MATLAB code: % File ce8_9c.m a=[0.3 0.6 1.0 1.5]; x=-10:0.001:10;

CHAPTER 8. NOISE IN MODULATION SYSTEMS

1 y1 y2 y3 y4

0.8 0.6

Output signal level

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -10

-8

-6

-4

-2 0 2 Input signal level

Figure 8.21: Compressor characteristic for Model 3. for j=1:6 for i=1:length(x) y1(i)=tanh(a(1)*x(i)); y2(i)=tanh(a(2)*x(i)); y3(i)=tanh(a(3)*x(i)); y4(i)=tanh(a(4)*x(i)); end end plot(x,y1,’k-’,x,y2,’k—’,x,y3,’k-.’,x,y4,’k:’) grid xlabel(’Input signal level’) ylabel(’Output signal level’) legend(’y1’,’y2’,’y3’,’y4’) % End of script file. The result is shown below.

Chapter 9

Principles of Digital Data Transmission in Noise 9.1

Problem Solutions

Problem 9.1 The signal-to-noise ratio is z=

A2 T A2 = N0 N0 R

Trial and error using the asymptotic expression Q (x) Q-function program in MATLAB) shows that PE = Q

exp

x2 =2 =

2z = 10 6 for z = 11:297 = 10:53 dB

Thus A2 = 11:297 N0 R or A =

11:297N0 R p = 11:297 10 6

20000

= 0:4753 V

A rough estimate of the required bandwidth is BW = R = 20000 Hz. 1

2 x (or use of a

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

Problem 9.2 The bandwidth should be equal to the data rate to pass the main lobe of the signal spectrum. By trial and error, solving the following equation for z, q PE jdesired = Q 2 zjrequired ; z = Eb =N0

we …nd that zjrequired = 6:78 dB (4.76 ratio); 8:39 dB (6.90 ratio), 9:58 dB (9.08 ratio), 10:53 dB (11.3 ratio) for PE jdesired = 10 3 , 10 4 , 10 5 , 10 6 , respectively. The required signal power is Ps, required = Eb, required =Tsymbol = Eb, required R = zjrequired N0 R For example, for PE = 10 3 ; N0 = 10 3 ; and R = 1000 bits/second, we get 10 3

Ps, required = zjrequired N0 R = 4:76

1000 = 4:76 W

Similar calculations allow the rest of the table to be …lled in, giving the following results. R, bps 1; 000 10; 000 100; 000

PE = 10 3 ; A2 and BW 4:76 W; 1 kHz 47:6 W; 10 kHz 476:0 W; 100 kHz

PE = 10 4 6:9 W 69 W 690:0 W

PE = 10 5 9:08 W 90:8 W 908:0 W

PE = 10 6 11:3 W 113:0 W 1130:0 W

Problem 9.3 a. For PE jdesired = 10 4 we have zjrequired = 8:39 dB (6.91 ratio). For R = B = 5; 000 bps, we have signal power = A2 = zjrequired N0 R = 6:91

10 6

= 6:91

5; 000

10 3 = 0:0346 W

b. For R = B = 10; 000 bps, the result is signal power = A2 = zjrequired N0 R = 6:91

10 6

10; 000 = 0:0691 W

c. For R = B = 100; 000 bps, we get signal power = A2 = 6:91

10 6

100; 000 = 0:69 W

9.1. PROBLEM SOLUTIONS

d. For R = B = 1; 000; 000 bps, the result is signal power = A2 = 6:91

10 6

1; 000; 000 = 6:91 W

Problem 9.4 The decision criterion is V

> ", choose + A

< ", choose

where V =

AT + N

N is a Gaussian random variable of mean zero and variance 2 = N0 T =2. Because P (A sent) = P ( A sent) = 1=2, it follows that 1 1 1 PE = P ( AT + N > " j A sent) + P (AT + N < " j A sent) = (P1 + P2 ) 2 2 2 where 3 2s s Z 1 2 =N T 2 2 0 e 2A T 2" 5 p P1 = d = Q4 + N N N0 T 0 0T AT +" = Q

2z +

2 =

;

= N0 T; z =

A2 T N0

Similarly AT +"

Z 1

2 =N T 0

e =N0 T p d = d N0 T N0 T 1 AT " 3 2s s 2 2 p p 2A T 2" 5 =Q 2z 2 = = Q4 N0 N0 T

P2 =

Use the approximation

e p

Q (u) = to evaluate

e u =2 p u 2

1 (P1 + P2 ) = 10 6 2 by solving iteratively for z for various values of "= . The results are given in the table below. "= zdB for PE = 10 6 "= zdB for PE = 10 6 0 10.53 0.6 11.34 0.2 10.69 0.8 11.67 0.4 11.01 1.0 11.98

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

Problem 9.5 Use the zjrequired values found in the solution of Problem 9.2 along with zjrequired = A2 =N0 R h i to get R = A2 = zjrequired N0 : Substitute N0 = 10 5 V2 /Hz, A = 40 mV to get R = 16 10 4 below. PE

105 = zjrequired = 160= zjrequired .

zjrequired , dB (ratio)

R, bps

8.39 (6.9) 9.58 (9.08) 10.53 (11.3)

23.19 17.62 14.16

10 10 5 10 6

This results in the values given in the table

Problem 9.6 a. Since the amplitudes A1 and A2 each occur 1/2 the time on average, the average probability of error is 0s

1 PE = Q @ 2

2A21 T A N0

1 + Q@ 2

2A22 T A N0

But the average signal energy is E=

1+ 2 2 1 + A22 =A21 2 A21 T + A22 T = A1 T = A1 T 2 2 2

Thus A21 T =

2 E 1+ 2

and A22 T =

A21 T =

2 2 E 1+ 2

b. Substitution of these into the average probability of error expression gives the stated result. A magni…ed plot of the probability of error is shown below. The curves starting at the lefthand curve are for 2 = 1; 0:8; 0:6; 0:4. The respective values of E=N0 for the crossing of the PE = 10 6 level are approximately 10.5, 10.8, 11.5, and 12.7 dB giving degradations of 0, 0.3, 1, and 2.2 dB for 2 = 1; 0:8; 0:6; 0:4, respectively..

9.1. PROBLEM SOLUTIONS

-3

-4

-5

-6

10.5

11.5 E/N0, dB

12.5

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

∆ T/T = 0 ∆ T/T = 0.1 ∆ T/T = 0.2 ∆ T/T = 0.3

-1

-2

-3

-4

-5

-6

10 z, dB

Problem 9.7

a. For the pulse sequences ( A; A) and (A; A), which occur half the time, no degrap dation results from timing error, so the error probability is Q 2Eb =N0 . The error probability for the sequences ( A; A) and hp (A; A), which occur ithe other half the time, result in the error probability Q 2Eb =N0 (1 2 j T j =T ) (sketches of the pulse sequences with the integration interval o¤set from the transition between pulses is helpful here). Thus, the average error probability is the given result in the problem statement. Plots are shown in Figure 9.2 for the giving timing errors.

b. Blowing up the plots of Figure 9.2 show that the intercepts of PE = 10 4 are about 8.4, 9.9, 12.4, and 15.9 dB, respectively, for timing errors of 0, 0.1, 0.2, and 0.3. The corresponding degradations are 1.5, 4, and 7.5 dB for timing errors of 0.1, 0.2, and 0.3, respectively

9.1. PROBLEM SOLUTIONS

Problem 9.8 As in the derivation in the text for antipodal signaling, 1 var [N ] = N0 T 2 The output of the integrator at the sampling time is V =

AT + N; A sent N; 0 sent

The probabilities of error, given these two signaling events, are 1 AT + N < AT 2 1 AT N< 2

P (error j A sent) = P = P Z

AT =2

e =N0 T p d = N0 T 1 2s 3 2T A 5 = Q4 2N0

Similarly,

P (error j 0 sent) = P =

1 N > AT 2

Z 1

e p

2 =N T 0

N0 T 2s 3 2 A T5 = Q4 2N0 AT =2

Thus

1 1 P (error j A sent) + P (error j 0 sent) 22 2 3 s A2 T 5 = Q4 2N0

PE =

Therefore, the average probability of error is "r PE = Q

Eave N0

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

Problem 9.9 a. The impulse response of the …lter is h (t) = (2 f3 ) e 2 f3 t u (t) and the step response is the integral of the impulse response, or a (t) = 1

e 2 f3 t u (t)

Thus, for a step of amplitude A, the signal out at t = T is s0 (T ) = A 1

e 2 f3 T

The power spectral density of the noise at the …lter output is Sn (f ) =

N0 =2 1 + (f =f3 )2

so that the variance of the noise component at the …lter output is Z 1 N0 f3 2 Sn (f ) df = = N0 BN E N = 2 1 where BN is the noise equivalent bandwidth of the …lter. Thus SNR =

2A2 1 e 2 f3 T s20 (T ) = E [N 2 ] N0 f3

b. To maximize the SNR, di¤erentiate with respect to f3 and set the result equal to 0. Solve the equation for f3 . This results in the equation 4 f3 T e 2 f3 T Let

1 + e 2 f3 T = 0

= 2 f3 T to get the equation (2 + 1) e

A numerical solution results in

1:25: Thus

f3, opt =

1:25 = 0:199=T = 0:199R 2 T

9.1. PROBLEM SOLUTIONS

Problem 9.10 Substitute (9.27) and (9.28) into PE = pP (E j s1 (t)) + (1 p) P (E j s2 (t)) h i Z 1 exp (v s01 )2 =2 20 p = p dv + (1 2 20 k

Z k exp 1

i (v s02 )2 =2 20 p dv 2 20

Use Leibnitz’s rule to di¤erentiate with respect to the threshold, k, set the result of the di¤erentiation equal to 0, and solve for k = kopt : The di¤erentiation gives exp p

or exp p or (kopt (s01

i (k s01 )2 =2 20 p + (1 2 20 h

exp p)

i (kopt s01 )2 =2 20 p = (1 2 20 s01 )2 + (kopt s02 ) kopt +

exp p)

s02 )2 = 2 20 ln

s202

s201 2

2 0 ln

(k s02 )2 =2 20 p 2 20 h

s01

=0 k=ko p t

(kopt s02 )2 =2 20 p 2 20

p 1 p p

2 0

kopt =

s02

p p

s02 + s01 2

Problem 9.11 a. Denote the output due to the signal as s0 (t) and the output due to noise as n0 (t). We wish to maximize s20 (t) 2 = 2 N0 E n0 (t)

j2 f t0 df 1 S (f ) Hm (f ) e R1 1 Hm (f ) Hm (f ) df

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

By Schwartz’s inequality, we have R1 R1 Z 1 S (f ) S (f ) df 2 2 s20 (t) 1 1 Hm (f ) Hm (f ) df R1 = jS (f )j2 df j2 f t0 df N0 N H (f ) H (f ) e E n20 (t) 0 m 1 m 1

where the maximum value on the right-hand side is attained if Hm (f ) = S (f ) e j2 f t0

b. The matched …lter impulse response is hm (t) = s (t0 t) by taking the inverse Fourier transform of Hm (f ) employing the time reversal and time delay theorems. c. The realizable matched …lter has zero impulse response for t < 0. d. By sketching the components of the integrand of the convolution integral, we …nd the following results: For t0 = 0; s0 (t0 ) = 0; For t0 = T =2; s0 (t0 ) = A2 T =2; For t0 = T; s0 (t0 ) = A2 T ; For t0 = 2T; s0 (t0 ) = A2 T: Problem 9.12 a. These would be x (t) and y (t) reversed and then shifted to the right so that they are nonzero for t > 0. p b. A = 7B: c. The outputs for the two cases are So1 (t) = A2 So2 (t) = 7B

[(t

t0 ) = ] = 7B 2

[(t

t0 ) =7 ]

[(t

t0 ) = ]

d. The shorter pulse gives the more accurate time delay measurement. e. Peak power is lower for y (t). Problem 9.13 a. The matched …lter impulse response is given by hm (t) = s2 (T

s1 (T

9.1. PROBLEM SOLUTIONS

b. Using (9.55) and a sketch of s2 (t) 2

s1 (t) for an arbitrary t0 , we obtain

Z 1 2 [s2 (t) s1 (t)]2 dt N0 1 2 T A2 t0 + 4A2 t0 + A2 N0 2 2 2 A T + 4A2 t0 ; 0 t0 N0 2

= = =

t0 T 2

This result increases linearly with t0 and has its maximum value for for t0 = T =2, which is 5A2 T =N0 : c. The probability of error is PE = Q

p 2 2

To minimize it, use the maximum value of , which is given in (b). d. The optimum receiver would have two correlators in parallel, one for s1 (t) and one for s2 (t). The outputs for the correlators at t = T are di¤erenced and compared with the threshold k calculated from (9.31), assuming the signals are equally likely.

Problem 9.14 Use 2max = 2Es =N0 for a matched …lter, where Es is the signal energy. The required values for signal energy are: (a) A2 T ; (b) 3A2 T =8; (c) A2 T =2; (d) A2 T =3 Problem 9.15 The required formulas are 1 (E1 + E2 ) 2 Z 1 T = s1 (t) s2 (t) dt E 0 1 = (E2 E1 ) 22 3 s (1 R12 ) E 5 = Q4 N0

E = R12 kopt PE

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

The energies for the three given signals are Z T A2 dt = A2 T EA = 0 Z T Z T t t B 2 sin2 B 2 cos2 dt = dt EB = T 2 T 0 0 Z B2 T 2 t B2T = 1 cos dt = 2 0 T 2 Z T 2 2 C 2 t EC = 1 + cos dt 4 T 0 Z T 2 2 t 2 C dt 1 cos = 4 T 0 Z C2 T 2 t 2 t = 1 2 cos + cos2 dt 4 0 T T Z C2 T 3 2 t 1 4 t 3C 2 T = 2 cos + cos dt = 4 0 2 T 2 T 8 a. For s1 = sA and s2 = sB , the average signal energy is E=

1 A2 T (EA + EB ) = 2 2

The correlation coe¢ cient is R12 = = =

1 E 1 E

Z T

AB cos

T =2) T

Z T

B2 2A2

t 2ABT dt = T E

AB sin

4AB

(A2 + B 2 =2)

The optimum threshold is kopt =

1 (EB 2

The probability of error is 2s 3 "s (1 R ) E 1 12 5=Q PE = Q 4 N0 N0 s " T 4AB = Q 1 2 2N0 (A + B 2 =2)

EA ) =

1 2

B2 2

A2 T

4AB A2 T B2 1 1 + (A2 + B 2 =2) 2 2A2 # "s 2 B T B2 A2 + =Q A2 + 2 2N0 2

# 4AB

9.1. PROBLEM SOLUTIONS

b. s1 = sA and s2 = sC : 1 1 2 (EA + EC ) = A + 3C 2 =8 T 2 2 AC A2 + 3C 2 =8

E = R12 = kopt =

1 (EC 2 "s

PE = Q

3C 2 8

A2 T

3C 2 A2 + 8

EA ) = T 2N0

1 2

c. s1 = sB and s2 = sC :

R12 = kopt =

1 (EC 2 "s

PE = Q d. s1 = sB and s2 =

B2 +

3C 2 4

B2 T

1 1 (EB + EC ) = 2 4 16BC 3 (B 2 + 3C 2 =4)

E =

EB ) =

1 4

3C 2 B2 + 4

T 4N0

16BC

sB : E = EB = B 2 T =2 R12 =

e. s1 = sC and s2 =

sC :

kopt = 0 2s 3 2T B 5 PE = Q 4 N0 E = EC = 3C 2 T =8 R12 =

1 kopt = 0

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE 2s

PE = Q 4

3 3C 2 T 5 8N0

Problem 9.16 a. The energies are EA =

Z T

A2 dt = A2 T

EB =

Z T =2

A dt +

( A)2 dt = A2 T

T =2

Z T =4

Z T

A dt +

Z T =2

( A) dt +

T =4

Z 3T =4

A dt +

T =2

Z T

( A)2 dt = A2 T

3T =4

b. Sketch the combinations (A; B) ; (B; C) ; and (A; C) to show that their respective products have as much area above the t-axis as below it. Therefore, RAB , RBC , and RAC are all zero because they are directly proportional to the integral of the products of the respective signals. The optimum threshold for each case is zero because it is proportional to the di¤erence of the respective signal energies. c. The probability of error for each case is 0s

PE = Q @

A2 T N0

1 A

Problem 9.17 Candidate modulation schemes are noncoherent FSK and DPSK. The bandwidth e¢ ciency for NFSK is 0.4 and for DPSK is 0.5, so either one will provide the desired data rate for the given bandwidth. For NFSK, the average probability of error is PE; NFSK =

1 1 exp ( z=2) = exp 2 2

Eb 2N0

where Pave = A2 is the average signal power. We want PE 1 exp 2

A2 T 4N0

= 10 6

1 exp 2

A2 T 4N0

10 6 , so solve

9.1. PROBLEM SOLUTIONS or

A2 T = 2N0

10 6

2 ln 2

or A2 2

2N0 R ln 2

10 6 ; R = 1=T = data rate

2 10 7

103 ln 2

10 6

= 0:0105 watts (normalized) = 10:5 mW For DPSK, the average probability of error is PE; DPSK =

1 1 exp ( z) = exp 2 2

Eb N0

1 exp 2

where Pave = A2 is the average signal power. We want PE

1 exp 2

A2 T 2N0

= 10 6

A2 T = 2N0

ln 2

10 6

A2 T 2N0

10 6 , so solve

or A2 2

N0 R ln 2

10 6 ; R = 1=T = data rate

10 7

103 ln 2

10 6

= 5:25 mW From the standpoint of simplicity, the NFSK receiver is simpler than the optimum DPSK receiver. If a suboptimum delay-and-multiply DPSK receiver is used then DPSK wins in terms of simplicity with some loss in performance. In this case, we must solve 0s 1 2T A A = 10 6 Q@ 2N0 s A2 T = 4:754 2N0 A2 2

= N0 R (4:754)2 =

10 7

= 9 mW

103 (4:754)2

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

which is still less than for NFSK. Problem 9.18 Numerical integration of (9.83) with (9.81) and (9.82) substituted shows that the degradation for 2 = 0:01 ( = 0:1 rad) is about 0.045 dB; for 2 = 0:05 ( = 0:224 rad) it is about 0.6 dB; for 2 = 0:1 ( = 0:316 rad) it is about 9.5 dB. For the constant phase error model, the degradation is Dconst =

20 log10 [cos ( const )]

As suggested in the problem statement, we set the constant phase error equal to the standard deviation of the Gaussian phase error for comparison purposes; thus, we consider constant phase errors of const = 0:1; 0:224; and 0.316 radians. The degradations corresponding to these phase errors are 0:044, 0:219, and 0:441 dB, respectively. The constant phase error degradations are much less serious than the Gaussian phase error degradations. Problem 9.19 p a. For ASK, PE = Q [ z] = 10 5 gives z = 4:2642 = 18:182 or z = 12:596 dB. b. For BPSK, PE = Q

2z = 10 5 gives z = 4:2642 =2 = 9:091 or z = 9:586 dB.

c. Binary FSK bit error probability is the same as ASK. d. The degradation of BPSK with a phase error of 5 degrees is Dconst = 20 log10 [cos (5o )] = 0:033 dB, so the required SNR to give a bit error probability of 10 5 is 9:586+0:033 = 9:619 dB. hp i p p 2 (1 1=2) z = Q [ z] = 10 5 gives z = 12:596 e. For PSK with m = 1= 2, PE = Q dB. f. Assuming that the separate e¤ects are additive, we add the degradation of part (d) to the SNR found in part (e) to get 12:596 + 0:033 = 12:629 dB.

Problem 9.20 a. Without frequency uncertainty, the bandwidth of the input …lters is BT ' 2=T Hz. With frequency uncertainty it must be BT + j f j Hz. Therefore, the noise power

9.1. PROBLEM SOLUTIONS

-1

-2

-3

-4

-5

-6

13 E/N0, dB

out of the …lters is (BT + j f j) N0 watts instead of BT N0 watts. The probability of error is therefore PE = = = =

1 exp 2 1 exp 2 1 exp 2 1 exp 2

1 A2 exp 2 4 (BT + j f j) N0 2 A 4 (1 + j f j =BT ) N0 BT 1 A2 2 (1 + j f j =BT ) 2N0 BT z 1 2 1 + j f j =BT

z 0 =2 =

b. The probability of error is plotted in Figure 9.3 with the leftmost curve being for no frequency error.The required value of z to provide a desired value of error probability, PE; 0 , is given by 1 z0 1 exp = PE; 0 2 2 1 + j f j =BT

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

or z0 =

2 (1 + j f j =BT ) ln (2PE; 0 )

In dB, this is z0; dB = 10 log10 [ 2 (1 + j f j =BT ) ln (2PE; 0 )] c. The degradation is the di¤erence in dB between z0; dB with frequency uncertainty and the value without it: Degradation = D = 10 log10 [ 2 (1 + j f j =BT ) ln (2PE; 0 )]

10 log10 [ 2 ln (2PE; 0 )]

= 10 log10 [(1 + j f j =BT )] dB

Thus, the degradation is seen to be independent of error probability. A table of degradation values is given below. j f j =BT D, dB 0 0 0.1 0.414 0.2 0.792 0.3 1.139 0.4 1.461 Problem 9.21 We have that 1

m2 =

Use this in PE = Q

% modulation 100

2 (1

Eb m2 ) N0

This results in the plot below: From this plot, we estimate the following values for the percent mudulation to produce a probabillity of error of 10 4 for each value of z = Eb =N0 speci…ed. Eb =N0 , dB % modulation to give PE = 10 4 8:5 62 9:0 70 9:5 78 10:0 87 10:5 98

9.1. PROBLEM SOLUTIONS

-2

-3

-4

Eb/N0 = 8.5

9.5

10.5dB

-5

-6

70 75 80 % modulation

100

Problem 9.22

a. For BPSK, an SNR of 10.53 dB is required to give an error probability of 10 6 (see the solution to Problem 9.2). For ASK and FSK it is 3.01 dB more, or 13.54 dB. For BPSK and ASK, the required bandwidth is 2R Hz where R is the data rate in bps. For FSK with minimum tone spacing of 0:5R Hz, the required bandwidth is 2:5R Hz. Hence, for BPSK and ASK, the required bandwidth is 100 kHz. For FSK, it is 125 kHz. b. For BPSK, the required SNR is 9.588 dB.for an error probability of 10 5 : For ASK and FSK it is 3.01 dB more or 12.598 dB. The required bandwidths are now 1 MHz for BPSK and ASK; it is 1.25 MHz for FSK.

Problem 9.23

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

The correlation coe¢ cient is

R12

Z 1 T = s1 (t) s2 (t) dt E 0 Z 1 T 2 = A cos (! c t) cos [(! c + E 0 = sinc (2 f T )

!) t] dt

where f is the frequency separation between the FSK signals in Hz. Using a calculator, we …nd that the sinc-function takes on its …rst minimum at an argument value of about 1.4, or f = 0:7=T . The minimum value at this argument value is about 0:216. The improvement in SNR is 10 log10 (1 R12 ) = 10 log10 (1 + 0:216) = 0:85 dB.

Problem 9.24

The encoded bit streams, assuming a reference 1 to start with, are: (a) 1 111 110 100 010; (b) 1 100 111 100 111; (c) 1 111 111 111 111; (d)1 010 101 010 101; (e) 1 111 111 010 101; (f) 1 110 000 011 011; (g) 1 100 101 010 110; (h) 1 100 001 000 010.

9.1. PROBLEM SOLUTIONS

Problem 9.25 The di¤erentially encoded bit stream is 1 000 011 001 111 (1 assumed for the starting reference bit). The phase of the transmitted carrier is 000 0 0 0 . Integrate the product of the received signal and a 1-bit delayed signal over a (0; T ) interval. Each time the integrated productRis greater than 0, decide 1; each time it is less than 0, decide 0. The T …rst integration gives 0 A2 cos (! 0 t + ) cos (! 0 t) dt = A2 T =2. The resulting decision is RT that a 0 was sent. The next integration gives 0 A2 cos (! 0 t) cos (! 0 t) dt = A2 T =2. The resulting decision is that a 1 was sent. Continuing in this manner, using the previous signal for a reference in the integration for the current signal and using the stated decision criterion, we obtain the original bit sequence. If the signal is inverted, the same demodulated signal is detected as if it weren’t inverted. Problem 9.26 Note that E [n1 ] = E

Z 0

n (t) cos (! c t) dt =

Z 0

E [n (t)] cos (! c t) dt = 0

which follows because E [n (t)] = 0: Similarly for n2 , n3 , and n4 . Consider the variance of each of these random variables: n21

var [n1 ] = E = =

Z 0 Z 0

T T Z 0 Z 0 T

N0 2

Z 0

Z 0 Z 0 T

E [n (t) n ( )] cos (! c t) cos (! c ) dtd

N0 (t T 2

n (t) n ( ) cos (! c t) cos (! c ) dtd

) cos (! c t) cos (! c ) dtd

cos2 (! c t) dt =

N0 T 4

where the fact that E [n (t) n ( )] = N20 (t ) has been used along with the sifting property of the -function to reduce the double integral to a single integral. Similarly for n2 , n3 , and n4 . Therefore, w1 has zero mean and variance 1 1 N0 T var [w1 ] = var [n1 ] + var [n2 ] = 4 4 8 Similarly for w2 , w3 , and w4 .

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

Problem 9.27 a. The noise power out of the input …lter with frequency uncertainty is (2R + j f j) N0 watts instead of 2RN0 watts. Thus, the probability of error is PE

A2 z0 = Q = Q 2 (2R + j f j) N0 s ! A2 = Q 2 (1 + j f j =2R) (2RN0 ) r z = Q 1 + j f j =2R

b. The required value of z to provide a desired value of error probability, PE; 0 , is given by r z0 Q = PE; 0 1 + j f j =2R or z0 = (1 + j f j =2R) Q 1 (2PE; 0 ) In dB, this is

n o 2 z0; dB = 10 log10 (1 + j f j =2R) Q 1 (2PE; 0 )

c. The degradation is the di¤erence in dB between z0; dB with frequency uncertainty and the value with out it: n o n o 2 2 Degradation = D = 10 log10 (1 + j f j =2R) Q 1 (2PE; 0 ) 10 log10 Q 1 (2PE; 0 ) = 10 log10 [(1 + j f j =2R)] dB

Thus, the degradation is seen to be independent of error probability. A table of degradation values is given below. j f j =R D, dB 0 0 0.2 0.414 0.4 0.792 0.6 1.139 0.8 1.461

9.1. PROBLEM SOLUTIONS

Problem 9.28

a. By plotting sequences of two successive bit waveforms, it is seen that no energy cancellation for the cases 11 or 00 (i.e., a cosinusoid followed by a cosinusoid of the same phases, or the negation of this situation). The probability of error for these cases is

PE; 1 = Q

Eb N0

However, for the cases 10 or 01, there is energy loss in the amount of 1 j T j =R (i.e., a cosinusoid followed by its negation or vice versa). The probability of error for these cases is s ! Eb j Tj PE; 1 = Q 1 N0 T For equally likely 1s and 0s in the data stream, each case occurs half the time, giving

PE; j T j =

1 Q 2

r r

Eb N0 Eb N0

! !

1 + Q 2 1 + Q 2

s r

Eb N0

Eb (1 N0

j Tj T

j T j R)

b. A plot of magni…ed error probability is given below for j T j R = 0; 0:1; 0:2; 0:3; 0:4 starting with the leftmost curve. The values of Eb =N0 giving PE; j T j = 10 6 can be estimated as 13.5, 13.8, 14.2, 14.8, and 15.5 dB for these values of j T j R, respectively. The value for 13.5 dB is for no delay error. The degradation values are 13:8 13:5 = 0:3 dB, 14:2 13:5 = 0:7 dB, 14:8 13:5 = 1:3 dB, and 15:5 13:5 = 2 dB for j T j =R = 0:1; 0:2; 0:3; 0:4, respectively.

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

-2

-3

-4

-5

-6

11.5

12.5

13.5

14.5

15.5

9.1. PROBLEM SOLUTIONS

Problem 9.29 Bandwidths are given by BBPSK = BDPSK = 2R; RBPSK = RDPSK = BBPSK, DPSK =2 = 50 kbps BCFSK = 2:5R; RCFSK = BCFSK =2:5 = 40 kbps BNCFSK = 4R; RNCFSK = BNCFSK =4 = 25 kbps

Problem 9.30 Consider the bandpass …lter, just wide enough to pass the ASK signal, followed by an envelope detector with a sampler and threshold comparator at its output. The output of the bandpass …lter is u (t) = Ak cos (! c t +

) + n (t)

= Ak cos (! c t +

) + nc (t) cos (! c t +

= x (t) cos (! c t +

)

ns (t) sin (! c t +

)

ns (t) sin (! c t +

)

where Ak = A if signal plus noise is present at the receiver input, and A = 0 if noise alone is present at the input. In the last equation x (t) = Ak + nc (t) Now u (t) can be written in envelope-phase form as u (t) = r (t) cos [! c t + where r (t) =

x2 (t) + n2s (t) and

+ (t)]

(t) = tan 1

ns (t) x (t)

The output of the envelope detector is r (t), which is sampled each T seconds and compared with the threshold. Thus, to compute the error probability, we need the pdf of the envelope for noise alone at the input and then with signal plus noise at the input. For noise alone, it was shown in Chapter 4 that the pdf of r (t) is Rayleigh, which is fR (r) =

r r2 =2N e ; r R

0; noise only

where N is the variance (mean-square value) of the noise at the …lter output. For signal plus noise at the receiver input, the pdf of the envelope is Ricean. Rather than use this complex expression, however, we observe that for large SNR at the receiver input r (t)

A + nc (t) ; large SNR, signal present

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

which follows by expanding the argument of the square root, dropping the squared terms, and approximating the square root as p

1 ; j j << 1 2

Thus, for large SNR conditions, the envelope detector output is approximately Gaussian for signal plus noise present. It has mean A and variance N (recall that the inphase and quadrature lowpass noise components have variance the same as the bandpass process). The pdf of the envelope detector output with signal plus noise present is approximately 2

e (r A) =2N p fR (r) = ; large SNR 2 N For large SNR, the threshold is approximately A=2. Thus, for noise alone present, the probability of error is exactly Z 1 r r2 =2N 2 P (E j 0) = e dr = e A =8N R A=2 For signal plus noise present, the probability of error is approximately Z A=2

P (E j S + N present) =

2 p e (r A) =2N p dr = Q A2 =2N 2 N

1 2 e A =4N

Now the average signal-to-noise ratio is z=

=N A

1 A2 =2 A2 = 2 N0 BT 4N

which follows because the signal is present only half the time. probability of error is 1 1 PE = P (E j S + N ) + P (E j 0) 2 2

e z 1 p + e z=2 4 z 2

Problem 9.31 The integral to be integrated is PE = e

Z 1 0

r r2 =N e I0 N

Therefore, the average

Ar N

9.1. PROBLEM SOLUTIONS

The Ricean pdf, from (7.150), is ( r

fR (r) =

2 exp

0; r < 0

n h

r2 +K 2 2

2K r ; r

Since it is a pdf, its integral over -1 < r < 1 is 1. The trick is to get the integrand of the expression for PE to look like a Ricean pdf. Obviously, a start is to identify N = 2 2 and p 2K A A2 A2 which can be solved to give K = 4N = z2 because z = 2N . The the integral for N = PE becomes Z p e z 1 r r2 =2 2 r e I dr PE = 2K 0 2 2 0 We multiply by 1 = eK e K to get Ke

PE = e

= eK

z Z 1 0

e (r =2 2

2 +K

1 1 e z = ez=2 e z = e z=2 2 2 2

where the integral is 1 because the integrand is a Ricean pdf. Problem 9.32 The binary and Gray encoded numbers are given in the table below: Decimal Binary Gray Decimal Binary Gray 0 00000 00000 16 10000 11000 1 00001 00001 17 10001 11001 2 00010 00011 18 10010 11011 3 00011 00010 19 10011 11110 4 00100 00110 20 10100 11110 5 00101 00111 21 10101 11111 6 00110 00101 22 10110 11101 7 00111 00100 23 10111 11100 8 01000 01100 24 11000 10100 9 01001 01101 25 11001 10101 10 01010 01111 26 11010 10111 11 01011 01110 27 11011 10110 12 01100 01010 28 11100 10010 13 01101 01011 29 11101 10011 14 01110 01001 30 11110 10001 15 01111 01000 31 11111 10000

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

Problem 9.33 Problem statement should be "Show that (9.138) is the average energy in terms of for M -ary antipodal PAM." Evaluation is facilitated by means of the following sums: m X

m X

1 = m;

k=1

m (m + 1) X 2 m (m + 1) (2m + 1) ; k = 2 6 k=1

The expression to be evaluated is M

Eave =

1 X 1 X (j Ej = M M j=1

= =

M 2

1) 1)2

1 2

j=1

M 2 X

j=1

M X1

k=1

j=1

1) +

M 2 X

1) (2M 6

4 1)2

1)2

(M 4 M2

1) (j

1)2

M 2 X

4 3

1) (M + 1) =

j=1

k=1

M X1

1 2

j=1

M 2 X

1 12

+ #

j=1

1)2 X 5 1 j=1

1)2

1) [2 (2M

Problem 9.34 a. Appy (9.142) since baseband is speci…ed: BP AM =

1) +

1)2

R 20 or 5 = ) M = 16 log2 M log2 M

b. Apply (9.141): 2 (M 1) Pb; antip. PAM = Q M log2 M

6 log2 M Eb M 2 1 N0

3 (M

1)]

9.1. PROBLEM SOLUTIONS

For M = 16 and a BEP of 10 6 this becomes 2 (16 1) Q 16 log2 16 Q

6 log2 16 Eb 162 1 N0 6 log2 16 Eb 162 1 N0

! !

= 10 6 =

32 15

10 6

Trial and error with a Q-function routine in MATLAB yields r

6 log2 16 Eb Eb 255 ' 4:598 ) = 162 1 N0 N0 24

4:5982 = 23:515 dB

For M = 16 and a BEP of 10 5 we have

6 log2 4 Eb 42 1 N 0

= ) )

Problem 9.35

a. BPSK or DPSK, 10.53 dB; b. 4-PAM (RF), 14.4 dB; c. 8-PAM (RF), 18.8 dB; d. 16-PAM (RF), 23.52 dB; e. 32-PAM (RF), 28.51 dB.

32 15 r

10 5

6 log2 16 Eb ' 4:0925 162 1 N0 Eb 255 4:09252 = 22:503 dB = N0 24

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE First 3 answers from Table 9.5. Solve equation (9.141) for 16-PAM and 32-PAM: ! ! r r 2 (16 1) 6 log2 16 Eb 30 24 E b = 10 6 =) Q = 10 6 Q 16 log2 16 162 1 N0 64 255 N0 ! r Eb =) Q 0:0941 = 2:13 10 6 N0 r Eb = 4:598 0:0941 =) N0 Eb =) 0:0941 = 21:14 N0 Eb = 224:67 = 23:52 dB =) N0

2 (32 1) Q 32 log2 32

6 log2 32 Eb 322 1 N0

= =) =) =) =)

! r 62 E 30 b 10 6 =) Q = 10 6 160 1023 N0 ! r Eb Q = 2:58 10 6 0:0293 N0 r Eb 0:0293 = 4:56 N0 Eb 0:0293 = 20:79 N0 Eb = 709:586 = 28:51 dB N0

Problem 9.36 For BPSK, the BEP is found from Pb; BPSK = Q

2Eb =N0 = 10 4 )

2Eb =N0 = 3:719; Eb =N0 = 8:4 dB

The data rate in terms of bandwidth for BPSK is RBPSK =

BRF 100 kHz = = 50 kbps 2 2

For DPSK 1 exp ( Eb =N0 ) = 10 4 ) Eb =N0 = 10 log10 [ ln 2 2

10 4 ] = 9:3 dB

9.1. PROBLEM SOLUTIONS

The data rate is the same as for BPSK. For coherent FSK, the required Eb =N0 is 3.01 dB more than for BPSK, or Eb =N0 = 8:4 + 3:01 = 11:41 dB. The data rate is RCFSK =

BRF 100 kHz = = 40 kbps 2:5 2:5

For noncoherent FSK 1 exp ( Eb =2N0 ) = 10 4 ) Eb =N0 = 10 log10 [ 2 ln 2 2 The data rate is RNCFSK =

10 4 ] = 12:31 dB

100 kHz BRF = = 25 kbps 4 4

For antipodal 4-ary PAM, 2 (4 1) Q 4 log2 4

6 log2 4 Eb 42 1 N 0

= 10 4 )

Eb = 11:3 dB N0

The data rate is: 1 1 R = BRF , PAM log2 M = 100 log2 4 = 100 kbps 2 2 For antipodal 8-ary PAM, 2 (8 1) Q 8 log2 8

6 log2 8 Eb 82 1 N 0

= 10 5 )

Eb = 16:5dB N0

The data rate in terms of bandwidth can be obtained from (9.141): 1 1 R = BRF , PAM log2 M = (100) log2 8 = 150 kbps 2 2 Problem 9.37 For = 0:25, P (f ) =

8 < T; :

0 T 2

T 1 + cos 0:25 jf j

0:75 2T

; 0:75 2T

The optimum transmitter and receiver transfer functions are 1=4

jHT (f )jopt =

A jP (f )j1=2 Gn (f ) jHC (f )j1=2

jf j

jf j 0:75 2T 1:25 Hz 2T jf j > 0:75 2T

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

and jHR (f )jopt =

K jP (f )j1=2

1=4

Gn (f ) jHC (f )j1=2

Setting all scale factors equal to 1, these become

1=2 1=4

jHT (f )jopt =

jP (f )j q

1=2 1=4

jHR (f )jopt = jP (f )j

1=4

1 + (f =fC )2

1 + (f =f3 )2 q

2 1=4

1 + (f =f3 )

1 + (f =fC )2

A MATL:AB program for computing the optimum transfer functions is given below. Plots for the various cases are then given in Figures 8.3 - 8.5. % Problem 9.37: Optimum transmit and receive …lters for raised-cosine pulse shaping, % …rst-order Butterworth channel …ltering, and …rst-order noise spectrum % A = char(’-’,’:’,’–’,’-.’); clf beta00 = input(’Enter starting value for beta: >=0 and <= 1 ’); del_beta = input(’Enter step in beta: starting value + 4*step <= 1 ’); R = 1; f_3 = R/2; f_C = R/2; delf = 0.01; HT_opt = []; HR_opt = []; for n = 1:4 beta = (n-1)*del_beta+beta00; beta0(n) = beta; f1 = 0:delf:(1-beta)*R/2; f2 = (1-beta)*R/2:delf:(1+beta)*R/2; f3 = (1+beta)*R/2:delf:R; f = [f1 f2 f3]; P1 = ones(size(f1)); P2 = 0.5*(1+cos(pi/(R*beta)*(f2-0.5*(1-beta)*R))); P3 = zeros(size(f3)); Gn = 1./(1+(f/f_3).^2); HC = 1./sqrt(1+(f/f_C).^2); P = [P1 P2 P3]; HT_opt = sqrt(P).*(Gn.^.25)./sqrt(HC);

9.1. PROBLEM SOLUTIONS

|HR,opt(f)||

R = 1 bps; channel filter 3-dB frequency = 0.5 Hz; noise 3-dB frequency = 0.5 Hz 1.5

0.5

0.1

0.2

0.3

0.4

0.5 f, Hz

0.6

0.7

0.8

0.9

|HT,opt(f)|

1.5

β = 0.2 β = 0.3

β = 0.4 β = 0.5

0.5

0.1

0.2

0.3

0.4

0.5 f, Hz

0.6

0.7

0.8

0.9

HR_opt = sqrt(P)./((Gn.^.25).*sqrt(HC)); subplot(2,1,1),plot(f,HR_opt,A(n,:)),xlabel(’f, Hz’),ylabel(’jH_R_,_o_p_t(f)jj’) title([’R = ’,num2str(R),’bps; channel …lter 3-dB frequency = ’,num2str(f_C),’ Hz; noise 3-dB frequency = ’,num2str(f_3),’Hz’]) if n == 1 hold on end subplot(2,1,2),plot(f,HT_opt,A(n,:)),xlabel(’f, Hz’),ylabel(’jH_T_,_o_p_t(f)j’) if n == 1 hold on end end legend([’nbeta = ’,num2str(beta0(1))],[’nbeta = ’,num2str(beta0(2))], [’nbeta = ’,num2str(beta0(3))],[’nbeta = ’,num2str(beta0(4))],1) a. f3 = fC = 1=2T

b. fC = 2f3 = 1=T

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

|HR,opt(f)||

R = 1 bps; channel filter 3-dB frequency = 1 Hz; noise 3-dB frequency = 0.5 Hz 1.5

0.5

0.1

0.2

0.3

0.4

0.5 f, Hz

0.6

0.7

0.8

0.9

|HT,opt(f)|

β = 0.2 β = 0.3 β = 0.4 β = 0.5

0.5

0.1

0.2

0.3

0.4

0.5 f, Hz

0.6

0.7

0.8

0.9

|HR,opt(f)||

R = 1 bps; channel filter 3-dB frequency = 0.5 Hz; noise 3-dB frequency = 1 Hz 1.5

0.5

0.1

0.2

0.3

0.4

0.5 f, Hz

0.6

0.7

0.8

0.9

1.5

|HT,opt(f)|

β = 0.2 β = 0.3

β = 0.4 β = 0.5

0.5

0.1

0.2

0.3

0.4

0.5 f, Hz

0.6

0.7

0.8

0.9

9.1. PROBLEM SOLUTIONS

c. f3 = 2fC = 1=T Problem 9.38 This a simple matter of sketching P (f ) = b b a (f =b) b a a (f =a) and P (f B), where B = 21 (a + b), side by side and showing that they add to a constant in the overlap region. Problem 9.39 a. The optimum transmitter and receiver transfer functions are 1=4

jHT (f )jopt =

A jP (f )j1=2 Gn (f ) jHC (f )j1=2

and jHR (f )jopt =

K jP (f )j1=2

1=4

Gn (f ) jHC (f )j1=2 where P (f ) is the signal spectrum and A and K are arbitrary constants. From the problem 1=4 1=4 statement, Gn (f ) = 10 11 = 10 11=4 and 1

jHC (f )j = q

1 + (f =4800)2

The signal spectrum for P (f ) = = = Thus,

= 1 and T1 = 9600 bps is, from (5.137),

T T 1 1 1 + cos = 0 jf j f ; 2 2T 2T jf j 1 1 + cos ; 0 jf j 9600 Hz 2 9600 9600 1 jf j cos2 ; 0 jf j 9600 Hz 9600 19200 s

= 9600 Hz

i1=4 jf j h 1 + (f =4800)2 ; 0 jf j 9600 Hz 19200 h i1=4 10 11=4 A jf j = cos 1 + (f =4800)2 ; 0 jf j 9600 Hz 97:98 19200 s i1=4 1 jf j h 11=4 = 10 K cos2 1 + (f =4800)2 ; 0 jf j 9600 Hz 9600 19200 h i1=4 1011=4 K jf j = cos 1 + (f =4800)2 ; 0 jf j 9600 Hz 97:98 19200

jHT (f )jopt = 10 11=4 A

jHR (f )jopt

1+ 2T

1 cos2 9600

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE 11=4

11=4

where A and K may be chosen so that 1097:98 A = 1097:98K = 1. b. From (9.149) Q (A= ) = 10 4 results in A= ' 3:72. c. Matching arguments of (9.151) and (9.155) we have 3:72 '

1=2

Gn (f ) P (f ) df jHC (f )j 1

# 1

Therefore, we need to evaluate the integral Z 1 1=2 1=2 Gn (f ) P (f ) Gn (f ) P (f ) df = 2 df I = jHC (f )j jHC (f )j 0 1 Z q 2 10 11=4 9600 f = cos2 1 + (f =4800)2 df; v = f =4800 9600 19200 0 Z 2 v p = 10 11=4 cos2 1 + v 2 dv ' 1:213 10 11=4 4 0 Z 1

Thus 3:72 =

ET ) ET = 6:44 1:213 10 11=4

10 5 J

Problem 9.40 The MATLAB program is given below and Figure 9.28 shows the probability of error curves. % Plots for Prob. 9.40 % clf A = char(’-’,’–’,’-.’,’:’,’–.’,’-..’); delta = input(’Enter value for delta: ’) taum_T0 = input(’Enter vector of values for tau_m/T: ’) L_taum = length(taum_T0); z0_dB = 0:.1:15; z0 = 10.^(z0_dB/10); for ll = 1:L_taum ll taum_T = taum_T0(ll) P_E = 0.5*qfn(sqrt(2*z0)*(1+delta))+0.5*qfn(sqrt(2*z0)*((1+delta)-2*delta*taum_T)); semilogy(z0_dB, P_E,A(ll,:)) if ll == 1

9.1. PROBLEM SOLUTIONS

37 PE versus z 0 in dB for δ = 0.5

-1

-2

-3

-4

τm/T = 0.2

-5

τm/T = 0.6 τm/T = 1

-6

z 0, dB

hold on; grid on; axis([0, inf, 10^(-6), 1]); xlabel(’z_0, dB’); ylabel(’P_E’) end end title([’P_E versus z_0 in dB for ndelta = ’, num2str(delta)]) if L_taum == 1 legend([’ntau_m/T = ’,num2str(taum_T0)],3) elseif L_taum == 2 legend([’ntau_m/T = ’,num2str(taum_T0(1))],[’ntau_m/T = ’,num2str(taum_T0(2))],3) elseif L_taum == 3 legend([’ntau_m/T = ’,num2str(taum_T0(1))],[’ntau_m/T = ’,num2str(taum_T0(2))], [’ntau_m/T = ’,num2str(taum_T0(3))],3) elseif L_taum == 4 legend([’ntau_m/T = ’,num2str(taum_T0(1))],[’ntau_m/T = ’,num2str(taum_T0(2))], [’ntau_m/T = ’,num2str(taum_T0(3))],[’ntau_m/T = ’,num2str(taum_T0(4))],3) end Problem 9.41 The program given in Problem 9.40 can be adapted for making the asked for plot.

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

Problem 9.42 The appropriate equations are: BPSK: (9.74) with m = 0 for nonfading; (9.205) for fading; DPSK: (9.112) for nonfading; (9.207) for fading; CFSK: (9.88) for nonfading; (9.206) for fading; NFSK: (9.122) for nonfading; (9.208) for fading. Degradations for PE = 10 4 are: Modulation type BPSK DPSK Coh. FSK Noncoh. FSK

Fading SNR, dB 33.99 37.00 37.00 40.01

Nonfaded SNR, dB 8.41 9.32 11.42 12.33

Fading Margin, dB 25.58 27.68 25.58 27.68

Problem 9.43 Comparing the exponents in (9.203) and (9.204) it is apparent that 1 2

2 w

=1+

1 1 Z or 2 2w = 1 = 1+ Z Z 1+Z

1 2

p 2 p

so that (9.203) becomes

PE = = =

2 w

Z 1

exp

w2 =2 2w

2 2w p p 2 2 1 1 1 p w = 2 2w 1 2 2 2 0 1 s 1@ Z A 1 2 1+Z 0

Problem 9.44 The expressions with Q-functions can be integrated by parts as shown in the text (BPSK and coherent FSK). The other two cases involve exponential integrals which are easily integrated (DPSK and noncoherent FSK).

9.1. PROBLEM SOLUTIONS Problem 9.45 The computer program given below is used for the solution. clf pc = input(’Enter sample values for channel pulse response (odd #): ’); L_pc = length(pc); disp(’Maximum number of zeros each side of decision sample: ’); disp((L_pc-1)/4) N = input(’Enter number of zeros each side of decision sample desired: ’); mid_samp = …x(L_pc/2)+1 Pc = []; for m = 1:2*N+1 row = []; for n = 2-m:2*N+2-m row = [row pc(mid_samp-1+n)]; end Pc(:,m) = row’; end disp(’’) disp(’Pc:’) disp(’’) disp(Pc) Pc_inv = inv(Pc); disp(’’) disp(’Inverse of Pc:’) disp(’’) disp(Pc_inv) coef = Pc_inv(:,N+1); disp(’’) disp(’Equalizer coe¢ cients:’) disp(’’) disp(coef) disp(’’) steps = (L_pc-length(coef))+1; p_eq = []; for m = mid_samp-‡oor(steps/2):mid_samp+‡oor(steps/2) p_eq(m) = sum(coef’.*‡iplr(pc(m-N:m+N))); end p_eq_p = p_eq(N+1:mid_samp+‡oor(steps/2)); disp(’’) disp(’Equalized pulse train’)

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE disp(’’) disp(p_eq_p) disp(’’) disp(’Noise enhancement factor, dB’) N_enh = coef’*coef; N_enh_dB = 10*log10(N_enh); disp(N_enh_dB) t=-5:.01:5; y=zeros(size(t)); subplot(2,1,1),stem(pc),ylabel(’Ch. output pulse’),... axis([1 L_pc -1 1.5]),... title([’Ch. pulse samp. = [’,num2str(pc),’]’]) subplot(2,1,2),stem(p_eq_p),xlabel(’n’),ylabel(’Equal. output pulse’),... axis([1 L_pc -1 1.5]),... title([’Output equalized with ’,num2str(N),’zero samples either side’]) pause % Equalizer freq response % clf alpha = coef’; T = 1; L = length(alpha); f = -0.5/T:.001:0.5/T; %Efreq = [exp(j*2*pi*2*f); exp(j*2*pi*f); exp(j*2*pi*0*f); exp(-j*2*pi*f); exp(-j*2*pi*2*f)]; nn = -(L-1)/2:1:(L-1)/2; Efreq = []; for n = 1:L Efreq = [Efreq; exp(j*2*pi*nn(n)*f)]; end freq_resp = alpha*Efreq; plot(f, abs(freq_resp)), xlabel(’{nitfT}’), ylabel(’{nitH}_e_q({nitf})’) title([’nalpha = [’, num2str(alpha), ’]’]) a. The given sample values give a singular sample matrix for a three-tap equalizer. b. Equalizer coe¢ cients: 1:7217 0:1565 2:0696

9.1. PROBLEM SOLUTIONS

Ch. output pulse

Ch. pulse samp. = [0 1.5

-0.11111

0.5

-1

0.5

-1

0.5

-0.11111

1 0.5 0 -0.5 -1

Output equalized with 2 zero samples either side Equal. output pulse

1.5 1 0.5 0 -0.5 -1

5 n

0:1565 1:7217 Plots of the unequalized and equalized pulses are shown below: The frequency response function and noise enhancement are given below: Problem 9.46 a. The equalizer coe¢ cients are: 1:3333 0:6667 0 b. The equalizer coe¢ cients are: 3:7877 1:5498 3:4334 0:0064 3:7906 Problem 9.47

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

Noise enhancement factor = 10.1118 dB 2 1

Eq. freq. response

0 -1 -2 -3 -4 -5 -6 -0.5

-0.4

Ch. output pulse

Ch. pulse samp. = [0 1.5

-0.3

-0.2

-0.1

0 fT

0.1

0.2

0.3

0.4

0.5

-0.11111

0.1

-1

0.5

-1

0.5

-0.11111

1 0.5 0 -0.5 -1

Output equalized with 1 zero samples either side 1.5 Equal. output pulse

1 0.5 0 -0.5 -1

5 n

9.1. PROBLEM SOLUTIONS

Noise enhancement factor = 3.4679 dB 1

Eq. freq. response

0.5

-0.5

-1

-1.5

-2 -0.5

-0.4

Ch. output pulse

Ch. pulse samp. = [0 1.5

-0.3

-0.2

-0.1

0 fT

0.1

0.2

0.3

0.4

0.5

-0.11111

0.1

-1

0.5

-1

0.5

-0.11111

1 0.5 0 -0.5 -1

Output equalized with 2 zero samples either side Equal. output pulse

1.5 1 0.5 0 -0.5 -1

5 n

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE Noise enhancement factor = 16.3252 dB 12 10

Eq. freq. response

8 6 4 2 0 -2 -4 -6 -0.5

-0.4

-0.3

-0.2

-0.1

0 fT

0.1

0.2

0.3

0.4

0.5

a. The solution is similar to Example 9.12 except that Snn (f ) =

N0 =2 1 + (f =f3 )2

so that f3 N0 exp ( 2 f3 j j) 2

Rnn ( ) =

The desired normalized autocorrelation matrix, with 2 1 + b2 z + 2 N0 4 bz + 2 e 2 [Ryy ] = T 4 2e

= Tm = T and f3 = 1=T , is

bz + 2 e 2 1 + b2 z + 2 bz + 2 e 2

5 bz + 2 e 2 2 1+b z+ 2

b. The following matrix is needed: 2

3 2 3 Ryd ( T ) bA [Ryd ] = 4 Ryd (0) 5 = 4 A 5 Ryd (T ) 0

9.1. PROBLEM SOLUTIONS

Multiply the matrix given in (a) by the column matrix of unknown weights to get the equations 2 32 0 3 2 3 1 + b2 z + 2 bz + 2 e 2 e 4 a 1 bA 2 N0 4 5 4 a00 5 = 4 A 5 1 + b2 z + 2 bz + 2 e 2 bz + 2 e 2 T 4 2 2 a01 0 bz + 2 e 1+b z+ 2 2e or 2

1 + b2 z + 2 4 bz + e 2 2 4 2e

bz + 2 e 2 1 + b2 z + 2 bz + 2 e 2

3 2 3 2 3 bA2 T =N0 bz a 1 5 4 a0 5 = 4 A2 T =N0 5 = 4 z 5 bz + 2 e 2 0 0 a1 1 + b2 z + 2 2e

where the factor N0 =T has been normalized out so that the right hand side is in terms of the signal-to-noise ratio, z (the extra factor of A needed mulitplies both sides of the matrix equation and on the left hand side is lumped into the new coe¢ cients a 1 , a0 , and a1 . Invert the matrix equation above to …nd the weights. A MATLAB program for doing so is given below. For z = 10 dB and b = 0:5, the weights are a 1 = 0:0773; a0 = 0:7821; and a1 =. 0:2781 c. According to (9.238) the minimum MSE is Normalized minimum MSE

[Ryd ]T [A]opt

= E d2 (t) = 1

1 z

= 1

1 10

bz z 0

0:5

a 1 4 a0 5 a1 2

10 10 0

= 0:1793

% Solution for Problem 9.47 % z_dB = input(’Enter the signal-to-noise ratio in dB ’); z = 10^(z_dB/10); b = input(’Enter multipath gain ’); R_yy(1,1)=(1+b^2)*z+pi/2;

(z is for normalization) 3 3 0:0783 4 0:7821 5 0:2781

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE R_yy(2,2)=R_yy(1,1); R_yy(3,3)=R_yy(1,1); R_yy(1,2)=b*z+(pi/2)*exp(-2*pi); R_yy(1,3)=(pi/2)*exp(-4*pi); R_yy(3,1)=R_yy(1,3); R_yy(2,1)=R_yy(1,2); R_yy(2,3)=R_yy(1,2); R_yy(3,2)=R_yy(1,2); disp(’R_yy’) disp(R_yy) B = inv(R_yy); disp(’R_yy^-1’) disp(B) R_yd = [b*z z 0]’ A = B*R_yd MSE = 1 - (1/z)*R_yd’*C

9.1. PROBLEM SOLUTIONS

Problem 9.48

In normalized form, from (9:245):

(2Eb =N0 ) + 1 [Ryy ] = 4 2 Eb =N0 0 2 3 2 Eb =N0 4 2Eb =N0 5 [Ryd ] = 0

2 Eb =N0 1 + 2 (2Eb =N0 ) + 1 2 Eb =N0

3 0 5 2 Eb =N0 2 1+ (2Eb =N0 ) + 1

The steepest descent tap weight adjustment algorithm is

[A](k+1) = [A](k) +

[Ryd ]

[Ryy ] [A](k)

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

3 2 k 3 ak+1 a 1 1 4 ak+1 5 = 4 ak0 5 0 ak1 ak+1 1 8 > > > > > > > > > > 2 > > 1 + 2 (2Eb =N0 ) + 1 < 4 + 2 Eb =N0 > > 0 > > > > > > > > > > : 2

3 2 Eb =N0 4 2Eb =N0 5 0 2 Eb =N0 1 + 2 (2Eb =N0 ) + 1 22 Eb =N30 ak 1 4 ak0 5 ak1

9 > > > > > > > > > 3 > > > 0 = 5 2 Eb =N0 > > 1 + 2 (2Eb =N0 ) + 1 > > > > > > > > > > ;

3 ak 1 = 4 ak0 5 ak1 2 3 8 9 2 Eb =N0 > > > > > > 4 5 > > 2E =N > > 0 b > > < = 0 3 2 + 1 + 2 (2Eb =N0 ) + 1 ak 1 + (2 Eb =N0 ) ak0 > > > > > > 4 (2 Eb =N0 ) ak + 1 + 2 (2Eb =N0 ) + 1 ak + (2 Eb =N0 ) ak 5 > > > > 1 0 1 > > : ; 2 k k (2 Eb =N0 ) a0 + 1 + (2Eb =N0 ) + 1 a1 2 3 Eb 2 N 6 Eb0 7 = 4 2 N0 5 0 2 k 3 8 9 a 1 > > > > > > 4 ak0 5 > > > > > > < = k a 1 2 3 + 1 + 2 (2Eb =N0 ) + 1 ak 1 + (2 Eb =N0 ) ak0 > > > > > > 2 k + k + (2 E =N ) ak 5 > > 4 > > (2 E =N ) a 1 + (2E =N ) + 1 a 0 0 0 b b b 1 0 1 > > : ; 2 k k (2 Eb =N0 ) a0 + 1 + (2Eb =N0 ) + 1 a1

9.1. PROBLEM SOLUTIONS

The separate update equations are ak+1 = 2 1

2Eb +1 N0

ak 1

ak+1 0

ak0

Eb ak 1 + ak1 N0

Eb N0

ak+1 1

Eb 1 ak0 + 1 1+ 2 N0 2Eb Eb + 1 1+ 2 +1 = 2 N0 N0 2Eb = 1 1+ 2 + 1 ak1 N0

ak0

Problem 9.49 From Example 9.12, the autocorrelation matrix equation with numerical values substituted, from (9.246), is 2 3 26 10 0 Ryy = 4 10 26 10 5 0 10 26 Using MATLAB its eigenvalues are

An acceptable range for

= 11:86

= 26:00

= 40:14

is therefore 0<

2 = 0:0498 40:14

a. Equation (9.246) scaled by 10 is 2

32 3 2 3 2:6 1 0 c 1 1 4 1 2:6 1 5 4 c0 5 = 4 2 5 0 1 2:6 c1 0 The inverse of the new Ryy is 2

0:4654 Ryy1 = 4 0:2101 0:0808

0:2101 0:5462 0:2101

3 0:0808 0:2101 5 0:4654

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE The recomputed weights are 2 3 2 c 1 0:4654 4 c0 5 = 4 0:2101 c1 0:0808

0:2101 0:5462 0:2101

32 3 2 3 0:0808 1 0:0452 0:2101 5 4 2 5 = 4 0:8824 5 0:4654 0 0:3394

which are the same as obtained in Example 9.12.

b. The eigenvalues of the scaled Ryy are 1.1858, 2.6000, and 4.0142 which are smaller by a factor of 10 than those found ifor the unscaled case.

Problem 9.50 The MATLAB program below computes the coe¢ cients among other quantities. output for an SNR of 20 dB and = 0:1 is also given. % Solution for Problem 9.50 % z_dB = input(’Enter the signal-to-noise ratio in dB ’); z = 10^(z_dB/10); beta = input(’Enter multipath gain ’); R_yy(1,1)=(1+beta^2)*2*z+1; R_yy(2,2)=R_yy(1,1); R_yy(3,3)=R_yy(1,1); R_yy(1,2)=2*beta*z; R_yy(1,3)=0; R_yy(3,1)=R_yy(1,3); R_yy(2,1)=R_yy(1,2); R_yy(2,3)=R_yy(1,2); R_yy(3,2)=R_yy(1,2); disp(’R_yy’) disp(R_yy) B = inv(R_yy); disp(’R_yy^-1’) disp(B) R_yd = [b*z z 0]’ C = B*R_yd Eigen = eig(R_yy) mu_max = 2/max(Eigen) >> pr9_50 Enter SNR in dB => 13.01

Its

9.2. COMPUTER EXERCISES Enter multipath gain, beta => 0.1 R_yy 41.3972 3.9997 0 3.9997 41.3972 3.9997 0 3.9997 41.3972 R_yy^-1 0.0244 -0.0024 0.0002 -0.0024 0.0246 -0.0024 0.0002 -0.0024 0.0244 R_yd = 1.9999 19.9986 0 C= 0.0012 0.4875 -0.0471 Eigen = 35.7407 41.3972 47.0537 mu_max = 0.0425

9.2

Computer Exercises

Computer Exercise 9.1 % ce9_1.m: Simulation of BEP for antipodal signaling % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clear all; clf n_sim = input(’Enter number of bits for simulation => ’); zdB = input(’Enter z = A^2*T/N_0 vector in dB => ’); z = 10.^(zdB/10); Lz = length(z); A = 1; T = 1;

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE N0 = (A^2*T)./z; S = A*T*sign(rand(1,n_sim)-.5); PE = []; ET = []; for k = 1:Lz N = sqrt(0.5*N0(k)*T)*randn(1,n_sim); Y = S+N; Z = sign(Y); V = 0.5*(1+Z); W = 0.5*(1+sign(S)); E = xor(V,W); ET(k) = sum(E); PE(k) = ET(k)/length(Y); end PEth = 0.5*erfc(sqrt(z)); disp(’’) disp(’SNR, dB Errors P_E est. P_E theory’) disp(’________________________________________’) disp(’’) disp([zdB’ET’PE’PEth’]) disp(’’) semilogy(zdB, PE, ’o’), xlabel(’z in dB’), ylabel(’P_E’), grid hold on semilogy(zdB, PEth) title([’BER simulated with ’, num2str(n_sim), ’bits per SNR value ’]) legend([’Simulated’], [’Theoretical’]) %End of script …le

A typical run is given below: >> ce9_1 Enter number of bits for simulation => 20000 Enter z = A^2*T/N_0 vector in dB => [2 4 6 8] SNR, dB Errors P_E est. P_E theory ______________________________________ 2.0000 704.0000 0.0352 0.0375 4.0000 238.0000 0.0119 0.0125 6.0000 44.0000 0.0022 0.0024 8.0000 4.0000 0.0002 0.0002

9.2. COMPUTER EXERCISES

A plot also appears showing the estimated and theoretical probabilities of error plotted versus signal-to-noise ratio.

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

Computer Exercise 9.2 % ce9_2: Compute degradation in SNR due to bit timing error % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clear all; clf A = char(’-’,’-.’,’:’,’–’); for m = 1:4 PE0 = 10^(-m); delT = []; deg = []; Eb1 = []; k = 0; for Y = 0:.025:.45; k = k+1; delT(k) = Y; Eb1(k) = fzero(@PE1, 2, [], delT(k), PE0); end Eb = fzero(@PE, 2, [], PE0); deg = Eb1 - Eb; plot(delT, deg, A(m,:)), ... if m == 1 hold on axis([0 .45 0 20]) xlabel(’Fractional timing error, nDeltaT/T’), ylabel(’Degradation in dB’),... title([’Degradation in SNR for bit synchronization timing error’]), grid end end legend([’P_E = 10^-^1’], [’= 10^-^2’], [’= 10^-^3’], [’= 10^-^4’], 2), %End of script …le

The output plot is shown in Fig. 9.7.

9.2. COMPUTER EXERCISES

Degradation in SNR for bit synchronization timing error 20 PE = 10-1

Degradation in dB

= 10-2 16

= 10-3

= 10-4

12 10 8 6 4 2 0

0.05

0.1

0.15 0.2 0.25 0.3 Fractional timing error, ∆ T/T

0.35

0.4

0.45

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

Computer Exercise 9.3

% ce9_3.m; Evaluates e¤ect of Gaussian % phase error in BPSK; …xed variance. % % Uses subprogram pb_phase_pdf % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clear all; clf sigma_psi2 = input(’Enter 3-vector of variances of phase error in radians^2 => ’); L_sig = length(sigma_psi2); Eb_N0_dB_min = input(’Enter minimum Eb/N0 in dB desired => ’); Eb_N0_dB_max = input(’Enter maximum Eb/N0 in dB desired => ’); A = char(’-.’,’:’,’–’,’-..’); a_mod = 0; for m = 1:L_sig Eb_N0_dB = []; Eb_N0 = []; PE = []; k = 1; for snr = Eb_N0_dB_min:.5:Eb_N0_dB_max Eb_N0_dB(k) = snr; Eb_N0(k) = 10.^(Eb_N0_dB(k)/10); sigma_psi = sqrt(sigma_psi2(m)); PE(k) = quadl(’pb_phase_pdf’,-pi,pi,[],[],Eb_N0(k),sigma_psi,a_mod); k = k+1; end semilogy(Eb_N0_dB, PE,A(m,:)),xlabel(’E_b/N_0, dB’),ylabel(’P_E’),... axis([Eb_N0_dB_min Eb_N0_dB_max 10^(-7) 1]),... if m == 1 hold on grid on end end P_ideal = .5*erfc(sqrt(Eb_N0)); semilogy(Eb_N0_dB, P_ideal) legend([’nsigma_ntheta_e^2 = ’,num2str(sigma_psi2(1))],[’nsigma_ntheta_e^2 = ’,num2str(sigma_psi2(2 = ’,num2str(sigma_psi2(3))],[’nsigma_ntheta_e^2 = 0’],3) title(’E¤ect of Gaussian phase error on detection of BPSK’)

9.2. COMPUTER EXERCISES

Effect of Gaussian phase error on detection of BPSK

-1

-2

-3

-4

σ θe = 0.05

-5

σ θe = 0.1 2

σ θe = 0.5

-6

σ θe = 0

-7

6 Eb/N0, dB

function XX = pb_phase_pdf(psi,Eb_N0,sigma_psi,a) arg = Eb_N0*(1-a^2)*(cos(psi)-a/sqrt(1-a^2)*sin(psi)).^2; T1 = .5*erfc(sqrt(arg)); T2 = exp(-psi.^2/(2*sigma_psi^2))/sqrt(2*pi*sigma_psi^2); XX = T1.*T2; %End of script …le A plot from the program is given in Fig. 9.8. Computer Exercise 9.4 % ce9_4: For a given data rate and error probability, …nd the required % bandwidth and Eb/N0 in dB; baseband, BPSK/DPSK, coherent FSK % and noncoherent FSK modulation may be speci…ed. % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clear all; clf I_mod = input(’Enter desired modulation: 0 = baseband; 1 = BPSK; 2 = DPSK; 3 = CFSK; 4 = NCFSK: ’);

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE I_BW_R = input(’Enter 1 if bandwidth speci…ed; 2 if data rate speci…ed: ’); if I_BW_R == 1 BkHz = input(’Enter vector of desired bandwidths in kHz => ’); elseif I_BW_R == 2 Rkbps = input(’Enter vector of desired data rates in kbps => ’); end PE = input(’Enter desired BEP => ’); N0 = 1; if I_BW_R == 1 if I_mod == 0 Rkpbs = BkHz; % Rate for baseband elseif I_mod == 1 j I_mod == 2 Rkbps = BkHz/2; % Rate in kbps for BPSK or DPSK elseif I_mod == 3 Rkbps = BkHz/3; % Rate in kbps for CFSK elseif I_mod == 4 Rkbps = BkHz/4; % Rate in kbps for NFSK end elseif I_BW_R == 2 if I_mod == 0 BkHz = Rkbps; % Transmission bandwidth for baseband in kHz elseif I_mod == 1 j I_mod == 2 BkHz = 2*Rkbps; % Transmission bandwidth for BPSK or DPSK in kHz elseif I_mod == 3 BkHz = 3*Rkbps; % Transmission bandwidth for CFSK in kHz elseif I_mod == 4 BkHz = 4*Rkbps; % Transmission bandwidth for NFSK in kHz end end R = Rkbps*1000; B = BkHz*1000; % PE = 0.5*erfc(sqrt(z)) if I_mod == 0 j I_mod == 1 sqrt_z = er…nv(1-2*PE); z = sqrt_z^2; elseif I_mod == 2 z = -log(2*PE); elseif I_mod == 3 sqrt_z_over_2 = er…nv(1-2*PE); z = 2*sqrt_z_over_2^2;

9.2. COMPUTER EXERCISES

elseif I_mod == 4 z = -2*log(2*PE); end Eb_N0_dB = 10*log10(z); % This is the required Eb/N0 in dB if I_mod == 0 A = sqrt(R*N0*z); % z = A^2/(R*N0) for baseband elseif I_mod == 1 j I_mod == 2 j I_mod == 3 j I_mod == 4 A = sqrt(2*R*N0*z); % z = A^2/2*R*N0 for BPSK, DPSK, CFSK, NFSK; A = modulated carrier amplitude end disp(’’) disp(’Desired P_E and required E_b/N_0 in dB: ’) format long disp([PE Eb_N0_dB]) format short disp(’’) disp(’R, kbps BW, Hz A, volts’) disp(’___________________________’) disp(’’) disp([Rkbps’BkHz’A’]) disp(’’) %End of script …le

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

A typical run is given below: >> ce9_4 Enter desired modulation: 0 = baseband; 1 = BPSK; 2 = DPSK; 3 = CFSK; 4 = NCFSK: 2 Enter 1 if bandwidth speci…ed; 2 if data rate speci…ed: 2 Enter vector of desired data rates in kbps => [5 10 50 100] Enter desired BEP => 1e-6 Desired P_E and required E_b/N_0 in dB: 0.000001000000000 11.180120598361940 R, kbps BW, Hz A, volts ___________________________ 1.0e+003 * 0.0050 0.0100 0.3622 0.0100 0.0200 0.5123 0.0500 0.1000 1.1455 0.1000 0.2000 1.6200 Computer Exercise 9.5 % …le: ce9_5.m % Simulation of suboptimum bandpass …lter/delay-and-multiply demodulator % with integrate-and-dump detection for DPSK; input bandpass …lter is % Butterworth with selectable bandwidth-bit period product % % User de…ned subprograms di¤_enc( ) and int_and_dump( ) called % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clear all; clf Eb_N0_dB_max = input(’Enter maximum Eb/N0 in dB ’); Eb_N0_dB_min = input(’Enter minimum Eb/N0 in dB ’); samp_bit = input(’Enter number of samples per bit used in simulation ’); n_order = input(’Enter order of Butterworth detection …lter ’); BWT_bit = input(’Enter …lter bandwidth normalized by bit rate ’); N_bits = input(’Enter total number of bits in simulation ’) clf ss = sign(rand(1,N_bits)-.5); % Generate sequence of random +-1s data = 0.5*(ss+1); % Logical data is sequence of 1s and 0s data_di¤_enc = di¤_enc(data);

% Di¤erentially encode data for DPSK

9.2. COMPUTER EXERCISES

s = 2*data_di¤_enc-1; % Generate bipolar data for modulation T_bit = 1; % Arbitrarily take bit time as 1 second BW = BWT_bit/T_bit; % Compute …lter bandwidth from BW*T_bit [num,den] = butter(n_order,2*BW/samp_bit); % Obtain …lter num/den coe¢ cients Eb_N0_dB = []; % Ensure that plotting arrays are empty Perror = []; k = 0; for kk = Eb_N0_dB_min:Eb_N0_dB_max % Loop simulates detector for each desired Eb/N0 k = k+1; Eb_N0_dB(k) = kk; Eb_N0 = 10^(Eb_N0_dB(k)/10); % Convert desired Eb/N0 from dB to ratio Eb = T_bit; % Bit energy is T_bit, assuming ampl = 1 N0 = Eb/Eb_N0; % Compute noise spectral density from Eb/N0 del_t = T_bit/samp_bit; % Compute sampling interval sigma_n = sqrt(N0/(2*del_t)); % Compute standard deviations of noise samples t=0:del_t:N_bits*T_bit-del_t; % Set up time axis sig = []; sig = s(ones(samp_bit,1),:); % Build array whose columns are samp_bit long sig = sig(:); % Convert matrix where bit samples occupy columns to vector zi = []; % Make sure various arrays are empty bits_out = []; y = []; y_det = []; y_ref = []; noise = sigma_n*randn(size(sig)); % Form sequence of Gaussian noise samples [y,zf] = …lter(num,den,sig+noise,zi); % Filter signal plus noise with chosen …lter zi = zf; % Save …nal values for future initial conditions in a block simulation n_tot = length(y);

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE

y_ref = [y(n_tot-samp_bit+1:n_tot,1); y(1:n_tot-samp_bit,1)]; % Reference signal for multiply is 1-bit delayed signal + noise y_mult = y.*y_ref; % Multiply received signal plus noise with 1 bit delayed version bits_out = int_and_dump(y_mult,samp_bit,N_bits+1); % di¤_enc appends a 1 bit to start of bit stream bits_out = bits_out(2:N_bits+1); error_array = abs(bits_out-data); % Compare detected bits with actual input data stream; errors = 1s error_array(1:5) = 0; % Don’t include …rst 5 bits due to transients ss = sum(error_array); % Sum to get total number of errors Perror(k) = ss/(N_bits-5); % Subtract 5 from denominator to account for initial errors set = 0 end % End of Eb/N0 loop disp(’E_b/N_0, dB; P_E’) % Display computed probability of error values versus Eb/N0 disp([Eb_N0_dB’Perror’]) % Plot simulated bit error probabilities versus Eb/N0 semilogy(Eb_N0_dB, Perror,’–’), grid, xlabel(’E_b/N_0; dB’), ylabel(’P_E’), hold title(’Simulation of BEP for delay-and-multiply detector with Butterworth pre…lter for DPSK’) % Plot theoretical bit error probability for optimum DPSK detector semilogy(Eb_N0_dB, 0.5*exp(-10.^(Eb_N0_dB/10))) % Plot approximate theoretical result for suboptimum detector semilogy(Eb_N0_dB, qfn(sqrt(10.^(Eb_N0_dB/10))),’-.’) legend([’Simulation; BT = ’, num2str(BWT_bit),’; ’, num2str(N_bits),’bits’],’Theory; optimum di¤erential detector’,’Theory; delay/multiply detector’, 3) % Functions output = di¤_enc(input) L_in = length(input) output = []; for k==1:L_in if k==1 output(k)=not(bitxor(input(k),1)); else output(k)=not(bitxor(input(k),output(k-1))); end end function y_out=int_and_dump(input, samp_bit, N_bits)

9.2. COMPUTER EXERCISES

samp_array=reshape(input, samp_bit, N_bits) % Reshape input array so that bits occupy columns integrate=sum(samp_array); bits_out=(sign(integrate)+1/2)/2; % End of script …le Computer Exercise 9.6 % ce9_6: program to plot Figure 9-28 % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clf z0dB = 0:1:30; z0 = 10.^(z0dB/10); for delta = -0.9:0.3:0.6; if delta > -eps & delta < eps delta = 0; end for taum_over_T = 0:1:1 T1 = 0.5*qfn(sqrt(2*z0)*(1+delta)); T2 = 0.5*qfn(sqrt(2*z0)*(1+delta)-2*delta*taum_over_T); PE = T1 + T2; if taum_over_T == 0 semilogy(z0dB, PE, ’-’) elseif taum_over_T == 1 semilogy(z0dB, PE, ’–’) end if delta <= -.3 text(z0dB(10), PE(10), [’ndelta = ’, num2str(delta)]) elseif delta > -0.3 & delta <= 0.3 text(z0dB(8), PE(8), [’ndelta = ’, num2str(delta)]) elseif delta > 0.3 text(z0dB(5), PE(5), [’ndelta = ’, num2str(delta)]) end if delta == -0.9 hold on ylabel(’P_E’), xlabel(’z_0 in dB’) axis([0 30 1E-5 1]) grid

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE Bit error probability for two-ray multipath channel

τm/T = 0

δ = -0.9 δ = -0.9

-1

τm/T = 1

δ = -0.6 δ = -0.6 -2

10 PE

δ = 0.6 -3

δ = -0.3 δ = -0.3 δ=0

δ = 0.6

-4

δ = 0.3

δ = 0.3 0

15 z 0 in dB

end end end legend([’ntau_m/T = 0’],[’ntau_m/T = 1’])

The output plot is shown in Fig. 9.9.

9.2. COMPUTER EXERCISES

Computer Exercise 9.7 Use Equations (9.74) (with m = 0), (9.88), (9.112), (9.122), and (9.205) - (9.208). Solve them for the SNR in terms of the error probability. Express the SNR in dB and subtract the nonfading result from the fading result. The MATLAB program below does this for the modulation cases of BPSK, coherent FSK (CFSK), DPSK, and noncoherent (NCFSK). % ce9_7.m: Program to evaluate degradation due to ‡at % Rayleigh fading for various modulation schemes % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % mod_type = input(’Enter type of modulation: 1 = BPSK; 2 = CFSK; 3 = DPSK; 4 = NCFSK: ’); PE = input(’Enter vector of desired probabilities of error: ’); disp(’’) if mod_type == 1 disp(’BPSK’) Z_bar = (0.25*(1 - 2*PE).^2)./(PE - PE.^2); z = (er…nv(1 - 2*PE)).^2; % MATLAB has an inv. error function elseif mod_type == 2 disp(’CFSK’) Z_bar = (0.5*(1 - 2*PE).^2)./(PE - PE.^2); z = 2*(er…nv(1 - 2*PE)).^2; elseif mod_type == 3 disp(’DPSK’) Z_bar = 1./(2*PE) -1; z = -log(2*PE); elseif mod_type == 4 disp(’NCFSK’) Z_bar = 1./PE -2; z = -2*log(2*PE); end Z_bar_dB = 10*log10(Z_bar); z_dB = 10*log10(z); deg_dB = Z_bar_dB - z_dB; disp(’’) disp(’P_E Degradation, dB’) disp(’_________________________’) disp(’’) disp([PE’deg_dB’])

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE disp(’’) %End of script …le

A typical run is given below: >> ce9_7 Enter type of modulation: 1 = BPSK; 2 = CFSK; 3 = DPSK; 4 = NCFSK: 3 Enter vector of desired probabilities of error: [1e-2 1e-3 1e-4] DPSK P_E Degradation, dB _________________________ 0.0100 10.9779 0.0010 19.0469 0.0001 27.6859 Computer Exercise 9.8 The solution to this exercise consists of two MATLAB programs, one for the zero-forcing case and one for the MMSE case. % ce9_8a.m: Plots unequalized and equalized sample values for ZF equalizer. The MATLAB program for the MMSE case is ce9_9b. % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition %: % clf pc = input(’Enter sample values for channel pulse response (odd #): ’); L_pc = length(pc); disp(’Maximum number of zeros each side of decision sample: ’); disp((L_pc-1)/4) N = input(’Enter number of zeros each side of decision sample desired: ’); mid_samp = …x(L_pc/2)+1; Pc = []; for m = 1:2*N+1 row = []; for n = 2-m:2*N+2-m row = [row pc(mid_samp-1+n)]; end Pc(:,m) = row’; end disp(’’)

9.2. COMPUTER EXERCISES disp(’Pc:’) disp(’’) disp(Pc) Pc_inv = inv(Pc); disp(’’) disp(’Inverse of Pc:’) disp(’’) disp(Pc_inv) coef = Pc_inv(:,N+1); disp(’’) disp(’Equalizer coe¢ cients:’) disp(’’) disp(coef) disp(’’) steps = (L_pc-length(coef))+1; p_eq = []; for m = mid_samp-‡oor(steps/2):mid_samp+‡oor(steps/2) p_eq(m) = sum(coef’.*‡iplr(pc(m-N:m+N))); end p_eq_p = p_eq(N+1:mid_samp+‡oor(steps/2)); disp(’’) disp(’Equalized pulse train’) disp(’’) disp(p_eq_p) disp(’’) t=-5:.01:5; y=zeros(size(t)); subplot(2,1,1),stem(pc),ylabel(’Ch. output pulse’),... axis([1 L_pc -.5 1.5]),... title([’Ch. pulse samp. = [’,num2str(pc),’]’]) subplot(2,1,2),stem(p_eq_p),xlabel(’n’),ylabel(’Equal. output pulse’),... axis([1 L_pc -.5 1.5]),... title([’Output equalized with ’,num2str(N),’zero samples either side’]) pause clf n = [-N:1:N]’; fT = -0.5:0.01:0.5; Heq = coef’*exp(-j*2*pi*n*fT); sigmaN2 = coef’*coef plot(fT, Heq), xlabel(’fT’), ylabel(’Eq. freq. response’), ...

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE title([’Noise enhancement factor = ’, num2str(10*log10(sigmaN2)), ’dB’]) %End of script …le

A typical run is given below: >> ce9_8a Enter sample values for channel pulse response (odd #): [.01 -.05 .2 -.3 1 -.2 .15 -.25 .07] Maximum number of zeros each side of decision sample: 2 Enter number of zeros each side of decision sample desired: 2 mid_samp = 5 Pc: 1.0000 -0.3000 0.2000 -0.0500 0.0100 -0.2000 1.0000 -0.3000 0.2000 -0.0500 0.1500 -0.2000 1.0000 -0.3000 0.2000 -0.2500 0.1500 -0.2000 1.0000 -0.3000 0.0700 -0.2500 0.1500 -0.2000 1.0000 Inverse of Pc: 1.0688 0.3050 -0.1338 -0.0441 0.0181 0.1482 1.1107 0.2824 -0.1388 -0.0441 -0.0631 0.1309 1.1285 0.2824 -0.1338 0.2381 0.0073 0.1309 1.1107 0.3050 0.0193 0.2381 -0.0631 0.1482 1.0688 Equalizer coe¢ cients: -0.1338 0.2824 1.1285 0.1309 -0.0631 Equalized pulse train -0.0000 -0.0000 1.0000 0.0000 0.0000 %The MATLAB program for the MMSE case is given below: % ce9.8b % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % z_dB = input(’Enter the signal-to-noise ratio in dB: ’); z = 10^(z_dB/10);

9.2. COMPUTER EXERCISES b = input(’Enter multipath gain: ’); R_yy(1,1)=(1+b^2)*z+pi/2; R_yy(2,2)=R_yy(1,1); R_yy(3,3)=R_yy(1,1); R_yy(1,2)=b*z+(pi/2)*exp(-2*pi); R_yy(1,3)=(pi/2)*exp(-4*pi); R_yy(3,1)=R_yy(1,3); R_yy(2,1)=R_yy(1,2); R_yy(2,3)=R_yy(1,2); R_yy(3,2)=R_yy(1,2); disp(’’) disp(’R_yy:’) disp(’’) disp(R_yy) B = inv(R_yy); disp(’’) disp(’R_yy^-1:’) disp(’’) disp(B) R_yd = [0 z b*z]’; disp(’’) disp(’R_yd:’) disp(’’) disp(R_yd) A = B*R_yd; disp(’’) disp(’Optimum coe¢ cients:’) disp(A) disp(’’) %End of script …le A typical run is given below: >> ce9_8b Enter the signal-to-noise ratio in dB: 7 Enter multipath gain: 0.6 R_yy: 8.3869 3.0101 0.0000 3.0101 8.3869 3.0101 0.0000 3.0101 8.3869 R_yy^-1:

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE 0.1399 -0.0576 -0.0576 0.1606 0.0207 -0.0576 R_yd: 0 5.0119 3.0071 Optimum coe¢ cients: -0.2267 0.6316 0.1319

0.0207 -0.0576 0.1399

Computer Exercise 9.9 % ce9_9.m: Simulation of multipath channel and adaptive MMSE equalizer % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clear all; clf Eb_N0_dB0 = input(’Enter starting Eb/N0 in dB => ’); %beta = input(’Enter multipath constant, beta => ’); N_bits = input(’Enter number of bits in simulation => ’); N_bits_train = input(’Enter number of training bits => ’); mu = input(’Enter feedback constant, mu => ’); samp_bit = 10; T = 1; A = 1; beta = 0.5; delta = T; delt = T/samp_bit; Eb = A^2*T; kk = 0; for Eb_N0_dB = Eb_N0_dB0:.5:Eb_N0_dB0+12 kk = kk + 1; Eb_N0plt(kk) = Eb_N0_dB; Eb_N0 = 10.^(Eb_N0_dB/10); N0 = Eb/Eb_N0; sigma_n = sqrt(N0/(2*delt)); data = zeros(size(N_bits)); data_10 = zeros(size(N_bits)); d = zeros(size(samp_bit, N_bits)); y = zeros(1, N_bits*samp_bit); data = sign(rand(1, N_bits)-0.5); data_10 = 0.5*(data +1);

9.2. COMPUTER EXERCISES d = data(ones(samp_bit, 1), :); d = d(:)’; t = 0:delt:N_bits*T-delt; tau = length(0:delt:delta); y = A*d + beta*delay1(d, tau); noise = sigma_n*randn(size(y)); z = y + noise; bits_out = int_and_dump(z, samp_bit, N_bits); bits_outs = 2*bits_out - 1; bits = bits_outs(ones(samp_bit, 1), :); bits = bits(:)’; error_array = abs(bits_out - data_10); error_array(1:10) = 0; errors = sum(error_array); PE_ue(kk) = errors/(N_bits - 10); % Equalizer z_eq = []; zd = []; zdd = []; a1 = []; a2 = []; a3 = []; epsilon = []; a1(1) = 1; a2(1) = 1; a3(1) = 1; zd = delay1(z, tau); zdd = delay1(z, 2*tau); z_eq(1) = a1(1)*z(1) + a2(1)*zd(1) + a3(1)*zdd(1); epsilon(1) = z_eq(1) - d(1); for k = 1:samp_bit*N_bits-1 a1(k+1) = a1(k) - mu*z(k)*epsilon(k); a2(k+1) = a2(k) - mu*zd(k)*epsilon(k); a3(k+1) = a3(k) - mu*zdd(k)*epsilon(k); z_eq(k+1) = a1(k+1)*z(k+1) + a2(k+1)*zd(k+1) + a3(k+1)*zdd(k+1); if k <= samp_bit*N_bits_train-1 epsilon(k+1) = z_eq(k+1) - d(k+1); elseif k > samp_bit*N_bits_train-1 epsilon(k+1) = z_eq(k+1) - bits(k+1); end end bits_out_eq = int_and_dump(z_eq, samp_bit, N_bits); error_array_eq = abs(bits_out_eq - data_10); error_array_eq(1:10) = 0; errors_eq = sum(error_array_eq); PE_e(kk) = errors_eq/(N_bits - 10); PE_gauss(kk) = qfn(sqrt(2*Eb_N0));

CHAPTER 9. PRINCIPLES OF DIGITAL DATA TRANSMISSION IN NOISE aa(:, kk) = [a1(k) a2(k) a3(k)] end semilogy(Eb_N0plt, PE_ue, ’–’, Eb_N0plt, PE_e, ’-.’, Eb_N0plt, PE_gauss, ’-’), grid,

... xlabel(’{nitE_b/N}_0, dB’), ylabel(’{nitP_b}’), axis([min(Eb_N0plt) max(Eb_N0plt) 1e-6 1]), ... legend([’Unequalized; ’, num2str(N_bits), ’bits’], [’Equalized, nmu = ’, num2str(mu)], ’Gauss noise, theory’, 1) pause %Enter to view tap weights %subplot(6,1,1), plot(d), axis([0 500 -1.1 1.1]), ylabel(’in data’), ... % legend([’T_b_i_t = ’, num2str(T),’; No. train bits = ’, num2str(N_bits_train)]) %subplot(6,1,2), plot(bits_out_eq), axis([0 500 -1.1 1.1]), ylabel(’eq_data’) subplot(4,1,1), plot(t, epsilon), axis([0 200 -2.1 2.1]), ylabel(’error’) subplot(4,1,2), plot(t, a1), axis([0 200 -1.1 1.1]), ylabel(’a_1’) subplot(4,1,3), plot(t, a2), axis([0 200 -1.1 1.1]), ylabel(’a_2’) subplot(4,1,4), plot(t, a3), axis([0 200 -1.1 1.1]), ylabel(’a_3’), xlabel(’t’) >> ce9_9 Enter starting Eb/N0 in dB => 4 Enter number of bits in simulation => 10000 Enter number of training bits => 500 Enter feedback constant, mu => .01

9.2. COMPUTER EXERCISES

Unequalized; 10000 bits Equalized, µ = 0.01 Gauss noise, theory

-1

-2

-3

-4

-5

-6

10 Eb /N0, dB

error

2 0 -2 0

100

120

140

160

180

200

100

120

140

160

180

200

100

120

140

160

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100 t

120

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1 0 -1

Chapter 10

Advanced Data Communications Topics 10.1

Problem Solutions

Problem 10.1 Use the relationship Rb = (log2 M ) Rs bps where Rb is the data rate, Rs is the symbol rate, and M is the number of possible signals per signaling interval. In this case, Rs = 2; 000 symbols per second. For M = 4, Rb = 2 2; 000 = 4; 000 bps, for M = 8, Rb = 6; 000 bps, for M = 16, Rb = 8; 000 bps, for M = 32, Rb = 10; 000 bps, and for M = 64, Rb = 12; 000 bps. The bit rare is linear as a function of log2 M: Problem 10.2 a. 5,000 symbols per second. b. Recall that y (t) = A [d1 (t) cos (! c t)

d2 (t) sin (! c t)] =

2A cos [! c t +

i (t)] ;

i (t) = tan

d2 (t) d1 (t)

Alternating every other bit of the given data sequence, 110110 010111 011011, between d1 (t) and d2 (t) gives d1 (t) = 1; 1; 1; 1; 1; 1; 1; 1; 1 and d2 (t) = 1; 1; 1; 1; 1; 1; 1; 1; 1 which results in the table below for the phases: d2 (t) 1 1 1 1 1 1 1 1 1 d1 (t) 1 1 1 1 1 1 1 1 1 3 7 3 3 3 7 i (t) 4 4 4 4 4 4 4 4 4 1

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

For QPSK the symbol switching points occur each Ts = 2Tb seconds. c. Now the switching instants are each Tb seconds. Start with d1 (t) = 1. Then d2 (t) is staggered, or o¤set, by Tb seconds. So the …rst phase shift is for d1 (t) = 1 and d2 (t) = 1 or 1 (t) = =4: After Ts = 2Tb seconds d1 (t) takes a 1-value, but d2 (t) is still 1, so 2 (t) = 3 =4: At 3Tb seconds, d2 (t) takes on a second 1-value while d1 (t) = 1, so 3 (t) = 3 =4: At 4Tb seconds, d1 t) = 1, so 4 (t) = =4, etc. The phase sequence is shown in the table below: d2 (t) d1 (t) i (t)

1 4

3 4

1 3 4

1 7 4

5 4

Problem 10.3 For QPSK, PE; symbol = 2Q

1 3 4

3 4

1 3 4

1 4

3 4

1 3 4

1 7 4

7 4

Es =N0 = 10 4

Trial and error using the asymptotic approximation for the Q-function gives Q (x) = 5 10 5 for x = 3:89 or [Es =N0 ]req’d 3:892 = 15:14 = 11:8 dB. If the quadrature-carrier amplitudes are A, then the amplitude of the envelope-phase-modulated q form of the carrier p p 2 2 is 2A, and Es =N0 = 2A T = (2N0 ) = A =N0 R: Hence, Areq’d = N0 R [Es =N0 ]req’d = p p (10 11 ) (15:14) R = 1:23 10 5 R: The answers are as follows: (a) 8:7 10 4 V; (b) 0.0012 V; (c) 0.0028 V; (d) 0.0039 V; (e) 0.0087 V; (f) 0.012 V. Problem 10.4 Take the expectation of the product after noting that the average values are 0 because E[n (t)] = 0: E [N1 N2 ] = E

Z Ts

n (t) cos (! c t) dt

= = =

E [n (t) n ( )] cos (! c t) sin (! c ) dtd

Z Ts Z Ts 0

n (t) sin (! c t) dt

Z Ts Z Ts

Z Ts

0 Z Ts

N0 (t 2

) cos (! c t) sin (! c ) dtd

N0 cos (! c t) sin (! c t) dt 2 0 Z N0 Ts sin (2! c t) dt = 0 4 0

10.1. PROBLEM SOLUTIONS

Problem 10.5 a. The integrator outputs, without noise, are (the polarities) V10 =

Z Ts

Z Ts 0

= V20 = = = =

xc (t) cos (2 fc t) dt

= A =

signs are determined by the data

cos 2 fc t +

Z A Ts cos sin 2 0 2 ATs cos sin 2 2 2 Z Ts xc (t) sin (2 fc t) dt 0 Z Ts A cos 2 fc t + 2 0 Z Ts A sin cos 2 0 2 ATs sin cos 2 2 2

sin 2 fc t

cos (2 fc t) dt

+ dbl freq terms dt

sin 2 fc t

sin (2 fc t) dt

+ dbl freq terms dt

b. This follows from the outline given in the problem statement. c. Note that Es = 2Eb and that the expression given in the problem statement reduces to (10.12) for = 0: A plot is given in Figure 10.1. The degradation at an error probability of 10 6 for 10 degrees phase imbalance is about 1.1 dB.

Problem 10.6 a. Both are equivalent in terms of symbol error probabilities. b. QPSK is 3 dB worse in terms of symbol error probability for equal transmission bandwidths, but it handles twice as many bits per second. c. Choose QPSK over BPSK in terms of performance; however, other factors might favor BPSK, such as simpler implementation.

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

-3

β = 0 deg β = 2.5 deg β = 5 deg β = 7.5 deg β = 10 deg

-4

PE, qaud ch

-5

-6

8.5

9.5

10 Eb/N0, dB

10.5

11.5

10.1. PROBLEM SOLUTIONS

Problem 10.7 For the data stream 101011 010010 100110 110011, the quadrature data streams are d1 (t) = 1, 1, 1, -1, -1, 1, 1, -1, 1, 1, -1, 1 d2 (t) = -1, -1, 1, 1, -1, -1, -1, 1, -1, 1, -1, 1 Each 1 above means a positive rectangular pulse Ts seconds in duration and each -1 above means a negative rectangular pulse Ts seconds in duration. For QPSK, these pulse sequences are aligned and for OQPSK, the one corresponding to d2 (t) is delayed by Ts =2 = Tb seconds with respect to d1 (t). Type I MSK corresponds to the OQPSK waveforms d1 (t) and d2 (t) multiplied, respectively, by cosine and sine waveforms with periods of 2Ts and type II MSK corresponds to d1 (t) and d2 (t) multiplied, respectively, by absolute-value cosine and sine waveforms with periods of Ts (one half cosine or sine per Ts pulse). Problem 10.8 For QPSK, the excess phase corresponds to a stepwise function that may change values each Ts seconds. The phase deviation is given by i (t) = tan

d2 (t) d1 (t)

so the sequence of phases for QPSK computed from d1 (t) and d2 (t) given in Problem 10.7 is 7 =4, 7 =4, =4, 3 =4, 5 =4, 7 =4, 7 =4, 3 =4, 7 =4, =4, 5 =4, =4 radians. For OQPSK, the phase can change each Ts =2 = Tb seconds because d2 (t) is delayed by Ts =2 seconds with respect to d1 (t). The maximum phase change is =2 radians. For MSK, the excess phase trajectories are straight lines of slopes =2Tb radians/second. Problem 10.9 Write the sinc-functions as sin(x) =x functions. Use appropriate trigonometric identities to reduce the product of the sinc-functions to the given form. For positive frequencies SM SK (f ) = jG (f )j2 SBP SK (f ) A2 T b = sinc2 [(f fc ) Tb 0:25] sinc2 [(f fc ) Tb + 0:25] 2 A2 Tb sin2 [(f fc ) Tb 0:25] sin2 [(f fc ) Tb + 0:25] = 2 f [(f fc ) Tb 0:25]g2 f [(f fc ) Tb + 0:25]g2 A2 Tb sin2 [(f fc ) Tb 0:25] sin2 [(f fc ) Tb + 0:25] = h i2 2 4 (f fc )2 Tb2 1=16

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS 0:25] = p12 sin (f

Use sin [(f

fc ) Tb

SM SK (f ) =

A2 Tb [sin (f 8 4

fc ) Tb

A2 Tb sin2 (f h 8 4

fc ) Tb

cos (f h

fc )2 Tb2

cos2 2 (f

16 (f

fc ) Tb to get

fc ) Tb ]2 [sin (f fc )2 Tb2

cos2 (f

A2 Tb cos2 2 (f fc ) Tb i2 h 8 4 (f fc )2 Tb2 1=16

32A2 Tb

p1 cos 2

fc ) Tb

fc ) Tb + cos (f i2

fc ) Tb ]2

1=16

fc ) Tb i2 fc )2 Tb2

Problem 10.10 If d (t) is a sequence of 1s or 0s, the instantaneous frequency is 1=4Tb Hz above the carrier (with the de…nition of i (t) given by (10.20) and (10.21)). If it is a sequence of alternating 1s and 0s, the instantaneous frequency is 1=4Tb Hz below the carrier. (a) The instantaneous frequency is 10; 000; 000 50; 000=4 = 9; 987; 500 Hz. (b) The instantaneous frequency is 10; 000; 000 + 50; 000=4 = 10; 012; 500 Hz.

10.1. PROBLEM SOLUTIONS

Problem 10.11 The Fourier transform of h (t) is Z 1 r 2 2 2B2 2 B H (f ) = exp t exp ( j2 f t) dt ln 2 ln 2 1 r Z 1 2 2 B 2 2 jf ln 2 2 exp t t dt; complete square in exponent = B ln 2 1 ln 2 B2 " !# r Z 1 2 2 2 B 2 2 jf ln 2 jf ln 2 2 jf ln 2 2 = B exp t t + dt ln 2 1 ln 2 B2 2 B2 2 B2 " # r Z 1 2 2B2 2 ln 2 2 jf ln 2 2 jf ln 2 exp f exp t dt; u = t = B 2 2 ln 2 2B ln 2 2 B 2 B2 1 " # r Z 1 2 ln 2 f 2 2 2B2 2 1 2 2B2 = B exp exp u du; = ln 2 2 B ln 2 2 2 ln 2 1 " # r Z 1 1 2 ln 2 f 2 p u2 p = B exp 2 2 du; integral is a Gauss pdf exp ln 2 2 B 2 2 2 2 1 " #r r ln 2 f 2 2 ln 2 exp ; substitute for 2 2 and simplify = B ln 2 2 B 2 2B2 " # ln 2 f 2 = exp 2 B

Problem 10.12 p a. The signal points lie on a circle of radius Es centered at the origin equally spaced at angular intervals of 22.5 degrees (360=16). The decision regions are pie wedges centered over each signal point. b. For M = 16, the upper bound (10.50) well approximates the symbol error probability. Thus "r # 2Es PE, 16-PSK ' 2Q sin N0 16 2s 3 2 log (16) E b 2 5 = 2Q 4 sin N0 16

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

Symbol error probability Bit error probability -1

-2

PE, Pb; 16-PSK

-3

-4

-5

-6

10 12 Eb/N0, dB

c. The bit error probability, for Gray bit encoding, is well approximated by PE, 16-PSK log2 16 "r # 1 8Eb Q sin 2 N0 16

Pb; 16-PSK = =

A plot comparing the two error probabilities is given in Figure 10.2. Problem 10.13 a. The bounds on symbol error probability are given by P1 where P1 = Q

PE; symbol

2P1

# 2Es sin ( =M ) N0

10.1. PROBLEM SOLUTIONS

For moderate signal-to-noise ratios and M 8, the actual error probability is very close to the upper bound. Assuming Gray encoding, the bit error probability is given in terms of symbol error probability as PE; symbol PE; bit ' log2 (M ) and the energy per bit-to-noise spectral density ratio is 1 Es Eb = N0 log2 (M ) N0 Thus, we solve 2 Q log2 (M )

# Eb 2 log2 (M ) sin ( =M ) = 10 4 N0

2 log2 (M )

Eb 10 5 sin ( =M ) = N0 2

8 1:5 10 4 , M = 8 > > < 2 10 4 , M = 16 log2 (M ) = 2:5 10 4 , M = 32 > > : 3 10 4 , M = 64

The arguments of the Q-function to give these various probabilities are found, approximately, by trial and error (either using MATLAB and a program for the Q-function or a calculator and the asymptotic expansion of the Q-function): 8 > > 3:62; M = 8 < 3:54; M = 16 Q-function argument = 3:48; M = 32 > > : 3:43; M = 64

Thus, for M = 8, we have r

Eb 2 log2 (8) sin ( =8) = 3:62 N0 r Eb 6 0:3827 = 4:17 N0 Eb N0

= 10 log10

1 6

3:62 0:3827

= 11:74 dB

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

For M = 16; we have r

Eb 2 log2 (16) sin ( =16) = 3:54 N0 r Eb 0:1951 = 3:54 8 N0

1 8

3:54 0:1951

Eb 2 log2 (32) sin ( =32) = 3:48 N r 0 Eb 10 0:098 = 3:48rQAM N0 " Eb 1 = 10 log10 N0 10

3:48 0:098

3:43 0:049

Eb N0

= 10 log10

= 16:14 dB

= 21:00 dB

= 26:11 dB

For M = 32; we have r

For M = 64; we have r

Eb 2 log2 (64) sin ( =64) = 3:43 N r 0 Eb 12 0:049 = 3:43 N0 Eb N0

= 10 log10

1 12

b. For QAM the values can be estimated from Figure 10.14(c): for M = 16 we get Eb =N0

12:4 dB to give Pb = 10 4 :

Problem 10.14 Let the coordinates of the received data vector, given signal labeled 1111 was sent, be (X; Y ) = (a + N1 ; a + N2 ) where N1 and N2 are zero-mean, uncorrelated Gaussian random variables with variances N0 =2. Since they are uncorrelated, they are also independent. Thus, P (C j I) = P (0 X 2a) P (0 Y 2a) Z 2a (x a)2 =N0 Z 2a (y a)2 =N0 e e p p dx dy = N0 N0 0 0

10.1. PROBLEM SOLUTIONS

Let

2 (x a) and v = N0

2 (y a) N0

This results in

Z p2a2 =N0 2 2 e u =2 e v =2 p p du p dv P (C j I) = p 2 2 2 2a =N0 2a2 =N0 Z p2a2 =N0 u2 =2 Z p2a2 =N0 v2 =2 e e p p = 4 du dv (by symmetry of integrands) 2 2 0 0 2 0s 132 2 2a A5 = 41 2Q @ N0 Z p2a2 =N0

Similar derivations can be carried out for the type II and type III regions.

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

Problem 10.15 The symbol error probability expression is 1 1 1 P (C j I) + P (C j II) + P (C j III) 4 2 4

Ps = 1 where

P (C j I) = [1

2Q (A)] ; A =

P (C j II) = [1

2Q (A)] [1

P (C j III) = [1

2a2 N0

Q (A)]

Thus PE = 1

= 1

1 = 1

1 M 1 M 1 M

(

2Q (A)]2 + 4

2 [1

Q (A)]2 p M ]+4

2Q (A)] [1

Q (A)]

+4 [1

(

4Q (A) +

2 [1

1 p M

4 M + 4 [1

4 M 0s Q@

2Q (A) + ] p 4Q (A)] + 4 M 2 [1

3Q (A) +

+4 [1

(

1 n M M

= 4 1

+4 [1 2Q (A)] o M Q (A) = 1 1 4 1 1

1 p M

3Q (A)]

)

]

)

; neglect Q2 (A)

Q (A)

2a2 A N0

Problem 10.16 The expression for a follows by computing the average symbol energy in terms of a. To accomplish this, the sums m X i=1

X m (m + 1) m (m + 1) (2m + 1) and i2 = 2 6 i=1

are convenient. The average symbol energy is (the overbar indicates an average over the signal constellation) ! Z Ts Bi2 2 A2i 2 Es = si (t) dt = + Ts = A2i + Bi2 T 2 2 s 0

10.1. PROBLEM SOLUTIONS

p M 1 a with M = 22n , n integer. where Ai ; Bi = a; 3a; ; value of Ai ; Bi equally likely, these averages become A2i

Assuming each

2 Bi2 = p

M =2 X

(2i

1)2 a2 (the factor of 2 accounts for negative amplitudes)

i=1

3 2 p p p M =2 M =2 M =2 2 X X X 2a 2a2 X (2i 1)2 = p 44 i2 4 i+ 15 = p M i=1 M i=1 i=1 i=1 2 p 3 p p p p M =2 M =2 + 1 M +1 M =2 M =2 + 1 p 2a2 4 = p 4 + M =25 4 6 2 M 3 2 p p p M =2 + 1 M +1 M =2 + 1 4 + 15 = a2 44 6 2 p

M =2

i p p a2 h p M +2 M +1 3 M +2 +3 3 p p a2 a2 M +3 M +2 3 M 6+3 = (M 1) = 3 3 Therefore 2a2 Es = A2i + Bi2 = (M 1) 3 and the given relationship follows. =

Problem 10.17 Use

M Es Eb PE; symbol and = log2 (M ) 2 (M 1) N0 N0 A tight bound for the symbol error probability for coherent M -ary FSK is r ! Es PE; symbol (M 1) Q N0 PE; bit =

so M PE; bit ' Q 2

Eb log2 (M ) N0

For M = 2 the exact expression is PE; binary = Q

Eb N0

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS a. Coherent FSK. Use the asymptotic expression for the Q-function or iteration on a calculator to get the following results: p For M = 2, PE; bit = Q Eb =N0 = 10 3 for Eb =N0 = 9:49 = 9:77 dB; p 2Eb =N0 = 10 3 for Eb =N0 = 5:42 = 7:34 dB; For M = 4, PE; bit = 2Q p For M = 8, PE; bit = 4Q 3Eb =N0 = 10 3 for Eb =N0 = 4:10 = 6:13 dB; p 4Eb =N0 = 10 3 for Eb =N0 = 3:35 = 5:25 dB; For M = 16, PE; bit = 8Q p 5Eb =N0 = 10 3 for Eb =N0 = 2:94 = 4:69 dB; For M = 32, PE; bit = 16Q p For M = 64, PE; bit = 32Q 6Eb =N0 = 10 3 for Eb =N0 = 2:67 = 4:27 dB.

b. Noncoherent FSK. Values can be estimated from Figure 10.15(b). For M = 2, PE; bit = 10 3 for Eb =N0 For M = 4, PE; bit = 10 3 for Eb =N0

10:8 dB; 8:4 dB; etc.

Problem 10.18 For M -ary noncoherent FSK, the symbol error probability was given in (10.68) as PE; symbol =

M X1

1 k

k=1

( 1)k+1 exp k+1

k Es k + 1 N0

The derived result, given by (10.89), is PE; symbol = 1 Take the exp

Es N0

M X1

Es N0

exp

k=0

M X1

k=1 M X1

k=1 M X1 k=1

Es (k + 1) N0

M k 1 k

( 1)k exp k+1

Es (k + 1) N0

( 1)k exp k+1

Es (k + 1) N0

Es N0

( 1)k exp k+1

Es N0

1 k

k=1

( 1)k exp k+1

inside the sum and break out the …rst term of the sum:

PE; symbol = 1

( 1)k+1 exp k+1

1 k+1

k Es k + 1 N0

exp

Es N0

10.1. PROBLEM SOLUTIONS

which shows they are equivalent. Problem 10.19 a. The bandwidth e¢ ciencies are given by 8 > 0:5 log2 (M ) ; M -PSK and M -QAM R < 2 log2 (M ) = M +3 ; coherent M -FSK B > : log2 (M ) ; noncoherent M -FSK 2M

For a data rate of 4 Mbps and a channel bandwidth of 1 MHz, the only candidate schemes are PSK, QAM or DPSK for which R = 4 = 0:5log2 (M ) or log2 (M ) = 8 or M = 28 = 256 B Eb 24 dB to give a bit error probability of 10 6 (Figure 10.14(c)). b. 256-QAM requires N 0 256-PSK and 256-DPSK require considerably more and are of the same order in complexity as 256-QAM.

c. Based purely of simplicity of implementation, the choice would probably be DPSK. However, QAM gives quite an advantage in terms of required power. It does not have constant envelope which might be another consideration.

Problem 10.20 The bandwidth e¢ ciencies of coherent and noncoherent FSK areve ( 2 log2 (M ) R M +3 ; coherent M -FSK = log (M ) 2 B 2M ; noncoherent M -FSK R 2 (M ) a. The required bandwidth e¢ ciency is B = 0:25 = 2 log M +3 . For M = [2 4 8 16 32 64] R we get B = [0:4000 0:5714 0:5455 0:4211 0:2857 0:1791] bits/s/Hz. Thus, M = 2 32 Eb will work. From Fig. 10.15(a) we get Pb = 10 4 for N = [11:5 8:6 7:2 6:4 5:7] dB 0

b. Figure 10.15(a) gives Eb =N0 = 5:7 dB for M = 32. R c. Solve B = 0:25

log2 (M ) 2M which gives M = [2 4].

d Solve 0:5 exp ( Eb =2N0 ) = 10 4 or Eb =N0 = 2ln 2 10 4 = 17:04 = 12:3 dB for M = 2: For M = 4 Fig. 10.15(b) gives Eb =N0 = 9:5 dB to give Pb = 10 4 .

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS e. In this case the cost of a noncoherent demodulator is 9:5 (over a factor of 2.4).

5:7 = 3:8 dB in power

Problem 10.21 Use Figure 10.16 noting that the abscissa is normalized baseband bandwidth. RF bandwidth is twice as large. The 90% power containment bandwidth is de…ned by the ordinate = -10 dB. (a) B90; BPSK = 1:6=Tb = 1:6Rb ; (b) B90; QPSK, OQPSK = 0:8=Tb = 0:8Rb ; (c) B90; MSK = 0:8=Tb = 0:8Rb ; (d) The di¤erence between (10.114) for BPSK and (10.118) for QAM as far as bandwidth is concerned is the factor of log2 M for the latter in the argument of the sinc-function. For 16-QAM, this is a factor of 4 so 16-QAM is a factor of 4 better than BPSK as far as bandwidth e¢ ciency is concerned. For a bit rate of 50 kbps, the resulting bandwidths are B90; BPSK = 80 kHz, B90; QPSK, OQPSK, MSK = 40 kHz, B90; 16-QAM = 20 kHz.

Problem 10.22 The baseband power spectrum is G (f ) = 2A2 Tb [log2 (M )] sinc2 [log2 (M ) Tb f ] where A2 = Es cos2 i = Es sin2 i M M 1 X 1) 1 X 2 2 (i 1) 2 2 (i = cos = sin M M M M i=1

i=1

The 10% power containment bandwidth is de…ned by (see Problem 10.21) log2 (M ) B90; MPSK = 0:8=Tb = 0:8Rb B90; MPSK

8 < 0:267Rb Hz; M = 8 0:25Rb Hz; M = 16 = 0:8Rb = log2 (M ) = : 0:16Rb Hz; M = 32

10.1. PROBLEM SOLUTIONS

Problem 10.23 a. From the given expression, B10% QAM, 4 = Rb = 20 kbps. b. B10% QAM, 16 = Rb =2 = 20 kHz; Rb = 40 kbps. c. B10% QAM, 64 = Rb =3 = 20 kHz; Rb = 60 kbps. d. B10% QAM, 256 = Rb =4 = 20 kHz; Rb = 80 kbps.

Problem 10.24 a. Find the autocorrelation function and Fourier transform it to obtain the power spectral density. By de…nition RASK ( ) = E [sASK (t) sASK (t + )] 1 2 = A E f[1 + d (t)] [1 + d (t + )] cos (! c t + ) cos [! c (t + ) + ]g 4 1 2 = A E f[1 + d (t)] [1 + d (t + )]g E fcos (! c t + ) cos [! c (t + ) + ]g 4 1 2 A E f[1 + d (t)] [1 + d (t + )]g cos (! c ) = 8 1 2 = A E f1 + d (t) + d (t + ) + d (t) d (t + )g cos (! c ) 8 1 2 A f1 + E [d (t) d (t + )]g cos (! c ) = 8 1 2 = A f1 + Rd ( )g cos (! c ) 8 The Fourier transform is SASK (f ) =

1 2 A [ (f 16

fc ) + (f + fc )] +

1 2 A [Sd (f 16

fc ) + Sd (f + fc )]

where Sd (f ) = F [Rd ( )] = T sinc2 (T f ) b. Write this as sP SK (t) = Am sin (! c t + ) + Ad (t)

m2 cos (! c t + )

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS Following a procedure similar to that used in (a), we have

RASK ( ) = E [sP SK (t) sP SK (t + )] 8 9 h i p 2 cos (! t + ) < = m sin (! t + ) + d (t) 1 m c c h i = A2 E p : m sin (! c (t + ) + ) + d (t + ) 1 m2 cos (! c (t + ) + ) ; m2 sin (! c t + ) sin (! c (t + ) + ) + cross terms in d (t) +d (t) d (t + ) 1 m2 cos (! c t + ) cos (! c (t + ) + ) (cross term average are 0) 1 2 2 1 = m A cos (! c ) + 1 m2 A2 E [d (t) d (t + )] cos (! c ) 2 2 = A2 E

(10.1)

The power spectral density is 1 SP SK (f ) = m2 A2 [ (f 4

fc ) + (f + fc )] +

1 1 4

m2 A2 [Sd (f

fc ) + Sd (f + fc )]

where Sd (f ) = T sinc2 (T f ) as in (a). Problem 10.25 To show:

t 2Tb

cos

t 2Tb

4Tb cos (2 Tb f ) 1 (4Tb f )2

Use the modulation theorem together with the Fourier transform of a rectangular pulse to get t t cos ! Tb sinc [2Tb (f 1=4Tb )] + Tb sinc [2Tb (f + 1=4Tb )] 2Tb 2Tb Rewrite the RHS as RHS = Tb

sin (2 Tb f 2 Tb f

=2) sin (2 Tb f + =2) + =2 2 Tb f + =2

Note that sin (2 Tb f

=2) =

cos (2 Tb f )

Then RHS = =

2Tb

1 1 + cos (2 Tb f ) 1 4Tb f 1 + 4Tb f 4Tb cos (2 Tb f ) 1 (4Tb f )2

10.1. PROBLEM SOLUTIONS

L = Rs /BL = 5

L = Rs /BL = 10

1.5 PLL Costas Data est.

PLL Costas Data est.

1 0

0.5

10 15 z, dB L = Rs /BL = 100

10 z, dB

0.2 PLL Costas Data est.

0.15 0.1 0.05 0

10 z, dB

Problem 10.26 This would involve an 8th power device which would produce a frequency component of 8 10 = 80 MHz at its output. This would be tracked by a PLL and then frequency divided to 10 MHz and used as the reference in a coherent demodulator for the 8-PSK signal.

Problem 10.27 A plot is given in Figure 10.3.

Problem 10.28 The dB di¤erence is 10 log10

2 2 ; SL = ; AV

= 10 log10

8 32

= 0:9691 dB

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

Problem 10.29 See the table below. As many shifts as will …t on the page are given. Shift 5 is within Hamming distance 1 HD Data Shft0 Shft1 Shft2 Shft3 Shft4 Shft5 Shft6 Shft7 Shft8 Shft9 Shft10 Shft11 Shft12 Shft13 Shft14

1 1

0 0 1

1 1 0 1

1 1 1 0 1

0 1 1 1 0 1

1 0 1 1 1 0 1

0 0 0 1 1 1 0 1

1 0 0 0 1 1 1 0 1

0 0 0 1 1 1 0 1

0 0 0 1 1 1 0

0 0 0 1 1 1

0 0 0 1 1

0 0 0 1

0 0 0

0 0

Problem 10.30 See Example 2.20 for the special case of N = 7 for more details. Problem 10.31 a. N = 2no. of stages

1 = 26

1 = 63.

b. Each clock pulse corresponds to one chip, so Tone seq period = 63 onds.

10 4 = 6:3 millisec-

c. The autocorrelation function consists of a periodic train of equilateral triangles with bases of 2 milliseconds between 1=64 and 1 unit high spaced by 6.3 milliseconds. d. The spectral lines are spaced by

f = 1/period = 1=6:3

10 3 = 158:73 Hz.

e. The weight of the impulse at f = 0 in the power spectral density is (1=63)2 = 2:52 10 4 V2 . The DC level of the m-sequence is 1=63 V. f. The …rst null in the envelope of the power spectrum is at 63 f = 10 kHz.

3 5 5 2 3 1 5 6 6 5 4 2 5 3 4

10.1. PROBLEM SOLUTIONS

Problem 10.32 a. The two boxes with arrows below are modulo-2 summed and the sum fed back to box 1. x1

x4 #

x7 #

b. Assuming an all-1s initial load the states are as follows. The output is the last column. 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0 0 0

1 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0 0

1 1 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 1 0 1 1 0 0 1 0 0 1 0 0 0

1 1 1 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 1 0 1 1 0 0 1 0 0 1 0 0

1 1 1 1 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 1 0 1 1 0 0 1 0 0 1 0

1 1 1 1 1 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 1 0 1 1 0 0 1 0 0 1

1 1 1 1 1 1 1 0 0 0 0 1 1 1 0 1 1 1 1 0 0 1 0 1 1 0 0 1 0 0

0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 0 1 1 0 1 1 0 0 0 0

0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 0 1 1 0 1 1 0 0 0

0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 0 1 1 0 1 1 0 0

0 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 0 1 1 0 1 1 0

0 0 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 0 1 1 0 1 1

0 0 0 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 0 1 1 0 1

1 0 0 0 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 1 1 1 0 1 0 1 1 0

0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1 1 0 1 0 0 0

0 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1 1 0 1 0 0

0 0 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1 1 0 1 0

0 0 0 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1 1 0 1

0 0 0 0 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1 1 0

1 0 0 0 0 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1 1

1 1 0 0 0 0 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 1 1 1 0 1

0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 0 1 0 1 0 0 0 1 1 0 1 1 1 0 0

0 0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 0 1 0 1 0 0 0 1 1 0 1 1 1 0

0 0 0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 0 1 0 1 0 0 0 1 1 0 1 1 1

0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 0 1 0 1 0 0 0 1 1 0 1 1

1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 0 1 0 1 0 0 0 1 1 0 1

0 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 0 1 0 1 0 0 0 1 1 0

1 0 1 0 0 0 0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 0 1 0 1 0 0 0 1 1

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS 0 1 1 1 1 1 1 1

0 0 1 1 1 1 1 1

0 0 0 1 1 1 1 1

1 0 0 0 1 1 1 1

1 1 0 0 0 1 1 1

1 1 1 0 0 0 1 1

0 1 1 1 0 0 0 1

c. The two boxes with arrows below are modulo-2 summed and the sum fed back to box 1.

x6 #

x7 #

d. Assuming an all-1s initial load the states are as follows. The output is the last column.

10.1. PROBLEM SOLUTIONS 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 1 1 1 0

1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 1 1 1

1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 1 1

1 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 1

1 1 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1

1 1 1 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0

1 1 1 1 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0

0 1 1 1 1 1 1 1

1 0 1 1 1 1 1 1

0 1 0 1 1 1 1 1

1 0 1 0 1 1 1 1

0 1 0 1 0 1 1 1

1 0 1 0 1 0 1 1

0 1 0 1 0 1 0 1

0 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 0 0 1 1 1 1 1 0 1 0 0 0

23 0 0 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 0 0 1 1 1 1 1 0 1 0 0

1 0 0 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 0 0 1 1 1 1 1 0 1 0

1 1 0 0 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 0 0 1 1 1 1 1 0 1

1 1 1 0 0 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 0 0 1 1 1 1 1 0

1 1 1 1 0 0 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 0 0 1 1 1 1 1

0 1 1 1 1 0 0 1 0 0 0 1 0 1 1 0 0 1 1 1 0 1 0 1 0 0 1 1 1 1

0 1 1 1 0 0 0 1 0 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 1 1 1 1 0 1

0 0 1 1 1 0 0 0 1 0 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 1 1 1 1 0

0 0 0 1 1 1 0 0 0 1 0 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 1 1 1 1

0 0 0 0 1 1 1 0 0 0 1 0 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 1 1 1

1 0 0 0 0 1 1 1 0 0 0 1 0 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 1 1

0 1 0 0 0 0 1 1 1 0 0 0 1 0 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0 1

1 0 1 0 0 0 0 1 1 1 0 0 0 1 0 0 1 0 0 1 1 0 1 1 0 1 0 1 1 0

1 0 0 0 1 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 0 1 0 1

1 1 0 0 0 1 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 0 1 0

0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 0 1

1 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 0

1 1 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1

1 1 1 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0

1 1 1 1 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

Problem 10.33

a. Lengths 11 and 13 are not included (reference should be to Table 10.12).

N =2 NA Barker code Delay = 0 Delay = 1 N =4

1 1

0 0 1

NA NU N

NU 2 -1

1 -1/2 NA

Barker code Delay = 0 Delay = 1 Delay = 2 Delay = 3 N =5

1 1

1 1 1

0 0 1 1

1 1 0 1 1

1 0 1

1 0

NA NU N

NU 4 1 0 1

1 0:25 0 0:25 NA

Barker code Delay = 0 Delay = 1 Delay = 2 Delay = 3 Delay = 5 N =7

1 1

1 1 1

1 1 1 1

0 0 1 1 1

1 1 0 1 1 1

1 0 1 1

1 0 1

1 0

NA N U N

NU 5 0 1 0 1

1 0 0:2 0 0:2 NA

Barker code Delay = 0 Delay = 1 Delay = 2 Delay = 3 Delay = 5 Delay = 6 Delay = 7

1 1

1 1 1

1 1 1 1

0 0 1 1 1

0 0 0 1 1 1

1 1 0 0 1 1 1

0 0 1 0 0 1 1 1

0 1 0 0 1 1

0 1 0 0 1

0 1 0 0

0 1 0

0 1

NU 7 0 1 0 1 0 1

N A NU N

1 0 0:143 0 0:143 0 0:143

b. The m-sequence is 111100010011010. Autocorrelation results for delays up to 9 are given.

10.1. PROBLEM SOLUTIONS

25 NA NU N

m-seq D=0 D=1 D=2 D=3 D=5 D=6 D=7 D=8

1 1

1 1 1

1 1 1 1

1 1 1 1 1

0 0 1 1 1 1

0 0 0 1 1 1 1

0 0 0 0 1 1 1 1

1 1 0 0 0 1 1 1 1

0 0 1 0 0 0 1 1 1

0 0 0 1 0 0 0 1 1

1 1 0 0 1 0 0 0 1

1 1 1 0 0 1 0 0 0

0 0 1 1 0 0 1 0 0

1 1 0 1 1 0 0 1 0

0 0 1 0 1 1 0 0 1

0 1 0 1 1 0 0

0 1 0 1 1 0

0 1 0 1 1

0 1 0 1

0 1 0

0 1

Problem 10.34 Note that the expectation of Ng is zero because n (t) has zero mean. Write the expectation of the square of Ng as an iterated integral. The expectation can be taken inside. Z Z E Ng2 = E 4 n (t) c (t) cos (! c t + ) dt n ( ) c ( ) cos (! c + ) d Tb Tb Z Z = E 4 n (t) n ( ) c (t) c ( ) cos (! c t + ) cos (! c + ) dtd Tb Tb Z Z = 4 E [n (t) n ( )] E [c (t) c ( )] E [cos (! c t + ) cos (! c + )] dtd ZTb ZTb N0 = 4 (t ) E [c (t) c ( )] E [cos (! c t + ) cos (! c + )] dtd Tb Tb 2 Z = 2N0 E c2 (t) E cos2 (! c t + ) dt = N0 Tb Tb | {z } 1 (10.2) Use has been made of the fact that

E [n (t) n ( )] =

N0 (t 2

)

to reduce the double integral to a single integral. The integral of the resulting cosine squared is Tb =2. The result for the variance (same as the mean square) is then found to be N0 Tb . Problem 10.35 The random variable to be considered is Z Tb NI = AI c (t) cos ( !t + 0

) dt

1 0 0:08 0 0:09 0:4 0:11 0:25

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

where is assumed uniform in[0; 2 ). Clearly, the expectation of NI is zero due to this random phase. The variance is the same as the mean square, which is given by Z Tb Z Tb var (NI ) = E A2I c (t) c ( ) cos ( !t + ) cos ( ! + ) dtd 0

Z Tb Z Tb

A2I E [c (t) c ( )] cos ( !t + ) cos ( ! + 0 0 Z Tb Z Tb 1 1 2 = AI T c [(t ) =Tc ] cos [ ! (t )] dtd Tc 2 0 0 Z Tb A2 T c T b 1 2 = AI T c dt = I 2 2 0 =

) dtd

Problem 10.36 a. The processing gain is Gp =

Tb Rc = or Rc = Gp Rb Tc Rb

For Gp = 1000 and a data rate of 10 kbps, Rc = 103

104 = 10 megachips per second;

b. BRF = 2Rc = 20 MHz; c. Use (10.155) and (10.158) to get the following: SNR For a JSR of 5 dB = 3:162 and SNR = 10 dB = 10, then 1+(SNR)(JSR)=G = p p 10 4 9:694 = Q (3:113) = 9:25 10 : 1+10 3:162=1000 = 9:694 and PE = Q SNR = For a JSR of 10 dB = 10 and SNR = 10 dB = 10, then 1+(SNR)(JSR)=G p p 10 3 9:091 = Q (3:015) = 1:3 10 : 1+10 10=1000 = 9:091 and PE = Q SNR For a JSR of 15 dB = 31:62 and SNR = 10 dB = 10, then 1+(SNR)(JSR)=G = p p 10 3 7:598 = Q (2:756) = 2:9 10 : 1+10 31:62=1000 = 7:598 and PE = Q SNR For a JSR of 30 dB = 1000 and SNR = 10 dB = 10, then 1+(SNR)(JSR)=G = p p 10 0:909 = Q (0:954) = 0:17: 1+10 1000=1000 = 0:909 and PE = Q

Problem 10.37 The limit of the arguement of the Q-function in (10.155) is s r Gp SNR lim = SNR!1 1 + (SNR) (JSR) =Gp JSR To get PE = Q (A) = 10 5 trial and error in MATLAB requires A = 4:264.

10.1. PROBLEM SOLUTIONS a. For JSR = 30 dB = 1000; 1:82 104 = 42:6 dB;

27 p

Gp =JSR =

Gp =1000 = 4:264 or Gp = 1000

4:2642 =

p p b. For JSR = 25 dB = 316:23; Gp =JSR = Gp =316:23 = 4:264 or Gp = 316:23 4:2642 = 5:75 103 = 37:6 dB; p p c. For JSR = 20 dB = 100; Gp =JSR = Gp =100 = 4:264 or Gp = 100 4:2642 = 1:82 103 = 32:6 dB Problem 10.38 p In the limit at SNR Eb =N0 ! 1, the argument of the Q-function for PE = Q SNR becomes r p 3N SNR = 3:08 to give PE = 10 3 K 1 Thus, 3 255 +1 (3:08)2 = 81:64

K =

Round this down to K = 81 users since the number of users must be an integer. Problem 10.39 With a propagation delay uncertainty of C = 2

2 1:5

1:5 ms the chip uncertainty becomes 10 3 s

106 chips/s

= 18; 000 half chips Equation (10.158) becomes Tacq = 17; 999 (1 + 1000Pfa ) a. Tacq = 27:197 s; b. Tacq = 18:114 s; c. Tacq = 14:112 s.

Pd 2Pd

1 Ti Pd

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

Problem 10.40 In this case T = 10 5 s= 5 so (10.161) becomes

10 5 s, Tb = 1=Rb = 1=500

103 = 2

10 6 s, and M = 4

10 5 = log2 (4) N 2 10 6 5 10 5 or N = = 12:5 subcarriers 4 10 6

We round this up to the next power of 2, or N = 16.

Problem 10.41 The equation for the e¤ective rate is

Re¤ =

Ndaata SC log2 M (Code rate) TFFT + Tguard

For example, the …rst set of values gives

Re¤ =

48 log2 2 4 s

1 = 6 Mbps 2

Taking all possible combinations of parameter values, we get the following results:

10.1. PROBLEM SOLUTIONS Mod. type BPSK BPSK BPSK BPSK BPSK BPSK QPSK QPSK QPSK QPSK QPSK QPSK 16-QAM 16-QAM 16-QAM 16-QAM 16-QAM 16-QAM 64-QAM 64-QAM 64-QAM 64-QAM 64-QAM 64-QAM

Ndaata SC 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48 48

Code rate 1/2 1/2 1/2 3/4 3/4 3/4 1/2 1/2 1/2 3/4 3/4 3/4 1/2 1/2 1/2 3/4 3/4 3/4 1/2 1/2 1/2 3/4 3/4 3/4

29 TFFT + Tguard , s 4 8 16 4 8 16 4 8 16 4 8 16 4 8 16 4 8 16 4 8 16 4 8 16

Re¤, Mbps 6 3 1.5 9 4.5 2.25 12 6 3 18 9 4.5 24 12 6 36 18 9 36 18 9 54 27 13.5

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

Problem 10.42 a. The number of users per reuse pattern remain the same at 120. For 20 dB minimum signal-to-interference ratio and a propagation power law of = 4, we have 20 = 10(4) log10

Dco dA

7:7815

or Dco dA

log10

20 + 7:7815 = 0:6945 40

Dco dA

= 100:6945 + 1 p = 5:9492 = 3N

or N = d11:798e = 12 (i = j = 2)

The e¢ ciency is v

120 12

6 MHz 10 5 = = voice circuits per base station per MHz 6 3 b. For 14 dB minimum signal-to-interference ratio and a propagation power law of we have Dco 14 = 10(4) log10 1 7:7815 dA 14 + 7:7815 Dco 1 = = 0:5445 log10 dA 40 Dco = 100:5445 + 1 dA p = 4:5038 = 3N or N = d6:7614e = 7 (i = 1; j = 2)

The e¢ ciency is v

= =

120 7

6 MHz 17 = 2:83 voice circuits per base station per MHz 6

= 4,

10.1. PROBLEM SOLUTIONS

Problem 10.43 The number of users per reuse pattern remain the same at 120. For 10 dB minimum signal-to-interference ratio and a propagation power law of = 3:5, we have 10 = 10(3:5) log10

Dco dA

7:7815

or log10

Dco dA

10 + 7:7815 = 0:508 35

Dco dA

= 100:508 + 1 p = 4:2214 = 3N

or N = d5:94e = 7 (i = 1; j = 2) The e¢ ciency is

= =

120 7

6 MHz 17 = 2:83 voice circuits per base station per MHz 6

Problem 10.44 a. Consider a hexigon made up of six equalateral triangles having sides d.. Pick one of these triangles; it consists of two right triangles each having hypoyenuse d and sides d=2 and 1=2. Thus 1 2 or

d 2

p d = 1= 3 b. This is obvious from (10:166).

= d2

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

10.2

Computer Exercises

Computer Exercise 10.1 % ce10_1.m: Bit error probabilities for coh. and noncoh. M-ary FSK % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clf z_dB = 3:.1:15; z = 10.^(z_dB/10); for j = 1:5 M=2^j; A = M/(2*(M-1)); if j == 1 Pbc = qfn(sqrt(z)); elseif j >= 2 Psc = (M-1)*qfn(sqrt(log2(M)*z)); Pbc = A*Psc; end Psnc = zeros(size(z)); for k = 1:M-1 B = (prod(1:(M-1))/(prod(1:k)*prod(1:(M-1-k))))*(-1)^(k+1)/(k+1); alpha=k*log2(M)/(k+1); Psnc = Psnc + B*exp(-alpha*z); end Pbnc = A*Psnc; subplot(1,2,1), semilogy(z_dB, Pbc, ’LineWidth’, 1.5), ... axis([min(z_dB) max(z_dB) 1e-6 1]), ylabel(’{nitP_b}’), xlabel(’{nitE_b/N}_0’), ... if j == 1 hold on; grid text(z_dB(45)+.2, Pbc(45), [’{nitM} = ’, num2str(M)]),... title(’{nitM}-ary CFSK’) else text(z_dB(45)+.2, Pbc(45), [’= ’, num2str(M)]) end subplot(1,2,2), semilogy(z_dB, Pbnc, ’LineWidth’, 1.5), ... axis([min(z_dB) max(z_dB) 1e-6 1]), ... ylabel(’{nitP_b}’), xlabel(’{nitE_b/N}_0’), ... if j == 1

10.2. COMPUTER EXERCISES

hold on; grid text(z_dB(45)+.2, Pbnc(45), [’{nitM} = ’, num2str(M)]),... title(’{nitM}-ary NCFSK’) else text(z_dB(45)+.2, Pbnc(45), [’= ’, num2str(M)]) end end % End of script …lw The curves obtained are identical to Fig. 10.15 Computer Exercise 10.2 % ce10_2.m: Out-of-band power for M-ary PSK, QPSK (OQPSK), and MSK % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clf A = char(’-’,’–’,’-.’,’:’,’–.’,’-..’); Tbf = 0:.025:7; LTbf = length(Tbf); A_BPSK = []; A_QPSK = []; A_MSK = []; for k = 1:LTbf A_BPSK(k) = quadl(’G_BPSK’, Tbf(k), 14); A_QPSK(k) = quadl(’G_QPSK’, Tbf(k), 7); A_MSK(k) = quadl(’G_MSK’, Tbf(k), 7); end OBP_BPSK = 10*log10(A_BPSK/A_BPSK(1)); OBP_QPSK = 10*log10(A_QPSK/A_QPSK(1)); OBP_MSK = 10*log10(A_MSK/A_MSK(1)); plot(Tbf, OBP_BPSK, A(1,:)), ylabel(’Relative out-of-band power, dB’), xlabel(’T_bf’),... axis([0 5 -40 0]),grid,... hold on plot(Tbf, OBP_QPSK, A(2,:)) plot(Tbf, OBP_MSK, A(3,:)) legend(’BPSK’, ’QPSK/OQPSK’, ’MSK’) % %

Function to compute area under BPSK spectrum

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

0 BPSK QPSK/OQPSK MSK

Relative out-of-band power, dB

-5 -10 -15 -20 -25 -30 -35 -40

0.5

1.5

2.5 Tbf

3.5

function y = G_BPSK(Tbf) y = 2*(sinc(Tbf)).^2; % Function to compute area under MSK spectrum % function y = G_MSK(Tbf) y = (32/pi^2)*((cos(2*pi*Tbf)).^2)./(1-(4*Tbf+eps).^2).^2; % Function to compute area under QPSK spectrum % function y = G_QPSK(Tbf) y = 4*(sinc(2*Tbf)).^2; % End of script …le A typical plot generated by the program is shown in Fig. 10.4.

4.5

10.2. COMPUTER EXERCISES Computer Exercise 10.3 % ce10_3.m: Computes bit error probability curves for jamming in DSSS. % For a desired P_E and given JSR and Gp, computes the required Eb/N0 % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clf A = char(’-’,’–’,’-.’,’:’,’–.’,’-.x’,’-.o’); Gp_dB = input(’Enter desired processing gain in dB ’); Gp = 10^(Gp_dB/10); z_dB = 0:.5:30; z = 10.^(z_dB/10); k = 1; JSR0 = []; for JSR_dB = 5:5:25; JSR0(k) = JSR_dB; JSR = 10^(JSR_dB/10); arg = z./(1+z*JSR/Gp); PE = 0.5*erfc(sqrt(arg)); semilogy(z_dB, PE, A(k+1,:)) if k == 1 hold on axis([0 30 1E-15 1]) grid on xlabel(’E_b/N_0, dB’), ylabel(’P_E’) title([’BEP for DS BPSK in jamming for proc. gain = ’, num2str(Gp_dB),’dB’]) end k = k+1; end PEG = 0.5*qfn(sqrt(2*z)); semilogy(z_dB, PEG), text(z_dB(30)-5, PEG(30), ’No jamming’) legend([’JSR = ’, num2str(JSR0(1))], [’JSR = ’, num2str(JSR0(2))], [’JSR = ’, num2str(JSR0(3))], [’JSR = ’, num2str(JSR0(4))], [’JSR = ’, num2str(JSR0(5))], 3) disp(’Strike ENTER to continue’); pause PE0 = input(’Enter desired value of P_E ’); JSR_dB_0 = input(’Enter given value of JSR in dB ’);

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS Gp_dB_0 = input(’Enter given value of processing gain in dB ’); JSR0 = 10^(JSR_dB_0/10); Gp0 = 10^(Gp_dB_0/10); PELIM = 0.5*erfc(sqrt(Gp0/JSR0)); disp(’’) disp(’In…nite Eb/N0 BEP limit’) disp(PELIM) if PELIM >= PE0 disp(’For given choices of G_p, JSR, & desired P_E, a solution is not possible ’); Gp0_over_JSR0 = (er…nv(1-2*PE0))^2; Gp0_over_JSR0_dB = 10*log10(Gp0_over_JSR0); disp(’The minimum required value of G_p over JSR in dB is:’) disp(Gp0_over_JSR0_dB) else arg0 = (er…nv(1-2*PE0))^2; SNR0 = 1/(1/arg0 - JSR0/Gp0); SNR0_dB = 10*log10(SNR0); end disp(’’) disp(’To give BEP of:’) disp(PE0) disp(’Requires GP, JSR, and Eb/N0 in dB of:’) disp([Gp_dB JSR_dB SNR0_dB])n % End of script …le

A typical run follows, with both a plot given and a speci…c output generated: >> ce10_3 Enter desired processing gain in dB: 30 Strike ENTER to continue Enter desired value of P_E: 1e-3 Enter given value of JSR in dB: 20 Enter given value of processing gain in dB: 30 In…nite Eb/N0 BEP limit 3.8721e-006 To give BEP of: 1.0000e-003 Requires GP, JSR, and Eb/N0 in dB of: 30.0000 25.0000 9.6085

10.2. COMPUTER EXERCISES

BEP for DS BPSK in jamming for processing gain = 30 dB

-5

-10

JSR = 5 JSR = 10 JSR = 15 JSR = 20 JSR = 25

-15

No jamming 10

15 Eb/N0, dB

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

Computer Exercise 10.4 % ce10_4.m: plots waveforms or spectra for gmsk and msk % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % A = char(’-’,’-.’,’–’,’:.’); samp_bit = input(’Enter number of samples per bit used in simulation: ’); N_bits = input(’Enter total number of bits: ’); GMSK = input(’Enter 1 for GMSK; 0 for normal MSK: ’); if GMSK == 1 BT_bit = input(’Enter vector of bandwidth X bit periods: ’); N_samp = 50; end I_plot = input(’Enter 1 to plot waveforms; 2 to plot spectra and out-of-band power ’); if I_plot == 1 f0 = 1; elseif I_plot == 2 f0 = 0; end clf kp = pi/2; T_bit = 1; LB = length(BT_bit); for k = 1:LB B = BT_bit(k)/T_bit; alpha = 1.1774/B; % Note that B is two-sided bandwidth del_t = T_bit/samp_bit; fs = 1/del_t; data = 0.5*(sign(rand(1,N_bits)-.5)+1); s = 2*data - 1; L = length(s); t=0:del_t:L*T_bit-del_t; s_t = s(ones(samp_bit,1),:); % Build array whose columns are samp_bit long s_t = s_t(:)’; % Convert matrix where bit samples occupy columns to vector L_s_t = length(s_t); tp=0:del_t:N_samp*T_bit-del_t; L_t = length(t); L_tp=length(tp); t_max = max(tp);

10.2. COMPUTER EXERCISES if GMSK == 1 hG = (sqrt(pi)/alpha)*exp(-pi^2*(tp-t_max/2).^2/alpha^2); freq1 = kp*del_t*conv(hG,s_t); L_hG = length(hG); freq = freq1(L_hG/2:length(freq1)-L_hG/2); elseif GMSK == 0 freq = kp*s_t; end phase = del_t*cumsum(freq); if GMSK == 0 y_mod = exp(j*2*pi*f0*t+j*phase); elseif GMSK == 1 y_mod = exp(j*2*pi*f0*t+j*phase); end if I_plot == 1 subplot(3,1,1),plot(t,s_t),axis([min(t), max(t), -1.2, 1.2]), ylabel(’Data’) if GMSK == 0 title(’MSK waveforms’) elseif GMSK == 1 title([’GMSK for BT_b = ’, num2str(BT_bit(k))]) end subplot(3,1,2),plot(t,phase), ylabel(’Excess phase, rad.’) for nn = 1:11 if nn == 1 hold on end subplot(3,1,2),plot(t,-nn*(pi/2)*ones(size(t)),’m’) subplot(3,1,2),plot(t,nn*(pi/2)*ones(size(t)),’m’) end subplot(3,1,3),plot(t,real(y_mod)), xlabel(’t’), ylabel(’Modulated signal’) elseif I_plot == 2 Z=PSD(real(y_mod),2048,fs); % MATLAB supplied function ZN = Z/max(Z); ET = sum(ZN); OBP = 1-cumsum(ZN)/ET; LZ = length(Z); del_FR = fs/2/LZ; FR = [0:del_FR:fs/2-del_FR]; subplot(1,2,1),semilogy(FR,ZN,A(k,:)), axis([0 3 10^(-8) 1]) if k == 1

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS hold on grid ylabel(’PSD, J/Hz’), xlabel(’fT_b_i_t’) title(’Power spectra and out-of-band power for GMSK’) end subplot(1,2,2),semilogy(FR,OBP,A(k,:)), axis([0 3 10^(-8) 1]) if k == 1 hold on grid ylabel(’Fraction of out-of-band power’), xlabel(’fT_b_i_t’) end if LB == 1 legend([’BT_b_i_t = ’,num2str(BT_bit(1))],1) elseif LB == 2 legend([’BT_b_i_t = ’,num2str(BT_bit(1))], [’BT_b_i_t = ’,num2str(BT_bit(2))],1) elseif LB == 3 legend([’BT_b_i_t = ’,num2str(BT_bit(1))], [’BT_b_i_t = ’,num2str(BT_bit(2))], [’BT_b_i_t = ’,num2str(BT_bit(3))],1) elseif LB == 4 legend([’BT_b_i_t = ’,num2str(BT_bit(1))], [’BT_b_i_t = ’,num2str(BT_bit(2))], [’BT_b_i_t = ’,num2str(BT_bit(3))], [’BT_b_i_t = ’,num2str(BT_bit(4))],1) end end end % End of script …le Two typical runs follow - one for plotting waveforms and one for plotting spectra. >> ce10_4 Enter number of samples per bit used in simulation: 20 Enter total number of bits: 50 Enter 1 for GMSK; 0 for normal MSK: 1 Enter vector of bandwidth X bit periods: .5 Enter 1 to plot waveforms; 2 to plot spectra and out-of-band power: 1 >> ce10_4 Enter number of samples per bit used in simulation: 10

10.2. COMPUTER EXERCISES

GMSK for BTb = 0.5

Data

1 0

Modulated signal

Excess phase, rad.

-1 0

25 t

-20 1

-1

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS Power0 spectra and out-of-band power for GMSK 0 10 10 -1

-2

Fraction of out-of-band power

-3

10 PSD, J/Hz

BTbit = 1 BTbit = 1.5

-2

-4

-5

-6

-7

-3

-4

-5

-6

-7

-8

BTbit = 0.5

-1

-8

2 fTbit

Enter total number of bits: 2000 Enter 1 for GMSK; 0 for normal MSK: 1 Enter vector of bandwidth X bit periods: [.5 1 1.5] Enter 1 to plot waveforms; 2 to plot spectra and out-of-band power: 2

Computer Exercise 10.5 % ce10_5.m: Mimo Alamouti system with receiver diversity % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clf; clear all sigma_n = 1; N_sim = input(’No. of bits in simulation => ’); SNRdB = 0:5:25; SNR = 10.^(SNRdB/10); k_max = length(SNRdB); for k = 1:k_max

10.2. COMPUTER EXERCISES sigma_n = 1/sqrt(SNR(k)); h0 = randn(1, N_sim) + j*randn(1, N_sim); h1 = randn(1, N_sim) + j*randn(1, N_sim); h2 = randn(1, N_sim) + j*randn(1, N_sim); h3 = randn(1, N_sim) + j*randn(1, N_sim); s0 = sign(rand(1, N_sim) - 0.5); data0 = (s0 + 1)/2; s1 = sign(rand(1, N_sim) - 0.5); data1 = (s1 + 1)/2; n0 = sigma_n*(randn(1, N_sim)+j*randn(1, N_sim)); n1 = sigma_n*(randn(1, N_sim)+j*randn(1, N_sim)); n2 = sigma_n*(randn(1, N_sim)+j*randn(1, N_sim)); n3 = sigma_n*(randn(1, N_sim)+j*randn(1, N_sim)); r0 = h0.*s0 +h1.*s1 + n0; r1 = -h0.*conj(s1) + h1.*conj(s0) + n1; r2 = h2.*s0 +h3.*s1 + n2; r3 = -h2.*conj(s1) + h3.*conj(s0) + n3; y0 = conj(h0).*r0 +h1.*conj(r1); y1 = conj(h1).*r0 - h0.*conj(r1); y2 = conj(h2).*r2 +h3.*conj(r3); y3 = conj(h3).*r2 - h2.*conj(r3); s0_hat2 = real(y0+y2); s1_hat2 = real(y1+y3); s0_hat1 = real(y0); s1_hat1 = real(y1); s0_out1 = (sign(s0_hat1) + 1)/2; s1_out1 = (sign(s1_hat1) + 1)/2; error01 = abs(s0_out1 - data0); error11 = abs(s1_out1 - data1); error_tot1 = sum(error01 + error11); Perror1(k) = error_tot1/(2*N_sim); s0_out2 = (sign(s0_hat2) + 1)/2; s1_out2 = (sign(s1_hat2) + 1)/2; error02 = abs(s0_out2 - data0); error12 = abs(s1_out2 - data1); error_tot2 = sum(error02 + error12); Perror2(k) = error_tot2/(2*N_sim); end SNRdB Perror1

CHAPTER 10. ADVANCED DATA COMMUNICATIONS TOPICS

Perror2 Perror_1rec_tr = 0.5*(1 - sqrt(SNR./(1 + SNR))); % semilogy(SNRdB, Perror1, ’LineWidth’, 1.5, SNRdB, Perror2, ’-.’, SNRdB, Perror_1rec_tr, ’–’); xlabel(’SNR; dB’),... semilogy(SNRdB, Perror1, ’LineWidth’, 1.5), xlabel(’SNR; dB’),... ylabel(’Bit error probability’), grid, hold on, ... title([’Alamouti-type diversity; ’, num2str(N_sim), ’bits simulated’]) semilogy(SNRdB, Perror2, ’-.’, ’LineWidth’, 1.5) semilogy(SNRdB, Perror_1rec_tr, ’–’, ’LineWidth’, 1.5) legend(’2 trans, 1 rec’, ’2 trans, 2 rec’, ’1 trans, 1 rec’) A run produces a plot like Figure 10.31.

Chapter 11

Optimum Receivers and Signal Space Concepts Problem 11.1 a. Given H1 , Z = N , so fZ (zjH1 ) = fN (n) jz=n = 10e 10z u (z) Given H2 , Z = S + N , where S and N are independent. Thus, the resulting pdf under H2 is the convolution of the separate pdfs of S and N : Z 1 fZ (zjH2 ) = fS (z ) fN ( ) d 1 Z 1 = 2e 2(z ) u (z ) 10e 10 u ( ) d 1 Z z 2z = 20e e 8 d ,z 0 0

1 8 e 8

= 20e 2z

= 2:5e 2z 1

e 8z u (z)

= 2:5 e 2z

e 10z u (z)

which is the result given in the problem statement. b. The likelihood ratio is (Z) =

2:5 e 2Z e 10Z fZ (ZjH2 ) = ; Z fZ (ZjH1 ) 10e 10Z

= 0:25 e8Z

1 , Z 1

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Note that an uppercase Z is used because we substitute data to carry out the test. c. The threshold is =

(1=3) (7 c11 ) = c22 ) (2=3) (7

Pr (H1 true) (c21 Pr (H2 true) (c12

0) 1 = 2 0)

d. The likelihood ratio test becomes

0:25 e8Z

H2 > 1 < 2 H1

This can be reduced to H2 > ln (4=2 + 1) Z = 0:1373 < 8 H1 e. The probability of detection is PD = 1

PM =

Z 1

2:5 e 2z

e 10z dz

0:1373

1 1 2z 1 = 2:5 e + e 10z 2 10 0:1373 1 0:2746 1 1:373 = 2:5 e e = 0:6991 2 10

The probability of false alarm is Z 1 PF = 10e 10z dz = 0:1373

e 10z 0:1373 = e 1:373 = 0:2533

Therefore, the risk is Risk

= Pr (H1 ) c21 + Pr (H2 ) c22 + Pr (H2 ) (c12 = =

c22 ) PM

Pr (H1 ) (c21 c11 ) (1 PF ) 1 2 2 1 7+ 0 + (7 0) (1 0:6991) (7 3 3 3 3 7 14 7 + 0:3009 0:7467 = 1:9952 3 3 3

0) (1

0:2533)

3 f. Consider the threshold set at . Then PF = e 10 to give PF

10 3 )

0:3 ln (10) = 0:6908

10 3 . Also, from part (e), Z 1 PD = 2:5 e 2z = 1:25e 2

e 10z dz

0:25e 10 ;

For values of 0, PD is a monotonically decreasing function with maximum value of 1 at = 0. For = 0:6908, PD = 0:3137. g. From part (f), PF = e 10 and PD = 1:25e 2 A plot of PD versus PF for various values of of the test.

0:25e 10

constitutes the operating characteristic

Problem 11.2 a. The likelihood ratio is e jZj =2 p = (Z) = e Z 2 =2 = 2

eZ =2 jZj

b. The decision regions are determined by the set of inequalities r H2 e jZj =2 2 p = (Z) = eZ =2 jZj ? 2 2 e Z =2 = 2 H1 or Z 2 =2

jZj ? ln H1

1 ln = 2 2

Thus, solve the equation Z 2 =2

jZj =

for Z > 0 and Z < 0 for the decision boundary points: Z > 0 : Z 2 =2 2

Z < 0 : Z =2 + Z

= 0 ) Z1;2 = 1 = 0 ) Z3;4 =

1+2 p 1+2

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS For example, for

= 0:1 p = 1 1 + 0:2 = 1 p = 1 1 + 0:2 =

Z1;2 Z3;4

1:0954 = 2.0954 ; 1

0:0954; Z > 0

1:0954 = 0:0954; -2.0954 ; Z < 0

So the decision regions in this particular case are R1 :

1<Z<

R2 :

2:0954

2:0954; 2:0954 < Z < 1 2:0954

Note that the equality sign can be associated with either region. Problem 11.3 Under both hypotheses, the statistic Z is zero mean and Gaussian. Under hypothesis H1 (noise alone) Z is zero-mean Gaussian with variance 2n and under hypothesis H2 (signal plus noise) it is zero-mean Gaussian with variance 2s + 2n : Thus, the likelihood ratio test is s 2 exp Z 2 =2 2s + 2n H2 Pr (H1 ) (c21 c11 ) n (Z) = ? = 2+ 2 2 =2 2 ] exp [ Z Pr (H2 ) (c12 c22 ) H1 s n n or Z 2 =2

exp

2 s +

or Z2

2 n

2+ s

= Z2

2 n

or H2

Z2 ?

2 n

2+ s 2 s

2 n

+ Z 2 =2 2n ?

2 s 2 ( 2+ n s 2 n

H2 4 H1 3

2+ 2 s n 2 n

H2 2) n

? 2 ln ( ) + ln

2 = 3; s

2+ 2 s n 2 n

2 ln ( ) + ln

For the rest of the problem, we assume that Z2 ?

2 = 1 so the LRT becomes n

[2 ln ( ) + ln (4)]

q 4 The boundaries of the decision regions are at Z1;2 = 3 [2 ln ( ) + ln (4)] which results in the decision regions r r 4 4 R1 : [2 ln ( ) + ln (4)] Z [2 ln ( ) + ln (4)] 3 3

5 and R2 :

1<Z<

4 [2 ln ( ) + ln (4)]; 3

4 [2 ln ( ) + ln (4)] < Z < 1 3

a. For c11 =qc22 = 0, c21 = c12 , and Pr (H1 ) = Pr (H2 ) = 1=2 we have Z1;2 =

4 3 ln (4) =

= 1 so that

1:3596.

b. For c11 = c22 = 0, c21 = c12 , p0 = Pr (H1 )q = 1=4, and q0 = 1 we have

1 4

4 3

p0 = Pr (H2 ) = 3=4 4 3 [2 ln (1=3) + ln (4)] which is imaginary

= 13 so that Z1;2 =

(i.e., Z 2 is being compared with a negative number in the LRT) which means that the decision is in favor of H2 (signal plus noise).

= 12 so c. For c11 = c22 = 0, c21 = c12 =2, p0 = Pr (H1 ) = Pr (H2 ) = 1=2 we have = cc21 12 q 4 that Z1;2 = 3 [2 ln (1=2) + ln (4)] = 0 so the decision is in favor of H2 .

d. For c11 = c22 = 0, c21 = 2c12 , p0 = Pr (H1 ) = Pr (H2 ) = 1=2 we have q 4 that Z1;2 = 3 [2 ln (2) + ln (4)] = 1:9227.

= cc21 = 2 so 12

Problem 11.4 2 = 9; n

For

2 = 16 the LRT is s H2 9 (25)

Z2 ?

2 ln ( ) + ln

25 9

= 28:125 ln ( ) + 14:367

p The boundaries of the decision regions are at Z1;2 = 28:125 ln ( ) + 14:367 which results in the decision regions p p R1 : 28:125 ln ( ) + 14:367 Z 28:125 ln ( ) + 14:367 and

R2 :

1<Z<

28:125 ln ( ) + 14:367;

28:125 ln ( ) + 14:367 < Z < 1

a. For c11 =pc22 = 0, c21 = c12 , and Pr (H1 ) = Pr (H2 ) = 1=2 we have Z1;2 = 14:367 = 3:7904 giving decision regions of R1 :

3:7904

R2 :

1<Z<

3:7904 3:7904; 3:7904 < Z < 1:

= 1 so that

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS The probability of false alarm is the probability that a decision is made in favor of signal plus noise when noise alone is present, and is given in this case by Z 3:7904 Z 3:7904 exp z 2 =2 9 exp z 2 =2 9 z p p PF = dz = 2 dz; u = 3 2 9 2 9 3:7904 "0 Z # Z 3:7904=3 1 exp u2 =2 exp u2 =2 1 p p du = 2 du = 2 2 2 2 3:7904=3 0 = 1

2Q (1:2635) = 0:7936

The probability of detection is the probability that a decision is made in favor of signal plus noise when signal plus noise present, and is given in this case by Z 3:7904 Z 1 exp z 2 =2 25 exp z 2 =2 25 p p PD = dz + dz 2 25 2 25 1 3:7904 Z 1 exp z 2 =2 25 z p = 2 dz; u = 5 2 25 3:7904 Z 1 2 exp u =2 p du = 2 2 3:7904=5 = 2Q (0:7581) = 0:4484 The risk is Risk

= Pr (H1 ) c21 + Pr (H2 ) c22 + Pr (H2 ) (c12 Pr (H1 ) (c21 c11 ) (1 1 1 1 = 1+ 0+ 1 2 2 2 = 0:6726

c22 ) PM

PF ) (1

0:4484)

1 2

0:7936)

b. For c11 = c22 = p0, c21 = c12 = 1, Pr (H1 ) = 1=4; and Pr (H2 ) = 3=42we have = 1=3 so that Z1;2 = 28:125 ln (1=3) + 14:367 = imaginary which says Z is being compared with a negative number in the LRT which means that the decision is in favor of H2 (signal plus noise). The probability of false alarm is the probability that a decision is made in favor of signal plus noise when noise alone is present, and is 0 in this case. Since the decision is always in favor of H2 the probability of detection is 1. The risk is Risk

= Pr (H1 ) c21 + Pr (H2 ) c22 + Pr (H2 ) (c12 Pr (H1 ) (c21 c11 ) (1 1 3 3 1+ 0+ 1 = 4 4 4 = 0

c22 ) PM

PF ) (1

1 4

7 c. For c11 = c22 = p 0, c21 = c12 =2 = 1=2, and Pr (H1 ) = Pr (H2 ) = 1=2 we have = 1=2 so that Z1;2 = 28:125 ln (1=2) + 14:367 =imaginary giving which gives PF = 0 and PD = 1. The risk is Risk

= Pr (H1 ) c21 + Pr (H2 ) c22 + Pr (H2 ) (c12 Pr (H1 ) (c21 c11 ) (1 1 1 1 1 = + 0+ 1 2 2 2 2 = 0

c22 ) PM

PF ) (1

1 2

d. For c11 =pc22 = 0, c21 = 2c12 = 2, and Pr (H1 ) = Pr (H2 ) = 1=2 we have 28:125 ln (2) + 14:367 = 5:8191 giving decision regions of Z1;2 = R1 :

5:8191

R2 :

1<Z<

= 2 so that

5:8191

5:8191; 5:8191 < Z < 1:

The probability of false alarm is the probability that a decision is made in favor of signal plus noise when noise alone is present, and is given in this case by Z 5:8191 Z 5:8191 exp z 2 =2 9 exp z 2 =2 9 z p p dz = 2 dz; u = PF = 3 2 9 2 9 5:8191 "0 Z # Z 5:8191=3 2 2 1 exp u =2 exp u =2 1 p p = 2 du = 2 du 2 2 2 0 5:8191=3 = 1

2Q (1:9397) = 0:9476

The probability of detection is the probability that a decision is made in favor of signal plus noise when signal plus noise present, and is given in this case by Z 5:8191 Z 1 exp z 2 =2 25 exp z 2 =2 25 p p PD = dz + dz 2 25 2 25 1 5:8191 Z 1 exp z 2 =2 25 z p = 2 dz; u = 5 2 25 5:8191 Z 1 2 exp u =2 p = 2 du 2 5:8191=5 = 2Q (1:1638) = 0:2445 The risk is Risk

= Pr (H1 ) c21 + Pr (H2 ) c22 + Pr (H2 ) (c12 c22 ) PM Pr (H1 ) (c21 1 1 1 1 = 2+ 0+ 1 (1 0:2445) 2 (1 0:9476) 2 2 2 2 = 1:3254

c11 ) (1

PF )

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS

Problem 11.5 De…ne A = (a1 ; a2; a3 ) and B = (b1 ; b2 ; b3 ). De…ne scalar multiplication by (a1 ; a2; a3 ) = ( a1 ; a2; a3 ) and vector addition by (a1 ; a2; a3 ) +(b1 ; b2 ; b3 ) = (a1 + b1 ; a2 + b2 ; a3 + b3 ). The properties listed under ”Structure of Signal Space” in the text then become: 1.

1A +

2 B = ( 1 a1 +

(A + B) =

2 b1 ;

1 a2 +

2 b2 ;

1 a3 +

2 b3 )

R3 ;

(a1 + b1; a2 + b2 ; a3 + b3 )

= ( a1 + b1; a2 + b2 ; a3 + b3 ) = ( a1 ; a2; a3 ) + ( b1; b2 ; b3 ) = A + B R3 ; 3.

1 ( 2 A) = ( 1 2 A)

R3 (follows by writing out in component form);

4. 1 A = A (follows by writing out in component form); 5. The unique element 0 is (0; 0; 0) so that A + 0 = A; 6.

A = ( a1 ; a2;

a3 ) so that A + ( A) = 0:

Problem 11.6 p p p p p p a. jAj = A A = 12 + 32 + 22 = 14; jBj = B B = 52 + 12 + 32 = 35; AB p p1+2 3 = 0:6325. cos = jAjjBj = 1 5+3 14 35 p p p p p p b. jAj = A A = 62 + 22 + 42 = 56; jBj = B B = 22 + 22 + 22 = 12; AB p p2+4 2 = 0:9258. cos = jAjjBj = 6 2+2 56 12 p p p p p p c. jAj = A A = 42 + 32 + 12 = 26; jBj = B B = 32 + 42 + 52 = 50; AB p p4+1 5 = 0:8043. cos = jAjjBj = 4 3+3 26 50 p

32 + 32 + 22 =

d. jAj = A A = 22; jBj = B B = p 3 ( 1)+3 ( 2)+2 3 AB p p 14; cos = jAjjBj = = 0:1709. 22 14

( 1)2 + ( 2)2 + 32 =

9 Problem 11.7 Consider (x; y) = lim

T !1

Then (x; y) = lim

T !1

Also ( x; y) = lim

T !1

Z T

y (t) x (t) dt =

Z T

x (t) y (t) dt

lim

T !1

x (t) y (t) dt =

Z T

lim

T !1

x (t) y (t) dt

= (x; y)

Z T

x (t) y (t) dt =

(x; y)

and (x + y; z) =

lim

T !1

lim

T !1

Z T

[x (t) + y (t)] z (t) dt

T Z T

x (t) z (t) dt + lim

T !1

= (x; z) + (y; z) Finally (x; x) = lim

T !1

Z T

x (t) x (t) dt = lim

T !1

Z T

y (t) z (t) dt

jx (t)j2 dt

The scalar product for power signals is considered in the same manner. Problem 11.8 a. This scalar product is (x1 ; x2 ) = lim

Z T

T !1 0

1 2

2e 4t dt =

b. Use the energy signal scalar product de…nition as in (a): (x1 ; x2 ) = lim

Z T

T !1 0

e (7 j2)t dt =

1 7

c. Use the power signal product de…nition: 1 (x1 ; x2 ) = lim T !1 2T

Z T

cos (2 t) cos (4 t) dt = 0

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS d. Use the power signal scalar product de…nition: 1 T !1 2T

(x1 ; x2 ) = lim

Z T

5 cos (2 t) dt = 0

Problem 11.9 Since the signals are assumed real, kx1 + x2 k

= =

lim

T !1

lim

T !1

Z T

[x1 (t) + x2 (t)]2 dt

Z T

x21 (t) dt + 2

lim

T !1

= kx1 k2 + 2 (x1 ; x2 ) + kx2 k2

Z T

x1 (t) x2 (t) dt + lim

T !1

Z T

x22 dt

From this it is seen that kx1 + x2 k2 = kx1 k2 + kx2 k2 if and only if (x1 ; x2 ) = 0.

Problem 11.10 By straight forward integration, it follows that

kx2 k2 kx3 k2

Z 1

p kx1 k = 2 1 r Z 1 2 2 2 = t dt = or kx2 k = 3 3 1 r Z 1 8 8 2 = (1 + t) dt = or kx3 k = 3 3 1

kx1 k2 =

12 dt = 2 or

Also (x1 ; x2 ) = 0 and (x3 ; x1 ) = 2 Since (x1 ; x2 ) = 0, they are orthogonal. Choose them as basis functions (not normalized). Clearly, x3 is their vector sum.

11 Problem 11.11 It follows that X

kx1 k2 =

an n ;

bm m

an bm ( n ;

n m N X

n=1 N X

and similarly kx2 k2 =

n=1

jan j2 because ( n ;

m) =

jbn j2

Also, in a similar fashion, (x1 ; x2 ) =

N X

an bn

n=1

Thus we must show that N X

N X

an bn

n=1

jan j

N X

n=1

jbn j2

Writing the both sides out we get XX n

an am bn bm

XX n

an an bm bm

This is equivalent to

XX n

(2an an bm bm

2an am bn bm )

2an am bn bm + am am bn bn )

(an an bm bm

The inequality is therefore equivalent to XX m

jan bm

am bn j2

which is seen to be true because each term in the sum is nonnegative.

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS

Problem 11.12 p p p p p p 2 + 12 + 32 = a. jAj = A A = 12 + 32 + 22 = 14; jBj = B B = 5 35; p p A B = 1 5 + 3 1 + 2 3 = 14; jA Bj = 14 jAj jBj = 14 35 = 22:1359 which is true. p p p p p p 2 + 22 + 22 = 12; b. jAj = A A = 62 + 22 + 42 = 56; jBj = B B =p 2p A B = 6 2 + 2 2 + 4 2 = 24; jA Bj = 24 jAj jBj = 56 12 = 25:923 which is true. p p p p p p 2 + 42 + 52 = c. jAj = A A = 42 + 32 + 12 = 26; jBj = B B = 3 50; p p A B = 4 3 + 3 4 + 1 5 = 29; jA Bj = 29 jAj jBj = 26 50 = 36:0555 which is true. q p p p p d. jAj = A A = 32 + 32 + 22 = 22; jBj = B B = ( 1)2 + ( 2)2 + 32 = p p p 14; A B = 3 ( 1) + 3 ( 2) + 2 3 = 3; jA Bj = 3 jAj jBj = 22 14 = 17:5499 which is true.

Problem 11.13 a. A set of normalized basis functions is 1 p s1 (t) 2 r 2 1 s2 (t) s1 (t) 2 (t) = 3 2 p 2 2 s1 (t) s2 (t) 3 s3 (t) 3 (t) = 3 3 1 (t)

b. Evaluate (s1 ;

1 ), (s2 ;

s1 =

1 ), and (s3 ;

1 2 1 ; s2 = p 2

1 ) to get

3 2

2 ; s3 =

2 1+

2 ( + 3 2

Note that a basis function set could have been obtained by inspection. For example, three unit-height, unit-width, nonoverlapping pulses spanning the interval could have been used.

13 Problem 11.14 Normalize the …rst vector to form the …rst orthonormal vector: 1

x1 3bi + 2b j b k =q jx1 j 32 + 22 + ( 1)2

1 p 3bi + 2b j b k 14 = 0:8018bi + 0:5345b j =

0:2673b k

Take the second vector and subtract its component along 1 p 3bi + 2b j 14 2:6429bi + 4:5714b j + 1:2143b k 2bi + 5b j+b k

v2 = =

b k

1 2bi + 5b j+b k p 3bi + 2b j 14

b k

Normalize v2 to form the second orthonormal vector:

= =

2:6429bi + 4:5714b j + 1:2143b k

( 2:6429)2 + (4:5714)2 + (1:2143)2

0:4878bi + 0:8437b j + 0:2241b k

Take the third vector and subtract its components along v3 =

6bi

2b j + 7b k

0:8018bi + 0:5345b j

0:2673b k

0:4878bi + 0:8437b j + 0:2241b k

= 3:0146bi

Normalize:

0:4307b j + 8:1825b k

6bi

2b j + 7b k

1 and

0:8018bi + 0:5345b j

0:2673b k

0:4878bi + 0:8437b j + 0:2241b k

3:0146bi 0:4307b j + 8:1825b k p 3:01462 + 0:43072 + 8:18252 = 0:3453bi 0:0493b j + 0:9372b k =

Finally, take the fourth vector and subtract components of the other three along it:

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS

v4 =

3bi + 8b j

3b k

0:2673b k

3bi + 8b j

3b k

0:8018bi + 0:5345b j

0:0493b j + 0:9372b k

3bi + 8b j

3b k

0:3453bi

0:8018bi + 0:5345b j

0:4878bi + 0:8437b j + 0:2241b k

0:3453bi

3bi + 8b j

= 0bi + 0b j + 0b k (actually of the order of 10 14 )

3b k

0:2673b k

0:4878bi + 0:8437b j + 0:2241b k 0:0493b j + 0:9372b k

Therefore there are only three orthornormal vectors. Problem 11.15 a. A basis set is r

2 cos (2 fc t) and T r 2 sin (2 fc t) for 0 2 (t) = T 1 (t)

b. The coordinates of the signal vectors are xi =

Z T

si (t)

1 (t) dt =

si (t)

2 (t) dt =

and yi =

Z T 0

E cos

E sin

i 4

where E = A2 T . Thus si (t) =

E cos

i 4

1 (t)

E sin

i 4

2 (t) ; i = 0; 1; 2; 3; 4; 5; 6; 7

p The signal point for i = 0 is at E; 0 with the rest at multiples of =4 radians on p a circle of radius E centered at the origen.

15 Problem 11.16

a. Normalize x1 (t) to produce 2

jjx1 jj

1 (t):

Z 1

t 2

dt =

Z 1

e 2t dt =

1 2t 1 1 = e 0 2 2

x1 (t) p p = 2e t u (t) 1 (t) = 1= 2

)

Take the second signal and subtract its component along 2t

v2 (t) = e

u (t)

= e 2t u (t)

Z 1p

p 2e t e 2t dt 2e t u (t) #1 "0 p p 2 3t e 2e t u (t) 3 0

= e 2t u (t) 2 (t):

Normalize v2 (t) to produce 2

jjv2 jj

Z 1

2 t e u (t) 3

2 t e 3

dt =

Z 1

e 4t

0 1

4 3t 4 2t e + e dt 3 9

1 1 4t 4 3t 2 2t = e + e e 4 9 9 36 0 v2 (t) = 6e 2t 4e t u (t) 2 (t) = 1=6

= )

Take the third signal and subtract its components along

v3 (t) = e

u (t)

Z 1

Z 1p

e 3t

p 2e t e 3t dt 2e t u (t)

6e 2t

4e t e 3t dt 6e 2t u (t)

1 and

3 6 2t e + e t u (t) 5 10

4e t u (t)

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Normalize v3 (t) to produce jjv3 jj2

= =

Z 1

3 (t):

e 3t

Z0 1

e 6t

1 600

)

3 (t) =

2 6 2t 3 e + e t dt 5 10 12 5t 51 4t 18 3t 9 e + e e + e 2t dt 5 25 25 100

p v3 (t) p = 6 10e 3t 1=10 6

12e 2t + 3e t u (t)

b. There does not appear to be a general pattern developing. Problem 11.17 a. Normalize x1 (t) to produce 2

jjx1 jj

= )

1 (t):

Z 1

1 31 2 t 1= 3 3 1 r x1 (t) 3 = t; 1 (t) = p 2 2=3 t2 dt =

Take the second signal and subtract its component along 1 : r Z 1r 3 2 3 2 v2 (t) = t tt dt t 2 2 1 = t

= t2 ; Normalize v2 (t) to produce 2

jjv2 jj

= = )

3 t4 2 4

t 1

dt =

Z 1

2 (t):

Z 1

2 2

1 1 5 t

t4 dt

= 1

2 5

x2 (t) = 2 (t) = p 2=5

5 2 t ; 2

17 Take the third signal and subtract its components along 1 and 2 : r r Z 1r Z 1r 3 3 3 5 2 3 5 2 3 v3 (t) = t tt dt t t t dt t 2 2 2 2 1 1 1

= t

3 t5 t 2 5 1 3 t; 1 t 5

= t3 Normalize v3 (t) to produce jjv3 jj2

3 t 5

1 7 t

6 t5

dt =

Z 1

1 9 t3

25 3

2 7

7 2

5 3 t 2

v3 (t) = 3 (t) = p 8=175

)

t2 1

3 (t):

Z 1

5 t6 2 6

6 4 9 t + t2 dt 5 25 62 9 2 8 + = 5 5 25 3 175 3 t ; 2

Take the fourth signal and subtract its components along 1 , r r Z 1r Z 1r 3 4 3 5 2 4 5 2 4 v4 (t) = t tt dt t t t dt t 2 2 2 2 1 1 r r Z 1 7 5 3 3 7 5 3 3 t t t4 dt t t 2 2 2 2 2 2 1 1

= t

Normalize v4 (t) to produce 2

jjv4 jj

= = )

3 t6 5 t7 t 2 6 1 2 7 5 2 t ; 1 t 1 7

= t4

Z 1

7 5 t8 2 28

t2 1

3 t6 26

2 , and

1 1

5 3 t 2

3 t 2

4 (t):

5 2 t 7

1 9 t

10 t7

7 7

dt =

1 25 t5

49 5

xv4 (t)

4 (t) = p

Z 1

8= (9

= 1

2 9

27 = p 2 2 49)

10 6 25 4 t + t dt 7 49 10 2 25 2 8 + = 7 7 49 5 9 49 t4

5 2 t ; 7

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS b. The pattern is not apparent.

Problem 11.18 Normalize s1 (t) to produce 2

jjs1 jj

1 (t):

Z 2

12 dt = 2

)

1 (t) =

s1 (t) 1 p =p ; 0 2 2

Take the second signal and subtract its component along 1 : Z 2 1 1 p cos ( t) dt p v2 (t) = cos ( t) 2 2 0 2 1 1 = cos ( t) sin ( t) 2 0 = cos ( t) ; 0 t 2 Normalize it: 2

jjv2 jj

= )

Z 2

1 1 + cos (2 t) dt = 1 2 2 0 0 t 2 2 (t) = s2 (t) = cos ( t) ; 0 2

cos ( t) dt =

A similar operation with the third signal shows that 3 (t) = s3 (t) = sin (

t) ; 0

Take the fourth signal and subtract its components along 1 and 3 : Z 2 Z 2 1 1 2 2 p p sin ( t) dt cos ( t) sin2 ( t) dt cos ( t) v4 (t) = sin ( t) 2 2 0 0 Z 2 sin ( t) sin2 ( t) dt sin ( t) 0

1 1 = 2 2

= sin ( t)

1 cos (2 t) ; 0 2

= Normalize it: 2

jjv4 jj

= )

Z 2

1 cos (2 t) 2

1 2

Z 1 1 2 1 1 1 2 cos (2 t) dt = + cos (4 t) dt = 4 4 2 2 4 0 0 v4 (t) = cos (2 t) ; 0 t 2 4 (t) = 1=2

19 From these results we …nd that p s1 (t) = 2 1 (t) s2 (t) =

2 (t)

s3 (t) =

3 (t)

s4 (t) =

1 2

1 1 cos (2 t) = p 2 2

1 (t) +

1 2

4 (t)

Problem 11.19 By the time delay and modulation theorems, (f ) =

sinc

1 2

+ sinc

1 2

e j2 k f

The total energy in the kth signal is E=

cos ( t= ) dt = =2

The energy for jf j EW

= =

Z W

W Z W

= 2

1 1 + cos (2 t= ) dt = 2 2 2

W is 2

1 2

sinc f

W 4 Z W 0

1 2

sinc v

Thus EW = E

sinc v Z W

+ sinc f + + sinc v +

1 2

sinc v

1 2

dv 1 2

dv (even integrand)

1 + sinc v + 2

The MATLAB program below computes EW =E: % pr11_19 % disp(’W_tau E_W/E’) for tau_W = 0.7:.1:1.3 v = 0:0.01:tau_W; y = (sinc(v-0.5)+sinc(v+0.5)).^2;

df 2

1 2

+ sinc v +

1 2

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS EW_E = trapz(v, y); disp([tau_W, EW_E]) end The results are given below: >> pr11_19 W_tau E_W/E 0.7000 0.8586 0.8000 0.9106 0.9000 0.9467 1.0000 0.9701 1.1000 0.9838 1.2000 0.9909 1.3000 0.9939 The product 0:8 W 0:9 for EW =E

11 12 = 0:9167.

Problem 11.20 a. By inspection of the given waveform set it follows that a satisfactory set of orthonormal functions is 1 (t)

2 (t)

3 (t)

4 (t)

2 cos (2 t) ; 0

2 sin (2 t) ; 0

2 cos (4 t) ; 0

2 sin (4 t) ; 0

For i = 0; 1; 2; 3 si (t) = A cos (i =2) cos (2 t) A sin (i =2) sin (2 t) p = E [cos (i =2) 1 (t) sin (i =2) 2 (t)] For i = 4; 5; 6; 7 si (t) = A cos ((i 4) =2) cos (4 t) A sin ((i 4) =2) sin (4 t) p E [cos ((i 4) =2) 3 (t) sin ((i 4) =2) 4 (t)] = b. The receiver looks like Figure 11.5 with four multipliers and integrators, one for each

21 orthonormal function. The signal coordinates are A = [Aij ; i = 0; 1; 2; 3; 4; 5; 6; 7; j = 1; 2; 3; 4] 8 hp i > E; 0; 0; 0; 0; 0; 0; 0 ; i = 0 > > > i h > p > > E; 0; 0; 0; 0; 0; 0 ; i=1 0; > > > h i > p > > 0; 0; E; 0; 0; 0; 0; 0 ; i = 2 > > > h i > p > < 0; 0; 0; E; 0; 0; 0; 0 ; i = 3 i h = p > E; 0; 0; 0 ; i=4 0; 0; 0; 0; > > > i h > p > > E; 0; 0 ; i = 5 0; 0; 0; 0; 0; > > > h i > p > > 0; 0; 0; 0; 0; 0; E; 0 ; i=6 > > > h > p i > : 0; 0; 0; 0; 0; 0; 0; E ; i = 7

From (11:96) the decision rule is

Choose H` so that d2 =

4 X

(Zj

A`j )2 = minumum

j=1

where Zj = Aij + Nj =

Z T =1 0

y (t)

j (t) dt =

Z 1

[si (t) + n (t)]

j (t) dt; j = 1; 2; 3; 4

This can be done suboptimally in two stages: (1) Correlate with two tones, one at frequency 1 Hz and the other at frequency 2 Hz and choose the output having the largest envelope (or squared envelope) (note that this will involve a correlation with both a sine and a cosine at each frequency); (2) Taking the pair of outputs from the correlators (one sine and one cosine) having the largest output envelope, determine in which =2 radian range the output pair lies (i.e., 0 to =2, =2 to , to 3 =2, or 3 =2 to 2 radians). c. As before E [Ni Nj ] = N20 ij . According to the suboptimum strategy given above, we make the comparison (y;

2 1 ) + (y;

2 2)

> (y; <

2 3 ) + (y;

2 4)

Suppose s0 (t) was really transmitted. Then the probability of a correct decision on

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS frequency is Pcf

= Pr

E + N1

= Pr

E + N1

+ N22 > N32 + N42 > N32 + N42

N22

The probability of correct decision on signal phase, from the analysis of quadriphase in Chapter 10 resulting in (10:15), is Pcp = 1

E N0

By symmetry, these expressions hold for either frequency or any phase. The overall probability of correct reception is Pc = Pcf Pcp for an overall probability of error of PE = 1

Problem 11.21 Write Pc =

Z 1

and let z = x

e y p

Pcf Pcp

E=N0

# 2 e x p dx dy

y. This gives Pc =

Z pE=N0 1

e z =2 p

1 1

# 2 e 2(y+z=2) p dy dz

Complete the square in the exponent p of the inside integral and use a table of de…nite integrals to show that is evaluates to 1= 2. The result then reduces to Pc = 1 which gives PE = Q

p E=N0

i E=N0 , the desired result.

23 Problem 11.22 a. The space is three-dimensional with signal points at the eight points p E=3 ;

E=3 ;

E=3

The optimum partitions are planes determined by the coordinate axes taken two at a time (three planes). Thus the optimum decision regions are the eight quadrants of the signal space. b. Consider S1 . Given the partitioning discussed in part (a), we make a correct decision only if kZ S1 k2 < kZ S2 k2 and kZ

S1 k2 < kZ

S4 k2

S1 k2 < kZ

S8 k2

and

where S2 , S4 , and S8 are the nearest-neighbor signal points topS1 . These are the most probable errors. Substituting Z = (Z1 ; Z2 ; Z3p ) and S1 = E=3 (1; 1; 1) ; S2 = p p E=3 (1; 1; 1; ) ; S4 = E=3 ( 1; 1; 1), and S8 = E=3 (1; 1; 1), the above conditions become Z1 + and

Z3 +

E=3 E=3

2 2

E=3

2 2

; Z2 +

E=3

< Z2

p 2 E=3 ;

by canceling like terms on each side. For S1 , these reduce to Z1 > 0; Z2 > 0; and Z3 > 0 to de…ne the decision region for S1 . Therefore, the probability of correct decision is P [correct dec.js1 (t)] = Pr (Z1 > 0; Z2 > 0; Z3 > 0) = [Pr (Zi > 0)]3 because the noises along each coordinate axis are independent. Note that E [Zi ] = (Es =3)1=2 , all i. The noise variances, and therefore the variances of the Zi s are all

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS

-1

-2

-3

n=4 n=3 n=2 n=1

-4

-5

-6

Eb/N0

N0 =2. Thus Z 1 3 p 2 1 1 exp Es =3 dy y N0 N0 0 " # r Z 1 1 u2 2 p = exp du ; u = y p 2 N0 2 2Es =3N0 h i3 p = 1 Q 2Es =3N0

P [correct dec.js1 (t)] =

Es =3

Since this is independent of the signal chosen, this is the average probability of correct detection. The symbol error probability is 1 Pc . Generalizing to n dimensions, we have in h in h p p PE = 1 1 Q 2Es =nN0 =1 1 Q 2Eb =N0

where n = log2 M . Note that Eb = Es =n since there are n bits per dimension and M = 2n . c. The symbol error probability is plotted in Fig. 11.1.

25 Problem 11.23 a. Integrate the product of the i

= = = '

1 and i signals over [0; Ts ]:

Z Ts

A2 cos f2 [fc + (i

A2 2

sin [2 (2fc + (2i 1) f ) t] sin [2 ( + 2 (2fc + (2i 1) f ) 2 (

0 Z A2 Ts

1) f ] tg cos [2 (fc + i f ) t] dt Z A2 Ts fcos [2 (2fc + (2i 1) f ) t]g dt + fcos [2 ( 2 0

sin [2 (2fc + (2i 1) f ) Ts ] sin [2 ( + 2 2 (2fc + (2i 1) f ) 2 ( 2 2 A Ts A Ts sin [2 f Ts ] = sinc (2 f Ts ) 2 2 f Ts 2

f ) t] f)

f ) t]g dt

Ts 0

f ) Ts ] f)

where the …rst term has been neglected because it is small for typical values of fc . The remaining term is 0 for its argument equal to a nonzero integer. The smallest such integer is 1, which gives 2 f Ts = 1 or

1 2Ts

b. We need 1=2Ts Hz of bandwidth per signal, so M=

M W = 2W Ts or W = 1=2Ts 2Ts

c. For vertices-of-a-hypercube signaling M = 2n where n is the dimensionality (number of orthogonal functions) which is 2W Ts , so log2 M = 2W Ts or W =

log2 M 2Ts

Problem 11.24 In the equation P (EjH1 ) =

Z 1 Z 1 0

fR2 (r2 jH1 ) dr2 fR1 (r1 jH1 ) dr1

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS

substitute fR1 (r1 jH1 ) =

2 2 2r1 e r1 =(2E +N0 ) ; r1 2 2E + N0

and fR2 (r2 jH1 ) =

2r2 r22 =N0 e ; r2 > 0 N0

The inside integral becomes

I = e r1 =N0 When substituted in the …rst equation, it can be reduced to Z 1 Z 1 2r1 2r1 2 r12 =(2E 2 +N0 ) r12 =N0 P (EjH1 ) = dr1 = e e r1 =K dr1 e 2+N 2+N 2E 2E 0 0 0 0 where K=

N0 2E 2 + N0 2 (E 2 + N0 )

The remaining integral is then easily integrated since its integrand is a perfect di¤erential if written as Z 1 K 2r1 r12 =K P (EjH1 ) = e dr1 2 2E + N0 0 K In the integral, let v = r12 =K to get Z 1 Z 1 2r1 r12 =K dr1 = e v dv = 1 e K 0 0 so that …nally P (EjH1 ) = =

N0 2E 2 + N0 K 1 = 2E 2 + N0 2E 2 + N0 2 (E 2 + N0 ) N0 1 = 2 2 (E 2 + N0 ) 2 1 + 1 2E 2 N0

which is the same as (10.142). Problem 11.25 a. By de…nition, the new signal set is M 1

s0i (t) = si (t)

1 X sj (t) M j=0

27 The signal energy for the ith signal in the new set is 2 32 Z Ts M 1 X 1 4si (t) Es0 = sj (t)5 dt M 0 j=0 2 3 Z Ts M M X1 X1 M X1 2 1 4s2i (t) si (t) = sj (t) + 2 sj (t) sk (t)5 dt M M 0 j=0

Z Ts 0

s2i (t) dt

2 M

M 1

= Es

j=0

M 1 M 1 Z Ts

1 X X si (t) sj (t) dt + 2 M j=0 k=0

= Es 1

jk = 0; j 6= k and 1; j = k

j=0 k=0

2 1 Es + 2 M M 1 M

sj (t) sk (t) dt

M 1M 1

2 X 1 X X Es ij + 2 Es jk ; M M j=0

= Es

j=0 k=0

M X1 Z Ts

M X1

Es = Es

j=0

= Es

1 2 Es + Es M M

M 1 M

b. The correlation coe¢ cient between two di¤erent signals is 2 3" # Z Ts M 1 M X X1 1 1 1 4sm (t) sj (t)5 sn (t) sk (t) dt; m 6= n mn = Es0 0 M M j=0 k=0 # P P 1 Z Ts " M 1 1 1 sm (t) sn (t) M sk (t) sn (t) j=0 sj (t) M sm (t) M 1 k=0 P 1 PM 1 = dt Es0 0 + M12 M j=0 k=0 sj (t) sk (t) 3 2 RT R s 1 PM 1 Ts s (t) s (t) dt s (t) s (t) dt n j m n j=0 0 RM 1 6 0 7 1 PM 1 Ts = s (t) s 5 4 m k (t) dt k=0 M 0 R Es0 P P M 1 Ts 1 + M12 M s (t) s (t) dt j k j=0 k=0 0 2 3 M 1 M 1 M 1M 1 1 4 1 X 1 X X 1 X = Es mn Es nj Es nk + 2 Es jk 5 Es0 M M M j=0

j=0 k=0

k=0

Es Es Es M 1 0 + = Es0 M M M Es (M 1 ; m 6= n M 1

Es M

c. Simply solve for Es in the above expression and substitute for Es in the orthogonal

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS signaling result. This gives Z 1( " Q

Pc; simplex = 1

2Es0 N0

M M 1

!#)M 1

e v =2 p dv 2

d. The union bound approximation for the symbol error probability for coherent orthogonal signaling is given by (10.67) as r ! Es Ps; orthog ' (M 1) Q N0 Using the substitution used in part c, we get s

Es0 N0

M M 1

The bit error probability can be approximated as s M log2 (M ) Eb Pb; simplex ' Q 2 N0

M M 1

Ps; simplex ' (M

1) Q

Plots are given in Fig. 11.2. Problem 11.26 a. Use the basis set r

2 cos ! i t; 0 T r 2 sin ! i t; 0 si (t) = T ci (t)

T; i = 1; 2;

; M

Base the decision on Z = (Zc1 ; Zs1 ; :::; ZcM ; ZsM ) where Zci = (y; ci ) and Zsi = (y; si ). Suppose that the ith hypothesis is true. Then the components of Z are p Ej Gcj + Ncj ; i = j Zci = Zsi

= Ncj ; i 6= j p = Ej Gsj + Nsj ; i = j = Nsj ; i 6= j

M-ary CFSK

M-ary Simplex

-1

-2

-3

M= 2

-2

M= 2

-4

= 16

=4 =8

-5

= 16 -6

-6

= 32

-7

= 32

-7

10 Eb /N0

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS where Gcj = Gj cos and Gsj = Gj sin are independent Gaussian random variables with variances 2 and Ncj ; Nsj are independent Gaussian random variables with variances N0 =2. The pdf of Z given hypothesis Hi is true is

fZjHi (zjHi ) = exp

2 + z2 zci si Ei 2 + N 0

exp @

1 N0

M X

j=1; j6=i

2 2 A zci + zsi =

Ei 2 + N0 N0M

Decide hypothesis i is true if

fZjHi (zjHi )

fZjHj ; all j

This constitutes the maximum likelihood decision rule since the hypotheses are equally probable. Alternatively, we can take the natural log of both sides and use that as a test. This reduces to computing

Ei 2 2 z 2 + zsi Ei 2 + N0 ci

and choosing the signal corresponding to the largest. If the signals have equal energy, then the optimum receiver is seen to be an M -signal replica of the one shown in Fig. 10.8a.

2 + Z 2 is Rayleigh distributed b. Assume that the Ei s are equal. Note that Zi = Zci si given Hi . Without loss of genarality assume that H1 is given. Then

fZ1 jH1 (z1 j H1 ) = fZi jH1 (zi j H1 ) =

z1 2+N

E zi exp N0

exp 0

z12 =2 E 2 + N0

zi2 =2N0 ; zi

0; i 6= 1

; z1

31 The probability of correct decision is Pc = E fPr [Zi < Z1 ; all i = 2; 3; :::; M ]g Z z1 Z 1 M 1 zi z1 2 2 2 = exp z =2 E + N exp z =2N dz dz1 0 0 i 1 i E 2 + N0 N0 0 0 Z 1 z1 exp z12 =2 E 2 + N0 M 1 = 1 exp z12 =2N0 dz1 , use binomial theorem 2 E + N0 0 Z 1 z1 exp z12 =2 E 2 + N0 = E 2 + N0 0 "M 1 #M 1 X M 1 (M 1 i) z12 =2N0 dz1 exp i i=0

M 1

X 1 E 2 + N0

1 2 E + N0

M X1 i=0 M X1 i=0

i=0 M X1 i=0

N0 E 2 + N0 N0 + (E 2 + N0 ) (M

1 i

N0 + (E

N0 2 + N ) (M 0

1 0 + 1) (M

1 i

1 + (E

where Ki ,

Ki z12 dz1

z1 exp

1 i

Z 1

2 =N

1 M 1 + 2 (E 2 + N0 ) 2N0

Problem 11.27 a. It can be shown that under hypothesis H1 ; Y1 is the sum of squares of 2N independent Gaussian random variables each having mean zero and variance E 2 11 = 2

N0 2

and Y2 is the sum of squares of 2N independent Gaussian random variables each having zero mean and variance N0 2 21 = 2

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Therefore, from Problem 5.37, the pdfs under hypothesis H1 are 2

y N 1 e y1 =2 11 ; y1 fY1 (y1 jH1 ) = 1 N N 2 11 (N ) and fY2 (y2 jH1 ) =

y2N 1 e y2 =N0 2N (N0 =2)N

(N )

; y2

b. The probability of error is PE = Pr (EjH1 ) = Pr (EjH2 ) = Pr (Y2 > Y1 jH1 ) Z 1 Z 1 = fY2 (y2 jH1 ) dy2 fY1 (y1 jH1 ) dy1 y1

Using integration by parts, it can be shown that I (x; a) =

Z 1

z e

dz = e

n X i=0

n! i!an+1 i

This can be used to get the formula given in the problem statement. c. Plots are given for N = 1

13 in steps of 2 in Fig. 11.3.

Problem 11.28 a. The characteristic function is found and used to obtain the moments. By de…nition, Z 1 m j ! m 1 (j!) = E e = ej ! e d (m) 0 To integrate, note that

Z 1 0

ej ! e

d =

1 j!

Di¤erentiate both sides m 1 times with respect to . This gives Z 1 ( 1)m 1 (m 1)! ej ! ( 1)m 1 m 1 e d = ( j!)m 0 Multiply both sides by m and cancel like terms to get Z 1 m m 1 ej ! e d = (m) ( 0

j!)m

PE for FSK signaling with diversity in Rayleigh fading: N = 1 - 13 paths in steps of 2 0

-1

-2

-3

-4

-5

-6

-7

10 12 SNR, dB

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS Thus

(j!) =

j!)m

(

Di¤erentiating this with respect to !, we get E[ ]=j

(0)

and E

= j2

(0) =

The variance is 2

Var ( ) = E

E2 ( ) =

m (m + 1)

m 2

b. Use Bayes’rule. But …rst we need fZ (z), which is obtained as follows: Z 1 fZ (z) = fZj (zj ) f ( ) d Z0 1 m m 1 = e z e d (m) 0 Z 1 m+1 m m ( +z) ( + z) m = e d (m + 1) ( + z)m+1 0 m m = ( + z)m+1 where the last integral is evaluated by noting that the integrand is a pdf and therefore integrates to unity. Using Bayes’rule, we …nd that m

f jZ ( jz) =

( + z)m+1 e ( +z) (m + 1)

Note that this is the same pdf as in part (a) except that has been replaced by + z and m has been replaced by m + 1. Therefore, we can infer the moments from part (a) as well. They are E [ jZ] =

m+1 m+1 and Var [ jZ] = +Z ( + Z)2

c. Assume the observations are independent. Then fz1 z2 j (z1; z2 j ) =

(z1 +z2 )

; z1 ; z2

The integration to …nd the joint pdf of Z1 and Z2 is similar to the procedure used to …nd the pdf of Z above with the result that fz1 z2 (z1; z2 ) =

m (m + 1) m ( + z1 + z2 )m+2

35 Again using Bayes’rule, we …nd that m+1

f jz1 z2 ( jz1 ; z2 ) =

+ z1 + z2 )m+2 (m + 2)

( +z1 +z2 ) (

Since this is of the same form as in parts (a) and (b), we can extend the results there for the moments to E [ jZ1 ; Z2 ] =

m+2 m+2 and V ar [ jZ1 ; Z2 ] = + Z1 + Z2 ( + Z1 + Z2 )2

d. By examining the pattern developed in parts. (a), (b), and (c), we conclude that m+1+K

f jz1 z2 :::zK ( jz1 ; z2; : : : zK ) =

+ z1 + z2 + : : : + zK )m+K (m + 2)

( +z1 +z2 +:::+zK ) (

and E [ jZ1 ; Z2 ; : : : ; ZK ] = var [ jZ1 ; Z2 ; : : : ; ZK ] =

m+2 + Z1 + Z2 + : : : + ZK m+K ( + Z1 + Z2 + : : : + ZK )2

Problem 11.29 The conditional mean and Bayes’estimates are the same if: 1. C (x) is symmetric; 2. C (x) is convex upward; 3. fAjZ is symmetric about the mean. Armed with this, we …nd the following to be true: a. Same - all conditions are satis…ed. b. Same - all conditions are satis…ed; c. Condition 3 is not satis…ed; d. Condition 2 is not satis…ed.

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS

Problem 11.30 The mean of the individual samples is A as is the mean of a ^ML (Z). Therefore 3 2 ! 2 K X 1 var [^ aML (Z)] = E 4 (Zk A) 5 K k=1 2 3 K X K X 1 4 = E (Zk A) (Zj A)5 K2 k=1 j=1

1 XX E [(Zk K2

A) (Zj

A)]

k=1 j=1

But, by the independence of samples and the fact that Zk E [(Zk

A) (Zj

A = Nk

0; k 6= j 2; k = j n

A)] =

Thus K

var [^ aML (Z)] =

1 XX K2

2 n kj

k=1 j=1 K

1 X K2

2 n

k=1

2 n

Problem 11.31 a. The conditional pdf of the noise samples is fZ z1 ; z2 ; : : : ; zK j 2n = 2

2 n

K=2

exp

K X i=1

zi2 =2 2n

The maximum likelihood estimate for 2n maximizes this expression. It can be found by di¤erentiating the pdf with respect to 2n and setting the result equal to zero. Solving for 2n , we …nd the maximum likelihood estimate to be K

b2n =

1 X 2 Zi K i=1

37 b. The variance of the estimate is

c. Yes. PK 2 d. i=1 Zi is a su¢ cient statistic.

var b2n =

2 4n K

Problem 11.32 We base our estimate on the observation Z0 , where Z0 = M0 + N where N is a Gaussian random variable of mean zero and variance N0 =2. The sample value M0 is Gaussian with mean m0 and variance 2m . The …gure of merit is mean-squared error. Thus, we use a MAP estimate. It can be shown that the estimate for m0 is the mean value of M0 conditioned on Z0 because the random variables involved are Gaussian. That is, m b 0 = E (M0 jZ0 = z0 ) = z0 where

E (M0 Z0 ) m p =q var (M0 ) var (Z0 ) 2 + N0 m

It is of interest to compare this with the ML estimate, which is c m b 0;ML = Z0

which results if we have no a priori knowledge of the parameter. Note that the MAP estimate reduces to the ML estimate when 2m ! 1. Note also that for N0 ! 1 (i.e., small signal-to-noise ratio) we don’t use the observed value of Z0 , but instead estimate m0 as zero (its mean). Problem 11.33 a. Use the basis function set 1 (t) =

2 cos ! c t and T

2 (t) =

2 sin ! c t T

and trigonometric expansion of the received data as H1 :

y(t) = A cos cos ! c t

A sin sin ! c t + n (t)

H2 :

y(t) = A cos cos ! c t + A sin sin ! c t + n (t)

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS to write the data in vector form as

H1 :

Z = (Z1 ; Z2 ) =

H2 :

Z = (Z1 ; Z2 ) =

where N1 =

Z T

n (t)

! r T T A cos + N1 ; A sin + N2 2 2 ! r r T T A cos + N1 ; A sin + N2 2 2

1 (t) dt and N2 =

Z T

n (t)

2 (t) dt

Note that E [N1 ] = [N2 ] = 0 and var[N1 ] = var[N2 ] = N0 =2. Thus r

T A cos , A1 2 r T E [Z2 j H1 ] = A sin , A2 2 r T E [Z1 j H2 ] = A cos , A1 2 r T E [Z1 j H2 ] = A sin , A2 2 E [Z1 j H1 ] =

Hence the conditional pdfs of the data, given fZj ;H1 (z1 ; z2 j ; H1 ) = fZj ;H2 (z1 ; z2 j ; H2 ) =

1 exp N0 1 exp N0

and the particular hypothesis, are i 1 h (z1 A1 )2 + (z2 + A2 )2 N0 i 1 h (z1 + A1 )2 + (z2 A2 )2 N0

b. Assume equally likely hypotheses so that fZj (z1 ; z2 j ) = =

1 1 fZj ;H1 (z1 ; z2 j ; H1 ) + fZj ;H2 (z1 ; z2 j ; H2 ) 2 2 2 n h io 3 2 2 1 exp (z A ) + (z + A ) 1 1 2 2 1 4 n N0 h io 5 1 2 N0 + exp (z1 + A1 )2 + (z2 A2 )2 N0

39 Expanding the exponents and taking out common terms results in fZj (z1 ; z2 j ) =

1 exp 2 N0

1 N0

T 2 A 2

z12 + z22 +

2 2 (A1 z1 A2 z2 ) + exp (A1 z1 A2 z2 ) N0 N0 1 1 T 2 exp z12 + z22 + A2 cosh (A1 z1 A2 z2 ) N0 N0 2 N0 exp

The maximum likelihood estimator for to signify actual data values)

satis…es (the data variables are capitalized

@ ln fZj (Z1 ; Z2 j ) @

= M L =0

From above ln fZj (Z1 ; Z2 j ) =

1 T Z12 + Z22 + A2 N0 2 2 + ln cosh (A1 Z1 A2 Z2 ) N0 ln ( N0 )

Taking the derivative with respect to and setting the result equal to 0 results in the condition "r #r T 2A T 2A tanh (Z1 cos Z2 sin ) ( Z1 sin Z2 cos ) = 0 2 N0 2 N0 Using de…nitions for Z1 and Z2 , which are r r Z T Z T 2 2 Z1 = y (t) cos ! c t dt and Z2 = y (t) sin ! c t dt T T 0 0 this can be put into the form 4z tanh AT

Z T 0

4z y (t) cos (! c t + ) dt AT

Z T

y (t) sin (! c t + ) dt = 0

where z = A2 T =2N0 . This implies a Costas loop type structure shown in the problem statement with integrators in place of the low pass …lters and a tanh function in the leg with the cosine multiplication. Note that for z small tanh (x) ' x and the phase estimator becomes a Costas loop if the integrators are viewed as low pass …lters.

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS c. The variance of the phase estimate is lower bounded by applying the Cramer-Rao bound which is " # 1 @ 2 fZj var ( ML ) E @ 2 The …rst derivative, from above, is Z Z T @fZj 2A T 2A tanh y (t) cos (! c t + ) dt = @ N0 N0 0 0

y (t) sin (! c t + ) dt

Assume low SNR so that tanh (x) ' x. Under this assumption the second derivative, after some work, can be shown to be Z Z @ 2 fZj 2A 2 T T = y (t) y t0 cos ! c t t0 dtdt0 N0 @ 2 0 0 After some development, it can be shown that E y (t) y t0

so that " # @ 2 fZj = E @ 2

= Pr [H1 ] E y (t) y t0 j H1 + Pr [H2 ] E y (t) y t0 j H2 1 E [A cos (! c t + ) + n (t)] A cos ! c t0 + + n t0 = 2 1 + E [ A cos (! c t + ) + n (t)] A cos ! c t0 + + n t0 2 A2 N0 = cos ! c t t0 + t t0 2 2

2A N0

2Z T Z T

2A N0

0 0 2Z T Z T 0

2A N0

0 2 A T2

4 2

cos ! c t

t0 dtdt0

A2 cos ! c t 2

t0 +

A2 cos2 ! c t 2

t0 dtdt0 +

2Z T Z T

E y (t) y t0

N0 T 2

T Since the SNR is z = A 2N0 this result becomes " # @ 2 fZj E = 4z (z + 1) @ 2

N0 2

t N0 T 2

cos ! c t

t0 dtdt0

41 Thus, the variance of the phase estimate, for low SNR, is bounded by 1 4z (z + 1)

var ( M L )

Reference should have been made to Table 10.7 where the tracking loop error variance is given as 2 = L 1 1=z + 0:5=z 2 = L 1 z+0:5 . the best that can be said of the comparison z2 is that both are proportional to 1=z 2 in the limit of small SNRs. Problem 11.34 a. Note that cos cos 1 m = m and sin cos 1 m = 1 s (t) =

2P m sin (! c t + )

Use the basis functions 1 (t) =

to write p s (t) = P T m sin

1=2

m2 cos (! c t + ) ;

2 cos ! c t and T

m2 cos

2 (t) =

1 (t)+

to write sign due to data

2 sin ! c t T

P T m cos

m2 sin

2 (t)

b. Use the above bases functions and base a decision on the vector 2 p 3 p P T m sin 1 m2 cos + N1 ; 5 Z = [Z1 ; Z2 ] = 4 p p 2 P T m cos 1 m sin + N2 , [S1 + N1 ; S2 + N2 ]

where N1 and N2 are independent Gaussian random variables with zero means and variances N0 =2 The pdf of Z conditioned on the signal sign and is fZj ;

(Z1 ; Z2 j ;

1 exp N0

To …nd the pdf conditioned only on to get

1 h (Z1 N0

S1 )2 + (Z2

S2 )2

use the fact the Pr(+ bit) = Pr (

1 1 fZj (Z1 ; Z2 j ) = fZj ; + (Z1 ; Z2 j ; +) + fZj ; 2 2

bit) = 1=2

(Z1 ; Z2 j ;

)

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS After considerable simpli…cation we obtain " p # 2m P T fZj (Z1 ; Z2 j ) = C exp (Z1 sin + Z2 cos ) N0 # " p p 2 1 m2 P T (Z1 cos Z2 sin ) cosh N0 where C includes all terms independent of . The log-likelihood function is p 2m P T L( ) = (Z1 sin + Z2 cos ) N0 " p # p 2 1 m2 P T + ln cosh (Z1 cos Z2 sin ) N0 Note that Z1 sin + Z2 cos

Z T

y (t)

Z1 cos

Z2 sin

Z T 0

y (t)

2 sin (! c t + ) dt T 2 cos (! c t + ) dt T

so that p Z 2m 2P T L( ) = y (t) sin (! c t + ) dt N0 0 " p # p Z 2 1 m2 2P T + ln cosh y (t) cos (! c t + ) dt N0 0 ) The maximum likelihood estimate satis…es @L( = 0. @ espression for L ( ) gives

@L ( ) @

Di¤erentiating the above

p Z 2m 2P T = 0= y (t) cos (! c t + ML ) dt N0 0 " p # p Z 2 1 m2 2P T tanh y (t) cos (! c t + ML ) dt N0 0 " p # p Z 2 1 m2 2P T y (t) sin (! c t + ML ) dt N0 0

11.1. COMPUTER EXERCISES

c. The block diagram follows the development of the one for Problem 11.33. The block diagram of the phase estimator has another arm that adds into the feedback to the VCO that acts as a phase-lock loop for the carrier component. Problem 11.35 Write the impulse response as h (t) =

1 T

T =2 T

and use the delay theorem to get H (f ) =

1 T sinc (T f ) e j f T = sinc (T f ) e j f T T

Thus the equivalent noise bandwidth is BN

11.1

Z 1

jH (f )j2 df 2 Hmax Z 1 0 = sinc2 (T f ) df 0 Z 1 1 = sinc2 (u) du T 0 1 = 2T =

Computer Exercises

Computer Exercise 11.1 % …le: ce11_1.m ROC; Pd near 1, PFA small % % PF = Q(A) < 1e-3 => A > 3.08; B = ln(eta)/d + d/2 % PD = Q(B) > 0.95 => B < -1.65; A = ln(eta)/d -d/2 % d = A - B > 4.73; % ln(eta) = (A^2 - B^2)/2 > 3.38 => eta > 29.37 % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS % clf clear all; clf d = [4.73 5 5.5 6]; log_eta = 3.38:0.01:9; lend = length(d); hold on for j = 1:lend dj = d(j); af = log_eta/dj + dj/2; ad = log_eta/dj - dj/2; pf = qfn(af); pd = qfn(ad); semilogy(pf, pd) text(pf(20), pd(20)-.01, [’d = ’, num2str(dj)]) end grid on hold o¤ xlabel(’Probability of False Alarm’) ylabel(’Probability of Detection’) title(’Receiver operating characteristic’) %

This function computes the Gaussian Q-function

% function Q=qfn(x) Q = 0.5*erfc(x/sqrt(2)); % % End of script …le

11.1. COMPUTER EXERCISES

Receiver operating characteristic 1 d=6

d = 5.5 d=5

0.95

Probability of Detection

d = 4.73 0.9

0.85

0.8

0.75

0.7

0.65

0.2

0.4 0.6 0.8 Probability of False Alarm

1.2 -3

x 10

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS

Computer Exercise 11.2 % …le: ce11_2.m % % From (11.171), sigmap^2/sigman^2 = (K + sigman^2/sigmaA^2)^-1 % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clf clear all; sigmaA2_sigman2 = 0.1:0.5:2.1; Ls = length(sigmaA2_sigman2); K = 1:25; hold on for l=1:Ls sigma_ratio = sigmaA2_sigman2(l); sigmap2_sigman2 = 1./(K + 1./sigma_ratio); plot(K, sigmap2_sigman2) text(K(1), sigmap2_sigman2(1),[’nsigma_A^2/nsigma_n^2 = ’, num2str(sigma_ratio)]) end hold o¤ xlabel(’K’),ylabel(’nsigma_p^2/nsigma_n^2’) % % End of script …le

Computer Exercise 11.3 % …le: ce11_3.m % Simulation of a digital PLL % % R. Ziemer & W. Tranter, Principles of Communications, 7th edition % clf clear all; AA = char(’-’,’:’,’–’,’-.’); K = 1000; epsilon = input(’Enter feedback constant epsilon => ’); theta0 = 0.5*pi; T = 1; A = 1;

11.1. COMPUTER EXERCISES

0.7

σ A /σ n = 2.1

0.6

σ A /σ n = 1.6

0.5

σ A /σ n = 1.1

0.4 2 2

σ p/σ n

σ A /σ n = 0.6

0.3

0.2

σ A /σ n = 0.1

0.1

15 K

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS

kk = 1:K; sigma_n = [0.1 0.5 1 2]; Ln = length(sigma_n); hold on for n = 1:Ln sigma = sigma_n(n); Z1(1) = sqrt(T/2)*A*cos(theta0) + sigma*randn(1); Z2(1) = -sqrt(T/2)*A*sin(theta0) + sigma*randn(1); theta(1) = theta0; for k = 1:K-1 theta(k+1) = theta(k) + epsilon*atan2(Z2(k), Z1(k)); Z1(k+1) = sqrt(T/2)*A*cos(theta(k+1)) + sigma*randn(1); Z2(k+1) = -sqrt(T/2)*A*sin(theta(k+1)) + sigma*randn(1); end plot(kk, theta, AA(n,:)) end hold o¤ xlabel(’k’),ylabel(’ntheta(k), rad’) legend([’nsigma_n = ’, num2str(sigma_n(1))], [’nsigma_n = ’, num2str(sigma_n(2))], [’nsigma_n = ’, num2str(sigma_n(3))], [’nsigma_n = ’, num2str(sigma_n(4))]) title([’First-order digital phase tracking device performance for nepsilon = ’, num2str(epsilon)]) % % End of script …le

11.1. COMPUTER EXERCISES

First-order digital phase tracking device performance for ε = 0.01 1.6

σ n = 0.1

1.4

σ n = 0.5

1.2

σn = 1

θ(k), rad

0.8 0.6 0.4 0.2 0 -0.2 -0.4

100

200

300

400

500 k

600

700

800

900

1000

CHAPTER 11. OPTIMUM RECEIVERS AND SIGNAL SPACE CONCEPTS

First-order digital phase tracking device performance for ε = 0.1 2

σ n = 0.1 σ n = 0.5

1.5

σn = 1 1

θ(k), rad

0.5

-0.5

-1

-1.5 0

100

200

300

400

500 k

600

700

800

900

1000

Chapter 12

Information Theory and Coding 12.1

Problem Solutions

Problem 12.1 The transition probabilities for the 3-hop system are (The first matrix represents the 2-hop system as defined by equations (12.19)-(12.22))  ( |) = This gives ∙ =

∙

11 12 21 22

∙

1  1 + 2  3 1  2 + 2  4 3  1 + 4  3 3  2 + 4  4

¸∙

1 2 3 4

1  1  1 + 2  3  1 + 1  2  3 + 2  4  3 1  1  2 + 2  3  2 + 1  2  2 + 2  4  2 3  1  3 + 4  3  1 + 3  2  3 + 4  4  3 3  1  2 + 4  3  2 + 3  2  4 + 4  4  4

Tracing all paths from the two inputs to the two outputs is consistent with the above. Note that there are four paths from a given input to a given output. Colored pencils help since there are a tptal of 16 paths. Problem 12.2 (a)  () = − log2 (52) = 57004 bits (b)  () = − log2 [(52)(52)] = 114009 bits (c)  () = − log2 [(52)(51)] = 113729 bits Problem 12.3 1

CHAPTER 12. INFORMATION THEORY AND CODING

The entropy is  () = −030 log2 (030) − 025 log2 (025) −020 log2 (020) − 015 log2 (015)

−010 log2 (010) = 22282 bits/symbol The maximum entropy is  () = log2 (5) = 23219 bits/symbol

Problem 12.4 The entropy is  () = −030 log2 (030) − 025 log2 (025) − 020 log2 (020) −010 log2 (010) − 010 log2 (010) − 005 log2 (005)

= 23660 bits/symbol

Problem 12.5 (a) The channel diagram is shown in Figure 12.1. (b) The output probabilities are  (1 ) =  (2 ) =  (3 ) =

1 1 (06 + 02 + 02) = 3 3 1 1 (03 + 05 + 02) = 3 3 1 1 (01 + 03 + 06) = 3 3

(c) Since all columns sum to one, equal input probabilities yield equal output probabilities. (d) The joint probability matrix is ⎤⎡ 6 ⎤ ⎡ 1 3 1 0 0 3 10 10 10 5 3 ⎦ 2 [ (;  )] = ⎣ 0 13 0 ⎦ ⎣ 10 10 10 2 6 2 1 0 0 3 10 10 10 which gives

⎡

[ (;  )] = ⎣

6 30 2 30 2 30

3 30 5 30 2 30

1 30 3 30 6 30

⎤ ⎦

12.1. PROBLEM SOLUTIONS

0.6 0.1 0.2 0.3 0.2

0.3 0.5

0.2 0.6

Figure 12.1: Channel diagram for Problem 12.5 Note that the column sum gives the output probabilities [ ( )], and the row sum gives the input probabilities [ ()]  Problem 12.6 This problem may be solved by raising the channel matrix ∙ ¸ 0993 0007 = 0007 0993 which corresponds to an error probability of 0007, to increasing powers  and seeing where the error probability reaches the critical value of 002. Consider the MATLAB program a = [0.993 0.007; 0.007 0.993]; n = 1; a1 = a; while a1(1,2)0.02 n=n+1; a1=a^n; end n-1

% channel matrix % initial value % save a

% display result

Executing the program yields  − 1 = 22. Thus we compute (MATLAB code is given) À a^2

CHAPTER 12. INFORMATION THEORY AND CODING ans = 0.9861 0.0139 0.0139 0.9861

This gives an error probability of 0.0139, which is less than 0.02. Next we compute  a^3 ans = 0.9793 0.0207 0.0207 0.9793 Thus 2 cascaded channels meets the specification for   002. However cascading 3 channels yields   002 and the specification is not satisfied. Problem 12.7 For the erasure channel we determine (;  ) = () − (| ) where (| ) = −

3 2 X X =1 =1

(   ) log2 ( | ) = −

3 2 X X =1 =1

( | )( ) log2

∙

( | )( ) ( )

Let (1 ) =  and (2 ) = 1 − . This gives (1 ) = (1 − )

(2 ) =  + (1 − ) =  (3 ) = (1 − )(1 − )

Thus ¸ ∙ ¸ (1 − )  −  log2 (| ) = −(1 − ) log2 (1 − )  ¸ ¸ ∙ ∙ (1 − ) (1 − )(1 − ) −(1 − ) log2 − (1 − )(1 − ) log2  (1 − )(1 − ) ∙

This is (| ) = − log2 () − (1 − ) log2 (1 − ) = − [ log2 () + (1 − ) log2 (1 − )] Since () = − log2 () − (1 − ) log2 (1 − )

12.1. PROBLEM SOLUTIONS

we have (;  ) = () − () = ()(1 − ) Therefore  =1−

bits/symbol

and capacity is achieved for equally likely inputs. Problem 12.8 The cascade of the two channels has the channel matrix ¸∙ ¸ ∙ 1 − 2 2 1 0 1 − 1  ( |) = 1 1 − 1 0 2 1 − 2 This gives  ( |) =

∙

(1 − 1 ) (1 − 2 ) (1 − 1 ) 2 + 1 2 1 (1 − 2 ) 1 2 + (1 − 1 ) 2 (1 − 1 ) (1 − 2 ) 1 (1 − 2 )

which can be written  ( |) =

∙

(1 − 1 ) (1 − 2 ) 2 1 (1 − 2 ) 2 (1 − 1 ) (1 − 2 ) 1 (1 − 2 )

The output probability for 2 is (2 ) = 2 Thus, the probability of 2 is independent of the input so that the cascade is a more general erasure channel in that (1 |2 ) and (3 |1 ) are not zero. We will encounter a channel like this when feedback channels are encountered. Problem 12.9 The capacity of the channel described by the transition probability matrix ⎡ ⎤   0 0 ⎢   0 0 ⎥ ⎢ ⎥ ⎣ 0 0   ⎦ 0 0   is easily computed by maximizing

 (;  ) =  ( ) −  ( |) The conditional entropy  ( |) can be written X XX  ( )  ( | ) log2  ( | ) =  ( )  ( | )  ( |) = − 





CHAPTER 12. INFORMATION THEORY AND CODING

where  ( | ) = −

X 

 ( | ) log2  ( | )

For the given channel  ( | ) = − log2  −  log2  =  so that  ( | ) is a constant independent of . This results since each row of the matrix contains the same set of probabilities although the terms are not in the same order. For such a channel X  ( )  =   ( |) = 

and

 (;  ) =  ( ) −  We see that capacity is achieved when each channel output occurs with equal probability. We now show that equally likely outputs result if the inputs are equally likely. Assume that  ( ) = 14 for all . Then X  ( ) =  ( )  ( | ) 

For each   ( ) =

1 1 1X 1  ( | ) = ( + ) = [ + (1 − )] = 4 4 4 4 

so that

1  ( ) =  4

all 

Since  ( ) = 14 for all ,  ( ) = 2, and the channel capacity is  = 2 +  log2  + (1 − ) log2 (1 − ) or  = 2 −  () This is shown in Figure 12.2. This channel is modeled as a pair of binary symmetric channels as shown. If  = 1 or  = 0, each input uniquely determines the output and the channel reduces to a noiseless channel with four inputs and four outputs. This gives a capacity of log2 4 = 2 bits/symbol. If  =  = 12 , then each of the two subchannels (Channel 1 and Channel 2) have zero capacity. However, 1 and 2 can be viewed as a single input and 3 and 4 can be viewed as a single input as shown in Figure 12.3. The result is equivalent to a noiseless channel with two inputs and two outputs. This gives a capacity of log2 2 = 1 bit/symbol.

12.1. PROBLEM SOLUTIONS

C 2

1/2

Figure 12.2: Channel capacity for Problem 12.9.

Channel 1

Channel 2

Figure 12.3: Additional illustration for Problem 12.9.

CHAPTER 12. INFORMATION THEORY AND CODING

Note that this channel is an example of a general symmetric channel. The capacity of such channels are easily computed. See Gallager (Information Theory and Reliable Communications, Wiley, 1968, pages 91-94) for a discussion of these channels. Problem 12.10 To show (12.30), write  (   ) in (12.29) as  ( | )  ( ) and recall that log2  (   ) = log2  ( | ) + log2  ( ) Evaluating the resulting sum yields (12.30). The same method is used to show (12.31) except that  (   ) is written  ( | )  ( )  Problem 12.11 The entropy is maximized when each quantizing region occurs with probability 025. Thus Z 1 1 −  = 1 − −1 = 4 0 which gives −1 = or

1 1 = ln 

In a similar manner Z 2 1

3 4

µ ¶ 4 = 02877 3

−  = −1 − −2 =

3 1 − −2 = 4 4

which gives −2 = or 2 = Also

1 2

1 ln (2) = 06931 

Z 3

−  =

2

gives

3 = 13863

Problem 12.12

1 4

12.1. PROBLEM SOLUTIONS

The thresholds are now parameterized in terms of . We have Z 1  −2 22 1 2 2   = 1 − −1  = 2  4 0 which gives

µ 2¶ ∙ ¸ 1 3 = ln(43) = − ln 2 4

Thus 1 = For 2 we have Z 2 1

p log(43) = 05364

 −2 22 3 1 2 2 2 2 2 2   = −1  − −2  = − −2  = 2  4 2

so that 2 =

p ln(2) = 06931

Finally, for 3 we have Z 3 1  −2 22 3 1 2 2 2 2 2 2   = −2  − −3  = −2  = − = 2  4 4 2 2 so that 3 = As a check we compute Z ∞ 3

p ln(2) = 11774

 −2 22 1 2 2 2   = −3  − 0 = −(11774) = 2 4

which is the correct result. Problem 12.13 Five thresholds are needed to define the six quantization regions. The five thresholds are −2 , −1 , 0, 1 , and 2 . The thresholds are selected so that the six quantization regions are equally likely. For 1 we have µ ¶ Z 1 Z ∞ 1 1 1 1 1 1 − 2 2 2 −2 22 =√   = − √   = −  6 2 2  2 0 2 1 so that 

1 

1 1 1 − = 2 6 3

CHAPTER 12. INFORMATION THEORY AND CODING

inding the inverse Q-function yields −1 For 2 we have

1 1 =√ 6 2

µ ¶ 1 1 = 04307 = 3  Z ∞

− 2 2 2



 = 

2

2 

Taking the inverse  function −1



µ ¶ 1 2 = 09674 = 6 

and 1 = 04307

and

2 = 09674

This gives the quantizing rule Quantizer Input −∞    −09674 −09674    −04307 −04307    0 0    04307 04307    09674 09674    ∞

Quantizer Output 0 1 2 3 4 5

Problem 12.14 The transition probabilities for the cascade of Channel 1 and Channel 2 are ∙ ¸ ∙ ¸∙ ¸ ∙ ¸ 11 12 09 01 075 025 07 03 = = 01 09 025 075 03 07 21 22 Therefore, the capacity of the overall channel is  = 1 + 07 log2 07 + 03 log2 03 = 01187 bits/symbol The capacity of Channel 1 is  = 1 + 09 log2 09 + 01 log2 01 = 05310 bits/symbol and the capacity of Channel 2 is  = 1 + 075 log2 075 + 025 log2 025 = 01887 bits/symbol

12.1. PROBLEM SOLUTIONS

It can be seen that the capacity of the cascade channel is less than the capacity of either channel taken alone, which is to be expected since each channel causes errors and the error probability on the cascade channel is greater than the error probability of either of the individual channels. Problem 12.15 The source entropy is 5 3 3 5  () = − log2 − log2 = 00944 8 8 8 8 and the entropy of the fourth-order extension is ¡ ¢   4 = 4 () = 4(09544) = 38177

The second way to determine the entropy of the fourth-order extension, is to determine the probability distribution of the extended source. We have µ ¶4 5 625 = 1 symbol with probability 8 4096 µ ¶3 3 5 375 4 symbols with probability = 8 8 4096 µ ¶2 µ ¶2 5 3 225 6 symbols with probability = 8 8 4096 µ ¶3 135 5 3 = 4 symbols with probability 8 8 4096 µ ¶4 81 3 = 1 symbol with probability 8 4096 Thus ¡

 

Problem 12.16

µ ¶ µ ¶ 375 225 625 375 225 625 log2 −4 log2 −6 log2 = − 4096 4096 4096 4096 4096 4096 µ ¶ 135 81 135 81 −4 log2 − log2 4096 4096 4096 4096 = 38177 bits/symbol

CHAPTER 12. INFORMATION THEORY AND CODING

For the fourth-order extension we have Source Symbol AAAA BAAA ABAA AABA AAAB AABB ABAB BAAB ABBA BABA BBAA ABBB BABB BBAB BBBA BBBB

P (·) 0.6561 0.0729 0.0729 0.0729 0.0729 0.0081 0.0081 0.0081 0.0081 0.0081 0.0081 0.0009 0.0009 0.0009 0.0009 0.0001

Codeword 0 100 101 110 1110 111100 1111010 1111011 1111100 1111101 1111110 111111100 111111101 111111110 1111111110 1111111111

The average wordlength:  = 19702 which is 4 = 197024 = 04926. The eﬃciencies are given in the following table:  1 2 3 4

 1 1.29 1.598 1.9702

 (  ) 0.4690 0.9380 1.4070 1.8760

Eﬃciency 46.90% 72.71% 88.05% 95.22%

Problem 12.17 For the Shannon-Fano code we have Source Symbol P( · ) Codeword 1/7 00 1 1/7 010 2 1/7 011 3 4 1/7 100 1/7 101 5 1/7 110 6 1/7 111 7

12.1. PROBLEM SOLUTIONS

11 000

001

010

011 100

1 0

101

2/7

2/7 2/7

1 0

1/7

0 1

Figure 12.4: Huﬀman coding for Problem 12.17. The entropy is  () = −7

1 1 log2 7 7

= log2 (7) = 28074

The average wordlength is µ ¶ 20 1 (2 + 3 + 3 + 3 + 3 + 3 + 3) = = 28571 = 7 7 The eﬃciency is therefore given by  () 28074 = 09826 = 9826% = 28571 

(12.1)

The diagram for determining the Huffman code is illustrated in Figure ??. This diagram yields the codewords shown. Note that we have six codewords of length three and one codeword of length two. This average wordlength is the same as for the Shannon-Fano code. Thus, the Huffman code and the Shannon-Fano code have the same average wordlength and the same efficiency. Problem 12.18 The codewords for the Shannon-Fano and Huffman codes are summarized in the following

CHAPTER 12. INFORMATION THEORY AND CODING

table: Source Symbol 1 2 3 4 5

Codewords Shannon-Fano Huﬀman 00 1 01 000 10 001 110 010 111 011

P (·) 0.40 0.20 0.17 0.13 0.10

The average wordlengths are:  = 225

(Shannon-Fano code)

 = 222

(Huﬀman code)

Note that the Huﬀman code gives the shorter average wordlength and therefore the higher eﬃciency. Problem 12.19 The entropy is  () = −085 log2 085 − 015 log2 015 = 060984 bits/symbol

We know that lim 

→∞

 ≤ 350 

and that  =  () →∞  lim

Therefore, for  large  () ≤ 350 or ≤

350 350 =  () 060984

which gives  ≤ 573921 symbols/s Since we cannot transmit partial symbols, the maximum symbol rate is 573 symbols/s. Problem 12.20 The Shannon-Fano and Huﬀman codes are as follows. (Note that the codes are not unique.)

12.1. PROBLEM SOLUTIONS

Source Symbol 1 2 3 4 5 6 7 8 9 10 11

Shannon-Fano Codeword 001 0000 0001 011 0100 0101 1000 1001 110 1110 1111

Huﬀman Codeword 010 0110 0111 0000 0001 0010 0011 100 101 110 111

The source entropy is  () = log2 11 = 34594 For the Shannon-Fano code the average wordlength =

1 [3(3) + 8(4)] = 37273 11

and the eﬃciency is =

 () 34594 = 9281% = 37273 

For the Huﬀman code the average wordlength =

1 [5(3) + 6(4)] = 35455 11

and the eﬃciency is =

 () 34594 = 9757% = 35455 

We see that the Huﬀman code is better in this case. Problem 12.21 The probability of message  is  ( ) =

Z 01

01(−1)

2  = 001 (2 − 1)

CHAPTER 12. INFORMATION THEORY AND CODING

A Huﬀman code yields the codewords in the following table: Source Symbol 10 9 8 7 6 5 4 3 2 1

P (·) 0.19 0.17 0.15 0.13 0.11 0.09 0.07 0.05 0.03 0.01

Codeword 11 000 010 011 100 101 0011 00100 001010 001011

The source entropy is given by  () = 30488 bits/source symbol and the average wordlength is  = 31000 binary symbols/source symbol A source symbol rate of 250 symbols (samples) per second yields a binary symbol rate of  = 250(31000) = 775 binary symbols/second and a source information rate of  =  () = 250(30488) = 752 bits/second

Problem 12.22 The source entropy is  () = −035 log2 035 − 025 log2 025 − 02 log2 02 − 015 log2 015 − 005 log2 005 = 21211 bits/source symbol

and the entropy of the second-order extension of the source is ¡ ¢   2 = 2 () = 42422

The second-order extension source symbols are shown in the following table along with the Shannon-Fano code.

12.1. PROBLEM SOLUTIONS

Source Symbol Probability 1 1 1 2 2 1 1 3 3 1 2 2 1 4 4 1 3 2 2 3 3 3 4 2 2 4 4 3 3 4 4 4 5 1 1 5 5 2 2 5 5 3 3 5 4 5 5 4 5 5

01225 00875 00875 00700 00700 00625 00525 00525 00500 00500 00400 00375 00375 00300 00300 00225 00175 00175 00125 00125 00100 00100 00075 00075 00025

Shannon-Fano Codeword 000 001 0100 0101 0110 0111 1000 1001 1010 10110 10110 11000 11001 11010 11011 111000 111001 111010 111011 111100 111101 1111100 1111101 1111110 1111111

The preceding table defnes the 25 codewords of the second-order extension of the source. Also given in the table are the probabilities of the 25 symbols of the second-order extension. This allows the average wordlength and efficiency to be easily computed. For the Huffman code, note that Computer Exercise 12.2 calls for the development of a MATLAB program to generate the Huffman code. After Computer Exercise 12.2 is worked, this problem offers an excellent set of source symbol probabilities for testing the problem generated in Computer Exercise 12.2. If this is done it will be found that the Huffman code offers a slight increase in efficency.

Problem 12.23

CHAPTER 12. INFORMATION THEORY AND CODING 6

x 10

8 7

Capacity

6 5 4 3 2 1 0 2 10

10 Bandwidth

Figure 12.5: Capacity for the AWGN channel given in Problem 12.24. The set of wordlengths from Table 12.3 is {1 3 3 3 5 5 5 5}. The Kraft inequality gives 8 X =1

¢ ¡ ¢ 1 3 ¡ 4 =1 2 = 2−1 + 3 2−3 + 4 2−5 = + + 2 8 32

Thus the Kraft inequality is satisfied. Problem 12.24 The Shannon-Hartley law with  = 60 and  = 10−5  gives Ã ¡ ¢! ¶ µ 60 105  =  log2 1 +  =  log2 1 +   This is illustrated in the following figure. It can be seen that as  → ∞ the capacity approaches a finite constant. For suﬃciently small , ln(1 + ) ≈ . Therefore lim  =

→∞

60(105 ) = 86562 × 106 ln(2)

as we see from the illustration that follows. Problem 12.25

12.1. PROBLEM SOLUTIONS

x 10

9 8

Capacity - bits/s

7 6 5 4 3 2 1 0

40 50 60 Signal Power - Watts

100

Figure 12.6: Capacity as a function of bandwidth for Problem 12.25. We now determine Ã ¡ ¢! ¡ 4¢ ¡ ¢  105  = 10 log2 1 + = 104 log2 (1 + 10 ) 4 10 The plot is given in Figure 12.6. For suﬃciently large  , ¡ ¢  ≈ 105 log2 ( ) and, therefore,  → ∞ as  → ∞.

Problem 12.26 Consider the ratio of the error probability of the coded system to the error probability of the uncoded system for high SNR. Since we are considering  À 1 we use the asympotic

CHAPTER 12. INFORMATION THEORY AND CODING

approximation for the Gaussian  function. Therefore, for the coded system Ãr ! r 2  1 −2 1 −(√2)2 1 √  √   =  ≈p =  2 2 2 2 For the uncoded system we simply let  = 1. Thus s ¶¸ ∙ µ 2 exp (−2) √ 1  =  exp 2 1 − =  2 exp (−2) 

√ Note that for large  the ratio becomes , where  is a constant defined by exp(2). For fixed , the loss in symbol error probability is therefore exponential with respect to the SNR , which is a greater loss than can be compensated by the error-correcting capability √ of the repetition code. Note that for fixed , the corresponding losss is proportional to . Problem 12.27 The parity-check matrix will have 15 − 11 = 4 rows and 15 columns. Since rows may be permuted, the parity-check matrix is not unique. One possible selection is ⎡ ⎤ 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 ⎢ 0 1 1 1 0 0 0 1 1 1 1 0 1 0 0 ⎥ ⎥ [] = ⎢ ⎣ 1 0 1 1 0 1 1 0 0 1 1 0 0 1 0 ⎦ 1 1 0 1 1 0 1 0 1 0 1 0 0 0 1 The corresponding generator matrix is ⎡ 1 0 0 0 0 ⎢ 0 1 0 0 0 ⎢ ⎢ 0 0 1 0 0 ⎢ ⎢ 0 0 0 1 0 ⎢ ⎢ 0 0 0 0 1 ⎢ ⎢ 0 0 0 0 0 ⎢ ⎢ ⎢ 0 0 0 0 0 ⎢ [] = ⎢ 0 0 0 0 0 ⎢ ⎢ 0 0 0 0 0 ⎢ ⎢ 0 0 0 0 0 ⎢ ⎢ 0 0 0 0 0 ⎢ ⎢ 0 0 0 0 1 ⎢ ⎢ 0 1 1 1 0 ⎢ ⎣ 1 0 1 1 0 1 1 0 1 1

0 0 0 0 0 1 0 0 0 0 0 1 0 1 0

0 0 0 0 0 0 1 0 0 0 0 1 0 1 1

0 0 0 0 0 0 0 1 0 0 0 1 1 0 0

0 0 0 0 0 0 0 0 1 0 0 1 1 0 1

0 0 0 0 0 0 0 0 0 1 0 1 1 1 0

⎤ 0 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ ⎥ ¸ 0 ⎥ ∙ 11 ⎥ 0 ⎥= 11 ⎥ 0 ⎥ ⎥ 0 ⎥ ⎥ 1 ⎥ ⎥ 1 ⎥ ⎥ 1 ⎥ ⎥ 1 ⎦ 1

12.1. PROBLEM SOLUTIONS

where 11 is the 11×11 identity matrix and 11 represents the first 11 columns of the paritycheck matrix. From the structure of the parity-check matrix, we see that each parity symbol is the sum of 7 information symbols. Since 7 is odd, the all-ones information sequence yields the parity sequence 1111. This gives the codeword [ ] = [111111111111111] A single error in the third position yields a syndrome which is the third column of the parity-check matrix. Thus, for the assumed parity-check matrix, the syndrome is ⎡ ⎤ 0 ⎢ 1 ⎥ ⎥ [] = ⎢ ⎣ 1 ⎦ 0 Problem 12.28 The generator matrix corresponding to the given parity-check matrix is ⎤ ⎡ 1 0 0 0 ⎢ 0 1 0 0 ⎥ ⎥ ⎢ ⎢ 0 0 1 0 ⎥ ⎥ ⎢ ⎥ [] = ⎢ ⎢ 0 0 0 1 ⎥ ⎢ 0 1 1 1 ⎥ ⎥ ⎢ ⎣ 1 0 1 1 ⎦ 1 1 0 1

The code words are of the form

1 2 3 4 1 2 3 ⎡ ⎤ 1 = 2 ⊕ 3 ⊕ 4 ⎣ 2 = 1 ⊕ 3 ⊕ 4 ⎦ 3 = 1 ⊕ 2 ⊕ 4

where

Thus, the 24 = 16 code words are (read veritcally) a1 a2 a3 a4

0 0 0 0

0 0 0 1

0 0 1 0

0 0 1 1

0 1 0 0

0 1 0 1

0 1 1 0

0 1 1 1

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1

c1 c2 c3

0 0 0

1 1 1

0 0 1

1 0 1

0 1 0

0 0 1

1 0 0

0 1 1

1 0 0

1 0 1

0 0 0

1 0 0

0 0 1

0 0 0

1 1 1

CHAPTER 12. INFORMATION THEORY AND CODING

Problem 12.29 We first determine [ ] = [] [ ] for  = 1 2. The generator matrix for the previous problem is used. For [1 ] we have ⎡ ⎤ 0 ⎡ ⎤ ⎢ 1 ⎥ ⎢ ⎥ 0 ⎢ ⎥ ⎢ 1 ⎥ ⎢ 1 ⎥ ⎥ ⎢ ⎥ [] [1 ] = [] ⎣ ⎦ = ⎢ ⎢ 1 ⎥ = [1 ] 1 ⎢ 1 ⎥ ⎢ ⎥ 1 ⎣ 0 ⎦ 0 and for [2 ] we have

⎤ 1 ⎡ ⎤ ⎢ 0 ⎥ ⎢ ⎥ 1 ⎢ ⎥ ⎢ 0 ⎥ ⎢ 1 ⎥ ⎥ ⎢ ⎥ [] [2 ] = [] ⎣ ⎦ = ⎢ ⎢ 0 ⎥ = [2 ] 1 ⎢ 1 ⎥ ⎢ ⎥ 0 ⎣ 0 ⎦ ⎡

We note that [1 ] ⊕ [2 ] 

Problem 12.30 For a Hamming code with  =  −  parity check symbols, the wordlength  is  = 2 − 1 and  = 2 −  − 1 The rate is

2 −  − 1  =  2 − 1 If  is large,  is large. As  → ∞ both  and  approach 2 . Thus =

lim  = lim  =

→∞

Problem 12.31

→∞

2 =1 2

12.1. PROBLEM SOLUTIONS

The codeword is of the form (1 2 3 4 1 2 3 ) where 1 = 1 ⊕ 2 ⊕ 3 2 = 2 ⊕ 3 ⊕ 4 3 = 1 ⊕ 2 ⊕ 4 This gives the generator matrix ⎡

1 ⎢ 0 ⎢ ⎢ 0 ⎢ [] = ⎢ ⎢ 0 ⎢ 1 ⎢ ⎣ 0 1

0 1 0 0 1 1 1

0 0 1 0 1 1 0

⎤ 0 0 ⎥ ⎥ 0 ⎥ ⎥ 1 ⎥ ⎥ 0 ⎥ ⎥ 1 ⎦ 1

Using the generator matrix, we can determine the codewords. The codewords are (read vertically) a1 a2 a3 a4

0 0 0 0

1 0 0 0

1 1 0 0

1 1 1 0

1 1 1 1

0 0 0 1

0 0 1 1

0 1 1 1

1 0 1 1

1 1 0 1

1 0 0 1

0 1 1 0

0 1 0 0

0 0 1 0

0 1 0 1

1 0 1 0

c1 c2 c3

0 0 0

1 0 1

0 1 0

1 0 0

1 1 1

0 1 1

1 0 1

0 1 0

0 0 0

0 0 1

1 1 0

0 0 1

1 1 1

1 1 0

1 0 0

0 1 1

Let  denote the -th code word and let  →  indicate that a cyclic shift of  yields  . The cyclic shifts are illustrated in the following figure. Problem 12.32 The parity-check matrix for the coder in the previous problem is ⎡

⎤ 1 1 1 0 1 0 0 [] = ⎣ 0 1 1 1 0 1 0 ⎦ 1 1 0 1 0 0 1 For the received sequence [] = [1101001]

CHAPTER 12. INFORMATION THEORY AND CODING

1

5

2

3

12

9

15

14

6

4

8

7

11

13

16

10

Figure 12.7: Illustration of cyclic property. the syndrome is

⎡

⎤ 0 [] = [] [] = ⎣ 0 ⎦ 0

This indicates no errors, which is in agreement with Figure 12.15. For the received sequence [] = [1101011] the syndrome is

⎡

⎤ 0 [] = [] [] = ⎣ 1 ⎦ 0

This syndrome indicates an error in the 6th position. Thus the assumed transmitted codeword is [ ] = [1101001] which is also in agreement with Figure 12.15. Problem 12.33 The probability of two errors in a 7 symbol codeword is µ ¶ 7  {2 errors} = (1 −  )5 2 = 212 (1 −  )5 2 and the probability of three errors in a 7 symbol codeword is µ ¶ 7  {3 errors} = (1 −  )4 3 = 353 (1 −  )4 3

12.1. PROBLEM SOLUTIONS

The ratio of  {3 errors} to  {2 errors} is =

353 (1 −  )4 5  Pr {3 errors} = = 5 2 Pr {2 errors} 3 1 −  21 (1 −  )

We now determine the coded error probability  . Since BPSK modulation is used Ãr ! 2  =  7 where  is  0 . This gives us the following table  8 dB 9 dB 10 dB 11 dB 12 dB 13 dB 14 dB

 00897 00660 00455 00290 00167 00085 00037

errors}  = Pr{3 Pr{2 errors} 01642 01177 00794 00497 00278 00141 00062

It is clear that as the SNR, , increases, the value of  becomes negligible. Problem 12.34 For the convolutional coder shown in Figure 12.24 we have the three outputs 1 = 1 ⊕ 2 ⊕ 3

2 = 1

2 = 1 ⊕ 2 Shifting in a 0 gives, in terms of the initial shift register contents 1 = 0 ⊕ 1 ⊕ 2 = 0 ⊕ 

2 = 0

3 = 0 ⊕ 1 and shifting in a binary 1 gives 1 = 1 ⊕ 1 ⊕ 2 = 1 ⊕ 

2 = 1

3 = 1 ⊕ 1

CHAPTER 12. INFORMATION THEORY AND CODING

where  = 1 ⊕ 2 . Clearly 1 = 0 ⊕  or 1 = 1 ⊕  are complements for  = 0 or 1. The output 2 for a 0 or 1 input are obviously complements. Also 1 = 0 ⊕ 1 or 1 = 1 ⊕ 1 are complements for 1 = 0 or 1

Problem 12.35 For the convolutional coder shown in Figure 12.24 we have the three outputs

1 = 1 ⊕ 2 ⊕ 3 ⊕ 4

2 = 1 ⊕ 2 ⊕ 4

Shifting in a 0 gives, in terms of the initial shift register contents

1 = 0 ⊕ 1 ⊕ 2 ⊕ 3 = 0 ⊕ 

2 = 0 ⊕ 1 ⊕ 3 = 0 ⊕  and shifting in a binary 1 gives

1 = 1 ⊕ 1 ⊕ 2 ⊕ 3 = 1 ⊕ 

2 = 1 ⊕ 1 ⊕ 3 = 1 ⊕ 

where  = 1 ⊕ 2 ⊕ 3 and  = 1 ⊕ 3 . Clearly 1 = 0 ⊕  or 1 = 1 ⊕  are complements for  = 0 or 1. In the same manner 2 = 0 ⊕  or 2 = 1 ⊕  are complements for  = 0 or 1.

Problem 12.36 For the encoder shown, the constraint span is  = 4. We first need to compute the states for the encoder. The output is 1 2 where 1 = 1 ⊕ 2 ⊕ 3 ⊕ 4 and 2 = 1 ⊕ 2 ⊕ 4 . We now compute the state transitions and the output using the state diagram shown in

12.1. PROBLEM SOLUTIONS

Figure 12.8. This gives the following table. State A

Previous 1 2 2 000

001

010

011

100

101

110

111

Input 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

1 2 3 4 0000 1000 0001 1001 0010 1010 0011 1011 0100 1100 0101 1101 0110 1110 0111 1111

Current State A E A E B F B F C G C G D H D H

Output 00 11 11 00 10 01 01 10 11 00 00 11 01 10 10 01

Problem 12.37 For the encoder shown, we have 1 = 1 ⊕ 2 ⊕ 3 and 2 = 1 ⊕ 3 . The trellis diagram is illustrated in Figure 12.9. The state transitions, and the output corresponding to each transition appears in the following table. State A

1 2 00

Input 0 1 0 1 0 1 0 1

1 2 3 000 100 001 101 010 110 011 111

State A C A C B D B D

Output 00 11 11 00 10 01 01 10

Problem 12.38 The source rate is  = 5000 symbols/second. The channel symbol rate is ³´   =  = 5000  

CHAPTER 12. INFORMATION THEORY AND CODING (0 0)

A (1 1) (1 1) (1 0) C B

(0 1)

(1 1)

(0 0)

(0 1) (0 0)

(0 0)

(0 1) (1 0) (1 1) G F (1 0) (1 0) H (0 1)

Figure 12.8: State diagram for Problem 12.36. If the burst duration is  , then

³´  = 5000   channel symbols can be affected. Assuming that  = 02 s, the burst length, expressed in symbols, is ³´ ³´  =   = (5000) (02) = 1000   symbols affected by the burst. The interleaving array contains  rows with each row representing a ( ) codeword. We now must now determine  and . Since   , it is clear that the burst affects more than 1000 code symbols. We also know that the rate of a Hamming code approaches 1 as  gets large, so that for large . /   =  

12.1. PROBLEM SOLUTIONS

(00)

(11)

(00)

(11)

(10)

(00) (10)

(01)

(00)

(11)

(10)

(01) (01)

(10) Termination

Steady-State Transitions

Figure 12.9: Trellis diagram for Problem 12.37. will exceed 1000 by only a small amount. We therefore try a (1023,1013) Hamming code having  −  = 10 parity symbols. The number of symbols aﬀected by the burst is  = 1000

1023 1013

= 10099

Since  must be an integer lower bounded by 1009.9, we let  = 1010. The full interleaving array contains  symbols, since  may be received in error (worst case) ( − 1) must be received correctly. Since the symbol rate on the channel is 5000 this gives a required error-free span of 1010 1022 ( − 1) = = 20443 s 5000 5000 10231013 This is probably not very practical but it illustrates the process and the problem of using long codes with low error-correcting capabilities for interleaving. Problem 12.39

CHAPTER 12. INFORMATION THEORY AND CODING

The source rate is  = 5000 symbols/second. The channel symbol rate is µ ¶ 23   =  = 5000  12 since we have a (23,12) code. If the burst duration is  , then µ ¶  23   =   = 5000  12 channel symbols can be aﬀected by the burst. The burst can aﬀect 3 columns, containing 3 symbols, of the interleaving table. Assuming that  = 02 s, the burst length, expressed in symbols, is µ ¶ 23 = 19167 symbols 3 =   = (5000) (02) 12 Thus

19167 = 63889 3 We therefore set  = 639, which gives an interleaving array with 639 rows and 23 columns. Since 3 columns can contain errors we must have 20 error-free columns representing 20(639) symbols. This represents an error-free span of =

20 20(639) = 13336 =  5000(2312)

We see that the more powerful code results in a much shorter required error-free span and a smaller interleaving array. Problem 12.40 The probability of  transmissions is the probability of  − 1 retransmissions. This is 2−1 . The expected value of  is therefore (∞ ) X −1 2 { } =  =1

From

∞ X

 =

=0

1  1−

||  1

we can obtain, by diﬀerentiating both sides of the above wrt , ∞ X

=1

−1 =

1 (1 − )2

12.1. PROBLEM SOLUTIONS

Figure 12.10: Detection gain characteristice for Problem 12.41. There the expected value of  is  { } =

1 (1 − 2 )2

Problem 12.41 The required plot is shown in Figure 12.10. Problem 12.42 The expected number of transmissions,  , is  = 1 Pr( = 1) + 2 Pr( = 2) + 3 Pr( = 3) · · ·  Pr( = ) · · ·

CHAPTER 12. INFORMATION THEORY AND CODING

or =

∞ X

2−1 

=1

where  is the probability of a decision on the   transmission, which is 1 ( − 1 erasures plus a decision). Thus, with  =  − 1 we have =

∞ X

2−1 =

=1

12.2

1 (1 − 2 )2

Computer Exercises

Computer Exercise 12.1 For two messages the problem is easy since there is only a single independent probability. The MATLAB program follows % File: ce12_1a.m a = 0.001:0.001:0.999; b =1-a; lna = length(a); H = zeros(1,lna); H=-(a.*log(a)+b.*log(b))/log(2); plot(a,H); xlabel(’a’) ylabel(’Entropy’) The plot is illustrated in Figure 12.11. For three source messages we have two independent probabilities. Two of them will be specified and the remaining probability can then be determined. The resulting MATLAB code follows: % File: ce12_1b.m n = 50; nn = n+1; H = zeros(nn,nn); z = (0:n)/n; z(1) = eps;

% number of points in vector % increment for zero % initialize entropy array % probability vector % prevent problems with log(0)

12.2. COMPUTER EXERCISES

1 0.9 0.8 0.7

E ntropy

0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.1

0.2

0.3

0.4

0.5 a

0.6

0.7

0.8

0.9

Figure 12.11: Entropy for binary source. for i=1:nn for j=1:nn c = 1-z(i)-z(j); % calculate third probability if c0 xx = [z(i) z(j) c]; H(i,j)= -xx*log(xx)’/log(2); % compute entropy end end end colormap([0 0 0]) % b&w colormap for mesh mesh(z,z,H); % plot entropy view(-45,30); % set view angle xlabel(’Probability of message a’) ylabel(’Probability of message b’)

CHAPTER 12. INFORMATION THEORY AND CODING

Figure 12.12: Entropy for three-symbol source. zlabel(’Entropy’) figure contour(z,z,H,10) xlabel(’Probability of message a’) ylabel(’Probability of message b’)

% new figure % plot contour map

The resulting plot is illustrated in Figure 12.12. One should experiment with various viewing locations. Next we draw a countour map with 10 countour lines. The result is illustrated in Figure 10.17. Note the peak at 0.33, 0.33, indicating that the maximum entropy occurs when all three source messages occur with equal probability Computer Exercise 12.2 The solution of this computer example is, in large part, taken from the book by Proakis, Salehi, and Bauch (Comtemporary Communication Systems using MATLAB, Thomson-

12.2. COMPUTER EXERCISES

Brooks/Cole, 2004, ISBN 0-534-40617-3).1 Modifications made to the ortiginally published program include changing the program from a function to a script, allowing the user to determine probability distributions and source codes for extended sources, allowing the user to enter the source distribution and desired extension order interactively and, in addition, the entropy of the source, the entropy of the extended source, the average word length, and the eﬃciency are displayed as the program is executed. This was done since a typical use of source coding is to continue incrementing the source extension until the eﬃciency reaches a point where transmission can be accomplished. The resulting MATLAB code follows. % File: ce12_2.m clear all msgprobs = input(’Enter message probabilities as a vector, ... e.g., [0.5 0.25 0.25]  ’); ext = input(’Enter desired extension (1 if no extension is desired)  ’); entp = -sum(msgprobs.*log(msgprobs))/log(2); % original source entropy dispa = [’The source entropy is ’, ... num2str(entp,’%15.5f’),’ bits/symbol.’]; disp(dispa) % % Determine probabilities and entropy of extended source. % if ext1 b = msgprobs; % working array for k=2:ext b = kron(b,msgprobs); % Kronecker tensor product end msgprobs = b; % probabilities for extended source entp = -sum(msgprobs.*log(msgprobs)) ... /log(2); % entropy of extended source dispaa = [’The extended source entropy is ’, ... um2str(entp,’%15.5f’),’ bits/symbol.’]; disp(dispaa) end % % Continue 1

This book is highly recommended as a supplemental text in basic communications courses. We have found it useful as a resource for in-class demonstrations and for homework assignments extending material covered in class. This particular computer example, as originally developed by Proakis, Salehi, and Bauch, provides an excellent example of several MATLAB programming techniques. It has been said that a key to understanding eﬃcient MATLAB programming techniques is to understand the use of the colon and the find command. This program makes eﬃcient use of both.

CHAPTER 12. INFORMATION THEORY AND CODING % n = length(msgprobs); % length of source (or extended source) q = msgprobs; % copy of msgprobs m = zeros(n-1,n); % initialize for i=1:n-1 [q,l] = sort(q); % order message probabilities m(i,:) = [l(1:n-i+1),zeros(1,i-1)]; q = [q(1)+q(2),q(3:n),1]; % combine messages in column vector end % % Code words are estabilished in the c array. % for i = 1:n-1 c(i,:) = blanks(n*n); % initialize character array end c(n-1,n) = ’0’; c(n-1,2*n) = ’1’; for i=2:n-1 c(n-i,1:n-1) = c(n-i+1,n*(find(m(n-i+1,:)==1)) ... (n-2):n*(find(m(n-i+1,:)==1))); c(n-i,n) = ’0’; c(n-i,n+1:2*n-1) = c(n-i,1:n-1); c(n-i,2*n) = ’1’; for j=1:i-1 c(n-i,(j+1)*n+1:(j+2)*n) = ... c(n-i+1,n*(find(m(n-i+1,:)==j+1)-1)+1:n*find(m(n-i+1,:)==j+1)); end end % % Establish code words, average word length, and entropy. % for i=1:n % determine codewords and wordlengths codewords(i,1:n) = c(1,n*(find(m(1,:)==i)-1)+1:find(m(1,:)==i)*n); wrdlengths(i) = length(find(abs(codewords(i,:))~=32)); end avewl = sum(msgprobs.*wrdlengths); % average word length codewords % display code words disp3 = [’The average wordlength is ’, ... num2str(avewl,’%15.5f’),’ symbols.’]; disp(disp3) disp4 = [’The efficiency is ’, ... num2str(entp*100/avewl,’%15.5f’),’ %.’]; disp(disp4)

12.2. COMPUTER EXERCISES

% End of function file. We will use the preceding MATLAB program to work Problem 12.28. The MATLAB dialog follows. Computer Exercise 12.3 The MATLAB code for generating the performance curves illustrated in Figure 11.28 in the text follows. % File: ce12_3.m zdB = 0:30; z = 10.^(zdB/10); qa = 0.5*exp(-0.5*z); qa3 = 0.5*exp(-0.5*z/3); qa7 = 0.5*exp(-0.5*z/7); pa7 = 35*((1-qa7).^3).*(qa7.^4)+21*((1-qa7).^2).*(qa7.^5)... +7*(1-qa7).*(qa7.^6)+(qa7.^7); pa3 = 3*(1-qa3).*(qa3.^2)+(qa3.^3); qr = 0.5./(1+z/2); qr3 = 0.5./(1+z/(2*3)); qr7 = 0.5./(1+z/(2*7)); pr7 = 35*((1-qr7).^3).*(qr7.^4)+21*((1-qr7).^2).*(qr7.^5)... +7*(1-qr7).*(qr7.^6)+(qr7.^7); pr3 = 3*(1-qr3).*(qr3.^2)+(qr3.^3); semilogy(zdB,qr,zdB,pr3,zdB,pr7,zdB,qa,zdB,pa3,zdB,pa7) axis([0 30 0.0001 1]) xlabel(’Signal-to-Noise Ratio, z - dB’) ylabel(’Probability ’) Executing the code yields the results illustrated in Figure 12.13 in the text. The curves can be identified from Figure 10.18 in the text. Computer Exercise 12.4 For the rate 0.5 BCH codes, we use the following MATLAB program. % File: ce12_4a.m zdB = 0:0.5:10; % set Eb/No in dB z = 10.^(zdB/10); % convert to linear scale ber1 = q(sqrt(4*2*z/7)); % SER for (7,4) BCH code ber2 = q(sqrt(7*2*z/15)); % SER for (15,7) BCH code ber3 = q(sqrt(16*2*z/31)); % SER for (31,16) BCH code

CHAPTER 12. INFORMATION THEORY AND CODING

-1

P robability

-2

-3

-4

10 15 20 Signal-to-Noise Ratio, z - dB

Figure 12.13: Results for Computer Exercise 12.3. ber4 = q(sqrt(30*2*z/63)); % SER for (63,30) BCH code berbch1 = ser2ber(2,7,3,1,ber1); % BER for (7,4) BCH code berbch2 = ser2ber(2,15,5,2,ber2); % BER for (15,7) BCH code berbch3 = ser2ber(2,31,7,3,ber3); % BER for (31,16) BCH code berbch4 = ser2ber(2,64,13,6,ber4); % BER for (63,30) BCH code semilogy(zdB,berbch1,’k’,zdB,berbch2,’k’,zdB,berbch3,’k’,zdB,berbch4,’k’) xlabel(’E_b/N_o in dB’) % label x axis ylabel(’Bit Error Probability’) % label y axis % End of script file. Executing the code for the rate 0.5 code yields the results shown in Figure 12.14. For SNRs exceeding 5 dB, the order of the plots is in order of word length , with  = being the top curve (least error correcting capability) and  = 63 being the bottom curve (greatest error correcting capability).

12.2. COMPUTER EXERCISES

-2

-4

Bit Error Probability

-6

-8

-10

-12

-14

5 6 Eb/No in dB

Figure 12.14: Rate 1/2 BCH code results for Computer Exercise 12.4.

CHAPTER 12. INFORMATION THEORY AND CODING

For the rate 0.75 BCH codes, we use the following MATLAB program. % File: ce12_4b.m zdB = 0:0.5:10; % set Eb/No in dB z = 10.^(zdB/10); % convert to linear scale ber1 = q(sqrt(11*2*z/15)); % SER for (15,11) BCH code ber2 = q(sqrt(21*2*z/31)); % SER for (31,21) BCH code ber3 = q(sqrt(45*2*z/63)); % SER for (63,45) BCH code ber4 = q(sqrt(99*2*z/127)); % SER for (127,99) BCH code berbch1 = ser2ber(2,15,3,1,ber1); % BER for (15,11) BCH code berbch2 = ser2ber(2,31,5,2,ber2); % BER for (31,21) BCH code berbch3 = ser2ber(2,63,7,3,ber3); % BER for (63,45) BCH code berbch4 = ser2ber(2,127,9,4,ber4); % BER for (127,99) BCH code semilogy(zdB,berbch1,zdB,berbch2,zdB,berbch3,zdB,berbch4) xlabel(’E_b/N_o in dB’) % label x axis ylabel(’Bit Error Probability’) % label y axis % End of script file. Executing the code for the rate 0.75 code yields the results shown in Figure 12.15. For SNRs exceeding 5 dB, the order of the plots is in order of word length , with  = 15 being the top curve (least error correcting capability) and  = 127 being the bottom curve (greatest error correcting capability). Computer Exercise 12.5 In order to illustrate the problems associated with using nchoosek with large arguments we compute » nchoosek(1000,500) Warning: Result may not be exact.  In C:\MATLABR11\toolbox\matlab\specfun\nchoosek.m at line 50 ans = 2.7029e+299 The warning indicates a possible with numerical percision. A possible way to mitigate this is to perform the calculation using logarithms as illustrated in the function below. %File: nkchoose.m Function out=nkchoose(n,k) % Computes n!/k!/(n-k)! a = sum(log(1:n));

% ln of n!

12.2. COMPUTER EXERCISES

Bit Error Probability

-5

-10

-15

5 6 Eb/No in dB

Figure 12.15: Rate 3/4 BCH code results for Computer Exercise 12.4.

CHAPTER 12. INFORMATION THEORY AND CODING b = sum(log(1:k)); c = sum(log(1:(n-k))); out = round(exp(a-b-c)); % End of function file.

% ln of k! % ln of (n-k)! % result

The main program for evaluating the performance of the (511,385) BCH code is shown below. % File: ce12_5.m zdB = 0:0.5:10; % set Eb/No in dB z = 10.^(zdB/10); % convert to linear scale ber1 = q(sqrt(z)); % FSK result ber2 = q(sqrt(385*z/511)); % SER for (511,385) BCH code ber3 = q(sqrt(768*z/1023)); % SER for (1023,768) BCH code berbch1 = ser2ber1(2,511,29,14,ber2); % BER for (511,385) BCH code berbch2 = ser2ber1(2,1023,53,26,ber3); % BER for (1023,768) BCH code semilogy(zdB,ber1,’k-’,zdB,berbch1,’k--’,zdB,berbch2,’k-.’) xlabel(’E_b/N_o in dB’) % label x axis ylabel(’Bit Error Probability’) % label y axis legend(’Uncoded’,’(511,385) BCH code’,’(1023,768) BCH code’,3) % End of script file. Note that this program calls ser2ber1 rather than ser2ber. In order to evaluate the performance of the (1023,768) BCH code, it is necessay to modify the routine for mapping the code symbol error rate so that the calculation is performed using logarithms as the following MATLAB code indicates. % File: ser2ber1.m function [ber] = ser2ber(q,n,d,t,ps) lnps = length(ps); % length of error vector ber = zeros(1,lnps); % initialize output vector for k=1:lnps % iterate error vector ser = ps(k); % channel symbol error rate sum1 = 0; sum2 = 0; % initialize sums for i=(t+1):d lnterm = log(nkchoose(n,i))+(i*log(ser))+(n-i)*log((1-ser)); term = exp(lnterm); sum1 = sum1+term; end for i=(d+1):n

12.2. COMPUTER EXERCISES

-5

Bit Error Probability

-10

-15

Uncoded (511,385) BCH code (1023,768) BCH code

-20

5 6 Eb/No in dB

Figure 12.16: Performance results for Computer Exercise 12.5. lnterm = log(i)+log(nkchoose(n,i))+(i*log(ser))+(n-i)*log((1-ser)); term = exp(lnterm); sum2 = sum2+term; end ber(k) = (q/(2*(q-1)))*((d/n)*sum1+(1/n)*sum2); end % End of function file. Executing the program yields the results shown in Figure 12.16. Computer Exercise 12.6 In order to work this computer exercise we write a MATLAB program to determine the outputs for each of the 16 tree branches illustrated in Figure 12.25. The inputs for each

CHAPTER 12. INFORMATION THEORY AND CODING

tree branch are illustrated in the right-hand column of Figure 12.25. The outputs for the portion of the tree are generated by inputting the appropriate binary vector corresponding the portion of the tree of interest. The MATLAB code follows. % File: ce12_6.m clear all invector = input(’Enter input as vector (in brackets with spaces)  ’); d = [’The input sequence is ’,num2str(invector),’.’]; disp(d) sr = [0 0 0]; for k=1:4 sr = [invector(1,k) sr(1,1) sr(1,2)]; v1 = dec2bin(rem(sr(1,1)+sr(1,2)+sr(1,3),2)); v2 = dec2bin(sr(1,1)); v3 = dec2bin(rem(sr(1,1)+sr(1,2),2)); c = [v1 v2 v3]; out(1,k) = bin2dec(c); end out = dec2bin(out) % End of script file. The program is illustrated by generating the outputs for the path labeled ’A’ in Figure 12.25., for which the input bit stream is 1010. The MATLAB dialog is as follows: » ce12_6 Enter input sequence as a vector (in brackets with spaces)  [1 0 1 0] The input sequence is 1 0 1 0. out = 111 101 011 101 The four outputs are represented as the rows following ‘out’ in the preceding dialog. This corresponds to the sequence 111 101 011 101, which is Path A as shown in Figure 12.25 in the text. (Note: If all outputs are 0, the sequence 000 is represented by a single 0.) Computer Exercise 12.7 This Computer Exercise is worked by modifying the MATLAB code in the text (c12ce6.m) to allow the calculation of results for  = 01 and  = 03 and plotting the results together so that they are easily compared. The revised MATLAB code follows.

12.2. COMPUTER EXERCISES

% File: ce12_7.m g = [0.1 0.3]; % gamma zdB = 0:0.1:10; % z in dB z = 10.^(zdB/10); % vector of z values qt = Q(sqrt(2*z)); % gamma=0 case for k=1:2 q1 = Q((1-g(k))*sqrt(2*z)); q2 = Q((1+g(k))*sqrt(2*z)); p2 = q1-q2; % P2 p1 = q2; % P1 pe(k,:) = p1./(1-p2); % probability of error N(k,:) = 1./((1-p2).*(1-p2)); end subplot(2,2,1) semilogy(zdB,pe(1,:),zdB,qt) title(’gamma = 0.1’) xlabel(’z - dB’); ylabel(’Error Probability’); grid subplot(2,2,2) semilogy(zdB,pe(2,:),zdB,qt) title(’gamma = 0.3’) xlabel(’z - dB’); ylabel(’Error Probability’); grid subplot(2,2,3) plot(zdB,N(1,:)) xlabel(’z - dB’); ylabel(’Ave. Transmissions/Symbol’); axis([0 10 1 1.4]); grid subplot(2,2,4) plot(zdB,N(2,:)) xlabel(’z - dB’); ylabel(’Ave. Transmissions/Symbol’); axis([0 10 1 1.4]); grid % End of script file. Executing the program gives the results shown in the illustration below. As expected, as  is increased, the error probability deceases at the cost of an increase in the average number of transmissions. The key, however, is that increased performance is achieved with only a moderate increase in the rate of retransmissions. This is especially true at higher SNRs.

Computer Exercise 12.8 We start with Computer Example 12.7 and add the required code to give the numbers of matching elments. The MATLAB code follows.

CHAPTER 12. INFORMATION THEORY AND CODING

gamma = 0.1

Error Probability

-5

-10

5 z - dB

1.4 1.3 1.2 1.1 1

-5

5 z - dB

Ave. Transmissions/Symbol

Error Probability

Ave. Transmissions/Symbol

gamma = 0.3

5 z - dB

1.4 1.3 1.2 1.1 1

Figure 12.17: Performance results for feedback channel.

12.2. COMPUTER EXERCISES

% File: ce12_8.m elements = 40; clear all; for k=1:elements z(k)=rand(1); if z(k)0.3 z(k)=’B’; else z(k)=’W’; end end mask = [’W’ ’B’]; t = [z(2:end) z(end)]; diff = z-t; % subtract string with its offset trans = [0 find(diff ~= 0)]; % vector with transition indexes if (trans(end) ~= numel(z)) trans = [trans numel(z)] % nasty case (no transition at end) end r = trans(2:end) - trans(1:end-1); % lengths of consecutive B’s and W’s count = [sum(r(1:2:end)) sum(r(2:2:end))]; % number of B’s and W’s if(z(1)==’W’) % if sequence starts with ’W’ count = count([2 1]); % swap count and mask mask = mask([2 1]); end disp(sprintf(’Sequence: %s’,z)); disp(sprintf(’B count: %i W count: %i’,count)); % Display results BWstr = []; for k=1:numel(r) BWstr = strcat(BWstr,num2str(r(k)),mask(mod(k,2)+1)); end disp(sprintf(’BW string: %s’,BWstr)); % End of script file. Executing the program gives (Keep in mind that separate executions will give separate results.):  CE12_8 Sequence: WWWWBWWWWWBWWBBWWWWBWBWWWWWBBBWBWBWWBBWW B count: 13 W count: 27

CHAPTER 12. INFORMATION THEORY AND CODING BW string:

4W1B5W1B2W2B4W1B1W1B5W3B1W1B1W1B2W2B2W

Acknowledgement: Thanks to Dr. John Tranter at the University of Minnesota for his help on this Computer Exercise.

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

Problem A.1 All parts of the problem are solved using the relation p Vrm s = 4kT RB where k B

1:38 10 23 J/K 30 MHz = 3 107 Hz

= =

a. For R = 10; 000 ohms and T = T0 = 290 K p Vrm s = 4 (1:38 10 23 ) (290) (104 ) (3 = 6:93 10 5 V rms = 69:3 V rms b. Vrm s is smaller than the result in part (a) by a factor of

107 )

10 = 3:16: Thus

100 = 10: Thus

Vrm s = 21:9 V rms c. Vrm s is smaller than the result in part (a) by a factor of Vrm s = 6:93 V rms d. Each answer becomes smaller by factors of 2; Problem A.2 Use

eV kT

10 = 3:16; and 10, respectively.

eV kT

I = Is exp We want I > 20Is or exp

1 > 20. 19

e 1:6 10 a. At T = 290 K, kT = 1:38 1023 290 = 40, so we have exp (40V ) > 21 giving

V i2rm s or

i2rm s B

ln (21) = 0:0761 volts 40 eV = 2eIB ' 2eBIs exp kT eV = 2eIs exp kT

= =

2 1:6

10 19

1:0075

1:5

10 5 exp (40

0:0761)

A /Hz

e b. If T = 90 K, then kT = 129, and for I > 20Is , we need exp(129V ) > 21 or

V >

ln (21) = 2:36 129

10 2 volts

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

Thus i2rm s B

2 1:6

10 19

1:0079

10 22 A2 /Hz

1:5

10 5 exp (129

0:0236)

approximately as before. Problem A.3 a. Use Nyquist’s formula to get the equivalent circuit of Req in parallel with RL , where Req is given by R3 (R1 + R2 ) Req = R 1 + R 2 + R3 The noise equivalent circuit consists of a rms noise voltage, Veq , in series with Req and a rms noise voltage, VL , in series with with RL with these series combinations being in parallel. The equivalent noise voltages are Veq

p 4kT Req B p = 4kT RL B =

The rms noise voltages across the parallel combination of Req and RL , by using superposition and voltage division, are Ve q RL VL Re q V01 = and V02 = Req + RL Req + RL Adding noise powers to get V02 we obtain V02

= = =

2 2 RL Veq

2 VL2 Req + 2 2 (Req + RL ) (Req + RL ) (4kT B) RL Req Req + RL RL R3 (R1 + R2 ) 4kT B R 1 R 3 + R2 R 3 + R1 RL + R 2 R L + R 3 R L

Note that we could have considered the parallel combination of R3 and RL as an equivalent load resistor and found the Thevenin equivalent. Let Rjj =

R 3 RL R3 + R L

The Thevenin equivalent resistance of the whole circuit is then R R

Req2

3 L (R1 + R2 ) Rjj (R1 + R2 ) L = RR33+R R L Rjj + R1 + R2 R +R + R1 + R2

RL R3 (R1 + R2 ) R 1 R 3 + R2 R3 + R1 RL + R2 RL + R 3 R L

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

and the mean-square output noise voltage is now V02 = 4kT BReq2 which is the same result as obtained before. b. With R1 = 2000 ; R2 = RL = 300 ; and R3 = 500 V02 B

= = =

, we have

RL R3 (R1 + R2 ) R1 R3 + R 2 R 3 + R 1 R L + R 2 R L + R3 RL 4 1:38 10 23 (290) (300) (500) (2000 + 300) 2000 (500) + 300 (500) + 2000 (300) + 300 (300) + 500 (300)

4kT B

2:775

10 18 V2 /Hz

Problem A.4 Find the equivalent resistance for the R1 ; R2 ; R3 combination and set RL equal to this to get RL =

R3 (R1 + R2 ) R1 + R2 + R 3

Problem A.5 Using Nyquist’s formula, we …nd the equivalent resistance looking back into the terminals with Vrm s across them. It is Req

= 50 k k 20 k k (5 k + 10 k + 5 k) = 50 k k 20 k k 20 k = 50 k k 10 k (50 k) (10 k) = 50 k + 10 k = 8; 333

Thus 2 Vrm s

= =

4kT Req B 4 1:38 10 23 (400) (8333) 2

3:68

10 10 V2

which gives Vrm s = 19:18 V rms

106

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

Problem A.6 To …nd the noise …gure, we …rst determine the noise power due to a source at the output, then due to the source and the network, and take the ratio of the latter to the former. Initally assume unmatched conditions. The results are V02 due to R , only = S

V02 due to R1 and R2

R2 k RL R S + R1 + R 2 k RL

R2 k RL R S + R 1 + R2 k RL

= +

V02 due to R ; R1 and R2 S

(4kT R1 B)

RL k (R1 + RS ) R2 + (R1 + RS ) k RL

R2 k RL RS + R1 + R 2 k R L

(4kT RS B)

(4kT R2 B)

[4kT (RS + R1 ) B]

RL k (R1 + RS ) R2 + (R1 + RS ) k RL

(4kT R2 B)

The noise …gure is (after some simpli…cation) F =1+

R1 + RS

RL k (R1 + RS ) R2 + (R1 + RS ) k RL

R S + R1 + R 2 k R L R2 k RL

R2 RS

In the above, Ra Rb Ra + Rb Note that the noise due to RL has been excluded because it belongs to the next stage. Since this is a matching circuit, we want the input matched to the source and the output matched to the load. Matching at the input requires that Ra k Rb =

RS = Rin = R1 + R2 k RL = R1 +

R2 RL R2 + RL

and matching at the output requires that RL = Rout = R2 k (R1 + RS ) =

R2 (R1 + RS ) R1 + R 2 + R S

Next, these expressions are substituted back into the expression for F . After some simpli…cation, this gives 2 2 R1 2RL RS (R1 + R2 + RS ) = (RS R1 ) R2 F =1+ + 2 2 RS R2 (R1 + RS + RL ) + RL (R1 + R2 + RS ) RS Note that if R1 >> R2 we then have matched conditions of RL = R2 and RS = R1 : Then, the noise …gure simpli…es to R1 F = 2 + 16 R2

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

Note that the simple resistive pad matching circuit is very poor from the standpoint of noise. The equivalent noise temperature is found by using Te

= T0 (F

= T0 1 + 16

R1 R2

Problem A.7 a. The important relationships are Fl = 1 + Tel = T0 (Fl T e0 = T e1 +

T el T0 1)

T e3 T e2 + Ga1 Ga1 Ga2

Completion of the table gives Ampl. No. 1 2 3

F not needed 6 dB 11 dB

T ei 300 K 864.5 K 3360.9 K

Gai 10 dB = 10 30 dB = 1000 30 dB = 1000

Therefore, T e0

864:5 3360:9 + 10 (10) (1000) 386:8 K

300 +

Hence,

T e0 T0 2:33 = 3:68 dB

b. With ampli…ers 1 and 2 interchanged T e0

300 3360:9 + 10 (10) (1000) 865:14 K

864:5 +

This gives a noise …gure of

865:14 290 3:98 = 6 dB 1+

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

c. See part (a) for the noise temperatures. d. For B = 50 kHz, TS = 1000 K, and an overall gain of Ga = 107 , we have, for the con…guration of part (a) Pna; out

= Ga k (T0 + Te0 ) B = 107 1:38 10 23 (1000 + 386:8) 5 =

We desire

9:57

104

10 9 watts

Psa; out Psa; out = 104 = Pna; out 9:57 10 9

which gives Psa; out = 9:57

10 5 watts

For part (b), we have Pna; out

107 1:38

1:29

10 23 (1000 + 865:14) 5

10 8 watts

Once again, we desire Psa; out Psa; out = 104 = Pna; out 1:29 10 8 which gives Psa; out = 1:29 and Psa; in =

10 4 watts

Psa; out = 1:29 Ga

10 11 watts

104

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

Problem A.8 a. The noise …gure of the cascade is Foverall = F1 +

F2 1 F 1 =L+ = LF Ga1 (1=L)

b. For two identical attenuator-ampli…er stages Foverall = L +

L 1 F 1 F 1 + + = 2LF (1=L) (1=L) L (1=L) L (1=L)

2LF; L >> 1

c. Generalizing, for N stages we have Foverall

NFL

Problem A.9 a. The data for this problem is Stage 1 (preamp) 2 (mixer) 3 (ampli…er)

Fi 2 dB = 1.58 8 dB = 6.31 5 dB = 3.16

Gi G1 1.5 dB = 1.41 30 dB = 1000

The overall noise …gure is F = F1 +

F2 1 F3 1 + G1 G1 G2

which gives 5 dB = 3:16 = 1:58 +

6:31 1 3:16 1 + G1 1:41G1

or 3:16

1:58

or G1

6:31 1 3:16 1 + G1 1:41G1 5:31 2:16 + = 4:33 = 6:37 dB 1:58 (1:41) (1:58)

b. First, assume that G1 = 15 dB = 31:62: Then F

= =

6:31 1 3:16 1 + 31:62 (1:41) (31:62) 1:8 = 2:54 dB 1:58 +

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

Then Tes

= T0 (F 1) = 290 (1:8 1) = 230:95 K

and Toverall

= Tes + Ta = 230:95 + 300 = 530:95 K

Now use G1 as found in part (a): F Tes Toverall

= 3:16 = 290 (3:16 1) = 626:4 K = 300 + 626:4 = 926:4 K

c. For G1 = 15 dB = 31:62; Ga = (31:62) (1:41) (1000) = 4:46 Pna; out

= Ga kToverall B = 4:46 104 1:38 =

3:27

10 23 (530:95) 107

10 6 watts

For G1 = 6:37 dB = 4:33; Ga = (4:33) (1:41) (1000) = 6:11 Pna; out

104 : Thus

6:11

103

7:81

10 7 watts

1:38

103 : Thus

10 23 (926:4) 107

Note that for the second case, we get less noise power out even wth a larger Toverall . This is due to the lower gain of stage 1, which more than compensates for the larger input noise power. d. A transmission line with loss L = 2 dB connects the antenna to the preamp. We …rst …nd TS for the transmission line/preamp/mixer/amp chain: FS = FTL +

F2 1 F3 1 F1 1 + + ; GTL GTL G1 GTL G1 G2

where GTL = 1=L = 10 2=10 = 0:631 and FTL = L = 102=10 = 1:58 Assume two cases for G1 : 15 dB and 6.37 dB. First, for G1 = 15 dB = 31:6, we have

1:58 +

2:84

6:31 1 3:16 1 1:58 1 + + 0:631 (0:631) (31:6) (0:631) (31:6) (1:41)

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

This gives TS = 290 (2:84

1) = 534 K

and Toverall = 534 + 300 = 834 K Now, for G1 = 6:37 dB = 4:33, we have FS

1:58 +

5:00

1:58 1 6:31 1 3:16 1 + + 0:631 (0:631) (4:33) (0:631) (4:33) (1:41)

This gives TS = 290 (5:00

1) = 1160 K

and Toverall = 1160 + 300 = 1460 K

Problem A.10 a. (a) Using Pna; out = Ga kTS B = 108

1:38

10 23 (1800) 3

with the given values yields Pna; out = 7:45 b. We want

10 5 watts

Psa; out = 105 Pna; out

or Psa; out = 105

7:45

10 5 = 7:45 watts

This gives Psa; in

= =

7:45 Psa; out = = 7:45 10 8 watts Ga 108 71:28 dBW = 41:28 dBm

106

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

Problem A.11 a. For

A = 1 dB, Y = 1:259 and the e¤ective noise temperature is Te =

For

(1:259) (300) = 858:3 K 1:259 1

A = 1:5 dB, Y = 1:413 and the e¤ective noise temperature is Te =

For

600

(1:413) (300) = 426:4 K 1:413 1

A = 2 dB, Y = 1:585 and the e¤ective noise temperature is Te =

600

(1:585) (300) = 212:8 K 1:585 1

b. These values can be converted to noise …gure using F =1+

Te T0

With T0 = 290 K, we get the following values: (1) For A = 1 dB; F = 5:98 dB; (2) For A = 1:5 dB; F = 3:938 dB; (3) For A = 2 dB; F = 2:39 dB. Problem A.12 a. Using the data given, we can determine the following: =

0:039 m

4 d GT PT GT

202:4 dB

= 39:2 dB = 74:2 dBW

This gives PS =

202:4 + 74:2 + 6

127:2 dBW

b. Using Pn = kTe B for the noise power, we get Pn

Te T0

10 log10 kT0

10 log10 [kT0 ] + 10 log10

174 + 10 log10

= =

108:6 dBm 138:6 dBW

1000 290

Te T0

+ 10 log10 (B)

+ 10 log10 106

APPENDIX A: PHYSICAL NOISE SOURCES AND NOISE CALCULATIONS

c. PS Pn

127:2

( 138:6)

= 11:4 dB = 101:14 = 13:8 ratio d. Assuming the SNR = z = Eb =N0 = 13:8; we get the results for various digital signaling techniques given in the table below: Modulation type BPSK DPSK Noncoh. FSK QPSK

Error probability p Q 2z = 7:4 10 8 1 z = 5:06 10 7 2e 1 z=2 = 5:03 10 4 2e Same as BPSK

APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS

Problem D.1 a. Consider z (t) = A cos (! 0 + ! d ) t + n (t) ; ! 0 = 2 f0 ; ! d = 2 fd = A cos (! 0 + ! d ) t + nc (t) cos ! 0 t with the PSD of n (t) given by Sn (t) = N0 =2 for jf f0 j also that the PSD of the quadrature noise components is

ns (t) sin ! 0 t B=2 and 0 otherwise. Note

N0 ; jf j B=2 0; otherwise

Snc (f ) = Sns (f ) =

Also, the quadrature noise components are uncorrelated. It follows that n0c (t) cos (! 0 + ! d ) t

n0s (t) sin (! 0 + ! d ) t = n0c (t) [cos ! 0 t cos ! d t =

sin ! 0 t sin ! d t] 0 ns (t) [sin ! 0 t cos ! d t + cos ! 0 t sin ! d t] n0c (t) cos ! d t n0s (t) sin ! d t cos ! 0 t n0c (t) sin ! d t + n0s (t) cos ! d t sin ! 0 t

nc (t) cos ! 0 t

ns (t) sin ! 0 t

Therefore nc (t) = n0c (t) cos ! d t

n0s (t) sin ! d t

ns (t) = n0c (t) sin ! d t + n0s (t) cos ! d t These two equations can be solved for n0c (t) and n0s (t). The results are n0c (t) = nc (t) cos ! d t + ns (t) sin ! d t n0s (t) =

nc (t) sin ! d t + ns (t) cos ! d t

To …nd the PSDs of n0c (t) and n0s (t) we …rst …nd their autocorrelation functions. de…nition, [nc (t) cos ! d t + ns (t) sin ! d t] [nc (t + ) cos ! d (t + ) + ns (t + ) sin ! d (t + )] = Rc ( ) cos ! d t cos ! d (t + ) + Rs ( ) sin ! d t sin ! d (t + )

Rn0c ( ) = E

Noting that Rc ( ) = Rs ( ) and using a trig identity, we …nd that Rn0c ( ) = Rc ( ) cos ! d

APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS

Similarly, it can be shown that Rn0s ( ) = Rs ( ) cos ! d The PSDs follow by applying the modulation theorem: 1 1 Sn0c (f ) = Sn0s (f ) = Snc (f + fd ) + Snc (f fd ) 2 2 8 jf j B=2 fd < N0 ; N0 =2; B=2 fd < jf j B=2 + fd = : 0; otherwise

b. To …nd the cross-spectral density of n0c (t) and n0s (t), we …rst …nd their cross-correlation function: Rn0c n0s ( ) = E n0c (t) n0s (t + ) = E =

[nc (t) cos ! d t + ns (t) sin ! d t] [ nc (t + ) sin ! d (t + ) + ns (t + ) cos ! d (t + )] Rc ( ) cos ! d t sin ! d (t + ) + Rs ( ) sin ! d t cos ! d (t + )

Again noting that Rc ( ) = Rs ( ) and using a trig identity, we …nd that Rn0c n0s ( ) =

Rs ( ) sin ! d =

Rs ( )

exp (j! d )

exp ( j! d ) 2j

Application of the frequency translation theorem results in the following for the crossspectral density: j [Sns (f fd ) Sns (f + fd )] 2 8 B=2 fd f B=2 + fd < jN0 =2; jN0 =2; (B=2 + fd ) f (B=2 fd ) = : 0; otherwise

Sn0c n0s (f ) =

For fd 6= 0 this cross-PSD is not identically zero as a sketch will show. Thus n0c (t) and n0s (t) are not uncorrelated. However, they are when sampled at the same instant because Rn0c n0s (0) = 0. Therefore, they are statistically independent when sampled at the same instant since they are jointly Gaussian. Problem D.2 The derivation here follows S. O. Rice, "Noise in FM Receivers," Proc. of the Symp. of Time Series Analysis, M. Rosenblatt, ed., New York: Wiley (1963), pp. 395-422.

APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS

a. We need the additional clicks due to modulation. Thus, consider z (t) = A cos (! 0 + ! d ) t + nc (t) cos ! 0 t = =

ns (t) sin ! 0 t 0 A cos (! 0 + ! d ) t + nc (t) cos (! 0 + ! d ) t n0s (t) sin (! 0 + ! d ) t A + n0c (t) cos (! 0 + ! d ) t n0s (t) sin (! 0 + ! d ) t

, R (t) cos [(! 0 + ! d ) t + where R (t) =

(t)]

[A + n0c (t)]2 + [n0s (t)]2 and tan

(t) =

n0s (t) A + n0c (t)

It follows from the cross-correlation function found in Problem D.1 that Rn0c n0s (0) = E n0c (t) n0s (t) = 0 From an appropriate sketch, it follows that the probability of a counter-clockwise origin encirclement in a small time interval is Pcc = Pr A + n0c (t) < 0 and n0s (t) undergoes a + to

zero crossing in [0;

]

The condition for n0s (t) to undergo a + to zero crossing in [0; ], from an appropriate 0 (t) 0 (t) s s sketch, is that n0s (0) > 0 and that n0s (0) + dndt < 0 for all dndt < 0. Therefore t=0

Pcc = Pr A + n0c (0) < 0; n0s (0) > 0; n0s (0) +

t=0

dn0s (t) dt t=0

< 0;

dn0s (t) <0 dt t=0

Thus we need the joint pdf of X , n0c (0) ; Y , n0s (0) ; and Z ,

dn0s (t) dt t=0

These are jointly Gaussian random variables (the derivative is a linear operation).

The

APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS

following expectations hold: E [X] = E [Y ] = E [Z] = 0 Z 1 Z 1 2 2 Sn0s (f ) df = N0 B , a Sn0c (f ) df = E X = E Y = 1 1 Z 1 jj2 f j2 Sn0s (f ) df E Z2 = 1

= E [XY ] = E [XZ] = E [Z j X] = =

1 2 N0 B 3 + 4 2 fd2 N0 B , b + a! 2d 3 E [Y Z] = 0 dRn0c n0s ( ) d ( Rs ( ) sin ! d ) = = a! d d d =0 =0 dn0s (t) dns (t) E j n0c (0) = E ! d nc (0) + j nc (0) dt t=0 dt t=0 ! d E [nc (0) j nc (0)] = ! d X

Therefore, the joint pdf of X; Y; Z is fXY Z (x; y; z) = fZjXY (z j x; y) fX (x) fY (y)

= fZjX (z j x) fX (x) fY (y) ( " #) x2 + y 2 (z + a! d )2 1 = + p exp 2a 2b (2 )3=2 a b

APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS

Thus, Pcc

= Pr [X < A; 0 < Y < Z ; all Z < 0] Z AZ 0 Z z Z AZ 0 Z z = fXY Z (x; y; z) dydzdx = fZj X (z j x) fX (x) fY (y) dydzdx 1 1 0 1 1 0 ( " #) Z AZ 0 Z z 1 x2 + y 2 (z + x! d )2 = dydzdx + p exp 2a 2b (2 )3=2 a b 1 1 0 8 # 9 " Z A Z = (z + x! d )2 =2b Z z exp y 2 =2a exp x2 =2a < 0 exp p p p dy dz dx; << 1 = ; : 1 2 a 2 a 2 b 1 0 8 9 Z A Z = exp (z + x! d )2 =2b exp x2 =2a < 0 z x z p p p = dz dx; = p ; & = p : ; a 2 a 2 a 2 b b 1 1 2 3 r Z r Z 1 exp 2 (& + )2 =2 =2 A2 b 1 exp a 4 5 p p & = p d& d ; = ! ; = d a p2 b 2a 2 2 2 0 "Z # r Z 2 1 exp u2 =2 =2 b 1 exp p p = p (u ) du d ; u = & + a p2 2 2 2 "Z # r Z Z 1 2 1 exp u2 =2 =2 exp u2 =2 b 1 exp p p p = p u du du d a p2 2 2 2 2 " # r Z 2 2 2 =2 =2 exp b 1 exp p p = p Q( ) d a p2 2 2 2 ) r (Z Z 1 1 exp p 1 + 2 2 =2 b 2 p d Q ( ) exp =2 d ; v = 1+ 2 = p p 2 a 2 2 2 r Z 1 p b 1 2 p 2 (1 + 2 ) = Q Q ( ) exp d p 2 a 1+ 2 2

To integrate the last integral, consider Z 1 xQ (Cx) exp

x =2

dx = uvj1 B

Z 1

dt; du =

vdu

integrated by parts with u = Q (Cx) =

Z 1

t2 =2

2 2 x =2 dx; v = Cx

dv = x exp

exp

x2 =2

exp

C 2 x2 2

APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS

Thus, by parts Z 1

xQ (Cx) exp

x2 2

dx =

Q (Cx) exp

Pcc

= = where

1 B

Z 1

exp

x2 2

exp

C 2 x2 =2 p dx 2

1 + C 2 x2 =2 p dx 2 D Z 1 exp w2 =2 D2 C p p = Q (CD) exp dw p 2 2 1 + C 2 D 1+C 2 p D2 C p Q D 1 + C2 = Q (CD) exp 2 1 + C2 p with the above integral substituted with C = and D = 2 , becomes r Z 1 p b 1 2 2 p Q Q ( ) exp d 2 (1 + ) p 2 a 1+ 2 2 9 p r 8 1 2) < = p Q 2 (1 + b 1+ 2 h i p p ; 2 a: Q 2 exp ( ) p1+ 2 Q 2 (1 + 2 ) r p p b 1+ 2 p Q 2 (1 + 2 ) Q 2 exp ( ) 2 a 1+ 2 np o p p 1 + 2Q 2 (1 + 2 ) Q 2 exp ( ) = Q (CD) exp

Therefore, Pcc

x2 2

= = =

D2 2

exp

!d fd = 2 s r 1 2 3 b 1 1 B 3 ( N0 B ) = = p 2 a 2 N0 B 2 3 A2 A2 = 2a 2N0 B

b. It follows that the probability of a clockwise origin encirclement is given by this

APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS

expression but with fd replaced by fd ( = ). Taking this into account, p p p Pc = 1 + 2Q 2 (1 + 2 ) + fd Q 2 exp ( ) i h p p p = 1 + 2Q 2 (1 + 2 ) + fd 1 Q 2 exp ( ) p p p 1 + 2Q 2 (1 + 2 ) fd Q 2 exp ( ) + fd exp ( = Pcc

Pcc

+ fd exp (

)

; fd > 0

Therefore 1

mod

= 2

(Pc + Pcc )

Pcc

A2 2N0 B

+ fd exp

; fd > 0

To make this applicable to square wave modulation, replace fd by jfd j to get h p i p p 1 + 2Q 2 (1 + 2 ) fd Q 2 exp ( ) + jfd j exp ( sw mod = 2 ,

where = 2

h p

1+ 2 s

= 24

)

B = 24 p 2 3

2 (1 + 0v u 2 u @ Q t2

1 + 12

fd B

0v u u Q @t2

fd Q fd

p 2 !1

1 + 12

fd B

exp ( fd Q 2

i ) (reduces to (D.23) for fd = 0) 3 fd p 2 exp ( )5

p p 2fd = 2 p Q 8 fd Q 6 exp ( ) ; B = 2fd 3 0 s 1 0s 1 2 2 A2 4fd A A A A = p Q @2 2fd Q @ 3 exp N0 B N0 B 2N0 B 3

= jfd j exp

A2 2N0 B

Typical results for v and

are shown below versus SNR.

fd p 24 B

fd Q

;

A2 2N0 B

exp (

APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS

Average encirclements per second, ν and δν, for fd = 5 Hz 0

-2

-4

ν, δν

-6

-8

ν -10

-15

δν -10

-5

A2/(2N0B) in dB

APPENDIX D: ZERO-CROSSING AND ORIGIN ENCIRCLEMENT STATISTICS

Average encirclements per second, ν and δν, for fd = 10 Hz 0

-2

ν, δν

-4

-6

-8

ν -10

-15

δν -10

-5

A2/(2N0B) in dB