Test Bank For Contemporary Communication Systems 1st Edition by M. Farooque Mesiya Chapter 1_15 by QuentinAxel

Chapter 2 2.1

Consider the signals displayed in Figure P2.1.Show that each of these signals can be expressed as the sum of rectangular Π (t ) and/or triangular Λ (t ) pulses.

Figure P2.1

Solution:

⎛t⎞ ⎛t⎞ x1 (t ) = Π ⎜ ⎟ + Π ⎜ ⎟ ⎝2⎠ ⎝4⎠

⎛t −3⎞ ⎛t −3⎞ b. x2 (t ) = 2Λ ⎜ ⎟− Λ⎜ ⎟ ⎝ 6 ⎠ ⎝ 2 ⎠ c.

⎛ 11 ⎞ ⎛ 7⎞ ⎛ 3⎞ ⎛ 1⎞ ⎛ 5⎞ ⎛ 9⎞ x3 (t ) = ... + Π ⎜ t + ⎟ + Π ⎜ t + ⎟ + Π ⎜ t + ⎟ + Π ⎜ t − ⎟ + Π ⎜ t − ⎟ + Π ⎜ t − ⎟ + ... 2⎠ ⎝ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ∞

= ∑ Π ⎡⎣t − ( 2n + 0.5 ) ⎤⎦ n =−∞

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⎛t+4⎞ ⎛t⎞ ⎛t−4⎞ d. x4 (t ) = ... + Λ ⎜ ⎟ + Λ⎜ ⎟ + Λ⎜ ⎟ + ... ⎝ 2 ⎠ ⎝2⎠ ⎝ 2 ⎠ ∞ ⎛ t − 4n ⎞ = ∑ Λ⎜ ⎟ ⎝ 2 ⎠ n =−∞ 2.2 For the signal x2 (t ) in Figure P2.1 (b) plot the following signals x2 (t − 3)

b. x2 (−t ) c.

x2 (2t )

d. x2 (3 − 2t ) Solution:

x2 (t − 3)

x2 (−t )

0 1 2 3 4 5 6 7 8 9

−6 −5 −4 −3 −2 −1

t 1 2 3

x2 (3 − 2t )

x2 (2t )

1 0

1 2 3 4 5 6

0 −2−1

1 2 3 4

2.3 Plot the following signals a.

x1 (t ) = 2Π (t / 2) cos(6π t )

⎡1 1 ⎤ b. x2 (t ) = 2 ⎢ + sgn ( t ) ⎥ ⎣2 2 ⎦

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x3 (t ) = x2 (−t + 2)

d. x4 (t ) = sinc ( 2t ) Π ( t / 2 ) Solution: x2 (t )

2 1.5

x1(t)

1 0.5

t 1 2 3 4 5 6

-0.5 -1 1 -1.5 0.8

-2 -1

-0.8

-0.6

-0.4

-0.2

0.2

0.4

0.6

0.8

1 0.6

xx4(t) (t)

0.4

x3 (t )

0.2

2 0

-0.2

-0.4 -1

1 2 3 4

-0.8

-0.6

-0.4

-0.2

0.2

0.4

0.6

0.8

2.4 Determine whether the following signals are periodic. For periodic signals, determine the fundamental period. a.

x1 (t ) = sin(π t ) + 5cos(4π t / 5)

Solution:

sin(π t ) is periodic with period T1 =

2π

= 2 . cos(4π t / 5) is periodic with period

T 2π 5 = . x1 (t ) is periodic if the ratio 1 can be written as ratio of 4π / 5 2 T2 integers. In the present case, T2 =

T1 2 × 2 4 = = T2 5 5

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Therefore, x1 (t ) is periodic with fundamental period To such that To = 5T1 = 4T2 = 10 b. x2 (t ) = e j 3t + e j 9t + cos(12t ) Solution:

2π . Similarly, e j 9t is periodic with period 3 2π 2π π T2 = . cos(12t ) is periodic with period T3 = = . x2 (t ) is periodic with 9 12 6 2π fundamental period To = LCM (T1 , T2 , T3 ) = . 3

e j 3t is periodic with period T1 =

x3 (t ) = sin(2π t ) + cos(10t )

Solution:

2π = 1 . cos(10t ) is periodic with period 2π T 2π π T2 = = . x3 (t ) is periodic if the ratio 1 can be written as ratio of integers. 10 5 T2 In the present case, sin(2π t ) is periodic with period T1 =

T1 1× 5 5 = = T2 π π Since π is an irrational number, the ratio is not rational. Therefore, x3 (t ) is not periodic.

π⎞ ⎛ d. x4 (t ) = cos ⎜ 2π t − ⎟ + sin(5π t ) 4⎠ ⎝ Solution:

π⎞ 2π ⎛ cos ⎜ 2π t − ⎟ is periodic with period T1 = = 1 . sin(5π t ) is periodic with 4⎠ 2π ⎝ T 2π 2 period T2 = = . x4 (t ) is periodic if the ratio 1 can be written as ratio of T2 5π 5 integers. In the present case,

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T1 1× 5 5 = = T2 2 2 Therefore, x4 (t ) is periodic with fundamental period To such that To = 2T1 = 5T2 = 2 2.5 Classify the following signals as odd or even or neither. a.

x(t ) = −4t

Solution:

x(−t ) = 4t = −(−4t ) = − x(t ) . So x(t ) is odd.

b. x(t ) = e

−t

Solution: e

−t

− −t

. So x(t ) is even.

x(t ) = 5cos(3t )

Solution:

Since cos(t ) is even, 5cos(3t ) is also even.

d. x(t ) = sin(3t − ) 2 Solution:

x(t ) = sin(3t − ) = − cos(3t ) which is even. 2 e. x(t ) = u (t ) Solution:

u (t ) is neither even nor odd; For example, u (1) = 1 but u (−1) = 0 ≠ −u (1) .

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x(t ) = sin(2t ) + cos(2t )

Solution:

x(t ) is neither even nor odd; For example, x(π / 8) = 2 but x(−π / 8) = 0 ≠ − x(π / 8) . 2.6 Determine whether the following signals are energy or power, or neither and calculate the corresponding energy or power in the signal: a. x1 (t ) = u (t ) Solution:

The normalized average power of a signal x1 (t ) is defined as T /2

T /2

1 1 1T 1 2 u (t ) dt = lim ∫ 1 dt = lim = ∫ T →∞ T T →∞ T T →∞ T 2 2 0 −T /2

Px1 = lim

Therefore, x1 (t ) is a power signal. b. x2 (t ) = 4 cos(2π t ) + 3cos ( 4π t ) Solution:

4 cos(2π t ) is a power signal. 3cos ( 4π t ) is also a power signal. Since x2 (t ) is sum of two power signals, it is a power signal. T /2

T /2

2 2 1 1 ⎡⎣ 4 cos(2π t ) + 3cos ( 4π t ) ⎤⎦ dt = lim ⎡⎣ 4 cos(2π t ) + 3cos ( 4π t ) ⎤⎦ dt Px2 = lim ∫ ∫ T →∞ T T →∞ T −T /2 −T /2 T /2

1 ⎡16 cos 2 (2π t ) + 9 cos 2 ( 4π t ) + 24 cos(2π t ) cos ( 4π t ) ⎤⎦dt T →∞ T ∫ ⎣ −T /2

= lim Now T /2

T /2

1 8 8T =8 16 cos 2 (2π t )dt = lim 1 + cos(4π t ) ]dt = lim [ ∫ ∫ T →∞ T T →∞ T T →∞ T −T /2 −T /2 lim

T /2

1 lim 9 cos 2 (4π t )dt = 4.5 T →∞ T ∫ −T /2

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T /2

24 24 cos(2π t ) cos ( 4π t )dt = lim ∫ ∫ ⎡⎣cos(6π t ) + cos ( 2π t )⎤⎦dt = 0 T →∞ T T →∞ T −T /2 −T /2 lim

Therefore, Px2 = 8 + 4.5 = 12.5 c.

x3 (t ) =

1 t

Solution: T /2 T /2 ⎛ 1 T /2 ⎞ 2 ⎛ −2 ⎞ Ex3 = lim ∫ 1/ t dt = lim ∫ t −2 dt = lim ⎜ − ⎟⎟ = Tlim ⎜ ⎟=0 T →∞ T →∞ T →∞ ⎜ →∞ t T / 2 ⎝ ⎠ T /2 − −T /2 −T /2 ⎝ ⎠ T /2 T /2 T /2 ⎞ 1 1 1⎛ 1 1 ⎛ −2 ⎞ 2 −2 Px3 = lim t dt t dt 1/ lim lim = = − = lim ⎜ ⎜ ⎟ ⎟=0 ⎟ →∞ T →∞ T ∫ T →∞ T ∫ T →∞ T ⎜ T T ⎝T / 2⎠ −T /2 −T /2 ⎝ t −T /2 ⎠

x3 (t ) is neither an energy nor a power signal. d. x4 (t ) = e−α t u (t ) Solution: T /2

T /2

Ex4 = lim ∫ e u (t ) dt = lim ∫ e T →∞

−α t

T →∞

−T /2

−2α t

⎛ e −2α t T /2 ⎞ 1 1 ⎟= dt = lim ⎜ − lim ( −e −α T + 1) = T →∞ ⎜ 2α 0 ⎟ 2α T →∞ 2α ⎝ ⎠

Thus x4 (t ) is an energy signal. e.

x5 (t ) = Π (t / 3) + Π (t )

Solution: x5 (t )

−3 / 2

3/ 2

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T /2

−1/2

1/ 2

−1/ 2

−3/ 2

−1/2

−3/2

Ex5 = lim ∫ Π (t / 3) + Π (t ) dt = ∫ 1dt + ∫ 4dt + ∫ 1dt 2

T →∞

−T /2

= 1+ 4 +1 = 6 Thus x5 (t ) is an energy signal. f.

x6 (t ) = 5e( −2t + j10π t )u (t )

Solution: T /2

T /2

Ex6 = lim ∫ 5e( −2t + j10π t )u (t ) dt = lim ∫ 5e −2t e j10π t ) dt = lim 25 ∫ e−4t dt T →∞

T →∞

−T /2

⎛ e = 25 lim ⎜ − T →∞ ⎜ 4 ⎝

−4 t T /2

T →∞

⎞ 25 25 ⎟= lim ( −e −2T + 1) = →∞ T ⎟ 4 4 ⎠

Thus x6 (t ) is an energy signal. ∞

g. x7 (t ) = ∑ Λ ⎡⎣( t − 4n ) / 2 ⎤⎦ n =−∞

Solution:

x7 (t ) is a periodic signal with period To = 4 . T /2

(

)

2 1 o 2 1 2 ⎡⎣ Λ ( t / 2 ) ⎤⎦ dt = ∫ (1 − t ) dt = ∫ 1 − 2t + t 2 dt Px7 = ∫ To −To / 2 To 0 20 1

t3 ⎞ 1⎛ 1⎛ 1⎞ 1 = ⎜ t − t 2 + ⎟ = ⎜1 − 1 + ⎟ = 2⎝ 3 ⎠0 2⎝ 3⎠ 6 2.7 Evaluate the following expressions by using the properties of the delta function: a.

x1 (t ) = δ (4t ) sin(2t )

Solution:

1 4

δ (4t ) = δ (t ) 1 1 x1 (t ) = δ (t ) sin(2t ) = δ (t ) sin(0) = 0 4 4

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π⎞ ⎛ b. x2 (t ) = δ (t ) cos ⎜ 30π t + ⎟ 4⎠ ⎝ Solution:

π⎞ π⎞ ⎛ ⎛ ⎛π ⎞ x2 (t ) = δ (t ) cos ⎜ 30π t + ⎟ = δ (t ) cos ⎜ 0 + ⎟ = cos ⎜ ⎟ δ (t ) 4⎠ 4⎠ ⎝ ⎝ ⎝4⎠ c. x3 (t ) = δ (t )sinc(t + 1) Solution:

x3 (t ) = δ (t )sinc(t + 1) = δ (t )sinc(0 + 1) = δ (t )sinc(1) = δ (t ) × 0 = 0 d. x4 (t ) = δ (t − 2)e − t sin(2.5π t ) Solution: x4 (t ) = δ (t − 2)e − t sin(2.5π t ) = δ (t − 2)e −2sin(2.5π × 2) = δ (t − 2)e −2sin(π ) = 0 ∞

x5 (t ) = ∫ δ (2t )sinc(t )dt −∞

Solution: ∞

∞

1 x5 (t ) = ∫ δ (2t )sinc(t )dt = ∫ δ (t )sinc(t ) dt 2 −∞ −∞ ∞

∞

1 1 1 δ (t )sinc(0)dt = ∫ δ (t )dt = ∫ 2 −∞ 2 −∞ 2 ∞

f. x6 (t ) = ∫ δ (t − 3) cos(t )dt −∞

Solution: ∞

∞

−∞

x6 (t ) = ∫ δ (t − 3) cos(t )dt = cos(3) ∫ δ (t − 3)dt = cos(3)

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∞

g. x7 (t ) = ∫ δ (2 − t ) −∞

1 dt 1− t3

Solution: ∞

x7 (t ) = ∫ δ (2 − t ) −∞

∞

1 1 dt = ∫ δ (t − 2) dt 3 1− t 1− t3 −∞

∞

1 1 1 dt = − ∫ δ (t − 2)dt = − = ∫ δ (t − 2) 3 1− 2 7 −∞ 7 −∞ ∞

h. x8 (t ) = ∫ δ (3t − 4)e −3t dt −∞

Solution: ∞

x8 (t ) = ∫ δ (3t − 4)e −3t dt = −∞

∞

1 ⎛ 4⎞ δ ⎜ t − ⎟ e −3t dt ∫ 3 −∞ ⎝ 3 ⎠

∞

4 1 e −4 e −4 ⎛ 4 ⎞ −3× ⎛ 4⎞ = ∫ δ ⎜ t − ⎟ e 3 dt = − = δ t dt ∫ ⎜⎝ 3 ⎟⎠ 3 −∞ ⎝ 3 ⎠ 3 −∞ 3

x9 (t ) = δ '(t ) ⊗ Π (t )

Solution: ∞

x9 (t ) = ∫ Π (t − τ )δ '(τ )dτ = (−1) −∞

d Π (t − τ ) dτ τ =0

= Π ' (t ) = δ ( t + 0.5 ) − δ ( t − 0.5 ) 2.8 For each of the following continuous-time systems, determine whether or not the system is (1) linear, (2) time-invariant, (3) memoryless, and (4) casual. a.

y (t ) = x(t − 1)

Solution:

The system is linear, time-invariant, causal, and has memory. The system has memory because current value of the output depends on the previous value of the input. The system is causal because current value of the output does not depend on future inputs. To prove linearity, let x(t ) = α x1 (t ) + β x2 (t ) . The response of the system to x(t ) is

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y (t ) = x(t − 1) = α x1 (t − 1) + β x2 (t − 1) = α y1 (t ) + β y2 (t ) T T where x1 (t ) ⎯⎯ → y1 (t ) and x2 (t ) ⎯⎯ → y2 (t ) .

To show that the system is time-invariant, let x1 (t ) = x(t − to ) be the system input. The corresponding output is y1 (t ) = x1 (t − 1) = x(t − 1 − to ) = y (t − to ) b. y (t ) = 3x(t ) − 2 Solution:

The system is nonlinear, time-invariant, causal, and memoryless. The system is memoryless because current value of the output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let x(t ) = α x1 (t ) + β x2 (t ) . The response of the system to x(t ) is y (t ) = 3x(t ) − 2 = 3 [α x1 (t ) + β x2 (t ) ] − 2 ≠ α y1 (t ) + β y2 (t ) = α [3 x1 (t ) − 2] + β [3 x2 (t ) − 2] To show that the system is time-invariant, let x1 (t ) = x(t − to ) be the system input. The corresponding output is y1 (t ) = 3x1 (t ) − 2 = 3x(t − to ) − 2 = y (t − to ) c.

y (t ) = x(t )

Solution:

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y (t ) = α x1 (t ) + β x2 (t ) ≤ α x1 (t ) + β x2 (t )

Because

α x1 (t ) + β x2 (t ) = α x1 (t ) + β x2 (t ) ≠ α y1 (t ) + β y2 (t ) for all α and β , the system is nonlinear. To show that the system is time-invariant, let x1 (t ) = x(t − to ) be the system input. The corresponding output is y1 (t ) = x1 (t ) = x(t − to ) = y (t − to )

d. y (t ) = [ cos(2t )] x(t ) Solution:

The system is linear, time-variant, causal, and memoryless. The system is memoryless because current value of the output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let x(t ) = α x1 (t ) + β x2 (t ) . The response of the system to x(t ) is y (t ) = [ cos(2t ) ] x(t ) = [ cos(2t ) ][α x1 (t ) + β x2 (t ) ] = α [ cos(2t ) ] x1 (t ) + β [ cos(2t ) ] x2 (t ) = α y1 (t ) + β y2 (t ) To prove time-variance, let x1 (t ) = x(t − to ) be the system input. The corresponding output is

y1 (t ) = [ cos(2t ) ] x1 (t ) = [ cos(2t ) ] x ( t − to ) ≠ y (t − to ) = cos ⎡⎣ 2 ( t − to ) ⎤⎦ x ( t − to ) e.

y (t ) = e x (t )

Solution:

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To prove nonlinearity, let x(t ) = α x1 (t ) + β x2 (t ) . The response of the system to x(t ) is y (t ) = eα x1 ( t ) + β x2 ( t ) = eα x1 ( t ) e β x2 (t ) ≠ α y1 (t ) + β y2 (t ) = α e x1 ( t ) + β e x2 ( t )

for all α and β To show that the system is time-invariant, let x1 (t ) = x(t − to ) be the system input. The corresponding output is y1 (t ) = e x1 (t ) = e x (t −to ) = y (t − to ) f. y (t ) = tx(t ) Solution:

The system is linear, time-variant, causal, and memoryless. The system is memoryless because current value of the output depends only on the current value of the input. The system is causal because current value of the output does not depend on future inputs. To prove nonlinearity, let x(t ) = α x1 (t ) + β x2 (t ) . The response of the system to x(t ) is y (t ) = t [α x1 (t ) + β x2 (t ) ] = α tx1 (t ) + β tx2 (t ) = α y1 (t ) + β y2 (t )

To prove time-variance, let x1 (t ) = x(t − to ) be the system input. The corresponding output is y1 (t ) = tx1 (t ) = tx ( t − to ) ≠ y (t − to ) = ( t − to ) x ( t − to ) t

g. y (t ) = ∫ x(2τ )dτ −∞

Solution:

The system is linear, time-invariant, causal, and has memory. The system has memory because current value of the output depends only on the past values of the input. The system is causal because current value of the output does not depend on future inputs. To prove linearity, let x(t ) = α x1 (t ) + β x2 (t ) . The response of the system to x(t ) is

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−∞

y (t ) = ∫ [α x1 (2τ ) + β x2 (2τ ) ] dτ = α ∫ x1 (2τ )dτ + β ∫ x2 (2τ )dτ = α y1 (t ) + β y2 (t ) To check for time-variance, let x1 (t ) = x(t − to ) be the system input. The corresponding output is t

t − to

−∞

y1 (t ) = ∫ x1 (2τ )dτ = ∫ x [ 2(τ − to ) ] dτ = ∫ x ( 2α ) dα = y ( t − to ) 2.9 Calculate the output y (t ) of the LTI system for the following cases: a.

x(t ) = e −2t u (t ) and h(t ) = u (t − 2) − u (t − 4)

Solution:

⎛ t −3⎞ h(t ) = u (t − 2) − u (t − 4) = Π ⎜ ⎟ ⎝ 2 ⎠ ∞

∞

⎛τ −3 ⎞ y (t ) = ∫ h(τ ) x(t − τ )dτ = ∫ e−2(t −τ )u (t − τ )Π ⎜ ⎟ dτ ⎝ 2 ⎠ −∞ −∞ For t < 2 , there is no overlap and y (t ) = 0. t

For 2 ≤ t < 4 , y (t ) = ∫ e

−2( t −τ )

t −2

dτ = ∫ e 0

−2τ

e −2τ 1 − e−2(t − 2) dτ = = −2 0 2

For t ≥ 4 , 4

y (t ) = ∫ e

−2( t −τ )

dτ = e

−2 t

⎧1 − e −2(t − 2) , ⎪ 2 ⎪ ⎪⎪ −2(t − 2) ( e 4 − 1) y (t ) = ⎨e , 2 ⎪ ⎪0, ⎪ ⎪⎩

( e − 1) e2τ e8 − e 4 = e −2t = e −2( t − 2) 2 2 2 2 4

∫ e dτ =e 2τ

−2 t

2≤t ≤4 t>4 otherwise

b. x(t ) = e − t u (t ) and h(t ) = e −2t u (t )

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Solution: ∞

y (t ) = ∫ e u (τ )e −τ

−2( t −τ )

−∞

u (t − τ )dτ = ∫ e −τ e−2(t −τ ) dτ 0

For t < 0 , there is no overlap and y (t ) = 0. y (t ) = e

−2 t

∫ e dτ =e τ

−2 t

eτ = e−2t ( et − 1) = e − t − e −2t , t ≥ 0 t

x(t ) = u (−t ) and h(t ) = δ (t ) − 3e−2t u (t )

Solution: ∞

∞

−∞

y (t ) = ∫ h(τ ) x(t − τ )dτ = ∫ ⎡⎣δ (τ ) − 3e −2τ u (τ ) ⎤⎦ u (τ − t )dτ Now ∞

∫ δ (τ )u(τ − t )dτ = u (−t )

−∞

∞

For t < 0 , ∫ 3e 0

∞

−2τ

3e −2τ 3 = dτ = −2 0 2 ∞

For t ≥ 0 , ∫ 3e u (τ )u (τ − t )dτ = ∫ 3e −2τ

−∞

5 ⎧3 ⎪⎪ 2 + 1 = 2 , y (t ) = ⎨ ⎪ 3 e −2t , ⎪⎩ 2

∞

−2τ

3e −2τ 3 = e −2t dτ = −2 t 2

t<0 t≥0

d. x(t ) = δ (t − 2) + 3e3t u (−t ) and h(t ) = u (t ) − u (t − 1) Solution:

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∞

−∞

y (t ) = ∫ h(τ ) x(t − τ )dτ = ∫ ⎡⎣δ (t − τ − 2) + 3e3(t −τ )u (−t + τ ) ⎤⎦ [u (τ ) − u (τ − 1)] dτ ∞

∞

−∞

= ∫ δ (t − τ − 2) [u (τ ) − u (τ − 1) ] dτ + ∫ 3e3(t −τ )u (−t + τ ) [u (τ ) − u (τ − 1) ] dτ

Now ∞

∫ δ (t − τ − 2) [u(τ ) − u(τ − 1)] dτ = u(t − 2) − u(t − 2 − 1) = Π ( t − 2.5)

−∞

For t < 0 , ∞

−∞

3( t −τ ) 3( t −τ ) 3t −3τ ∫ 3e u(−t + τ ) [u(τ ) − u(τ − 1)] dτ = ∫ 3e dτ = 3e ∫ e dτ

= 3e3t

−3τ 1

e 1⎞ ⎛ = e 3t ⎜ 1 − 3 ⎟ −3 0 ⎝ e ⎠

For 0 < t ≤ 1 , ∞

∫ 3e

3( t −τ )

−∞

u (−t + τ ) [u (τ ) − u (τ − 1) ] dτ = ∫ 3e t

= 3e3t ⎧ 3t ⎛ 1⎞ ⎪e ⎜ 1 − e3 ⎟ , ⎠ ⎪ ⎝ ⎪ ⎛ 1⎞ y (t ) = ⎨e3t ⎜ e −3t − 3 ⎟ , e ⎠ ⎪ ⎝ ⎪1, ⎪ ⎩0,

3( t −τ )

dτ = 3e ∫ e−3τ dτ 3t

−3τ 1

e 1⎞ ⎛ = e3t ⎜ e −3t − 3 ⎟ −3 t e ⎠ ⎝

t≤0 0 < t ≤1 2≤t ≤3 otherwise

2.10 The impulse response function of a continuous-time LTI is displayed in Figure P2.2(b). Assuming the input x(t ) to the system is waveform illustrated in Figure P2.2(a), determine the system output waveform y (t ) and sketch it. Solution:

For t < 1 , there is no overlap and y (t ) = 0.

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Figure P2.2 x(τ )

(a)

(b)

h(t − τ ) 1

1< t < 2

h(t − τ )

(c)

5<t <6

3 t -1

As shown in Figure (b), y (t ) = ∫ dτ = t − 1 for 1 ≤ t < 2 0

For 2 ≤ t < 3 , y (t ) = ∫ dτ = 1 0 1

For 3 ≤ t < 4 , y (t ) = ∫ dτ =1 − t + 3 = 4 − t t −3

Referring to Figure (c), y (t ) = ∫ dτ = t − 5 for 5 ≤ t < 7 5

For 7 ≤ t < 8 , y (t ) = ∫ dτ = t − t + 2 = 2 t −2 7

For 8 ≤ t < 10 , y (t ) = ∫ dτ = 7 − t + 3 = 10 − t t −3

y(t ) 2

t 10 17

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2.11 An LTI system has the impulse response h(t ) = e −0.5(t − 2)u (t − 2) . a.

Is the system casual?

Solution:

Yes. The system is casual because h(t ) = 0 for t < 0 . b. Is the system stable? Solution: ∞

∞

−∞

The system is stable because ∫ h(t ) dt = ∫ e

∞

−0.5 x

e −0.5 x =2 dx = −0.5 0

c. Repeat parts (a) and (b) for h(t ) = e −0.5(t + 2)u (t + 2) . Solution:

The system is not causal but stable. 2.12 Write down the exponential Fourier series coefficients of the signal x(t ) = 5sin(40π t ) + 7 cos ( 80π t − π / 2 ) − cos (160π t + π / 4 )

Solution:

Applying the Euler’s formula, we get the following terms: ⎡ e j 40π t − e − j 40π t ⎤ ⎡ e j 80π t e− jπ /2 + e− j 80π t e jπ /2 ⎤ ⎡ e j160π t e jπ / 4 + e− j160π t e− jπ /4 ⎤ + x(t ) = 5 ⎢ 7 ⎥ ⎢ ⎥−⎢ ⎥ 2j 2 2 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ j 40π t − j 40π t j 80π t − j 80π t jπ /4 j160π t − jπ /4 − j160π t = − j 2.5e + j 2.5e − j 3.5e + j 3.5e − 0.5e e − 0.5e e

The Fourier coefficients are

C1 = − j 2.5, C−1 = j 2.5 C2 = − j 3.5, C−2 = j 3.5 C3 = −0.5e jπ /4 , C−3 = −0.5e − jπ /4 a. Is x(t) periodic? If so, what is its period? Solution:

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Yes. ωo = 40π ⇒ f o = 20, To =

1 1 . The period is = 0.05sec . 20 20

2.13 A signal has the two-sided spectrum representation shown in Figure P2.3. Figure P2.3

a. Write the equation for x(t). Solution:

π⎞ π⎞ ⎛ ⎛ x(t ) = 14 cos ⎜100π t − ⎟ + 10 + 8cos ⎜ 300π t − ⎟ 3⎠ 2⎠ ⎝ ⎝ b. Is the signal periodic? If so, what is its period? Solution:

Yes. It is periodic with fundamental period To =

1 = 0.02 . 50

c. Does the signal have energy at DC? Solution:

Yes as indicated by the presence of DC term C0 = 10 . 2.14 Write down the complex exponential Fourier series for each of the periodic signals shown in Figure P2.4. Use odd or even symmetry whenever possible. Solution:

a. To = 3, f o = 1/ 3

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2 3 ⎤ 1 1⎡ C0 = ⎢ ∫ 2dt − ∫ 1dt ⎥ = ( 4 − 1) = 1 3 ⎣0 2 ⎦ 3 2 3 2π nt 2π nt −j −j ⎤ 1⎡ Cn = ⎢ ∫ 2e 3 dt − ∫ e 3 dt ⎥ 3 ⎣0 2 ⎦ 2 2π nt 3 ⎤ −j j 3 ⎡ − j 2π3nt ⎢ 2e = −e 3 ⎥ 3 × 2π n ⎢ 0 2⎥ ⎣ ⎦ j ⎡⎣ 2e − j 4π n /3 − 2 − e − j 2π n + e − j 4π n /3 ⎤⎦ = 2π n 3e − j 2π n /3 ⎛ e + j 2π n /3 − e − j 2π n /3 ⎞ j 3 − j 4π n /3 = − = 1 e ( ) πn ⎜ ⎟ 2π n 2j ⎝ ⎠

3e − j 2π n /3 ⎛ 2π n ⎞ = sin ⎜ ⎟ , n = ±1, ±2,..... πn ⎝ 3 ⎠

b. To = 2, f o = 1/ 2 1

1 ⎤ 1 t2 1⎡ 1 C0 = ⎢ ∫ tdt ⎥ = = (1 − 1) = 0 2 ⎣ −1 ⎦ 2 2 −1 4

Cn =

1 1 ⎤ 1 ⎡ − jπ nt ⎤ 1 ⎡ − jπ nt = − te dt ) ⎢∫ ⎥ ⎢ ∫ td ( e ⎥ 2 ⎣ −1 j 2π n ⎣ −1 ⎦ ⎦

1 ⎤ j ⎡ − jπ nt 1 − jπ nt − te e dt ⎢ ⎥ ∫ −1 2π n ⎣ −1 ⎦

1 − jπ nt 1 ⎤ j ⎡ − jπ n + jπ n +e + e e ⎢ −1 ⎥ 2π n ⎣ jπ n ⎦

j ⎡ e − jπ n e + jπ n e − jπ n e jπ n ⎤ = ⎢ + + 2 2 − 2 2⎥ 2 ⎣ πn jπ n jπ n ⎦ πn je − jπ n j ( −1) = = , n = ±1, ±2,..... πn πn n

c. To = 2, f o = 1/ 2 1

1 ⎛ t2 ⎞ ⎤ 2A ⎡ 1 A 1 C0 = − t dt = A ) ⎥ ⎜ t − ⎟ = A ⎜⎛1 − ⎟⎞ = ⎢∫ ( 2 ⎣0 2 ⎠0 ⎝ 2⎠ 2 ⎦ ⎝

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Since x3 (t ) is an even function of time, Cn =

1 1 ⎤ An ⎡ 2 A 1 π π 1 cos( ) cos( ) t nt dt A nt dt A =⎢ − = − ( ) ⎥ ∫0 ∫0 t cos(π nt )dt 2 ⎣ 2 ∫0 ⎦

= 0− =

1 1 ⎤ A A ⎡ 1 π π sin( ) sin( ) sin(π nt )dt ⎥ td nt dt t nt = − + ( ) ⎢ ∫ ∫ 0 πn 0 πn ⎣ 0 ⎦

1 A ⎡ cos(π nt ) ⎤ A A 0 − ⎢ ⎥ = − 2 2 [ cos(π n) − 1] = 2 2 [1 − cos(π n)] π n ⎢⎣ π n 0 ⎥⎦ π n π n

⎧0, ⎪ = ⎨ 2A ⎪⎩ π 2 n 2 ,

n even n odd

d. To = 3, f o = 1/ 3 1 1 3/ 2 ⎤ 1 ⎡1 1 ⎤ 2A ⎤ 2A ⎡t2 2A ⎡ 3/2 C0 = +t1 ⎥ = ⎢ + ⎥ = ⎢ ⎢ ∫ tdt + ∫ 1dt ⎥ = 3 ⎣0 ⎥⎦ 3 ⎣ 2 2 ⎦ 3 1 ⎦ 3 ⎢⎣ 2 0

Since x4 (t ) is an even function of time, Cn =

1.5 3/2 ⎤ 2A ⎡1 ⎤ An ⎡ 2 A =⎢ = + x t nt dt t nt dt ( ) cos(2 / 3) cos(2 / 3) cos(2π nt / 3)dt ⎥ π π ⎥ ⎢∫ 4 ∫ ∫ 2 ⎣ 3 0 1 ⎦ 3 ⎣0 ⎦

Now 1

∫ t cos(2π nt / 3)dt = 0

1 1 ⎤ ⎤ 3 ⎡ 3 ⎡ 1 π π td sin(2 nt / 3) t sin(2 nt / 3) sin(2π nt / 3)dt ⎥ = − ( ) ⎢∫ ⎥ ⎢ ∫ 0 2π n ⎣ 0 0 ⎦ 2π n ⎣ ⎦

3 ⎡ 3 1⎤ π sin(2 n / 3) cos(2π nt / 3) 0 ⎥ + ⎢ 2π n ⎣ 2π n ⎦ 3 ⎡ 3 sin(2π n / 3) + = [cos(2π n / 3) − 1]⎤⎥ ⎢ 2π n ⎣ 2π n ⎦ =

3/2

∫ cos(2π nt / 3)dt = 2π n sin(2π nt / 3) 1

3/2 1

=−

3 2π n

sin(2π n / 3)

Substituting yields

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2 A ⎛ 3 ⎞ ⎡ ⎛ 2π n ⎞ ⎤ 3 A ⎡ ⎛ 2π n ⎞ ⎤ ⎜ ⎟ ⎢cos ⎜ ⎟ − 1⎥ = 2 2 ⎢cos ⎜ ⎟ −1 3 ⎝ 2π n ⎠ ⎣ ⎝ 3 ⎠ ⎦ 2π n ⎣ ⎝ 3 ⎠ ⎥⎦ 2

Cn =

∞

x5 (t ) = ∑ p ( t − 8n ) n =−∞

where p ( t ) = Π ⎡⎣( t − 2 ) / 2 ⎤⎦ + Π ⎡⎣( t − 2 ) / 4 ⎤⎦ − Π ⎡⎣( t − 6 ) / 2 ⎤⎦ − Π ⎡⎣( t − 6 ) / 4⎤⎦ The FT of the pulse shape p ( t ) over [ 0, To ] is given by To

P ( f ) = ∫ p (t )e − j 2π ft dt 0

The FS coefficients of a periodic signal with basic pulse shape p ( t ) are given by T

Cn =

1 o p (t )e − j 2π nfot dt ∫ To 0

Comparing yields Cn =

1 P ( f ) f = nf o To

Now ℑ Π ⎡⎣( t − 2 ) / 2 ⎤⎦ ←⎯ → 2sinc ( 2 f ) e − j 4π f ℑ Π ⎡⎣( t − 2 ) / 4 ⎤⎦ ←⎯ → 4sinc ( 4 f ) e − j 4π f ℑ Π ⎡⎣( t − 6 ) / 2 ⎤⎦ ←⎯ → 2sinc ( 2 f ) e − j12π f ℑ Π ⎡⎣( t − 6 ) / 4 ⎤⎦ ←⎯ → 4sinc ( 4 f ) e − j12π f

Therefore,

P ( f ) = 2sinc ( 2 f ) e− j 4π f ⎡⎣1 − e− j 8π f ⎤⎦ + 4sinc ( 4 f ) e− j 4π f ⎡⎣1 − e− j 8π f ⎤⎦ The FS coefficients of a periodic signal x5 (t ) are now obtained as

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Cn = =

{

}

1 2sinc ( 2nf o ) e− j 4π nfo ⎡⎣1 − e − j 8π nfo ⎤⎦ + 4sinc ( 4nf o ) e− j 4π nfo ⎡⎣1 − e− j 8π nfo ⎤⎦ To

{

}

1 sinc ( n / 4 ) e − jπ n /2 ⎡⎣1 − e − jπ n ⎤⎦ + 2sinc ( n / 2 ) e − jπ n /2 ⎡⎣1 − e − jπ n ⎤⎦ 4

2.15 For the rectangular pulse train in Figure 2.23, compute the Fourier coefficients of the new periodic signal y(t) given by a.

y (t ) = x(t − 0.5To )

Solution: ∞ ⎡ ( t − nTo ) ⎤ The FS expansion of a periodic pulse train x(t ) = ∑ Π ⎢ ⎥ of n =−∞ ⎣ τ ⎦ rectangular pulses is given by

∞

x(t ) = ∑ Cnx e j 2π nfot n =−∞

where the exponential FS coefficients are Cnx =

τ To

sinc ( nf oτ )

Let the FS expansion of a periodic pulse train y (t ) = x(t − 0.5To ) be expressed as ∞

y (t ) = ∑ Cny e j 2π nfot n =−∞

where Cny = =

1 1 y (t )e − j 2π nfot dt = ∫ x(t − 0.5To )e − j 2π nfot dt ∫ To To To To 1 − j 2π nf o (ν + 0.5To ) − j 2π nf o ( 0.5To ) 1 x(ν )e dν = e x(ν )e − j 2π nfoν dν ∫ ∫ To To To To Cnx

= e − jπ nCnx

Time Shifting introduces a linear phase shift in the FS coefficients; their magnitudes are not changed.

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b. y (t ) = x(t )e j 2π t / To Solution:

Cny = =

1 1 y (t )e − j 2π nfot dt = ∫ x(t )e j 2π tfo e− j 2π nfot dt ∫ To To To To 1 − j 2π tf o ( n −1) x(t )e dt ∫ To To

= Cnx−1 c.

y (t ) = x(α t )

Solution: Cny =

1 1 y (t )e − j 2π nfot dt = ∫ x (α t )e − j 2π nfot dt ∫ To To To To ⎛f ⎞

− j 2π n ⎜ o ⎟ν 1 ⎝α ⎠ ( ) = x ν e dν α To α∫To

= Cnx

Time Scaling does not change FS coefficients but the FS itself has changed as the f 2f 3f harmonic components are now at the frequencies ± o , ± o , ± o ,…

2.16 Draw the one-sided power spectrum for the square wave in Figure P2.5 with duty cycle 50%. Figure P2.5

Solution:

The FS expansion for the square wave is given by

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∞

x(t ) = ∑ Cn e j 2π nfot n =−∞

where 1 Cn = ∫ x(t )e − j 2π nfot dt To To T /2 To ⎫⎪ 1 ⎧⎪ o − j 2π nfot = ⎨A ∫ e dt − A ∫ e− j 2π nfot dt ⎬ To ⎪⎩ 0 To / 2 ⎪⎭

= =

{

To / 2 To A e − j 2π nfot − e− j 2π nfot 0 To /2 To ( − j 2π nf o )

}

jA − jπ n e − 1 − e− j 2π n + e− jπ n ) ( 2π n

Therefore, ⎧ j2 A , ⎪− Cn = ⎨ π n ⎪⎩0,

n odd otherwise

Average power in the frequency component at f = nf o equals Cn . Figure 2

displays the one-sided power spectrum for the square wave.

One-sided Power Spectrum of Square wave 0.9 0.8

One-sided Power Spectrum

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

10 15 Frequency (xfo)

a. Calculate the normalized average power. Solution: 25 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

The normalized average power for a periodic signal x (t) is given by T /2

A2To 1 o 2 = = A2 ( ) Px = x t dt To −T∫o /2 To b. Determine the 98% power bandwidth of the pulse train. Solution:

1 3 5 7 9 11 13 15 17 19 21 2.17

Total Power including current Fourier coefficient 0.8106 0.9006 0.9331 0.9496 0.9596 0.9663 0.9711 0.9747 0.9775 0.9798 0.9816

98% Power bandwidth = 21× f o

Determine the Fourier transforms of the signals shown in Figure P2.6.

Figure P2.6

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Solution (a)

The pulse x1 (t ) can be expressed as x1 (t ) = A ⎡⎣Π ( t + 0.5 ) − Π ( t − 0.5 ) ⎤⎦ Now ℑ Π ( t ) ←⎯ → sinc ( f )

Applying the time-shifting property of the FT, we obtain ℑ Π ( t + 0.5 ) ←⎯ → sinc ( f ) e jπ f ℑ Π ( t − 0.5 ) ←⎯ → sinc ( f ) e − jπ f

Adding

X 1 ( f ) = A ⎡⎣sinc ( f ) e jπ f − sinc ( f ) e− jπ f ⎤⎦ = Asinc ( f ) ⎡⎣e jπ f − e− jπ f ⎤⎦ = Aj 2sinc ( f ) sin (π f ) = j 2π fA ⎡⎣sinc 2 ( f ) ⎤⎦ Solution (b)

The pulse x2 (t ) can be expressed as x2 (t ) = Π ( 2t ) cos ( 2π t )

Now 1 ℑ Π (2t ) ←⎯ → sinc( f / 2) 2 1 ℑ cos(2π t ) ←⎯ → ⎡⎣δ ( f − 1) + δ ( f + 1) ⎤⎦ 2 1 1 X 2 ( f ) = ℑ{Π (2t )} ⊗ ℑ{cos(2π t )} = sinc( f / 2) ⊗ ⎡⎣δ ( f − 1) + δ ( f + 1) ⎤⎦ 2 2 1 = ⎡⎣sinc ⎡⎣0.5 ( f − 1) ⎤⎦ + sinc ⎡⎣0.5 ( f + 1) ⎤⎦ ⎤⎦ 4

Solution (c)

The pulse x3 (t ) can be expressed as

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x3 (t ) = p ( t ) + p (−t )

where p ( t ) = e − t ⎡⎣u ( t ) − u ( t − 1) ⎤⎦ From Table 2.2, we have ℑ e− t u (t ) ←⎯ →

1 1 + j 2π f

Using the time-shifting property of the FT, we obtain e u ( t − 1) = e e −t

−1 − ( t −1)

e − j 2π f e −(1+ j 2π f ) = u ( t − 1) ←⎯→ e 1 + j 2π f 1 + j 2π f ℑ

−1

Combining

1 e −(1+ j 2π f ) 1 ⎡1 − e−(1+ j 2π f ) ⎤ p(t ) ←⎯→ P ( f ) = − = ⎦ 1 + j 2π f 1 + j 2π f 1 + j 2π f ⎣ ℑ

By time-reversal property, ℑ p ( −t ) ←⎯ → P (− f ) =

1 ⎡1 − e −(1− j 2π f ) ⎤ ⎦ 1 − j 2π f ⎣

X3 ( f ) = P ( f ) + P (− f ) = = =

1 1 −(1− j 2π f ) ⎡1 − e−(1+ j 2π f ) ⎤ + ⎡ ⎤ ⎦ 1 − j 2π f ⎣1 − e ⎦ 1 + j 2π f ⎣

{(1 − j 2π f ) ⎡⎣1 − e 1 + ( 2π f ) 1

2 1 + ( 2π f )

− (1+ j 2π f )

}

⎤ + (1 + j 2π f ) ⎡1 − e−(1− j 2π f ) ⎤ ⎦ ⎣ ⎦

⎡⎣1 − e−1 cos ( 2π f ) + 2π fe−1 sin ( 2π f ) ⎤⎦

Solution (d)

The pulse x4 (t ) can be expressed as x4 (t ) = Π ( t / 2 ) + Π ( t / 4 )

Now

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ℑ Π ( t / 2 ) ←⎯ → 2sinc ( 2 f ) ℑ Π ( t / 4 ) ←⎯ → 4sinc ( 4 f )

Adding X 4 ( f ) = 2sinc ( 2 f ) + 4sinc ( 4 f )

2.18 Use properties of the Fourier transform to compute the Fourier transform of following signals. a. sinc 2 (Wt ) Solution:

τ ⎛ fτ ⎞ ℑ Λ (t / τ ) ←⎯ → sinc 2 ⎜ ⎟ 2 ⎝ 2 ⎠ Using the duality property, we obtain ℑ Wsinc 2 (Wt ) ←⎯ → Λ ( f / 2W )

ℑ → sinc 2 (Wt ) ←⎯

1 Λ ( f / 2W ) W

Thus the Fourier transform of a sinc2 pulse is a triangular function in frequency. b. Π (t / T ) cos(2π f c t ) Solution: ℑ Π ( t / T ) ←⎯ → Tsinc ( fT )

1 ℑ cos(2π f ct ) ←⎯ → ⎡⎣δ ( f − f c ) + δ ( f + f c ) ⎤⎦ 2 1 X ( f ) = ℑ{Π ( t / T )} ⊗ ℑ{cos(2π f c t )} = Tsinc ( fT ) ⊗ ⎡⎣δ ( f − f c ) + δ ( f + f c ) ⎤⎦ 2 T sin c ⎡⎣T ( f − f c ) ⎤⎦ + sinc ⎡⎣T ( f + f c ) ⎤⎦ = 2

{

}

( e cos10π t ) u (t ) −t

Solution:

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From Table 2.2, ℑ e− t u (t ) ←⎯ →

1 1 + j 2π f

Now

{

} {

}

ℑ ( e − t u (t ) ) cos10π t = ℑ ( e −t u (t ) ) ⊗ ℑ{cos10π t} =

1 1 ⊗ ⎡⎣δ ( f − 5 ) + δ ( f + 5 ) ⎤⎦ 1 + j 2π f 2

⎤ 1⎡ 1 1 + ⎢ ⎥ 2 ⎣1 + j 2π ( f − 5 ) 1 + j 2π ( f + 5 ) ⎦

1 ⎡1 + j 2π ( f + 5 ) + 1 + j 2π ( f − 5 ) ⎤ ⎢ ⎥ 2 2 2 ⎢⎣ (1 + j 2π f ) − ( j 2π 5) ⎥⎦

⎡ ⎤ 1 + j 2π f =⎢ ⎥ 2 2 ⎢⎣ (1 + j 2π f ) + 100π ⎥⎦ ⎡ ⎤ 1 + j 2π f ⎥ =⎢ 2 2 2 ⎢⎣ (1 + 100π ) + j 4π f − 4π f ⎥⎦ d.

te− t u (t )

Solution:

Applying the differentiation in frequency domain property in (2.85), we obtain ℑ → te −t u (t ) ←⎯

j d ℑ{e − t u (t )} 2π df

That is, −1 ⎡ ⎤ j d ⎣(1 + j 2π f ) ⎦ j 1 j 2π ℑ{te u (t )} = =− 2 2π df 2π (1 + j 2π f ) −t

e. e −π t

(1 + j 2π f )

Solution:

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∞

X(f ) = ∫ e

−π t 2

− j 2π ft

−∞

∞

dt = ∫ e

(

−π t 2 + j 2 ft

) dt

−∞

Multiplying the right hand side by e −π f eπ f yields 2

X(f ) =e

−π f 2

∞

( ∫e

−π t 2 + j 2 ft − f 2

) dt = e −π f

−∞

∞ 2

−π ( t + jf ) dt ∫e 2

−∞

Substituting t + jf = ν , we obtain X(f ) =e

−π f 2

∞

−πν ∫ e dν 2

−∞

4sinc (t ) cos (100π t ) 2

Solution:

⎛ f ⎞ ℑ 4sinc 2 ( t ) ←⎯ → 4Λ ⎜ ⎟ ⎝2⎠ 1 ℑ cos (100π t ) ←⎯ → ⎡⎣δ ( f − 50 ) + δ ( f + 50 ) ⎤⎦ 2 ⎛ f ⎞ 1 X ( f ) = ℑ{4sinc 2 ( t )} ⊗ ℑ{cos (100π t )} = 4Λ ⎜ ⎟ ⊗ ⎡⎣δ ( f − 50 ) + δ ( f + 50 ) ⎤⎦ ⎝2⎠ 2

{

}

= 2 Λ ⎡⎣0.5 ( f − 50 ) ⎤⎦ + Λ ⎡⎣ 0.5 ( f − 50 ) ⎤⎦ 2.19 Find the following convolutions: a. sinc(Wt ) ⊗ sinc(2Wt ) Solution:

We use the convolution property of Fourier transform in (2.79). ℑ x(t ) ⊗ y (t ) ←⎯→ X ( f )Y ( f )

Now ℑ → sinc(2Wt ) ←⎯

1 Π ( f / 2W ) 2W

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Therefore, 1 1 Π( f / W ) × Π ( f / 2W ) 2W W 1 = Π( f / W ) 2W 2

ℑ → sinc(Wt ) ⊗ sinc(2Wt ) ←⎯

b. sinc 2 (Wt ) ⊗ sinc(2Wt ) Solution:

Again using the convolution property of Fourier transform 1 1 Λ ( f / 2W ) × Π ( f / 2W ) 2W W 1 = Λ ( f / 2W ) 2W 2

ℑ → sinc 2 (Wt ) ⊗ sinc(2Wt ) ←⎯

2.20 The FT of a signal x(t ) is described by X(f ) =

1 5 + j 2π f

Determine the FT V ( f ) of the following signals: a. v(t ) = x(5t − 1) Solution:

⎡ ⎛ 1 ⎞⎤ v(t ) = x(5t − 1) = x ⎢5 ⎜ t − ⎟ ⎥ ⎣ ⎝ 5 ⎠⎦ Let ℑ y (t ) = x(5t ) ←⎯ →Y ( f ) =

1 ⎛ f ⎞ 1 1 X⎜ ⎟= 5 ⎝ 5 ⎠ 5 ( j 2π f / 5 ) + 5 1 j 2π f + 25

− j 2π f − j 2π f 1 ⎛ 1⎞ ℑ 5 = v(t ) = y ⎜ t − ⎟ ←⎯→ V ( f ) = Y ( f )e e 5 j 2π f + 25 ⎝ 5⎠

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b. v(t ) = x(t ) cos(100π t ) Solution:

⎡⎣ x(t )e j100π t + x(t )e− j100π t ⎤⎦ v(t ) = x(t ) cos(100π t ) = 2 Now ⎡⎣ x(t )e j100π t + x(t )e− j100π t ⎤⎦ ℑ 1 ←⎯→V ( f ) = ⎡⎣ X ( f − 50 ) + X ( f + 50 ) ⎤⎦ 2 2 1 1 = + j 2π ( f − 50 ) + 5 j 2π ( f + 50 ) + 5 After simplification, we get V( f ) =

j 2π f + 5 j 20π f + (π 2104 + 25 − 4π 2 f 2 )

c. v(t ) = x(t )e j10t Solution: 5⎞ 1 ⎛ ℑ →V ( f ) = X ⎜ f − ⎟ = v(t ) = x(t )e j10t ←⎯ 5⎞ π⎠ ⎛ ⎝ j 2π ⎜ f − ⎟ + 5 π⎠ ⎝ dx(t ) d. v(t ) = dt

Solution:

Using the differentiation property of FT V ( f ) = j 2π fX ( f ) =

d ℑ x(t ) ←⎯→ j 2πfX ( f ) , we obtain dt

j 2π f 5 + j 2π f

e. v(t ) = x(t ) ⊗ u (t ) Solution: ℑ Using the convolution property of FT x(t ) ⊗ y (t ) ←⎯→ X ( f )Y ( f ) , we can write

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V ( f ) = X ( f )U ( f ) =

⎛1 1 1 ⎞ ⎜ δ( f )+ ⎟ j 2π f ⎠ 5 + j 2π f ⎝ 2

1 1 1 = δ(f ) + 2 5 + j 2π f j 2π f ( 5 + j 2π f ) =

1 1 δ(f )+ 10 j 2π f ( 5 + j 2π f )

2.21 Consider the delay element y (t ) = x(t − 3) . a. What is the impulse response h(t)? Solution:

Since y (t ) = h(t ) ⊗ x(t ) = x(t − 3) , the impulse response of the delay element is given from (2.16) as h(t ) = δ (t − 3) b. What is the magnitude and phase response function of the system? Solution:

H ( f ) = ℑ{δ (t − 3)} = e − j 6π f H( f ) =1 H ( f ) = −6π f 2.22 The periodic input x(t ) to an LTI system is displayed in Figure P2.7. The frequency response function of the system is given by H( f ) =

2 2 + j 2π f

Figure P2.7

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a. Write the complex exponential FS of input x(t ) . Solution:

The input x(t ) is rectangular pulse train in Example 2.24 shifted by To / 4 and duty cycle

= 0.5 . That is,

∞ ⎡ ( t − nTo − To / 4 ) ⎤ x(t ) = gTo ( t − To / 4 ) = ∑ Π ⎢ ⎥ 0.5To n =−∞ ⎣ ⎦

The complex exponential FS of gTo (t ) from Example 2.24 with To = 2 ( f o = 0.5 ) and

= 0.5 is given by

∞

gTo (t ) = 0.5 ∑ sinc ( 0.5n )e jπ nt n =−∞

In Exercise 2.15(a), we showed that time shifting introduces a linear phase shift in the FS coefficients; their magnitudes are not changed. The phase shift is equal to e − j 2π nfou for a time shift of u. The exponential FS coefficients x(t ) are Cn = 0.5sinc ( 0.5n ) e

− j 2π nf o ( 0.25To )

= e − jπ n /2 0.5sinc ( 0.5n )

∞

x(t ) = 0.5 ∑ ⎡⎣e − jπ n /2sinc ( 0.5n ) ⎤⎦e jπ nt n =−∞

b. Plot the magnitude and phase response functions for H ( f ) . Solution:

H( f ) =

2 1 = 2 + j 2π f 1 + jπ f

H ( nf o ) =

1 1 = 1 + jπ f 1 + jπ nf o

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0 -10

-5 Phase Response(degrees)

Magnitude Response(dB)

-20 -10

-15

-20

-30 -40 -50 -60 -70

-25 -80 -30

4 6 Frequency (f)

-90

4 6 Frequency (f)

c. Compute the complex exponential FS of the output y (t ) . Solution:

Using (2.114), the output of an LTI system to the input ∞

x(t ) = ∑ ⎡⎣0.5e− jπ n / 2sinc ( 0.5n ) ⎤⎦e jπ nt n =−∞

is given by ⎡ ⎤ − jπ n /2 ⎢ ⎥ 0.5 e y (t ) = ∑ Cn H (0.5n)e jπ nt = ∑ ⎢ sinc ( 0.5n ) ⎥ e jπ nt n =−∞ n =−∞ ⎢1 + j 05π n ⎥ FS coefficient of y ( t ) ⎣⎢ ⎦⎥ ∞

∞

2.23 The frequency response of an ideal LP filter is given by ⎧⎪5e − j 0.0025π f , H( f ) = ⎨ ⎪⎩0,

f < 1000 Hz f > 1000 Hz

Determine the output signal in each of the following cases: a.

x(t ) = 5sin ( 400π t ) + 2 cos (1200π t − π / 2 ) − cos ( 2200π t + π / 4 )

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Solution:

⎧⎪5, H( f ) = ⎨ ⎪⎩0, ⎧− ⎪ 0.0025π f , H( f ) = ⎨ ⎪⎩0, Since H (200) = 5 and

f < 1000 Hz f > 1000 Hz f < 1000 Hz f > 1000 Hz

H (200) = −π / 2 , the output of the system for an input

5sin(400πt) can now be expressed using (2.117) as 25sin(400πt − π/2). Next H (600) = 5 and

H (600) = −3π / 2 = π / 2 , the output of the system for

an input 2 cos (1200π t − π / 2 ) can now be expressed using (2.117) as 10 cos (1200π t − π / 2 + π / 2 ) = 10 cos (1200π t ) .

Now H (1100) = 0. So the LP filter doesn’t pass cos ( 2200π t + π / 4 ) . The output of the LP filter is therefore given by y (t ) = 25sin ( 400π t − π / 2 ) + 10 cos (1200π t )

b. x(t ) = 2sin(400π t ) +

sin ( 2200π t ) πt

Solution:

Since H (200) = 5 and

H (200) = −π / 2 , the output of the system for an input

2sin(400πt) is 10sin(400πt − π/2). To calculate the response to

sin ( 2200π t ) , we note that πt

sin ( 2200π t ) ⎛ f ⎞ ℑ = 2200 × sinc ( 2200t ) ←⎯ →Π⎜ ⎟ πt ⎝ 2200 ⎠

In frequency domain, the output of LP filter to

sin ( 2200π t ) is πt

⎛ f ⎞ − j 0.0025π f ⎛ f ⎞ ⎛ f ⎞ − j 0.0025π f Π⎜ Π⎜ ⎟ 5e ⎟ = 5Π ⎜ ⎟e ⎝ 2200 ⎠ ⎝ 2000 ⎠ ⎝ 2000 ⎠

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Now ⎛ f ⎞ − j 0.0025π f ℑ Π⎜ ←⎯→ 2000sinc ⎡⎣ 2000 ( t − 0.00125 ) ⎤⎦ ⎟e ⎝ 2000 ⎠ The output of the LP filter is therefore given by y (t ) = 10sin ( 400π t − π / 2 ) + 10000sinc ⎡⎣ 2000 ( t − 0.00125 ) ⎤⎦ c.

x(t ) = cos(400π t ) +

sin (1000π t ) πt

Solution:

Since H (200) = 5 and

H (200) = −π / 2 , the output of the system for an input

cos(400π t ) is 5cos ( 400π t − π / 2 ) . To calculate the response to

sin (1000π t ) , we note that πt

sin (1000π t ) ⎛ f ⎞ ℑ = 1000 × sinc (1000t ) ←⎯ →Π ⎜ ⎟ πt ⎝ 1000 ⎠

In frequency domain, the output of LP filter to

sin (1000π t ) is πt

⎛ f ⎞ ⎛ f ⎞ − j 0.0025π f 5e− j 0.0025π f Π ⎜ ⎟ = 5Π ⎜ ⎟e ⎝ 1000 ⎠ ⎝ 1000 ⎠ Now ⎛ f ⎞ − j 0.0025π f ℑ Π⎜ ←⎯→1000sinc ⎡⎣1000 ( t − 0.00125 ) ⎤⎦ ⎟e ⎝ 1000 ⎠ The output of the LP filter is therefore given by y (t ) = 5cos ( 400π t − π / 2 ) + 5000sinc ⎡⎣1000 ( t − 0.00125 ) ⎤⎦ d. x(t ) = 5cos(800π t ) + 2δ (t ) Solution:

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Since H (400) = 5 and

H (400) = −π , the output of the system for an input

5cos ( 800π t ) is 25cos ( 800π t − π ) . The response of the system to δ (t ) is h(t ) .

The impulse response of the ideal LP filter is 10000sinc ⎡⎣ 2000 ( t − 0.00125 ) ⎤⎦ . Combining y (t ) = 25cos ( 800π t − π ) + 2h(t ) = 25cos ( 800π t − π ) + 20 × 103 sinc ⎡⎣ 2000 ( t − 0.00125 ) ⎤⎦ 2.24 The frequency response of an ideal HP filter is given by ⎧⎪4, H( f ) = ⎨ ⎪⎩0,

f > 20 Hz, f < 20 Hz

Determine the output signal y (t ) for the input a.

x(t ) = 5 + 2 cos ( 50π t − π / 2 ) − cos ( 75π t + π / 4 )

Solution: y (t ) = 8cos ( 50π t − π / 2 ) − 4 cos ( 75π t + π / 4 )

b. x(t ) = cos ( 20π t − 3π / 4 ) + 3cos (100π t + π / 4 ) Solution: y (t ) = 12 cos (100π t + π / 4 )

2.25 The frequency response of an ideal BP filter is given by ⎧⎪2e− j 0.0005π f , 900< f < 1000 Hz, H( f ) = ⎨ otherwise ⎪⎩0, Determine the output signal y (t ) for the input a.

x(t ) = 2 cos (1850π t − π / 2 ) − cos (1900π t + π / 4 )

Solution: H (925) = 2 and H (950) = 2 .

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H ( f ) = −0.0005π f . Therefore, H (925) = −0.0005π f = −0.0005π × 925 = −0.462π H (950) = −0.0005π f = −0.0005π × 950 = −0.475π y (t ) = 4 cos (1850π t − π / 2 − 0.462π ) − 2 cos (1900π t + π / 4 − 0.475π )

b. x(t ) = sinc ( 60t ) cos(1900π t ) Solution:

X(f ) = =

1 ⎛ f ⎞ 1 Π ⎜ ⎟ ⊗ [δ ( f − 950) + δ ( f + 950) ] 60 ⎝ 60 ⎠ 2

1 ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ Π⎜ ⎟+Π⎜ ⎟⎥ ⎢ 120 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦

Y( f )= H( f )X ( f ) = 2e − j 0.0005π f ×

1 ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ × ⎢Π ⎜ ⎟+Π⎜ ⎟⎥ 120 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦

Now 1 ⎛ f − 950 ⎞ ℑ j1900π t ×Π⎜ ⎟ ←⎯→ sinc ( 60t ) e 60 ⎝ 60 ⎠ e − j 0.0005π f

1 ⎛ f − 950 ⎞ ℑ j1900π ( t − 0.00025 ) ×Π⎜ ⎟ ←⎯→ sinc ⎡⎣60 ( t − 0.00025 ) ⎤⎦ e 60 ⎝ 60 ⎠

Similarly 1 ⎛ f + 950 ⎞ ℑ − j1900π t ×Π⎜ ⎟ ←⎯→ sinc ( 60t ) e 60 ⎝ 60 ⎠ e− j 0.0005π f

1 ⎛ f + 950 ⎞ ℑ − j1900π ( t − 0.00025 ) ×Π⎜ ⎟ ←⎯→ sinc ⎡⎣ 60 ( t − 0.00025 ) ⎤⎦ e 60 ⎝ 60 ⎠

Therefore,

y (t ) = sinc ⎡⎣60 ( t − 0.00025 ) ⎤⎦ ⎡⎣e j1900π ( t −0.00025) + e − j1900π ( t −0.00025) ⎤⎦ = 2sinc ⎡⎣60 ( t − 0.00025 ) ⎤⎦ cos ⎡⎣1900π ( t − 0.00025 ) ⎤⎦

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x(t ) = sinc 2 ( 30t ) cos(1900π t )

Solution:

X(f ) = =

1 ⎛ f ⎞ 1 Λ ⎜ ⎟ ⊗ [δ ( f − 950) + δ ( f + 950) ] 30 ⎝ 60 ⎠ 2 1 ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ Λ⎜ ⎟ + Λ⎜ ⎟⎥ ⎢ 60 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦

Y( f )= H( f )X ( f ) = 2e − j 0.0005π f ×

1 ⎡ ⎛ f − 950 ⎞ ⎛ f + 950 ⎞ ⎤ Λ⎜ ⎟+ Λ⎜ ⎟⎥ ⎢ 60 ⎣ ⎝ 60 ⎠ ⎝ 60 ⎠ ⎦

Now 1 ⎛ f − 950 ⎞ ℑ 2 j1900π t ×Λ⎜ ⎟ ←⎯→ sinc ( 30t ) e 30 ⎝ 60 ⎠ e − j 0.0005π f

1 ⎛ f − 950 ⎞ ℑ j1900π ( t − 0.00025) 2 ×Λ⎜ ⎟ ←⎯→ sinc ⎡⎣30 ( t − 0.00025 ) ⎤⎦ e 30 ⎝ 60 ⎠

Similarly 1 ⎛ f + 950 ⎞ ℑ 2 − j1900π t ×Π⎜ ⎟ ←⎯→ sinc ( 30t ) e 30 ⎝ 60 ⎠ e − j 0.0005π f

1 ⎛ f + 950 ⎞ ℑ − j1900π ( t − 0.00025 ) 2 ×Π⎜ ⎟ ←⎯→ sinc ⎡⎣30 ( t − 0.00025 ) ⎤⎦ e 30 ⎝ 60 ⎠

Therefore, y (t ) = sinc 2 ⎡⎣30 ( t − 0.00025 ) ⎤⎦ ⎡⎣ e j1900π ( t −0.00025) + e− j1900π ( t −0.00025) ⎤⎦ = 2sinc 2 ⎡⎣30 ( t − 0.00025 ) ⎤⎦ cos ⎡⎣1900π ( t − 0.00025 ) ⎤⎦ 2.26 The signal 2e −2t u (t ) is input to an ideal LP filter with passband edge frequency equal to 5 Hz. Find the energy density spectrum of the output of the filter. Calculate the energy of the input signal and the output signal. Solution:

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∞

e −4t =1 Ex = ∫ 2e u (t ) dt = ∫ 4e dt = 4 −4 0 −∞ 0 2

−2 t

ℑ x(t ) = 2e −2t u (t ) ←⎯ →

−4 t

2 2 + j 2π f 2

The energy density spectrum of the output y(t), Y ( f ) , is related to the energy density spectrum of the input x(t) 2

2 ⎛ f ⎞ Y( f ) = H( f ) X( f ) = Π⎜ ⎟ ⎝ 10 ⎠ 2 + j 2π f 2

∞

2 ⎛ f ⎞ E y = ∫ y (t ) dt = ∫ Y ( f ) df = ∫ Π ⎜ ⎟ df ⎝ 10 ⎠ 2 + j 2π f −∞ −∞ −∞ 2

1 1 df = 2∫ df =∫ 1 + jπ f 1+ π 2 f 2 0 −5

Making change of variables π f = u ⇒ df = Ey =

5π

du = tan ( u ) π ∫ 1+ u π 2

−1

5π 0

du , we obtain

(1.507 ) = 0.9594

Thus output of the LP filter contains 96% of the input signal energy. 2.27 Calculate and sketch the power spectral density of the following signals. Calculate the normalized average power of the signal in each case. a. x(t ) = 2 cos (1000π t − π / 2 ) − cos (1850π t + π / 4 ) x1 ( t )

x2 ( t )

Solution: T /2

1 Rx (τ ) = lim x(t ) x(t − τ )dt = Rx1 (τ ) + Rx2 (τ ) − Rx1x2 (τ ) − Rx2 x1 (τ ) T →∞ T ∫ −T /2 Now

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T /2

1 Rx1 (τ ) = lim x1 (t ) x1 (t − τ )dt T →∞ T ∫ −T /2 1 π⎞ π⎤ ⎛ ⎡ 2 cos ⎜1000π t − ⎟ 2 cos ⎢1000π ( t − τ ) − ⎥ dt T →∞ T ∫ 2⎠ 2⎦ ⎝ ⎣ −T /2 T /2

= lim

T /2

{

}

1 cos (1000πτ ) + cos ⎡⎣1000π ( 2t − τ ) − π ⎤⎦ dt T →∞ T ∫ −T /2

= 2 lim

= 2 cos (1000πτ )

The second term is zero because it integrates a sinusoidal function over a period. Similarly, T /2

1 1 x2 (t ) x2 (t − τ )dt = cos (1850πτ ) ∫ T →∞ T 2 −T / 2

Rx2 (τ ) = lim

The cross-correlation term T /2

T /2

1 1 x1 (t ) x2 (t − τ ) = lim 2 cos (1000π t − π / 2 ) cos ⎡⎣1850π ( t − τ ) + π / 4 ⎤⎦dt ∫ T →∞ T T →∞ T ∫ −T / 2 −T / 2

Rx1 x2 (τ ) = lim

T /2

1 = lim {cos (850π t − 1850πτ + 3π / 4 ) + sin ( 2850π t − 1850πτ − π / 4 )}dt T →∞ T ∫ −T / 2 is zero because it integrates a sinusoidal function over a period in each case. Similarly, it can be shown that all other cross-correlation terms are zero. Therefore, 1 Rx (τ ) = Rx1 (τ ) + Rx2 (τ ) = 2 cos (1000πτ ) + cos (1850πτ ) 2

1 ⎧ ⎫ Gx ( f ) = ℑ{Rx (τ )} = ℑ ⎨2 cos (1000πτ ) + cos (1850πτ ) ⎬ 2 ⎩ ⎭ 1 = [δ ( f − 500) + δ ( f + 500) ] + [δ ( f − 925) + δ ( f + 925) ] 4 The normalized average power is obtained by using (2.172) as

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∞

−∞

Px = ∫ Gx ( f )df = ∫ [δ ( f − 500) + δ ( f + 500) ] df +

∞

1 ∫ [δ ( f − 925) + δ ( f + 925)] df 4 −∞

1 1 5 = 1+1+ + = 4 4 2

Power Spectral Density

0.8

0.6

0.4

0.2

0 -1

-0.8

-0.6

-0.4

-0.2 0 0.2 Frequency (kHz)

0.4

0.6

0.8

b. x(t ) = [1 + sin(200π t ) ] cos(2000π t ) Solution:

x(t ) = [1 + sin(200π t ) ] cos(2000π t ) = cos(2000π t ) + sin(200π t ) cos(2000π t ) 1 1 = cos(2000π t ) + sin ( 2200π t ) − sin (1800π t ) 2 2 x (t ) 1

x2 ( t )

x3 ( t )

Now T /2

1 Rx (τ ) = lim x(t ) x(t − τ )dt = Rx1 (τ ) + Rx2 (τ ) + Rx3 (τ ) T →∞ T ∫ −T /2 where 1 cos ( 2000πτ ) 2 1 Rx2 (τ ) = cos ( 2200πτ ) 8 1 Rx3 (τ ) = cos (1800πτ ) 8

Rx1 (τ ) =

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Therefore, 1 1 1 Rx (τ ) = cos ( 2000πτ ) + cos ( 2200πτ ) + cos (1800πτ ) 2 8 8 1 1 ⎧1 ⎫ Gx ( f ) = ℑ{Rx (τ )} = ℑ ⎨ cos ( 2000πτ ) + cos ( 2200πτ ) + cos (1800πτ ) ⎬ 8 8 ⎩2 ⎭ 1 1 = [δ ( f − 1000) + δ ( f + 1000) ] + [δ ( f − 1100) + δ ( f + 1100) ] 4 16 1 + [δ ( f − 900) + δ ( f + 900) ] 16

The normalized average power is obtained by using (2.172) as ∞

∞

1 1 Px = ∫ Gx ( f )df = ∫ [δ ( f − 1000) + δ ( f + 1000) ] df + ∫ [δ ( f − 1100) + δ ( f + 1100)] df 4 −∞ 16 −∞ −∞ ∞

1 1 1 1 3 [δ ( f − 900) + δ ( f + 900)] df = + + = ∫ 16 −∞ 2 8 8 4

Power Spectral Density

0.25

0.2

0.15

0.1

0.05

-1

-0.8

-0.6

-0.4

-0.2 0 0.2 Frequency (kHz)

0.4

0.6

0.8

x(t ) = cos 2 ( 200π t ) sin (1800π t )

Solution:

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1 1 1 ⎡⎣1 + cos ( 400π t ) ⎤⎦ sin (1800π t ) = sin (1800π t ) + sin (1800π t ) cos ( 400π t ) 2 2 2 1 1 1 = sin (1800π t ) + sin (1400π t ) + sin ( 2200π t ) 2 4 4

x(t ) =

Now

Rx (τ ) = Rx1 (τ ) + Rx2 (τ ) + Rx3 (τ ) where 1 Rx1 (τ ) = cos (1800πτ ) 8

Rx2 (τ ) =

1 cos (1400πτ ) 32

Rx3 (τ ) =

1 cos ( 2200πτ ) 32

Therefore, 1 1 1 Rx (τ ) = cos (1800πτ ) + cos (1400πτ ) + cos ( 2200πτ ) 8 32 32

1 1 ⎧1 ⎫ Gx ( f ) = ℑ{Rx (τ )} = ℑ ⎨ cos (1800πτ ) + cos (1400πτ ) + cos ( 2200πτ ) ⎬ 32 32 ⎩8 ⎭ 1 1 = [δ ( f − 900) + δ ( f + 900) ] + [δ ( f − 700) + δ ( f + 700) ] 16 64 1 + [δ ( f − 1100) + δ ( f + 1100) ] 64 The normalized average power is obtained by using (2.172) as ∞

∞

1 1 Px = ∫ Gx ( f )df = [δ ( f − 900) + δ ( f + 900)]df + ∫ [δ ( f − 700) + δ ( f + 700)] df ∫ 16 −∞ 64 −∞ −∞ ∞

1 1 1 1 3 [δ ( f − 1100) + δ ( f + 1100)] df = + + = ∫ 64 −∞ 8 32 32 16

46 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

0.1 0.09

Power Spectral Density

0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0

-1

-0.8

-0.6

-0.4

-0.2

0.2

0.4

0.6

0.8

47 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Chapter 4 4.1

The message signal s1 (t ) = sin(100π t ) modulates the carrier c(t ) = cos(36000π t ) to generate the DSB-SC AM signal. a. Sketch the spectrum of message signal. Solution:

S1 ( f ) =

1 ⎡δ ( f − 50 ) − δ ( f + 50 ) ⎤⎦ 2j ⎣ S1 ( f ) 1/ 2

−50

(S1 ( f ) π /2 50 −50

−π / 2

b. Determine and sketch the spectrum of the resulting DSB-SC AM signal. Identify the upper and lower sidebands in the spectrum. Solution: x1 (t ) = s1 (t )cos(36000π t ) = sin(100π t )cos(36000π t )

1 [sin(36100π t ) − sin(35900π t )] 2

Taking the FT of both sides, we obtain

X1 ( f ) =

1 ⎡δ ( f − 18050 ) − δ ( f + 18050 ) − δ ( f − 17950 ) + δ ( f + 17950 ) ⎤⎦ 4j ⎣

X1 ( f ) USB

LSB

USB

LSB 1/ 4

−18.05

f (kHz)

−17.95

17.95

18.05

(X 1 ( f )

π /2 18.05

−17.95 −18.05

f (kHz)

17.95

π /2

c. Repeat (a) and (b) for s2 (t ) = cos(600π t ) cos(1800π t ) . Solution:

s2 (t ) = cos(600π t )cos(1800π t ) =

1 [cos(1200π t ) + cos(2400π t )] 2

1 S2 ( f ) = ⎡⎣δ ( f − 600 ) + δ ( f + 600 ) + δ ( f − 1200 ) + δ ( f + 1200 ) ⎤⎦ 4

S2 ( f ) 1/ 4

−1200

−600

600

1200

The resulting DSB-SC AM signal is given by

x2 (t ) = s2 (t ) cos(36000π t ) = =

1 [cos(1200π t ) + cos(2400π t )] cos(36000π t ) 2

1 [cos(34800π t ) + cos(37200π t ) + cos(33600π t ) + cos(38400π t )] 4

Taking the FT of both sides, we obtain

1 X 2 ( f ) = ⎡⎣δ ( f − 17400) + δ ( f + 17400) + δ ( f − 18600) + δ ( f + 18600) 4 +δ ( f − 16800 ) + δ ( f + 16800 ) + δ ( f − 19200) + δ ( f + 19200 ) ⎤⎦

X2 ( f ) USB

LSB

USB

1/ 4

−19.2 −18.6 −17.4 −16.8

f (kHz) 16.8 17.4

18.6 19.2

d. Calculate the PEP supplied by the DSB-SC modulator in each case. Solution: Since max s1 (t ) = max s2 (t ) = 1 , PEPDSB is obtained from (4.7) as t

Ac2 ⎡ max s1 (t ) ⎤ Ac2 ⎡ max s1 (t ) ⎤ t ⎣ ⎦ ⎣ t ⎦ =1 PEPDSB = = 2 2 2

4.2

because Ac = 1

The message signal s(t ) = 2cos(6π t ) + 0.5sin(10π t ) modulates the carrier signal

c(t ) = Ac cos(100π t ) using DSB-SC AM scheme. a. Write an expression for the modulated signal x(t ) . Solution:

x(t ) = s (t )c(t ) = Ac [ 2 cos(6π t ) + 0.5sin(10π t )] cos(100π t ) = Ac [ 2 cos(6π t ) cos(100π t ) + 0.5sin(10π t ) cos(100π t ) ] = Ac [ cos(94π t ) + cos(106π t ) − 0.25sin(90π t ) + 0.25sin(110π t ) ] b. Determine and sketch the spectrum of the modulated signal x(t ) . Solution: Taking the FT of both sides, we obtain

X( f ) =

Ac ⎡δ ( f − 47 ) + δ ( f + 47 ) + δ ( f − 53) + δ ( f + 53) 2 ⎣ ⎤ 0.25 0.25 0.25 0.25 δ ( f − 45) + δ ( f + 45) + δ ( f − 55) − δ ( f + 55) ⎥ − j j j j ⎦

X(f)

USB

Ac 2

LSB

USB

Ac 8 -55 -53

-50

45 47 50

-47 -45

53 55

f (Hz)

(X ( f )

π /2 -45 -55 -53 -50 -47

55 0

45 47

f (Hz)

−π / 2 c. Calculate the power contained in the modulated signal x(t ) .

Solution: ⎡ 1 1 ( 0.25 )2 ( 0.25 )2 ⎤ 17 A2 c Px = A ⎢ + + + ⎥= 2 2 ⎥⎦ 16 ⎢⎣ 2 2 2 c

A modulating signal s(t ) = 2cos(6π t ) + 5sin(10π t ) modulates the carrier signal c(t ) = 100cos(1000π t ) using convention AM scheme. The modulator operates with a modulation index of 0.7.

4.3

a. Write an expression for the modulated signal x(t ) . Solution: The message signal s(t ) = 2cos(6π t ) + 5sin(10π t ) is displayed in Figure (a). Using MATLAB, the minimum value of s (t ) is −6.9412.. The normalized message signal sn (t ) is now obtained as

sn (t ) =

s(t ) min s(t )

2cos(6π t ) + 5sin(10π t ) = 0.288cos(6π t ) + 0.72sin(10π t ) 6.9412

The conventional AM signal is given by x AM (t ) = 100 {1 + 0.7 [ 0.288 cos(6π t ) + 0.72 sin(10π t ) ]} cos(1000π t )

1.5

Conventional AM signal xAM(t)

Message signal s(t)

Figure (b) displays the signal x AM (t ) .

2 0 -2 -4 -6 -8

0.5 0 -0.5 -1 -1.5

0.1

0.2

0.3

0.4

0.5 t

(a)

0.6

0.7

0.8

0.9

-2

0.1

0.2

0.3

0.4

0.5 t

0.6

0.7

0.8

0.9

(b)

b. Determine and sketch spectrum of the modulated signal x(t ) . Solution:

X ( f ) = 50 ⎡⎣δ ( f − 500 ) + δ ( f + 500 ) ⎤⎦ + 10.08 ⎡⎣δ ( f − 503) + δ ( f + 503) + δ ( f − 497 ) + δ ( f + 497 ) ⎤⎦ +

25.2 ⎡δ ( f − 505) − δ ( f + 505) − δ ( f − 495) + δ ( f + 495) ⎤⎦ j ⎣

X(f) Carrier USB

Carrier

LSB

USB

25 f (Hz) 0

-505 -503 -500 -497 -495

495 497 500 503 505

(X ( f )

π /2 497 500 503 505

-503 -500 -497 -495 -505

495

f (Hz)

−π / 2 c. Compare the power contained in the sidebands to the total power in the modulated signal. What is the power efficiency this AM scheme? Solution:

1 1 2 2 ( 0.288 ) + ( 0.72 ) 2 2 = 0.04147 + 0.2592 = 0.301 W .

Power in the normalized message signal Psn =

Ac2 Ac2 ma2 + Psn Power in the conventional AM signal Px = 2 2 1002 2 = 1 + ( 0.7 ) × 0.301 = 5.737 kW 2

(

)

(

)

Ac2 ma2 1002 2 Power in the sidebands = Psn = ( 0.7 ) × 0.301 = 737 W 2 2 The power efficiency of this AM scheme is

η=

737 = 12.85% 5737

d. Calculate the PEP supplied by the AM transmitter. Solution:

Using MATLAB, max sn (t ) = 1 . The PEPAM supplied by the transmitter is t

obtained from (4.34) as PEPAM =

1002 264.45 2 = 14.45 kW [1 + 0.7 ×1.0] = 2 2

Note that PEPAM is approximately 2.5 times the average power supplied by the AM signal. 4.4 The output of an AM modulator is given by x(t ) = Ac cos(2000π t ) + 50 cos(2100π t ) + 50 cos(1900π t ) The carrier power is 5 kW. a. Determine the modulation index ma . Solution:

Since the carrier power is 5 kW, we have

1 2 Ac = 5000 2 Therefore, Ac = 100 Substituting

x(t ) = 100 cos(2000π t ) + 50 cos(2100π t ) + 50 cos(1900π t ) = 100 [1 + cos(100π t ) ] cos(2000π t ) Therefore, ma = 1.0 . b. Write an expression for the normalized modulating signal. Solution:

sn (t ) = cos(100π t ) c. Calculate the power efficiency. Solution:

The power efficiency is given from (4.30) as

η=

ma2 Psn 1 + ma2 Psn

Psn =

η=

1 2

1.02 × 0.5 = 0.333 or 33.3% 1 + 1.02 × 0.5

d. Calculate the PEP supplied by the AM transmitter and tabulate it along with the normalized average power and the carrier power. Solution:

PEPAM =

2 Ac2 ⎡ 1002 2 1 + ma × max sn (t ) ⎤ = [1 + 1] = 20 kW t ⎣ ⎦ 2 2

Normalized average power Px = Carrier Power Average Power Px 5 kW 7 kW

Ac2 1002 ⎡ 1 ⎤ ⎡⎣1 + ma2 Psn ⎤⎦ = 1 + = 7.5 kW 2 2 ⎢⎣ 2 ⎥⎦

Peak Envelope Power PEPAM 20 kW

4.5 The envelope detector shown in Figure 4.16 (a) is used to demodulate the message signal s(t ) from the conventional AM signal x(t ) = Ac [1 + s(t )] cos(2π fct ) where s(t ) is a square wave taking on 1 and -1 and having a period To 1/ f c . Sketch the demodulated signal if RL C = To / 40 and To / 4 . Solution:

The demodulated signal was calculated using the following MATLAB script: %Exercise 4.5 Envelope detection of square wave clc Fs = 1000 %Sampling rate Ts=1/Fs fc = 100; %carrier freq. fm=1. To = 1 t = [0:1/Fs:2]; A =1.25 m = square(2*pi*fm*t, 50); figure(1) plot(t,m) axis([0 2 -1.25 1.25]) smin=min(m) mn=m/abs(smin); %Modulator c = cos(2*pi*fc*t); ma = abs(smin)/A % AM = A*(1+ma*mn).*c; figure(2) plot(t,AM,'b') N=4; TRC =To/4 [bLP,aLP]=butter(N,2*fm*10*Ts); figure(3) BB_sig = Envelope_Detector(AM,bLP,aLP,Ts,TRC);%clean baseband signal plot(t(1:2000),BB_sig(1:2000),'k'); hold on TRC =To/40 BB_sig = Envelope_Detector(AM,bLP,aLP,Ts,TRC);%clean baseband signal plot(t(1:2000),BB_sig(1:2000),'r--'); xlabel('\itt'); ylabel('Envelope detector output signal y(\itt)');hold off legend( 'Detector RC time constant= T_o/4','Detector RC time constant= T_o/40')

1.5 Detector RC time constant= To/4 Detector RC time constant= To/40

Envelope detector output signal y(t)

0.5

-0.5

-1

-1.5

0.2

0.4

0.6

0.8

1 t

1.2

1.4

1.6

1.8

4.6 Consider the demodulation of conventional AM signals using a square law detector 2 (t ) , where are k1 and k2 constants with transfer characteristic y (t ) = k1 xAM (t ) + k2 xAM with k2 k1 . a. Calculate the detector output y(t ) . Solution:

The conventional AM signal is given by xAM (t ) = Ac [1 + ma sn (t )]cos(2π f c t ) The detector output y(t ) is given by y (t ) = k1 Ac [1 + ma sn (t )]cos(2π f c t ) + k2 Ac2 [1 + ma sn (t )]2 cos 2 (2π f c t ) = k1 Ac [1 + ma sn (t )]cos(2π f c t ) +

k2 Ac2 [1 + 2ma sn (t ) + ma2 sn2 (t )] [1 + cos(4π f c t )] 2

b. Calculate the output yD (t ) of the LP filter. Note that it contains the desired signal as well as distortion term. Show that if Ac s (t ) , the distortion can be neglected. Solution:

The message signal sn (t ) can be recovered by passing it through a LPF. The output yD (t ) of the LP filter is yD (t ) =

k2 Ac2 [1 + 2ma sn (t ) + ma2 sn2 (t ) ] 2 signal term

distortion term

k A2 m2 s 2 (t ) The output contains the distortion term 2 c a n . Distortion can be kept 2 within an acceptable level if ma sn (t ) 1. 4.7 The spectrum of signal x(t ) is illustrated in Figure P4.1. Derive an expression for the spectrum of the signal y(t ) = x(t ) cos(2π f c t ) + xˆ (t ) sin(2π f c t ) Assume B << f c . Sketch the spectrum Y ( f ) . Figure P4.1

X(f)

−B

Solution:

We express y (t ) in terms of the signal x − (t ) = x(t ) − jxˆ (t ) as

{

y (t ) = Re x − (t )e j 2 πfct

{

}

* 1 − x (t )e j 2 πfct + ⎡⎣ x − (t ) ⎤⎦ e − j 2 πfct 2 1 = x − (t )e j 2 πfct + x + (t )e − j 2 πfct 2

{

}

Taking FT of both sides, the spectrum Y ( f ) is obtained as

Y( f ) =

1 X − ( f − f c ) + X + ( f + f c )} { 2

X − ( f − fc ) represents the negative frequency portion of X ( f ) shifted in 2 X + ( f + fc ) represents frequency by the carrier frequency fc. On the other hand, 2 the positive frequency portion of X ( f ) shifted in frequency by the amount –fc. Thus, it is spectrum of an lower sideband SSB-AM signal with modulating signal spectrum X ( f ) . Figure displays the spectra of y(t ) .

Y( f ) +

X − ( f − fc ) 2

X ( f + fc ) 2

− fc

− fc + B

fc − B

4.8 Find the Hilbert transform of the following time signals a.

x1 (t ) = sinc(t )

Solution:

sinc(t ) =

sin (π t ) 2sin (π t / 2) cos (π t / 2) = = sinc(t / 2) cos (π t / 2) πt πt

Using the low-pass high-pass product property, we can write H sinc(t / 2) cos (π t / 2 ) ←⎯ → sinc(t / 2) sin (π t / 2 )

Therefore, xˆ1 (t ) = sinc(t / 2)sin (π t / 2) b. x2 (t ) =

sin(4π t ) cos(300π t ) t

Solution: sin(4π t ) H cos(300π t ) ←⎯ → 4π sinc(4t ) sin ( 300π t ) t

Therefore, xˆ2 (t ) = 4π sinc(4t )sin ( 300π t ) c. x3 (t ) = cos(300π t ) + 3sin(600π t ) Solution: H → sin(300π t ) − 3cos(600π t ) cos(300π t ) + 3sin(600π t ) ←⎯

Therefore, xˆ3 (t ) = sin(300π t ) − 3cos(600π t ) 4.9 If x(t) and y(t) are signals with nonoverlapping spectra, where x(t) is LP and y(t) is BP signal, then H x(t ) y (t ) ←⎯→ x(t ) yˆ (t )

Solution:

Let g (t ) = x(t ) y(t ) . Then

G( f ) = X ( f ) ⊗ Y ( f ) = X ( f ) ⊗ [Y ( f )u( f ) + Y ( f )u(− f )] where u ( f ) denotes the unit step function. Since X ( f ) and Y ( f ) are nonoverlapping , X ( f ) ⊗ (Y ( f )u ( f ) ) is zero for f < 0. Similarly, X ( f ) ⊗ (Y ( f )u (− f ) ) is zero for f > 0. Thus

Gˆ ( f ) = − jsgn( f )G( f ) = − jsgn( f ) [ X ( f ) ⊗ Y ( f )] = − jsgn( f ) X ( f ) ⊗ [Y ( f )u( f ) + Y ( f )u(− f )] = − jX ( f ) ⊗ Y ( f )u( f ) + jX ( f ) ⊗ Y ( f )u(− f ) = X ( f ) ⊗ [ − jY ( f )u( f ) + jY ( f )u(− f )] = X ( f ) ⊗ [ − jsgn( f )Y ( f )] = X ( f ) ⊗ Yˆ ( f ) 4.10 Consider the signal x(t ) = ⎡⎣ 2Wsinc ( 2Wt ) + Wsinc2 (Wt ) ⎤⎦ cos(2π f ct ) .

a. Determine and sketch the spectrum of analytic signal x + (t ) . Solution:

The analytic signal x+(t) is given from (4.60) as x + (t ) = x(t ) + jxˆ (t ) Now H 2Wsinc(2Wt ) cos(2π f ct ) ←⎯ → 2Wsinc(2Wt )sin(2π f ct ) H Wsinc2 (Wt ) cos(2π fct ) ←⎯ →Wsinc2 (Wt )sin(2π fct )

Therefore, x + (t ) = ⎡⎣ 2Wsinc ( 2Wt ) + Wsinc 2 (Wt ) ⎤⎦ [ cos(2π f c t ) + j sin(2π f c t ) ] = ⎡⎣ 2Wsinc ( 2Wt ) + Wsinc 2 (Wt ) ⎤⎦ e j 2π fct X(f )

(a) 1

− fc − W

− fc

− fc + W

fc − W

fc + W

fc − W

fc + W

X +( f )

(b)

X ( f ) 2

(c)

f −W

⎡ ⎛ f ⎞ ⎛ f ⎞⎤ X + ( f ) = ⎢Π ⎜ ⎟+ Λ⎜ ⎟⎥ ⊗ δ ( f − fc ) ⎝ 2W ⎠ ⎦ ⎣ ⎝ 2W ⎠ ⎡ ⎛ f − fc ⎞ ⎛ f − fc ⎞ ⎤ = ⎢Π ⎜ + Λ⎜ ⎟ ⎟⎥ ⎝ 2W ⎠ ⎦ ⎣ ⎝ 2W ⎠

b. Determine and sketch the spectrum of the complex envelope x (t ) . Solution:

x (t ) = x + (t )e − j 2π fct = ⎡⎣ 2Wsinc ( 2Wt ) + Wsinc 2 (Wt ) ⎤⎦ e j 2π fc t e − j 2π fc t = 2Wsinc ( 2Wt ) + Wsinc 2 (Wt )

⎛ f ⎞ ⎛ f ⎞ X ( f ) = Π ⎜ ⎟+ Λ⎜ ⎟ ⎝ 2W ⎠ ⎝ 2W ⎠ +

The spectra of x (t ) and x (t ) are displayed in the Figure. 4.11 An LSB-AM signal is generated by modulating 1 kHz carrier by the message signal s(t ) = 2cos(6π t ) + 0.5sin(10π t ) . The carrier power is 5 kW. a. Determine the signal sˆ(t ) . Solution:

sˆ(t ) = 2sin(6π t ) − 0.5cos(10π t ) b. Write an expression for the LSB-AM signal. Solution:

Since the carrier power is 5 kW, we have

1 2 Ac = 5000 2 Therefore, Ac = 100 The LSB-AM signal can be expressed using (4.80) as

xLSB (t ) = 50 {[ 2 cos(6π t ) + 0.5sin(10π t ) ] cos(2000π t ) + [ 2sin(6π t ) − 0.5cos(10π t ) ] sin(2000π t )} = 50 {[ 2 cos(6π t ) cos(2000π t ) + 2sin(6π t ) sin(2000π t ) ] + [ 0.5sin(10π t ) cos(2000π t ) − 0.5cos(10π t ) sin(2000π t )]} = 50 [ 2 cos(1994π t ) − 0.5sin(1990π t )] c. Determine and sketch the magnitude spectrum of the LSB-AM signal. Solution:

X LSB ( f ) = 50 {δ ( f − 997) + δ ( f + 997) + j 0.25 [δ ( f − 995) − δ ( f + 995)]}

X LSB ( f ) 50

12.5 -1000 -997 -995

f (Hz)

995 997 1000

d. Compare the PEP and the normalized average power of the LSB-AM signal. Solution:

PEPLSB − AM = Power supplied by sinsusoidal waveform with amplitude 50 × max t

Now s 2 (t ) + sˆ 2 (t ) = 4 + 0.25 + 2sin(4π t ) . Therefore, max t

s 2 (t ) + sˆ 2 (t )

s 2 (t ) + sˆ2 (t ) = 2.5 .

( 50 × 2.5) = 7.81 kW = 2

PEPLSB − AM

Average power of the message signal Ps =

22 0.52 + = 2.125 W 2 2

Ac2 Ps 1002 × 2.125 Average power of the LSB-AM signal Px = = = 5.312 kW 4 4 4.12 Suppose the message signal s(t ) = sinc(2t ) modulates the carrier signal c(t ) = 10cos(50π t ) . a. Plot the spectrum of the USB-AM signal and compare it with that of the DSBSC signal. Solution:

The spectra of the DSB-SC and USB-AM signals are displayed Figures (b) and (c) respectively.

S( f ) 0.5 (a)

−1

X DSB ( f ) (b) 2.5

−24

−25

−26

f 24

X USB ( f ) (c) 2.5

−26 −25.5 −25

f 0

25 25.5 26

b. From the plot of the Fourier transform of the USB signal, show that the USB signal is given by xUSB (t ) = 5cos(51π t )sinc(t )

17 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Solution:

From Figure (c), we can write

5 XUSB ( f ) = ⎡⎣Π ( f − 25.5) + Π ( f + 25.5) ⎤⎦ 2 Taking the inverse FT of both sides, we obtain 5 5 xUSB (t ) = sinc(t )e j 51π t + sinc(t )e− j 51π t 2 2 j 51π t ⎡e + e− j 51π t ⎤ = 5sinc(t ) ⎢ ⎥ = 5sinc(t ) cos(51π t ) 2 ⎣ ⎦ c. Now show that 5 [ s (t ) cos(50π t ) − sˆ(t ) sin(50π t ) ] equals xUSB (t ) of part (b). Solution:

We can express 5 [ s (t ) cos(50π t ) − sˆ(t ) sin(50π t ) ] as 5 [ s (t ) cos(50π t ) − sˆ(t ) sin(50π t )] = 5 Re {[ s (t ) + jsˆ(t )] e j 50π t } = 5 Re {s + (t )e j 50π t } = =

{

∗ 5 + s (t )e j 50π t + ( s + (t ) ) e− j 50π t 2

}

5 + s (t )e j 50π t + s − (t )e − j 50π t } { 2

Now 5 + 5 ℑ s (t )e j 50π t + s − (t )e − j 50π t } ←⎯ → {S + ( f − 25) + S − ( f + 25)} { 2 2 5 = {S + ( f − 25) + S − ( f + 25)} 2 5 = {Π ( f − 25.5 ) + Π ( f + 25.5 )} 2 The right-hand side is identical to the expression for X USB ( f ) from part (b) above.

Therefore, 5 [ s (t ) cos(50π t ) − sˆ(t ) sin(50π t ) ] equals xUSB (t ) = 5sinc(t ) cos(51π t ) .

Ac [ s(t)cos(2π fct) + sˆ(t)sin(2π fct )] . Assume 2 that the LO at the demodulator generates reference carrier with a fixed phase offset φ . That is, the receiver multiplies x(t ) with cos(2π f c t + φ ) and LP filters the output to produce the demodulated signal yD (t ) .

4.13 Consider the LSB-AM signal x(t ) =

a. Find yD (t ) in terms of s(t ) and sˆ(t ) . Comment on the distortion in the demodulated signal introduced by the phase offset. Solution:

y (t ) = xLSB (t ) × 2cos(2πf ct + φ ) =

Ac {s(t ) cos(2πfct ) + sˆ(t )sin(2πfct )} × 2 cos(2πfct + φ ) 2

Ac {2s(t ) cos(2πfct ) cos(2πfct + φ ) + 2sˆ(t )sin(2πfct ) cos(2πfct + φ )} 2 ⎧ ⎡ ⎤ ⎡ ⎤ ⎫⎪ Ac ⎪ ⎢ ⎥ ⎢ = ⎨ s(t ) cos(4πfc t + φ ) + cos(φ ) + sˆ(t ) sin(4πf c t + φ ) − sin(φ ) ⎥ ⎬ 2 ⎪ ⎢ Filtered ⎥ ⎢ Filtered ⎥⎪ out by LP filter out by LP filter ⎣ ⎦ ⎣ ⎦⎭ ⎩ =

By passing y (t ) through a LP filter, the double-frequency components are eliminated leaving us with y D (t ) =

Ac {s(t ) cos(φ ) − sˆ(t ) sin(φ )} 2

The demodulated signal is attenuated by the factor cos(φ ) . Further, it contains a residue of sˆ(t ) . b. Calculate the output spectrum YD ( f ) . Solution: Ac A ℑ {s (t ) cos(φ ) − sˆ(t ) sin(φ )} = c {S ( f ) cos(φ ) + j sgn( f ) S ( f ) sin(φ )} 2 2 AS( f ) = c {cos(φ ) + j sgn( f ) sin(φ )} 2 ⎧ Ac S ( f ) A S ( f ) jφ e , f ≥0 {cos(φ ) + j sin(φ )} = c ⎪⎪ 2 2 =⎨ ⎪ Ac S ( f ) cos(φ ) − j sin(φ ) = Ac S ( f ) e − jφ , f <0 } ⎪⎩ 2 { 2

YD ( f ) =

4.14

The main technical difficulty in the phasing method of generating SSB-AM signal (Figure 4.22) implementing the Hilbert transformer over a wide-range of frequencies. An alternative method of generating SSB-AM signal proposed by Weaver is shown in Figure P4.2. Although, it is a more complex design, it avoids the use of wide-band phase shifter. We assume that the message signal s(t ) is band-limited to B Hz. The cutoff frequency of ideal LP filters is B/2 Hz. Prove that x(t ) is an USB-AM signal. Figure P4.2 Weaver’s SSB-AM modulator

s (t )

υU (t )

LPF

xU (t )

cos π Bt

cos 2π ( f c + 0.5 B ) t

x (t )

+ −90o

90o

− sin π Bt

× υ (t ) L

LPF

xL (t )

sin 2π ( f c + 0.5 B ) t

Solution:

Let the message signal s (t ) = cos ( 2π f m t ) , where 0 < f m < B . The input to upper LPF is

υU (t ) = cos(2π fmt )cos(π Bt ) =

{

}

1 cos ⎡⎣2π ( fm + B / 2) t ⎤⎦ + cos ⎡⎣2π ( fm − B / 2) t ⎤⎦ 2

The input to lower LPF is

υL (t ) = − cos ( 2π fmt ) sin (π Bt ) =

{

}

1 sin ⎡⎣2π ( fm − B / 2) t ⎤⎦ − sin ⎡⎣2π ( fm + B / 2) t ⎤⎦ 2

The outputs of LPFs are

1 xU (t ) = cos ⎡⎣2π ( fm − B / 2) t ⎤⎦ 2 1 xL (t ) = sin ⎡⎣2π ( fm − B / 2) t ⎤⎦ 2 The output of the Weaver USB-AM modulator is x(t ) = xU (t ) cos ⎡⎣ 2π ( f c + B / 2 ) t ⎤⎦ + xL (t ) sin ⎡⎣ 2π ( f c + B / 2 ) t ⎤⎦ 1 = cos ⎡⎣ 2π ( f m − B / 2 ) t ⎤⎦ cos ⎡⎣ 2π ( f c + B / 2 ) t ⎤⎦ 2 1 + sin ⎡⎣ 2π ( f m − B / 2 ) t ⎤⎦ sin ⎡⎣ 2π ( f c + B / 2 ) t ⎤⎦ 2 cos ⎡⎣ 2π ( f c + f m ) t ) ⎤⎦ cos ⎡⎣ 2π ( f c − f m + B ) t ) ⎤⎦ = + 4 4 cos ⎡⎣ 2π ( f c + f m ) t ) ⎤⎦ cos ⎡⎣ 2π ( f c − f m + B ) t ) ⎤⎦ + − 4 4 1 = cos ⎡⎣ 2π ( f c + f m ) t ) ⎤⎦ 2 It can be shown that a LSB-AM signal is generated instead if outputs of upper and lower LPFs are multiplied by cos ⎡⎣ 2π ( f c − B / 2 ) t ⎤⎦ and

sin ⎡⎣ 2π ( f c − B / 2 ) t ⎤⎦ , respectively. 4.15 One disadvantage of synchronous demodulation of SSB-AM signals is the interference caused by the presence of residual frequency components from the removed sideband. By using a demodulation scheme shown in Figure P4.3, it is possible to demodulate either upper or lower sideband and cancel the other sideband. Prove that by choosing the sign of 90 phase shifter, either the upper or lower sideband component is demodulated while suppressing the other. Figure P4.3

Solution:

The output of the multiplier in the upper branch is given by yU (t ) = { Au cos [ 2π ( f c + f m )t ] + AA cos [ 2π ( f c − f m )t ]} × Ac cos 2π f c t =

Ac Au AA cos ( 2π f mt ) + cos [ 2π (2 f c + f m )t ]} + c l {cos ( 2π f mt ) + cos [ 2π (2 f c − f m )t ]} { 2 2

The output of the LP filter in the upper branch can now be expressed as

yU (t ) =

Ac Au AA cos ( 2π f mt )} + c l {cos ( 2π f mt )} { 2 2

The output of the multiplier in the lower branch is given by yL (t ) = { Au cos [ 2π ( f c + f m )t ] + AA cos [ 2π ( f c − f m )t ]} × − Ac sin ( 2π f c t ) =

Ac Au AA sin ( 2π f m t ) − sin [ 2π (2 f c + f m )t ]} + c l {− sin ( 2π f m t ) − sin [ 2π (2 f c − f m )t ]} { 2 2

The output of the LP filter in the lower branch can now be expressed as

yL (t ) =

Ac Au AA ⎡⎣sin ( 2π f mt ) ⎤⎦ − c l ⎡⎣sin ( 2π f mt ) ⎤⎦ 2 2

After 90o phase shift, the output of the lower branch is given as

{ (

Ac Au sin 2π f mt + 90o 2

)} − A2A {sin ( 2π f t + 90 )} = A2A {cos ( 2π f t )} − A2A {cos ( 2π f t )} c

The demodulated output is now obtained by adding the output of upper and lower branches as Ac Au AA AA AA cos ( 2π f mt ) + c l cos ( 2π f mt ) + c u cos ( 2π f mt ) − c l cos ( 2π f mt ) 2 2 2 2 = Ac Au cos ( 2π f mt )

yD (t ) =

If we instead choose 90o phase shift, the output of the lower branch is given as

{ (

Ac Au sin 2π f mt − 90o 2

)} − A2A {sin ( 2π f t − 90 )} = − A2A {cos ( 2π f t )} + A2A {cos ( 2π f t )} c

The demodulated output is now obtained by adding the output of upper and lower branches as Ac Au AA AA AA cos ( 2π f mt ) + c l cos ( 2π f mt ) − c u cos ( 2π f mt ) + c l cos ( 2π f mt ) 2 2 2 2 = Ac Al cos ( 2π f mt )

yD (t ) =

4.16 Consider the phasing method illustrated in Figure 4.22 to generate USB-AM signal. The modulating signal is a single tone of frequency f m . The 90o phase-shift network introduces a constant phase offset of φo between the in-phase and quadrature carriers. a. Calculate the output of the USB-AM modulator. Solution:

The output of USB-AM modulator in Figure 4.22 can be expressed as

Ac {cos ( 2πfmt ) cos ( 2πfct ) − sin ( 2πfmt ) sin ( 2πfct + φo )} 2 A = c cos ( 2πf mt ) cos ( 2πfc t ) − sin ( 2πf mt ) ⎡⎣sin ( 2πfct ) cos (φo ) + cos ( 2πfct ) sin (φo ) ⎤⎦ 2 A = c {cos ( 2πf mt ) cos ( 2πfc t ) − sin ( 2πf mt ) sin ( 2πfc t ) cos (φo )} 2 A − c sin ( 2πf mt ) cos ( 2πfc t ) sin (φo ) 2

x(t ) =

{

}

Assuming phase small offset φo , cos (φo ) ≈ 1. Therefore, x(t ) =

A sin (φo ) Ac cos ⎡⎣ 2π ( f c + f m ) t ⎤⎦ − c sin ( 2πf mt ) cos ( 2πf ct ) 2 2

Now Ac sin (φo ) A sin (φo ) sin ( 2πf mt ) cos ( 2πf c t ) = c sin ⎡⎣ 2π ( f c + f m ) t ⎤⎦ − sin ⎡⎣ 2π ( f c − f m ) t ⎤⎦ 2 4

{

}

Substituting x(t ) =

A sin (φo ) Ac cos ⎡⎣ 2π ( f c + f m ) t ⎤⎦ − c sin ⎡⎣ 2π ( f c + f m ) t ⎤⎦ − sin ⎡⎣ 2π ( f c − f m ) t ⎤⎦ 2 4

{

}

For phase small offset φo ,

x(t ) ≈

Ac sin (φo ) A c . Therefore, 4 2

A sin (φo ) Ac cos ⎡⎣ 2π ( f c + f m ) t ⎤⎦ + c sin ⎡⎣ 2π ( f c − f m ) t ⎤⎦ 2 4

b. Calculate the residual power in the lower sideband as a percentage of the power in the desired sideband.

Ac2 sin 2 (φo ) sin 2 (φo ) Power in the undesired sideband 32 = = Ac2 Power in the desired sideband 4 8 c. Evaluate the magnitude of undesired sideband rejection for phase offset φo = 1.5o . 2 D Power in the undesired sideband sin (1.5 ) 42.83 = = × 10−6 = −37.7 dB Power in the desired sideband 4 4

Thus, the power in the undesired sideband is approximately 40 dB below that in the desired sideband. 4.17. A VSB-AM signal is generated from an input s (t ) by first generating a conventional AM signal (ma = 0.875) with a carrier frequency f c = 45.75 MHz and then passing this signal through a filter whose frequency response is given by

⎧ f − fc + fv , ⎪ 2 fv ⎪⎪ H ( f ) = ⎨1, ⎪0, ⎪ ⎪⎩

fc − fv ≤ f < fc + fv f c + f v ≤ f < f c + 4.5 MHz otherwise

where f v = 0.75 MHz . a. Determine an expression for the resulting VSB signal if s1 (t ) = cos(30π t ×103 ) . Solution: x AM (t ) = ⎡⎣1 + 0.875 cos ( 2π f m t ) ⎤⎦ cos ( 2π f c t ) = cos ( 2π f c t ) + 0.4375 cos ⎡⎣ 2π ( f c + f m ) t ⎤⎦ + 0.4375 cos ⎡⎣ 2π ( f c − f m ) t ⎤⎦

The VSB filter output is

{

}

xVSB (t ) = cos ( 2π f c t ) + 0.4375cos ⎡⎣ 2π ( f c + f m ) t ⎤⎦ + 0.4375cos ⎡⎣ 2π ( f c − f m ) t ⎤⎦ ⊗ h(t )

⎧δ ( f + fc ) + δ ( f − fc ) ⎫ ⎪ ⎪⎪ H( f ) ⎪ ⎡ ⎤ δ δ f f f f f f + + + − + ( ) ( ) XVSB ( f ) = ( ) ( ) ⎨ ⎬ c m c m 2 ⎪+0.4375 ⎢ ⎥⎪ ⎢⎣+δ ( f + ( fc − f m ) ) + δ ( f − ( fc − f m ) ) ⎥⎦ ⎭⎪ ⎩⎪ Now f c = 45.75 MHz and f m = 15 kHz . From Figure H ( 45.75 × 106 ) = 0.5 H ( 45.765 × 106 ) = 0.51 H ( 45.735 × 106 ) = 0.49

0.5 ⎡δ ( f + 45.75 ×106 ) + δ ( f − 45.75 ×106 ) ⎤ ⎦ 2 ⎣ 0.223 ⎡δ ( f + 45.765 ×106 ) + δ ( f − 45.765 ×106 ) ⎤ + ⎦ 2 ⎣ 0.214 ⎡δ ( f + 45.735 ×106 ) + δ ( f − 45.735 ×106 ) ⎤ + ⎦ 2 ⎣

XVSB ( f ) =

This corresponds to

(

)

(

)

(

xVSB (t ) = 0.5cos 91.5 ×106 π t + 0.223cos 91.53 × 106 π t + 0.214 cos 91.47 × 106 π t

)

b. Determine the spectrum of the VSB signal. X VSB ( f ) Carrier

Carrier

0.5

0.25 f (MHz -45.765 - 45.75

-45.735

45.735

45.75 45.765

c. Calculate the ratio of the PEP to the average power.

25 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

( 0.5) + ( 0.223) + ( 0.214 ) = 0.172 Power in the VSB-AM signal P = 2

PEPVSB − AM =

1 0.878 2 = 0.439 ( 0.5 + 0.223 + 0.214 ) = 2 2

PEPVSB − AM 0.439 = = 2.55 0.172 Px d. Repeat (a) – (c) for s2 (t ) = cos(π t ×106 ) and s3 (t ) = cos(4π t ×106 ) . Solution:

For s2 (t ) = cos(π t ×106 ) , we have H ( 45.75 × 106 ) = 0.5 H ( 46.25 × 106 ) = 0.833 H ( 45.25 × 106 ) = 0.167

0.5 ⎡δ ( f + 45.75 ×106 ) + δ ( f − 45.75 ×106 ) ⎤ ⎦ 2 ⎣ 0.364 ⎡δ ( f + 46.25 ×106 ) + δ ( f − 46.25 ×106 ) ⎤ + ⎦ 2 ⎣ 0.073 ⎡δ ( f + 45.25 ×106 ) + δ ( f − 45.25 ×106 ) ⎤ + ⎦ 2 ⎣

XVSB ( f ) =

(

)

(

)

(

xVSB (t ) = 0.5cos 91.5 ×106 π t + 0.364 cos 92.5 ×106 π t + 0.073cos 90.5 × 106 π t

)

X VSB ( f ) Carrier

Carrier

0.5

0.25 f (MHz) -46.25 - 45.75

-45.25

45.25

45.75 46.25

( 0.5) + ( 0.364 ) + ( 0.073) = 0.194 Power in the VSB-AM signal P = 2

PEPVSB − AM =

1 0.878 2 = 0.439 ( 0.5 + 0.364 + 0.073) = 2 2

PEPVSB − AM 0.439 = = 2.26 0.194 Px For s3 (t ) = cos(4π t ×106 ) , we have H ( 45.75 ×106 ) = 0.5 H ( 47.75 ×106 ) = 1 H ( 43.75 ×106 ) = 0

0.5 ⎡δ ( f + 45.75 ×106 ) + δ ( f − 45.75 ×106 ) ⎤ ⎦ 2 ⎣ 0.4375 ⎡δ ( f + 47.75 ×106 ) + δ ( f − 47.75 ×106 ) ⎤ + ⎦ 2 ⎣

XVSB ( f ) =

(

)

(

xVSB (t ) = 0.5cos 91.5 × 106 π t + 0.4375cos 95.5 ×106 π t

)

X VSB ( f ) Carrier

Carrier

0.5

0.25 f (MHz) 0

45.75 47.75

( 0.5) + ( 0.4375) = 0.220 Power in the VSB-AM signal P = 2

PEPVSB − AM =

1 0.878 2 = 0.439 ( 0.5 + 0.4375) = 2 2

PEPVSB − AM 0.439 = =2 0.22 Px

4.18 A superheterodyne receiver is designed to cover the RF frequency range of 45 860 MHz, with channel spacings of 8 MHz. . Assume high-side injection. a. If the receiver down-converts the RF signals to an IF of 40 MHz, calculate the range of frequencies for the LO. Solution:

The LO minimum and maximum frequencies are given by

( f LO )min = ( fc )min + f IF = 45 + 40 = 85 MHz ( f LO )max = ( fc )max + f IF = 860 + 40 = 900 MHz b. Determine the range of image frequencies. Solution:

The range of image frequencies is given by

(f (f

) )

image min

= ( f c )min + 2 f IF = 45 + 80 = 125 MHz

image max

= ( f c )max + 2 f IF = 860 + 80 = 940 MHz

Note that band of image frequencies and the signal band overlap. This is undesirable. So we consider up-conversion to a higher IF of 1.2 GHz. c. Calculate the range of frequencies for the LO. Solution:

( f LO )min = ( fc )min + f IF = 45 + 1200 = 1245 MHz ( f LO )max = ( fc )max + f IF = 860 + 1200 = 2060 MHz d. Determine the range of image frequencies. Solution:

(f (f

) )

image min

= ( f c )min + 2 f IF = 45 + 2400 = 2445 MHz

image max

= ( f c )max + 2 f IF = 860 + 2400 = 3260 MHz

e. Is there any overlap between image and signal bands? Solution: 28 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

By choosing higher vs lower IF (i.e., up-conversion vs down-conversion), the overlap between signal and image frequency bands is eliminated. f. Since it is difficult to realize highly selective filtering at high IF, a dual conversion superheterodyne design is desirable choice for wide bandwidth receiver spanning several octaves. Draw the block diagram of a dual conversion superheterodyne receiver and specify the reasons for your choice of first and second IF frequencies. Solution:

The block diagram of a dual conversion superheterodyne receiver is shown. The first IF of 1.2 GHz was selected so that the VCO leakage is out of IF band. Further, the band of image frequencies doesn’t overlap with the signal band. The second IF frequency of 45.75 MHz was chosen because it is easy to realize highly selective filtering in that frequency range using the SAW filter technology.

Mixer RF Filter

LNA

Mixer BP Filter

IF Filter

1200 MHz

45.75 MHz

∼

VCO: 1245-2060 MHz

1154.25 MHz

LNA Low-noise amplifier VGA Variable gain amplifier

VGA To Demodulator

Chapter 5 5.1

A message signal s (t ) is shown in Figure P5.1. Sketch the FM and PM waveforms if the carrier frequency is 50 Hz, k f = 30 Hz/V and k p = 10π radians/V. Figure P5.1

s (t ) 1 0

… t (msec)

−1

s(t), x(t)

Solution:

t (msec)

s(t), x(t)

t (msec) 5.2 An angle-modulated signal is described by θ (t ) ⎡ ⎤ x(t ) = 100 cos ⎢ 2π × 105 t + 0.01cos(4π × 103 t ) ⎥ . ⎢ ⎥ φi ( t ) ⎣ ⎦

a. Write an expression for the instantaneous frequency of x(t ) . Solution:

The instantaneous frequency of the angle-modulated signal is given by 1 dφi (t ) 1 d ⎡⎣ 2π × 10 t + 0.01cos(4π × 10 t ) ⎤⎦ fi (t ) = = 2π dt 2π dt 5 3 3 = 10 − 0.01× 2 × 10 sin(4π ×10 t ) 5

= 105 − 20sin(4π × 103 t ) By comparing with (5.16), we conclude that ∆f max = 20 Hz. b. If x(t ) is an FM signal, what is the message signal sn (t ) ? 2 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Solution:

If x(t ) is an FM signal, the excess phase θ (t ) is given by t

θ (t ) = 2π ∆f max ∫ sn (t )dt = 0.01cos(4π × 103 t ) −∞

Taking the derivative of both sides yields 2π∆f max sn (t ) = −40π sin(4π ×103 t ) Therefore, the normalized message signal is sn (t ) = − sin(4π × 103 t ) c. If x(t ) is a PM signal, what is the message signal sn (t ) ? Solution:

If x(t ) is a PM signal, the excess phase θ (t ) is given by

θ (t ) = ∆φmax sn (t ) = 0.01cos(4π ×103 t ) Therefore, the normalized message signal is sn (t ) = cos(4π ×103 t ) 5.3 An angle-modulated signal is given by ⎡ ⎤ x(t ) = 10 cos ⎢ 2π × 106 t + 2sin(4π × 103 t ) ⎥ . ⎢⎣ ⎥⎦ θ (t ) a. Determine the peak phase deviation. Solution:

The excess phase θ (t ) is

θ (t ) = ∆φmax sn (t ) = 2sin(4π ×103 t ) Therefore, the peak phase deviation is ∆φmax = 2 radians b. Determine the peak frequency deviation.

Solution: 1 dφi (t ) 1 d ⎡⎣ 2π ×10 t + 2sin(4π ×10 t ) ⎤⎦ fi (t ) = = 2π dt 2π dt 6 3 3 = 10 + 4 ×10 cos(4π ×10 t ) 6

By comparing with (5.16), we conclude that ∆f max = 4 ×103 Hz. c. Calculate the average power in the modulated signal x(t ) . Solution:

The power in the angle-modulated signal x (t ) is Ac2 102 Px = = = 50 W 2 2 c. Is this a frequency- or phase-modulated signal? Solution:

The angle-modulated signal x(t ) can be interpreted as either a PM or an FM signal. It is a PM signal with the peak phase deviation ∆φmax = 2 radians and the normalized message signal sn (t ) = sin(4π × 103 t ) . It is a FM signal with the peak frequency deviation ∆f max = 4 ×103 and the normalized message signal sn (t ) = cos(4π × 103 t ) . 5.4 Consider an FM signal generated by modulating a 10 MHz carrier with a 1 kHz sinusoidal waveform such that peak frequency deviation is 2.5 kHz. a. Determine the bandwidth of the modulated signal. Solution:

BT = 2(∆f max + f m ) Substituting f m = 1 kHz and ∆f max = 2.5 kHz , we obtain BT = 2(2.5 + 1) = 7 kHz b. If the modulating signal amplitude is doubled, determine the bandwidth of the modulated signal. 4 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Solution:

Doubling the amplitude doubles the peak frequency deviation ∆f max . Therefore, ∆f max = 5 kHz Substituting yields BT = 2(5 + 1) = 12 kHz c. Determine the bandwidth of the modulated signal if the modulating signal frequency is doubled. Solution:

Now f m = 2 kHz and ∆f max = 2.5 kHz . Substituting BT = 2(2.5 + 2) = 9 kHz d. Determine the bandwidth of the modulated signal if both the amplitude and frequency of the modulating signal are doubled. Solution:

Now f m = 2 kHz and ∆f max = 5 kHz . Substituting BT = 2(5 + 2) = 14 kHz 5.5 Let the modulating signal be s (t ) = 10 cos(2π × 300t ) + 25cos(2π × 600t ) a. Write an expression for the FM waveform xFM (t ) when Ac = 100, f c = 5 MHz, and k f = 200 Hz/V . Solution: t ⎡ ⎤ xFM (t ) = Ac cos ⎢ 2πf c t + 2π ∆f max ∫ sn (α )dα ⎥ −∞ ⎣ ⎦

In the present case, max s (t ) = 35 . Therefore,

sn (t ) =

1 [10 cos(2π × 300t ) + 25cos(2π × 600t )] 35

∆f max = k f max s (t ) = 200 × 35 = 7 × 103

Substituting t ⎡ ⎤ 2π × 7 × 103 xFM (t ) = 100 cos ⎢10π × 106 t + [10 cos(2π × 300α ) + 25cos(2π × 600α )]dα ⎥ ∫ 35 −∞ ⎣ ⎦

⎡ 2π × 7 × 103 ⎡ 10 25 ⎤⎤ sin(2π × 300t ) + sin(2π × 600t ) ⎥ ⎥ = 100 cos ⎢10π × 106 t + ⎢ 35 2π × 600 ⎣ 2π × 300 ⎦⎦ ⎣ 20 25 ⎡ ⎤ = 100 cos ⎢10π × 106 t + sin(2π × 300t ) + sin(2π × 600t ) ⎥ 3 3 ⎣ ⎦

b. Determine maximum frequency deviation ∆f max , maximum phase deviation ∆φmax , and deviation ratio D of the modulated signal. ∆f max = 7 ×103 Hz The maximum phase deviation ∆φmax is the maximum value of the angle

θ (t ) =

20 25 sin(2π × 300t ) + sin(2π × 600t ) , which equals 15 radians. 3 3

Deviation ratio D =

∆f max 7 ×103 70 = = B 600 6

c. Determine bandwidth of the modulated signal. BT = 2(∆f max + B) = 2(7 ×103 + 600) = 15.2 kHz 5.6 Consider a PM signal generated by modulating a 10 MHz carrier with a 1 kHz sinusoidal waveform with peak phase deviation of 1 radian. a. Write an expression for the PM signal and determine its Carson bandwidth. Solution:

The PM signal is

xPM (t ) = Ac cos ⎡⎣ 2π × 107 t + cos(2π × 103 t ) ⎤⎦ For a sinusoidal modulating signal with β = ∆φmax = 1 kHz , the bandwidth of an

angle-modulated signal is given by BT = 2(∆φmax + 1) f m = 2(1 + 1) × 1 = 4 kHz b. Plot the spectrum of the PM signal. Solution:

Consider a PM signal with maximum phase deviation β produced by a sinusoidal modulating signal. That is, xPM (t ) = Ac cos [ 2πf c t + β cos(2πf mt ) ]

The PM signal can be expressed as

{

xPM (t ) = Ac Re e j 2 πfct e j β cos(2 πfmt )

}

( m) The function e is periodic with period 1/ fm, and therefore it can be expanded in a complex Fourier series: j β cos 2 πf t

j β cos ( 2 πf m t )

∞

= ∑ Cn e j 2π nfmt n =−∞

where Cn = f m ∫

1/ f m

j β cos ( 2 πf m t ) − j 2π nf m t

dt =

1 2π j ( β cos z − nz ) e dz 2π ∫0

π⎞ ⎛ Substituting cos ( z ) = sin ⎜ z + ⎟ , we obtain 2⎠ ⎝ ⎡

⎛

π⎞

⎤

1 2π j ⎢⎣ β sin ⎜⎝ z + 2 ⎟⎠ − nz ⎥⎦ Cn = e dz 2π ∫0 Making change of variable z + ⎛

Cn =

π 2

= y yields

nπ ⎞

jnπ jnπ 1 5π /2 j ⎜⎝ β sin y − ny + 2 ⎟⎠ 1 2π j ( β sin y − ny ) ⎫ 2 ⎧ 2 e dy e e dy e Jn ( β ) = = ⎨ ∫0 ⎬ 2π ∫π /2 ⎩ 2π ⎭

Since the integrand is periodic function of y with period 2π, the limits of integral can be changed to any period of length 2π to obtain

Cn = e

jnπ 2

jnπ

⎧ 1 2π j ( β sin y − ny ) ⎫ dy ⎬ = e 2 J n ( β ) ⎨ ∫0 e ⎩ 2π ⎭

The PM signal can, therefore, be expressed as nπ ⎞ ⎛ ∞ j ⎜ n 2 πf m t + ⎟ ⎪ ⎫ ⎪⎧ j 2 πf c t 2 ⎠ ⎝ xPM (t ) = Re ⎨ Ac e J n ( β )e ⎬ ∑ n =−∞ ⎪⎩ ⎪⎭ ∞ nπ ⎤ ⎡ = Ac ∑ J n ( β ) cos ⎢ 2π ( f c + nf m ) t + 2 ⎥⎦ ⎣ n =−∞

In the present case, the PM signal is given by ∞ nπ ⎤ ⎡ xPM (t ) = Ac ∑ J n (1) cos ⎢ 2π (107 + n ×103 ) t + 2 ⎥⎦ ⎣ n =−∞

The spectrum of the PM signal is displayed in the Figure. PM Spectrum for Ac = 1,

fm(kHz) = 1

& Max phase deviation = 1

-10

-2 0 2 Frequency(kHz)

Magnitude spectrum Magnitude

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

-8

-6

-4

f − f c ( kHz )

c. Determine the normalized average power in the PM signal. Solution:

The power in the PM signal x(t ) is Px =

Ac2 2

d. Calculate the percentage of the total power at the carrier frequency. Solution:

Ac2 2 The average power in the component at the carrier frequency is J 0 (β ) 2 where β = ∆φmax for a PM signal. For β = ∆φmax = 1, J 0 (1) = 0.765 . Therefore, percentage of the average power at the carrier frequency = 58.5%.

e. Repeat parts (a) and (d), if the amplitude of the modulating signal is doubled. Solution:

If the amplitude of the modulating signal is doubled, the peak phase deviation is also doubled. That is, β = ∆φmax = 2 . Therefore, the PM signal becomes nπ ⎤ ⎡ , xPM (t ) = Ac cos ⎢ 2π × 106 t + 2 cos(2π × 103 t ) + 2 ⎥⎦ ⎣

and the bandwidth of the signal is given by BT = 2(∆φmax + 1) f m = 2(2 + 1) × 1 = 6 kHz

The spectrum of the PM signal is shown below. PM Spectrum for Ac = 1,

fm(kHz) = 1

& Max phase deviation = 2

-10

-2 0 2 Frequency(kHz)

Magnitude spectrum Magnitude

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

-8

-6

-4

f − f c ( kHz )

For β = 2, J 0 (2) = 0.224 . The percentage of the average power at the carrier frequency = J 02 (2) = ( 0.224 ) = 0.0502 or 5.02% . 2

5.7 A single-tone modulated FM signal is given by xFM (t ) = 10 cos ⎡⎣ 2π × 106 t + 4sin(4π × 103 t ) ⎤⎦

a. Plot the spectrum of the FM signal. Solution:

The FM signal can be expressed as ∞

xFM (t ) = 10 ∑ J n (4) cos ⎡⎣ 2π (106 + 2n × 103 ) t ⎤⎦ n =−∞

FM Spectrum for A c = 1,

fm(kHz) = 2

& beta = 4

0.5

Magnitude spectrum Magnitude

0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0

-10

-8

-6

-4

-2 0 2 Frequency(kHz)

f − f c ( kHz )

b. Determine Carson’s bandwidth of the FM signal. Solution:

β = 4, f m = 2 × 103 Hz BT = 2( β + 1) f m = 2(4 + 1) × 2 ×103 = 20 × 103 Hz

c. Determine the normalized average power of the FM signal. Solution:

The normalized average power of the FM signal x(t ) is 102 Px = = 50 W 2 d. Calculate the percentage of the total power at the carrier frequency. How does the magnitude of power contained in the carrier frequency component changes if the amplitude of the modulating signal is doubled?

Solution:

The average power in the component at the carrier frequency is Ac2 2 J 0 (β ) 2

For β = 4, J 0 (4) = −0.3971 . Therefore, percentage of the average power at the carrier frequency = 15.77%. If the amplitude of the modulating signal is doubled, β = 8 and J 0 (8) = 0.1717 . Therefore, percentage of the average power at the carrier frequency = 2.95%. 5.8

The message signal s (t ) modulates a carrier at 1 MHz to produce an anglemodulated signal. s (t ) is periodic waveform with period To as shown in Figure P5.2. Assuming Am = 1 and To = 1 sec , determine Figure P5.2

s (t )

…

Am −To

T − o 2

T − o 4

… 0

− Am

To 4

To 2

a. Determine the 97% bandwidth of the message signal. Solution:

The Fourier series coefficients are given by C0 = 0 because x (t ) is an odd function.

Cn =

1 x(t )e − j 2π nfot dt ∫ To To −T /4

T /4

T /2

1 o 1 o 4t − j 2π nfot 1 o − j 2π nfot − j 2π nf o t = −e dt + e dt + e dt To −T∫o /2 To −T∫o / 4 To To To∫/4 To /2 −To /4 − je − j 2π nfot je − j 2π nfot 4 ⎡ jT te − j 2π nfot To2 e − j 2π nfot ⎤ = + + 2⎢ o + 2⎥ To 2π nf o −T /2 To 2π nf o T /4 To ⎢⎣ To 2π nf o (To 2π nf o ) ⎥⎦ o

To / 4

− To /4

⎛ je+ jπ n je+ jπ n /2 ⎞ ⎛ je − jπ n je− jπ n / 2 ⎞ =⎜ − − ⎟+⎜ ⎟ 2π n ⎠ ⎝ 2π n 2π n ⎠ ⎝ 2π n +

⎤ To2 4 ⎡ jTo2 − jπ n /2 + jπ n /2 − jπ n / 2 − jπ n /2 e e e e + + − ⎢ ⎥ ( ) ( ) 2 To2 ⎢⎣ 8π n ( 2π n ) ⎥⎦

j ⎡ ⎛ n ⎞⎤ ⎛πn ⎞ ⎛πn ⎞ cos (π n ) − cos ⎜ ⎟ + cos ⎜ ⎟ + sinc ⎜ ⎟ ⎥ ⎢ πn ⎣ ⎝ 2 ⎠⎦ ⎝ 2 ⎠ ⎝ 2 ⎠

j ⎡ n ⎤ n ( −1) + sinc ⎛⎜ ⎞⎟ ⎥ , n = ±1, ±2,...... ⎢ πn ⎣ ⎝ 2 ⎠⎦

Using the procedure illustrated in Example 2.27, the 97% bandwidth of the 10 signal is calculated as B = . To b. Calculate the bandwidth of an FM signal with k f = 100 Hz/V . Solution:

The peak frequency deviation, ∆f max , of the FM signal is given from (5.15) by ∆f max = k f max s (t ) = 100 × 1 = 100 Hz

BT = 2(∆f max + B) = 2(100 + 10) = 220 Hz c. Determine the bandwidth of an PM signal with k p = π / 2 . Solution:

The peak phase deviation, ∆φmax , of the PM signal is given from (5.9) by ∆φmax = k p max s (t ) = π / 2

BT = 2(∆φmax + 1) B = 2(π / 2 + 1)10 = 51.41 Hz d. Calculate the bandwidth of an FM signal with k f = 1000 Hz/V . Solution: ∆f max = k f max s (t ) = 1000 × 1 = 1000 Hz

BT = 2(∆f max + B) = 2(1000 + 10) = 2020 Hz e. How do answers to parts (a) − (c) change for the case where Am = 2 and To = 1 msec . Solution:

For To = 1 msec , the message bandwidth B = 10 kHz For Am = 2 , ∆f max = 200 , BT = 2(∆f max + B) = 2(200 + 10000) = 20.4 kHz For Am = 2 , ∆φmax = π and BT = 2(∆φmax + 1) B = 2(π + 1)10000 = 82.83 kHz f. How do answers to parts (a) − (c) change for the case where Am = 1 and To = 0.5 msec . Solution:

For To = 0.5 msec , the message bandwidth B = 20 kHz For Am = 1 , ∆f max = 100 , BT = 2(∆f max + B) = 2(100 + 20000) = 40.2 kHz For Am = 1 , ∆φmax = π / 2 , BT = 2(∆φmax + 1) B = 2(π / 2 + 1)20000 = 102.83 kHz 5.9

A 1 MHz carrier is frequency-modulated by a single tone of frequency 2 kHz, resulting in the peak frequency deviation of 10 kHz. a. What is the bandwidth occupied by the modulated signal? Plot the spectrum of the FM signal (only sidebands in the Carson’s bandwidth). Solution:

∆f max = 10 kHz, B = f m = 2 kHz

BT = 2(∆f max + B) = 2(10 + 2) = 24 kHz

β=

∆f max 10 = =5 2 fm

The spectrum of the FM signal contains 2(β +1) = 12 sidebands within the Carson’s bandwidth of 24 kHz. FM Spectrum for Ac = 1,

fm(kHz) = 2

& beta = 5

0.5

Magnitude spectrum Magnitude

0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0

-10

-5

0 Frequency(kHz)

f − f c ( kHz )

b. If the amplitude of the modulating sinusoidal signal is increased by a factor of 3 and its frequency is decreased to 1 kHz, how is the bandwidth of the modulated signal modified? Plot the spectrum of the FM signal. Solution:

If the amplitude of the modulating sinusoidal signal is increased by a factor of 3, the peak FM deviation is increased by a factor of 3 to ∆f max = 30 kHz . ∆f max = 30 kHz, B = f m = 1 kHz BT = 2(∆f max + B) = 2(30 + 1) = 62 kHz

β=

∆f max 30 = = 30 1 fm

The spectrum of the FM signal contains 2(β +1) = 62 sidebands within the Carson’s bandwidth of 62 kHz.

FM Spectrum for Ac = 1,

fm(kHz) = 1

& beta = 30

0.5

Magnitude spectrum Magnitude

0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 -30

-20

-10

0 Frequency(kHz)

f − f c ( kHz )

c. Repeat part (b) with frequency of the modulating sinusoidal signal increased to 3 kHz. Solution:

∆f max = 30 kHz, B = f m = 3kHz BT = 2(∆f max + B) = 2(30 + 3) = 66 kHz ∆f 30 β = max = = 10 3 fm The spectrum of the FM signal contains 2(10 + 1) = 22 sidebands within the Carson’s bandwidth of 66 kHz. FM Spectrum for Ac = 1,

fm(kHz) = 3

& beta = 10

0.5

Magnitude spectrum Magnitude

0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0

-30

-20

-10

0 10 Frequency(kHz)

f − f c ( kHz )

5.10 Consider the parallel RLC tuned circuit in Figure P5.3 used as a slope detector.

Figure P5.3 Parallel resonant circuit

I ( jω )

X ( jω )

V ( jω )

a. Show that the transfer function of the tuned circuit is given by H ( jω ) =

jωωo / Q V ( jω ) = 2 Comment: X ( jω ) ωo − ω 2 + jωωo / Q

where

ωo = Q=

1 1 , ω3dB = RC LC

ωo C =R L ω3dB

Solution:

⎛ 1 ⎞ + jωC ⎟ I = V ( jω ) ⎜ ⎝ jω L ⎠ ⎛ 1 ⎞ + jωC ⎟ + V ( jω ) X ( jω ) = RI + V ( jω ) = RV ( jω ) ⎜ ⎝ jω L ⎠ ⎛ R ⎞ = V ( jω ) ⎜ + RjωC + 1⎟ ⎝ jω L ⎠ The transfer function H ( jω ) of the parallel RLC tuned circuit can be expressed as H ( jω ) =

V ( jω ) 1/ R = X ( jω ) ⎛ 1 1⎞ ⎜ jω L + jωC + R ⎟ ⎝ ⎠

Substituting ωo =

1 C , we obtain ,Q = R L LC

jωωo2 L R = = H ( jω ) = 2 jω L ⎞ ⎛ ω ⎛ jω L ⎞ ⎛ 2 2 jωωo ωo − ω 2 + ⎜ 1 − ω LC + ⎟ ⎜1 − 2 + ⎟ ⎜ R ⎠ ⎝ ωo ⎝ R ⎠ ⎝ R jω L R

jω L R

jωωo R

L C

⎛ 2 jωωo 2 ⎜ ωo − ω + R ⎝

L⎞ ⎟ C⎠

jωωo / Q ω − ω 2 + jωωo / Q 2 o

b. Determine appropriate values for R, L and C for ωo = 2π ×106 and Q = 20 . Solution:

We choose C = 0.01 µ F and L = 0.1 mH so that f o = 106 =

1 −3

0.1× 10 × 0.01× 10−6

Next we calculate R using Q = 20 . 20 = R

C 0.01× 10−6 ⇒ 400 = R 2 = R 210−4 −3 L 0.1× 10

R = 400 × 104 = 2000 ohms

c. Plot the magnitude frequency response of the parallel RLC tuned circuit from 940 kHz to 1 MHz. Select the discriminator center frequency, discriminator constant K FD , and permissible peak frequency deviation for the input signal. Solution:

The frequency response H ( jω ) of the tuned circuit from 940 kHz to 1 MHz is plotted in the Figure. We choose the discriminator center frequency as 970 kHz. The discriminator constant K FD is now calculated from the figure as K FD =

0.7778 − 0.5222 0.2556 = = 12.78 µ V/Hz 6 ( 0.98 − 0.96 ) ×10 0.02 ×106

The determination of permissible peak frequency deviation is subject to the amount of acceptable distortion in the demodulated signal. From the amplitude vs frequency characteristic above, we observe that ∆f max = 20 kHz is a good choice. Frequency Response of Parallel Resonant Circuit 1

0.9

|H(j ω)| Magnitude

0.8

0.7

0.6

0.5

0.4

940

950

960

970 Frequency(kHz)

980

990

1000

5.11 Delay-line FM discriminator Consider the FM signal xFM (t ) = Ac cos[2πf c t + θ (t )] t

where θ (t ) = 2π ∆f max ∫ sn (t )dt . We use the FM demodulation scheme illustrated in −∞

Figure P5.4. The input FM signal is passed through a delay line that produces a delay of π / 2 radians at the carrier frequency f c . That is, 2π f c td = π / 2 . The output of delay-line is subtracted from the incoming FM signal, and the resulting difference signal is envelope detected. Let y (t ) = xFM (t ) − xFM (t − td ) Assume θ (t ) − θ (t − td ) 1 . Figure P5.4

xFM (t ) Delay t = 1 d

y (t )

−

4 fc

Envelope detector

yD (t )

a. Show that the envelope of y (t ) is given by ⎡ π θ (t ) − θ (t − td ) ⎤ v(t ) = 2 Ac sin ⎢ + ⎥ 2 ⎣4 ⎦ Solution: y (t ) = xFM (t ) − xFM (t − td )

= Ac cos [ 2πf c t + θ (t ) ] − Ac cos [ 2πf c (t − td ) + θ (t − td )]

θ (t ) + θ (t − td ) θ (t ) − θ (t − td ) ⎤ ⎡ ⎤ ⎡ − π ⎥ 2 Ac sin ⎢ πf c td + y (t ) = sin ⎢ 2πf ct − πf ctd + ⎥ 2 2 ⎦ ⎣ ⎦ ⎣ Hig-frequency term

Low-frequency envelope

The first term on the right-hand side is modulating signal dependent carrier term. The second-term is modulating signal dependent low-frequency envelope. Since θ (t ) − θ (t − td ) 1 , the argument of the second term exhibits

. This implies that envelope always remains positive 4 to assure proper detection. small variations around

The envelope detector tracks the envelope of y (t ) which is given by

θ (t ) − θ (t − td ) ⎤ ⎡ v(t ) = 2 Ac sin ⎢ πf c td + ⎥ 2 ⎣ ⎦ ⎡ π θ (t ) − θ (t − td ) ⎤ = 2 Ac sin ⎢ + ⎥ 2 ⎣4 ⎦ b. Show that the envelope detector output can be expressed as yD (t ) =

1 ⎡ 1 dθ ⎤ 1 1+ td = [1 + π ∆f max td sn (t )] 2 ⎢⎣ 2 dt ⎥⎦ 2

Solution: ⎡ π θ (t ) − θ (t − td ) ⎤ v(t ) = 2 Ac sin ⎢ + ⎥ 2 ⎣4 ⎦ ⎡ ⎛π⎞ ⎛ θ (t ) − θ (t − td ) ⎞ ⎛ π ⎞ ⎛ θ (t ) − θ (t − td ) ⎞ ⎤ = 2 Ac ⎢sin ⎜ ⎟ cos ⎜ ⎟ + cos ⎜ ⎟ sin ⎜ ⎟⎥ 2 2 ⎝4⎠ ⎝ ⎝ ⎠ ⎠⎦ ⎣ ⎝4⎠ ⎡ ⎛ θ (t ) − θ (t − td ) ⎞ ⎛ θ (t ) − θ (t − td ) ⎞ ⎤ = 2 Ac ⎢ cos ⎜ ⎟ + sin ⎜ ⎟⎥ 2 2 ⎠ ⎝ ⎠⎦ ⎣ ⎝

Since θ (t ) − θ (t − td ) 1 , the output of the envelope detector can be approximated as ⎡ θ (t ) − θ (t − td ) ⎤ yD (t ) = 2 Ac ⎢1 + ⎥ 2 ⎣ ⎦ ⎡ 1 dθ (t ) ⎤ td ⎥ = 2 Ac ⎢1 + ⎣ 2 dt ⎦ Now dθ (t ) = dt

t d ⎡ 2π ∆f max ∫ sn (t )dt ⎤ ⎢⎣ ⎥⎦ −∞ = 2π ∆f max sn (t ) dt

Substituting ⎡ θ (t ) − θ (t − td ) ⎤ yD (t ) = 2 Ac ⎢1 + ⎥ 2 ⎣ ⎦

(*)

= 2 Ac [1 + π ∆f max td sn (t ) ]

c. Calculate the output yD (t ) for a single-tone modulated FM signal xFM (t ) = Ac cos[2πf c t + β sin 2πf mt ] . Assume f c f m so that f mtd 1 . Solution:

For a single-tone modulating signal t

θ (t ) = 2π ∆f max ∫ sn (t )dt = β sin 2πf mt −∞

Taking the derivative of both sides 2π ∆f max sn (t ) = β 2π f m cos ( 2πf mt ) ⇒ sn (t ) = cos ( 2πf mt )

Substituting into equation (*), we obtain yD (t ) = 2 Ac [1 + π ∆f max td sn (t ) ] = 2 Ac ⎡⎣1 + π ∆f max td cos ( 2π f mt ) ⎤⎦ 5.12

A superheterodyne FM receiver operates in the frequency range of 88-108 MHz. Assuming that an IF of 10.7 MHz is selected, determine the range of variation of

the local oscillator frequency f LO . Does the range of image frequencies fall outside of the 88-108 MHz band? Solution: We will assume high-side injection. So the LO frequency is f LO = f c + f IF . The

range of LO frequencies is obtained as [88 + 10.7 , 108 + 10.7 ] = [98.7 , 118.7 ] . The image frequencies are located at fimage = f c + 2 f IF . So the range of image frequencies is obtained as [88 + 21.4 , 108 + 21.4] = [109.4 , 129.4] . Image frequencies fall outside the signal band. 5.13

A first-order PLL has phase detector characteristic shown in Figure P5.5. Assume that the phase detector output voltage swing is ±1.5 V. Determine a. Phase detector gain constant Solution:

Phase detector gain constant K PD =

V/radians

b. Hold-in range of the PLL. Solution:

The magnitude of the hold-in range ∆ωH is calculated by finding the frequency offset of the input that causes a phase error of ±π/2 radians. The phase detector output voltage swing is ±1.5 V for this range of phase error variation. We can obtain the PLL hold-in range by multiplying the change in phase detector voltage by KVCO F (0) . Since F (0) = 1 for a first-order PLL, ∆ωH is given by ∆ωH = 3KVCO Figure P5.5

υe (t )

…

−π

π 2

2π

… Phase error θ e

5.14 A first-order PLL is operating in phase lock when a frequency step ∆ω is applied. Assume that the loop gain K = 500π . Assume K PD = 0.8V/radians a. Determine the steady-state phase error, in degrees, for ∆ω = 100π , 200π , and 800π . Solution:

For a first-order PLL, the steady-state phase-error due to frequency-step ∆ω is given by

θ e (∞ ) =

∆ω K

∆ω

θ e (∞ )

100π 200π 800π

0.2 0.4 1.6

b. Plot the control voltage in each case. What is the steady-state control voltage after the initial transients have died? Solution:

The control voltage υe (t ) is given from (5.110) as

υe (t ) =

K PD ∆ω 1 − e− K t K

(

)

1.4

1.2

1 Freq. offset = 100*pi Freq. offset = 200*pi Freq. offset = 800*pi

ve(t)

0.8

0.6

0.4

0.2

0.5

1.5

2.5 t(sec)

3.5

4.5

5 -3

x 10

∆ω

υ e (∞ )

100π 200π 800π

0.16 0.32 1.28

c. What is the 3-dB bandwidth of the loop in each case? Comment on the tradeoff between the noise performance of the first-order PLL versus its steady-state phase error in part(a). Solution:

The 3-dB bandwidth of the PLL is K = 500π (1/sec). The shortcoming of the first-order PLL is that steady-state phase error θ e (∞) is inversely proportional to the loop gain K. Larger the loop gain, smaller the steady-state phase error θ e (∞) . However, the larger loop gain implies larger loop bandwidth and more output noise. 5.15 Consider two-pole, second-order PLL with loop filter F ( s ) =

1 . 1 + sτ 1

a. Derive closed-loop transfer function of the PLL in the standard form. Solution:

The open-loop transfer function G ( s ) for the loop is given by G ( s) =

K s (1 + sτ 1 )

We can now write the closed-loop transfer function H ( s ) as

H (s) =

K ⎛ ⎞ K s (1 + sτ 1 ) ⎜⎜1 + ⎟⎟ ⎝ s (1 + sτ 1 ) ⎠

K / τ1 s s 2 + + K / τ1

τ1

We can express H ( s ) in the standard form as H (s) =

ωn2 s 2 + 2sζωn + ωn2

where

ωn =

ζ =

τ1

1 2 Kτ 1

b. Determine and compare the 3-dB bandwidth and noise equivalent bandwidth BN of the PLL. Solution:

H ( jω ) =

ωn2

ωn2 − ω 2 + 2ζ jωnω

H ( jω ) =

ωn4

(ω − ω ) + ( 2ζω ω ) 2 n

Now H ( jω3dB ) =

ωn4

1 2

(ω − ω ) + 4ζ ω ω

(ω − ω ) + ( 2ζω ω ) 2 n

2 3dB

3dB

Solving for ω3dB, we have

⎡ ⎣

(

ω ⎡ B3dB = n ⎢1 − 2ς 2 + 2π ⎣

(1 − 2ς ) + 1⎤⎥⎦

∞

1 − 2ς 2

∞

BN = ∫ H ( f ) df = ∫ 0

1/2

ωn4

)

1/2

⎤ + 1⎥ ⎦

ω3dB = ωn ⎢1 − 2ς 2 +

2 n

2 2

2 n

∞

ωn4 df 4 2 2 2 2 4 0 ( 2π f ) + 8π ωn f ( 2ζ − 1) + ωn

=∫ =

ωn 8ζ

B3dB ωn ⎡ = 1 − 2ς 2 + ⎢ 2π ⎣ BN

(1 − 2ς )

8ζ ⎡ = 1 − 2ς 2 + 2π ⎢⎣

(1 − 2ς )

8ζ ⎤ + 1⎥ × ωn ⎦ 1/2

1/2

⎤ + 1⎥ ⎦

8ζ ⎡ 1 − 2ζ 2 + It can be shown that ⎢ 2π ⎣ B3dB < BN .

(1 − 2ζ ) 2

1/2

⎤ + 1⎥ ⎦

< 1 for ζ > 0. Therefore,

5.16 A frequency step of ∆ω is applied to the second-order PLL with loop active PI loop filter analyzed in Section 5.5.4. a. Derive an expression for the control voltage applied to the VCO. What is the steady-state control voltage after the initial transients have died? Solution:

The closed-loop transfer function H ( s ) the second-order PLL with loop active PI loop filter is given by H (s) =

Θout ( s ) 2sζωn + ωn2 = 2 Θin ( s ) s + 2 sζωn + ωn2

The control voltage υe (t ) in s-domain is now obtained as Ve ( s ) =

(

)

sΘin 2 sζωn + ωn2 sΘout ( s ) = K VCO K VCO s 2 + 2 sζωn + ωn2

(

)

For a step change ∆f in frequency, Θin ( s ) =

Ve ( s ) =

(

2π∆f 2 sζωn + ωn2

(

)

sK VCO s + 2 sζωn + ωn2 2

∆ω . Substituting yields s2

)

The control voltage υe (t ) is now obtained by taking the inverse Laplace transform as

υe (t ) =

⎞ 2π∆f ⎛ ς e−ςωnt ⎜1 − e −ςωnt cos( 1 − ς 2 ωnt ) + sin( 1 − ς 2 ωnt ) ⎟ u (t ), ς < 1 ⎟ K VCO ⎜⎝ 1− ς 2 ⎠

Therefore, the steady-state control voltage after the initial transients have died is given by lim υe (t ) = t →∞

2π∆f K VCO

The above result can also be derived by applying the final value theorem of the Laplace transform. b. Calculate the peak phase error for the step frequency change ∆ω . Solution:

The phase error θ e (t ) , for the step frequency change ∆ω , is given from (5.130) as

θ e (t ) =

∆ω e −ζωnt ⎛ 1 ⎜ sin ωn ⎜⎝ 1 − ζ 2

(

⎞ 1 − ζ 2 ωn t ⎟ , ζ < 1 ⎟ ⎠

)

The peak phase error θ p is obtained by setting

θp =

∆ω

ωn

(

dθ e (t ) = 0 and solving to give dt

)

− tan −1 κ / κ

where

κ = 1− ζ 2 / ζ 5.17 Consider the second-order PLL with an active PI loop filter. A single tone modulated FM signal with maximum frequency deviation ∆f max is applied to the PLL, where the frequency of the modulating signal is f m . a. Calculate the magnitude of the phase error. Solution:

For the second-order active PI loop, the phase error in the s-domain is given from (5.124) as

Θin ( s ) s 2 Θe ( s ) = 2 s + 2 sζωn + ωn2 The magnitude of the phase error in the frequency domain can be expressed as Θe ( jω ) =

Θin ( jω ) ω 2

(ω − ω ) + 4ζ ω ω 2 n

2 n

For an FM signal with maximum frequency deviation ∆f max caused by a sinusoidal tone of frequency ωm , the maximum phase deviation from (5.26) is

β=

∆f max ∆ωmax = ωm fm

The magnitude of the PLL phase error in tracking the FM signal with the phase deviation β is now obtained as

Θe ( jω ) =

βω 2

(ω − ω ) + 4ζ ω ω 2 n

2 n

b. For fixed ∆f max , calculate the value of modulating signal frequency f m for which the peak phase error occurs. What is the corresponding value of the peak phase error? Solution:

The phase error at ω = ωm = 2π f m is given by Θe ( jωm ) =

βωm2

(ω − ω ) + 4ζ ω ω 2 n

2 2 m

For fixed ∆f max , we calculate

2 n

= 2 m

d Θe ( jωm )

d ωm occurs for ωm = ωn . Substituting yields

β 2

2 ⎛ ωn2 ⎞ 2 ωn ζ − 1 + 4 ⎜ 2 ⎟ ωm2 ⎝ ωm ⎠

= 0 . We find that the maximum error

⎛ ∆ωmax ⎞ 1 ⎟ ⎝ ωn ⎠ 2ζ

ε p = max Θe ( jωm ) = ⎜ fm

c. Calculate the parameters of PLL ( K ,τ 1 ,τ 2 ) for a FM signal with ∆f max = 75 kHz and f m = 15 kHz . The PLL uses the phase detector with characteristic in Figure P5.5. Assume ζ = 0.707 , and τ 1 = τ 2 . Solution:

The PLL must demodulate the signal with uniform gain and stay within the linear portion of the phase detector characteristic. To assure uniform gain, we choose

ωn > ωm = 15 kHz In order that the PLL operates within the linear portion of the phase detector characteristic, the peak phase error ε p < π / 2 . That is, ⎛ ∆ωmax ⎞ 1 π < ⎟ ⎝ ωn ⎠ 2ζ 2

εp =⎜

or 75 π < fn 2 2 or fn >

2 × 75

= 33.76 kHz

Therefore, we choose f n = 34 kHz . Now from Table 5.4, we obtain parameters of the second-order PLL with an active PI loop filter as follows:

ωn =

ζ =

τ1

τ 2ωn 2

and ζ =

⇒τ2 =

Also, ωn × ζ =

2ζ

ωn

τ2 2

τ1

2 = 6.62 µ sec 2π × 34 × 103

τ2 K . Assuming τ 2 = τ 1 , 2 τ1

K = 2 × ωn × ζ = 2 × 0.707 × 2π × 34 × 103 = 3.02 × 105 (1/ sec)

Chapter 6 6.1

An unfair coin is tossed 3 times. Assume that the probability of a head p = 0.35 and the probability of a tail 1− p = 0.65. a. What is the sample space? Solution: Ω = { HHH , HHT , HTT , HTH , THH , THT , TTT , TTH }

b. Calculate P { HHT } and P {1 tail} . Solution: P { HHT } = 0.35 × 0.35 × 0.65 ≈ 0.08 P {1 tail} = P { HHT , HTH , THH } = 3 × 0.08 = 0.24

c. Determine P {1st toss is not head 1 head} . Solution:

New sample space = { HTT , THT , TTH } P {1st toss is not head 1 head} = P {THT , TTH } =

2 3

Alternatively, ⎧⎪ ⎫⎪ P { A ∩ B} P ⎨1st toss is not head 1 head ⎬ = P { B} A ⎪⎩ ⎪⎭ B where A = {THH , THT , TTT , TTH } B = { HTT , THT , TTH } A ∩ B = {THH , THT , TTT , TTH } ∩ { HTT , THT , TTH } = {THT , TTH } P {THT } = P {TTH } = P { HTT } = 0.35 × 0.65 × 0.65 ≈ 0.148

P {1st toss is not head 1 head} = P { A B} = =

P { A ∩ B} P { B}

P {THT , TTH }

P { HTT , THT , TTH }

0.148 + 0.148 2 = 0.148 + 0.148 + 0.148 3

6.2 Three tetrahedral dice are rolled. Let A be the event that at least one 4 appears and B be the event that no two dice show the same value. a. Find P ( A ) , P ( B ) , and P ( AB ) . Solution:

There are 43 = 64 possible outcomes. From the tree diagram, we observe that 4 appears at least once in 37 outcomes. Therefore, P ( A) =

37 64

From the tree diagram, we observe that no two dice show the same value in 24 outcomes. Therefore,

P ( B) =

24 3 = 64 8

A ∩ B = {124,134,142,143, 214, 234, 241, 243,314,324,341,342, 412, 413, 421, 423, 431, 432} P ( AB ) =

18 9 = 64 32

b. Are A and B statistically independent? Prove your answer. A and B are not statistically independent because P ( AB ) =

9 37 3 111 ≠ P ( A) P ( B ) = × = 32 64 8 512

Alternatively, P ( A ) = P {At least one 4 appers} = P {One 4 appers ∪ Two 4's apper ∪ Three 4's apper} ⎛ 3⎞ ⎛3⎞ 37 2 2 3 = ⎜ ⎟ 0.25 × ( 0.75 ) + ⎜ ⎟ ( 0.25 ) × 0.75 + ( 0.25 ) = 64 ⎝1 ⎠ ⎝ 2⎠

n! 4! = = 24 (n − k )! (4 − 3)! 24 3 P ( B ) = P {One of the ordered pairs w/o replacement occurs} = = 64 8

Number of ordered pairs w/o replacement = ( n )k =

4 2

3 1

12 34

2 3

1 2

1 2 3 4

3 4 1 2

1 2

3 4 1 2

3 1 2 34 1 4 2

3 4

6.3 Internet traffic from Los Angeles is routed to New York via Chicago with probability 0.8. In case of congestion, the traffic is routed via Denver with conditional probability of packet being dropped 0.3. Assuming that the conditional probability of packet being dropped via Chicago route is 0.2, determine a. The probability that a packet is dropped. Solution:

Let A = {Packet is routed through Chicago} B = {Packet is dropped }

Now

{ }

P { A} = 0.8 , P { B A} = 0.2 , and

( ) (

P B A = 0.3

)

P ( B ) = P ( A) P ( B A ) + P A P B A = 0.8 × 0.2 + 0.2 × 0.3 = 0.16 + 0.06 = 0.22

b. The conditional probability that a packet is routed via Chicago given that it is not dropped. Solution:

( ) = P ( B A) P ( A) = P ( B A) P ( A) ( ) P ( B) P ( B) P ( B) P ( B A ) = 1 − P ( B A ) = 1 − 0.2 = 0.8 P AB =

P AB

× 0.8 0.64 = = 0.82 ( ) 0.8 1 − 0.22 0.78

P AB =

6.4 The PMF of random variable x is given by ⎧ K / xi , xi = 3, 6,9,12 p x ( xi ) = ⎨ otherwise ⎩0,

a. What is the value of K?

Solution:

∑ p ( x ) = 3 + 6 + 9 + 12 + = 1 x

25 ⎡12 + 6 + 4 + 3 ⎤ K⎢ =K =1 ⎥ 36 36 ⎣ ⎦ 36 K= 25 b. Find P { x > 6} . Solution:

P { x > 6} = P { x = 9 or 12} = P { x = 9 } + P { x = 12} =

36 36 7 + = 25 × 9 25 × 12 25

c. Find P {6 ≤ x ≤ 12} . Solution:

P {6 ≤ x ≤ 12} = P { x = 6 or 9 or 12} = P { x = 6 } + P { x = 9 } + P { x = 12} = 6.5

36 36 36 13 + + = 25 × 6 25 × 9 25 × 12 25

The CDF of the random variable x is given by ⎧ ⎪0, ⎪ ⎪ (3 + 2 x) Fx ( x) = ⎨ , ⎪ 6 ⎪ ⎪1, ⎩

3 2 3 3 − ≤x≤ 2 2 3 x> 2 x<−

a. Plot the CDF and PDF. Solution:

Fx (x) fx (x)

1.0

1/3

0.5

−1.5

x −1.5

1.5

x 1.5

b. Find P { x ≤ −1.5} . Solution: P { x ≤ −1.5} = Fx (−1.5) = 0

c. Find P {−1 < x ≤ 1} . Solution:

P {−1 < x ≤ 1} = Fx (1) − Fx (−1) =

5 1 4 2 − = = 6 6 6 3

d. Find P { x ≤ 2} . Solution: P { x ≤ 2} = Fx (2) = 1

e. Find P { x > 3} . Solution: P { x > 3} = 1 − P { x ≤ 3} = 1 − Fx (3) = 0

6.6 The PDF of the random variable x is given by f x ( x) = K Λ ⎡⎣( x − 2 ) / 2 ⎤⎦ . a. What is the value of K?

Solution: Area of the traingle = 1× K = 1 ∴K =1

b. Plot the PDF.

1.2

fx (x)

1.0 Fx (x)

Fx(x)

0.8

0.6

0.4

0.2

x 1

0.5

1.5

2 x

2.5

3.5

c. Determine and plot the CDF. Solution:

For x < 1, Fx ( x) = 0 x

t2 x2 For 1 ≤ x ≤ 2, Fx ( x) = ∫ ( t − 1)dt = − t = − x + 0.5 2 2 1 1 For 2 ≤ x ≤ 3, x

⎛ t2 ⎞ ⎛ x2 ⎞ x2 Fx ( x) = 0.5 + ∫ ( −t + 3)dt = 0.5 − ⎜ − 3t ⎟ = 0.5 − ⎜ − 3 x − 2 + 6 ⎟ = − + 3 x − 3.5 2 ⎝2 ⎠2 ⎝ 2 ⎠ 2 x

For x ≥ 3, Fx ( x) = 1 d. P { x ≤ 2} . Solution: P { x ≤ 2} = Fx (2) = 0.5

e. Find P {−1 ≤ x − 2 ≤ 1} . Solution:

P {−1 ≤ x − 2 ≤ 1} = P {1 < x ≤ 3} = Fx (3) − Fx (1) = 1− 0 = 1 f. Find P { x > 2} Solution: P { x > 2} = 1 − P { x ≤ 2} = 1 − 0.5 = 0.5

6.7 The PDF of random variable x is given by f x ( x) = 0.5e

−α x

−∞ < x < ∞

a. Determine α so that f x ( x) is a PDF. Solution: ∞

∫ 0.5e

−α x

−∞

∞

dx = 2 ∫ 0.5e

−α x

⎛ e −α x ⎞ 1 1 dx = ⎜ ( 0 − 1) = = 1 ⎟ = α −α ⎝ −α ⎠ 0

α = 1.0 b. Find its CDF. Solution: x

−∞

Fx ( x) = ∫ f x (t )dt = ∫ 0.5e dt x

−t

For x < 0, Fx ( x) = ∫ 0.5et dt = 0.5 et −∞

−∞

= 0.5 ( e x − 0 ) = 0.5e x

For x ≥ 0, x

e−t = 0.5 − 0.5 ( e− x − 1) = 1 − 0.5e− x Fx ( x) = 0.5 + ∫ 0.5e dt = 0.5 + 0.5 −1 0 0 x

−t

⎧⎪0.5e x , Fx ( x) = ⎨ −x ⎪⎩1 − 0.5e , c.

x≤0 x≥0

Find P {−1 ≤ x ≤ 1}

Solution:

P {−1 ≤ x ≤ 1} = Fx (1) − Fx (−1) = 1 − 0.5e −1 − 0.5e −1 = 1 − e−1 = 0.632 d. Determine E { x} and Var ( x ) E { x} = 0

{ }

Var ( x ) = E x 2 − E 2 { x} 0

∞

−∞

= ∫ x 2 f x ( x)dx = 2 × 0.5∫ x 2 e− x dx =

2! =2 1

6.8 The CDF of random variable x is shown in Figure P6.1. Figure P6.1

Fx ( x ) 1 5/8

3/8

−1

a. Sketch the PDF of x, f x ( x ) .

Solution:

fx (x) 3/8 3/8 (a) Let y = 2x −1. Compute the values of P {−2 < y ≤ 1} . 1/4

x 0

-1

b. Let y = 2x −1. Compute the values of P {−2 < y ≤ 1} . Solution:

⎧ 1 ⎫ ⎛ 1⎞ 5 4 1 P {−2 < y ≤ 1} = P ⎨− < x ≤ 1⎬ = Fx (1) − Fx ⎜ − ⎟ = − = ⎩ 2 ⎭ ⎝ 2⎠ 8 8 8 c. Find the value of E { x x > 0} . Solution: ∞

E { x x > 0} = ∫ x f x ( x x > 0 ) dx . −∞

Fx ( x x > 0 ) = P ( x ≤ x x > 0 ) =

P ( x ≤ x, x > 0 ) P ( x > 0)

Fx ( x )

P ( x > 0)

,x>0

Differentiating both sides, we obtain f x ( x x > 0) =

fx ( x)

P ( x > 0)

,x>0

Since P ( x > 0 ) = 3 / 8 , the conditional PDF f x ( x x > 0 ) is a rectangular pulse of amplitude 1 over the interval (1,2] centered at x = 1.5. Therefore, E { x x > 0} = 1.5 .

6.9 The random variable x is passed through a soft limiter with transfer characteristic y = g ( x ) is shown in Figure P6.2. Figure P6.2 y 1 0

a −1

a. Find an expression for the mean and variance of the output random variable y for an arbitrary continuous random variable x . Solution: ∞

−a

∞

x E { y} = ∫ g ( x) f x ( x ) dx = ∫ f x ( x ) dx − ∫ f x ( x ) dx + ∫ f x ( x ) dx a −∞ −a −∞ a a

1 = ∫ x f x ( x ) dx − Fx ( −a ) + ⎡⎣1 − Fx ( a ) ⎤⎦ a −a ∞

−a

∞

⎛x⎞ E y 2 = ∫ g ( x) f x ( x ) dx = ∫ ⎜ ⎟ f x ( x ) dx + ∫ f x ( x ) dx + ∫ f x ( x ) dx a⎠ −∞ −a ⎝ −∞ a

{ }

1 = 2 ∫ x 2 f x ( x ) dx + Fx ( − a ) + ⎡⎣1 − Fx ( a ) ⎤⎦ a −a

{ }

Var ( y ) = E y 2 − E 2 { y} b. If x is a Gaussian random variable, determine the PDF of the output random variable y . Solution: Fy ( y ) is zero for y < −1 . For y > 1 , Fy ( y ) = P { y ≤ y} = 1 .

For −1 < y < 1 , Fy ( y ) = P { y ≤ y} = P { x ≤ ay} = Fx ( ay )

The CDF Fy ( y ) of random variable y can now be expressed as

⎧0, ⎪ Fy ( y ) = ⎨ Fx ( ay ) , ⎪1, ⎩

y < −1 y ≤1 y >1

Note that at the points y = ±1 , Fy ( y ) is discontinuous and the steps are given by Fy (−1+ ) − Fy (−1− ) = Fx (− a ) Fy (1+ ) − Fy (1− ) = 1 − Fx (a )

For −1 < y < 1 , y = g ( x) =

f y ( y) =

x . Therefore a

f x ( x) = af x ( x ) x = y = af x ( ay ) g '( x) x = g −1 ( y )

The PDF f y ( y ) of random variable y is given by f y ( y ) = Fx (− a )δ ( y + 1) + [1 − Fx (a ) ] δ ( y − 1) + af x (ay ),

y ≤1

If x is N (0, σ x2 ) , f y ( y ) is given by ⎡ 2 2 2 ⎛ a ⎞ ⎛ −a ⎞ ⎤ a f y ( y) = Q ⎜ e− a y /2σ x ⎟ δ ( y + 1) + ⎢1 − Q ⎜ ⎟ ⎥ δ ( y − 1) + ⎢⎣ 2πσ x2 ⎝σx ⎠ ⎝ σ x ⎠ ⎥⎦ 2 2 2 ⎛ a ⎞ a e − a y /2σ x , = Q⎜ ⎟ ⎡⎣δ ( y + 1) + δ ( y − 1) ⎤⎦ + 2 2πσ x ⎝σx ⎠

y ≤1

c. Determine the mean and variance of the output random variable y if x is a N (0, σ x2 ) . Solution:

∞ ⎛ a ⎞1 E { y} = ∫ y f y ( x ) dy = Q ⎜ ⎟ ∫ y ⎡⎣δ ( y + 1) + δ ( y − 1) ⎤⎦ dy ⎝ σ x ⎠ −1 −∞ ∞

+∫y

2 2

2πσ

−1

e − a y /2σ x dy

2 x

1 2 2 2 ⎛ a ⎞ a = Q⎜ − + 1 1 y e − a y /2σ x dy = 0 ( ) ⎟ ∫ 2πσ x2 ⎝σx ⎠ −1

{ }

E y

∞

⎛ a ⎞1 2 = ∫ y f y ( y ) dy = Q ⎜ ⎟ ∫ y ⎡⎣δ ( y + 1) + δ ( y − 1) ⎤⎦ dy ⎝ σ x ⎠ −1 −∞ 2

∞

+ ∫ y2 −∞

a 2πσ x2

e − a y /2σ x dy 2 2

2 2 2 ⎛ a ⎞ 1 2 a = 2Q ⎜ e − a y /2σ x dy ⎟ + 2∫ y 2 2πσ x ⎝σx ⎠ 0

2 2 2 ⎛ a ⎞ 1 2 a Var ( y ) = E y 2 − E 2 { y} = E y 2 = 2Q ⎜ e − a y /2σ x dy ⎟ + 2∫ y 2πσ x2 ⎝σx ⎠ 0

{ }

6.10 Let y = x be the output of a full-wave rectifier. a. Determine the CDF of the output random variable y for an arbitrary continuous random variable x . Solution:

The equivalent event for { y ≤ y} is { x ≤ y} . Therefore, for y ≥ 0 Fy ( y ) = P { x ≤ y} = P {− y < x ≤ y} = Fx ( y ) − Fx (− y )

b. Find the PDF of y for a Gaussian random variable x . Solution:

Assuming x is a continuous random variable f y ( y ) = f x ( y ) + f x ( − y ),

y≥0

If f x ( x) is even, then f x ( x) = f x (− x) and

f y ( y ) = 2 f x ( y ),

y≥0

If x is N (0, σ x2 ) , f y ( y ) is given by 2 e − y / 2σ x 2

f y ( y) =

σx

y≥0

6.11 x is a Gaussian random variable with mean mx = 2 and variance σ x2 = 16 a. Calculate P { x ≤ 2} , P { x ≥ 3} , P {6 < x ≤ 10} , P { x − 2 ≤ 4} and P { x > 16} . Solution:

(

)

If x is N mx , σ x2 , we have. ⎛m −x⎞ P{ x ≤ x} = Fx ( x) = Q ⎜ x ⎟ ⎝ σx ⎠ ⎛ x − mx ⎞ P { x > x} = Q ⎜ ⎟ ⎝ σx ⎠

Now ⎛ 2−2⎞ P{ x ≤ 2} = Fx (2) = Q ⎜ ⎟ = Q ( 0 ) = 0.5 ⎝ 4 ⎠ ⎛ 3− 2 ⎞ ⎛1⎞ P { x > 3} = Q ⎜ ⎟ = Q ⎜ ⎟ = 0.4013 ⎝ 4 ⎠ ⎝4⎠ .

P {6 < x ≤ 10} = Fx (10) − Fx (6) ⎛ 2 − 10 ⎞ ⎛ 2−6⎞ = Q⎜ ⎟ −Q⎜ ⎟ = Q ( −2 ) − Q ( −1) ⎝ 4 ⎠ ⎝ 4 ⎠ = −Q ( 2 ) + Q (1) = −0.02275 + 0.1586 = 0.1358 P { x − 2 ≤ 4} = P {−4 ≤ x − 2 ≤ 4} = P {−2 ≤ x ≤ 6} = Fx (6) − Fx (−2) ⎛ 2−6⎞ ⎛ 2+2⎞ = Q⎜ ⎟ −Q⎜ ⎟ = Q ( −1) − Q (1) ⎝ 4 ⎠ ⎝ 4 ⎠ = 1 − Q (1) − Q (1) = 1 − 2Q (1) = 1 − 2 × 0.1586 = 0.6828

P { x > 16} = P { x < −16 or x > 16} = P { x < −16 } + P { x > 16} ⎛ 2 + 16 ⎞ ⎛ 16 − 2 ⎞ = Q⎜ ⎟ +Q⎜ ⎟ = Q ( 4.5 ) + Q ( 3.5 ) ⎝ 4 ⎠ ⎝ 4 ⎠ = 0.0000034 + 0.00023 = 0.0002334

b. P { x > α } = 10−3 , find α. Solution:

⎛α −2⎞ −3 P{x > α} = Q ⎜ ⎟ = 10 4 ⎝ ⎠ α −2 ⇒ = 3.09 4 α = 14.36 c. P { x ≤ α } = 0.995 , find α. Solution:

⎛α −2⎞ P{x ≤ α} = 1− P{x > α} = 1− Q ⎜ ⎟ = 0.995 ⎝ 4 ⎠ ⎛α −2⎞ ∴Q ⎜ ⎟ = 1 − 0.995 = 0.005 ⎝ 4 ⎠ α −2 = 2.575 ⇒ α = 12.3 4 d. Find P { x − mx > 6σ x } . Solution:

P { x − mx > 6σ x } = P { x − mx < −24 or x − mx > 24} = P { x < −22 } + P { x > 26} ⎛ 2 + 22 ⎞ ⎛ 26 − 2 ⎞ = Q⎜ ⎟ +1− Q ⎜ ⎟ ⎝ 4 ⎠ ⎝ 4 ⎠ = 2 × Q(6) = 0.00198 × 10−6

6.12 x and y have joint PDF f x , y ( x, y ) = K ( x 2 + y 2 ) ,

− 1 ≤ x ≤ 1, − 1 ≤ y ≤ 1

a. Find the value of K? Solution: 1 1

∫ ∫ K (x + y ) 2

−1 −1

⎛ x3 ⎞ dxdy = K ∫ dy ⎜ + y 2 x ⎟ ⎝ 3 ⎠ −1 −1 1

⎛ y y3 ⎞ ⎛1 ⎞ ⎛4⎞ = 2 K ∫ dy ⎜ + y 2 ⎟ = 2 K ⎜ + ⎟ = 2 K ⎜ ⎟ = 1 ⎝3 ⎠ ⎝3⎠ ⎝ 3 3 ⎠ −1 −1 1

⇒K=

3 8

b. Determine the joint CDF Fx , y ( x, y ) . Solution: x

⎛ u3 ⎞ 3 y x 3 y Fxy ( x, y ) = ∫ ∫ u 2 + v 2 dudv = ∫ dv ⎜ + v 2u ⎟ 8 −1 −1 8 −1 ⎝ 3 ⎠ −1

(

)

⎛ x3 ⎞ 3 ⎛ x 3v v 3 3 y 1 v v3 ⎞ = ∫ dv ⎜ + v 2 x + + v 2 ⎟ = ⎜ + x+ + ⎟ 8 −1 ⎝ 3 3 3 3 3 ⎠ −1 ⎠ 8⎝ 3 3 ⎛ x3 y xy 3 y y 3 ⎞ 3 ⎛ x 3 x 1 1 ⎞ = ⎜ + + + ⎟+ ⎜ + + + ⎟ 8⎝ 3 3 3 3 ⎠ 8⎝ 3 3 3 3⎠ 3 y 3 y 1 = x + 1 + ( x + 1) + x 3 + x + 2 8 8 8

(

)

(

)

That is, Fxy ( x, y ) =

y 3 y3 1 x + 1 + ( x + 1) + x3 + x + 2 , 8 8 8

(

)

(

)

− 1 ≤ x ≤ 1; − 1 ≤ y ≤ 1

c. Calculate the marginal PDFs of x and y . Solution:

3 1 3⎛ y3 ⎞ f x ( x) = ∫ x 2 + y 2 dy = ⎜ x 2 y + ⎟ 8 −1 8⎝ 3 ⎠ −1

(

)

3⎛ 1 1⎞ 3⎛ 1⎞ = ⎜ x2 + + x2 + ⎟ = ⎜ x2 + ⎟ , 8⎝ 3 3⎠ 4⎝ 3⎠

−1 ≤ x ≤ 1

⎞ 3 1 3 ⎛ x3 f y ( y ) = ∫ x 2 + y 2 dx = ⎜ + y 2 x ⎟ 8 −1 8⎝ 3 ⎠ −1

(

)

3⎛1 1 1⎞ ⎞ 3⎛ = ⎜ + y2 + + y2 ⎟ = ⎜ y2 + ⎟ , 8⎝3 3 3⎠ ⎠ 4⎝

−1 ≤ y ≤ 1

6.13 x and y have joint PDF f x , y ( x, y ) = xye −( x + y ) ,

x ≥ 0; y ≥ 0

a. Determine f x ( x) and f y ( y ) . Solution: ∞

∞

f x ( x) = ∫ xye−( x + y ) dy = xe− x ∫ ye− y dy −x

= xe ,

x≥0

∞

f y ( y ) = ∫ xye −( x + y ) dx = ye− y xe− x ∫ xe− x dx 0

−y

= ye ,

y≥0

b. Are x and y statistically independent? Solution: f y ( y)

f xy ( x, y ) = xye

−( x + y )

= xe

−x

ye− y = f x ( x) f y ( y )

fx ( x)

Therefore, x and y statistically independent. c. Find f y ( y x) and f x ( x y ) . Solution:

Since x and y statistically independent,

f y ( y x) = f y ( y ) = ye− y ,

y≥0

To prove this, f y ( y x) =

f xy ( x, y ) f x ( x)

xye −( x + y ) = ye− y = f y ( y ) −x xe

Similarly, f x ( x y ) = f x ( x) = xe − x ,

x≥0

d. Find E ( y x = x) and E ( x y = y ) . Solution: ∞

∞

−∞

∞

E ( y x = x) = ∫ y f y ( y x)dy = ∫ y f y ( y )dy = ∫ y 2 e− y dy = 2 0

E( y) ∞

∞

−∞

∞

E ( x y = y ) = ∫ x f x ( x y )dx = ∫ x f x ( x) dx = ∫ x 2 e − x dx = 2 0

E( x)

e. Find E ( x ) and E ( y ) . Solution: ∞

∞

−∞

∞

−∞

E ( x ) = ∫ x f x ( x)dx = ∫ x 2 e− x dx = 2 E ( y ) = ∫ y f y ( y )dy = ∫ y 2 e− y dy =2 6.14 Show that the correlation coefficient ρ xy between random variable x and y satisfies −1 ≤ ρ xy ≤ 1 . Solution:

Let w = x - ay . Then

{

}

Var ( w ) = E ( x - ay ) − ⎡⎣ E ( x - ay ) ⎤⎦ 2

{ } { } = E { x } − m + E {a y } − a m − E {2axy} + 2am m = E x 2 − 2axy + a 2 y 2 − mx2 − 2amx m y + a 2 m 2y 2

2 x

2 y

= Var ( x ) + a Var ( y ) − 2aCov ( xy ) 2

Since Var ( w ) ≥ 0 for any a, 2aCov ( xy ) ≤ Var ( x ) + a 2Var ( y ) . Now if we choose a =

σx , σy 2

⎛σ ⎞ σ 2 x Cov ( xy ) ≤ Var ( x ) + ⎜ x ⎟ Var ( y ) ⎜σy ⎟ σy ⎝ ⎠ or 2Cov ( xy ) ≤ σ xσ y + σ xσ y or Cov ( xy ) ≤ σ xσ y ⇒ ρ xy ≤ 1 Similarly by choosing a = −

σx , we can show that σy

Cov ( xy ) ≥ −σ xσ y ⇒ ρ xy ≥ −1 .

6.15 Random variables x and y have joint PDF f x , y ( x, y ) = C ( x + y ) ,

0 ≤ x ≤ 2; 0 ≤ y ≤ 1

a. Calculate C. ∞ ∞

2 1

−∞ −∞

0 0

∫ ∫ f xy ( x, y)dxdy = ∫ ∫ C ( x + y)dxdy 1

2 ⎛ 1⎞ y2 ⎞ ⎛ = C ∫ dx ⎜ xy + ⎟ = C ∫ dx ⎜ x + ⎟ 2 ⎠0 2⎠ ⎝ ⎝ 0 0 2

⎛ x2 x ⎞ = C ⎜ + ⎟ = 3C = 1 ⇒ C = 1/ 3 ⎝ 2 2⎠0 b. Determine E ( x ) and E ( y ) . Solution:

∞

x+ y xy y 2 1⎛ 1⎞ f x ( x) = ∫ f xy ( x, y )dy = ∫ dy = + = ⎜ x + ⎟, 0 ≤ x ≤ 2 3 3 6 0 3⎝ 2⎠ 0 −∞ 2

x+ y xy x 2 2 f y ( y ) = ∫ f xy ( x, y )dx = ∫ dx = + = ( y + 1) , 0 ≤ y ≤ 1 3 3 6 0 3 −∞ 0 ∞

∞

x⎛ 1⎞ 1 ⎛ x3 x 2 ⎞ 1 ⎛ 8 4 ⎞ 11 E { x} = ∫ xf x ( x)dx = ∫ ⎜ x + ⎟dx = ⎜ + ⎟ = ⎜ + ⎟ = 3⎝ 2⎠ 3⎝ 3 4 ⎠0 3⎝3 4⎠ 9 −∞ 0 1

2y 2 ⎛ y3 y 2 ⎞ 2 1 1 5 E { y} = ∫ yf y ( y )dy = ∫ ( y + 1)dy = ⎜ + ⎟ = ⎛⎜ + ⎞⎟ = 3 3⎝ 3 2 ⎠0 3⎝3 2⎠ 9 −∞ 0 b. Find Var ( x ) and Var ( y ) .

{ }

E x

{ }

E y

∞

x2 ⎛ 1⎞ 1 ⎛ x 4 x3 ⎞ 1 ⎛ 16 8 ⎞ 16 = ∫ x f x ( x)dx = ∫ ⎜ x + ⎟dx = ⎜ + ⎟ = ⎜ + ⎟ = 3 ⎝ 2⎠ 3⎝ 4 6 ⎠0 3⎝ 4 6⎠ 9 −∞ 0 2

∞

2 y2 2 ⎛ y 4 y3 ⎞ 2 1 1 7 = ∫ y f y ( y )dy = ∫ ( y + 1)dy = ⎜ + ⎟ = ⎛⎜ + ⎞⎟ = 3 3⎝ 4 3 ⎠ 0 3 ⎝ 4 3 ⎠ 18 −∞ 0 1

{ }

16 121 144 − 121 23 − = = 9 81 81 81

{ }

7 25 13 − = 18 81 162

Var ( x ) = E x 2 − E 2 { x} = Var ( y ) = E y 2 − E 2 { y} =

c. Calculate Cov( x, y ) and ρ xy . Solution: COV ( x , y ) = E { xy} − E { x} E { y} ∞

∞

−∞

E { xy} = ∫ dy ∫ xyf xy ( x, y )dx = ∫ dx 0

x y ( x + y )dy 3 ∫0 2

2 ⎡ x x 2 ⎤ 1 ⎡ x 2 x3 ⎤ 1 ⎡4 8⎤ 2 x⎡y y x⎤ 1 = ∫ dx ⎢ + = dx ⎥ ⎢ + ⎥= ⎢ + ⎥ = ⎢ + ⎥= ∫ 3⎣ 3 2 ⎦0 3 0 ⎣ 3 2 ⎦ 3 ⎣ 6 6 ⎦0 3 ⎣ 6 6 ⎦ 3 0 2

COV ( x, y ) =

2 5 11 2 55 1 − = − =− 3 9 9 3 81 81

d. What is E ( x + y ) and Var ( x + y ) ? Solution:

E { x +y} = E { x} + E { y} =

11 5 16 + = 9 9 9

Var { x +y} = Var { x} + Var { y} + 2COV ( x, y ) =

23 13 2 13 + 46 − 4 55 + − = = 81 162 81 162 162

6.16 x and y are two N (0,1) random variables. Determine the PDF of z = x + y if a.

x and y are statistically independent.

Solution:

Since x and y are Gaussian random variables, their sum z = x + y is also Gaussian with mean and variance E { z} = E { x} + E { y} = 0 + 0 = 0 Var ( z ) = Var ( x ) + Var ( y ) = 1 + 1 = 2

The PDF of z can now be expressed as f z ( z) =

1 − z 2 /4 e 4π

b. The correlation coefficient ρ xy = −1/ 2 . Solution:

Since x and y are Gaussian random variables, their sum z = x + y is also Gaussian with mean and variance E { z} = E { x} + E { y} = 0 + 0 = 0

Var ( z ) = Var ( x ) + Var ( y ) + 2Cov( x, y )

= Var ( x ) + Var ( y ) + 2 ρ xy Var ( x ) Var ( y ) = 1 + 1 + 2 ( −1/ 2 ) = 1 The PDF of z can now be expressed as f z ( z) =

1 − z 2 /2 e 2π

6.17 Let y = e- ax . a. Find the CDF and PDF of y if x is a uniformly distributed random variable in [0,1] . Assume a > 0. Solution:

For x ∼ U [ 0,1] , y assumes values in [ e− a , 1]. For y ≤ 0, P { y ≤ y} = 0 . For y in [ e − a , 1], we have

{

}

P { y ≤ y} = P e − ax ≤ y = P {− ax ≤ log e y}

⎧ ⎧ log e (1/ y ) ⎫ log e (1/ y ) ⎫ = P ⎨x > ⎬ = 1− P ⎨ x ≤ ⎬ a a ⎩ ⎭ ⎩ ⎭ ⎛ log e (1/ y ) ⎞ = 1 − Fx ⎜ ⎟ a ⎝ ⎠ ⎧0, y ≤ e− a ⎪ ⎛ log e (1/ y ) ⎞ ⎪ Fy ( y ) = ⎨1 − Fx ⎜ e− a ≤ y ≤ 1 ⎟, a ⎝ ⎠ ⎪ ⎪1, y ≥1 ⎩ Now ⎧0, ⎪ Fx ( x) = ⎨ x, ⎪1 ⎩

x≤0 0 ≤ x ≤1 x >1

Substituting yields

⎧0, ⎪ ⎪ log e ( y ) Fy ( y ) = ⎨1 + , a ⎪ ⎪1, ⎩

y ≤ e− a e− a ≤ y ≤ 1 y >1

⎧1 ⎪ , f y ( y ) = ⎨ ya ⎪0, ⎩

e− a ≤ y ≤ 1 elsewhere

b. Find the CDF and PDF of y if x is a Gaussian random variable. For the case a = −1, y is called lognormal random variable. Solution:

We consider the case a = −1. For y ≤ 0, P { y ≤ y} = 0 . For y > 0,

{

}

P { y ≤ y} = P e x ≤ y = P { x ≤ log e y} = Fx ( log e y ) y≤0

⎧⎪0, Fy ( y ) = ⎨ ⎪⎩ Fx ( log e y ) ,

y>0

For y > 0 dF ( log e y ) d log e y 1 fy ( y) = x = f x ( log e y ) = f x ( log e y ) dy dy y Therefore, y≤0

⎧0, ⎪ f y ( y) = ⎨ 1 ⎪ y f x ( log e y ) , ⎩ For f x ( x) =

f y ( y) =

1 y

2πσ x2

e − ( x − mx ) /2σ x , we have 2

2πσ x2 1

y>0

e − (loge y − mx ) /2σ x , 2

y>0

6.18 The PDF of Rayleigh distributed random variable x is given by

f x ( x) =

e − x /2σ , 2

0≤ x<∞

π⎞ ⎛ , E x 2 = 2σ 2 and Var ( x ) = ⎜ 2 − ⎟ σ 2 2⎠ 2 ⎝

a. Show that E { x} = σ

( )

Solution: ∞

E { x} = ∫

e − x /2σ dx = 2 2

∞

xe σ ∫

2 − x 2 /2σ 2

Now for the N (0, σ 2 ) random variable, we have ∞

1 2πσ

∫ xe

2 − x 2 /2σ 2

dx = σ 2

−∞

Since the integrand is even, we obtain ∞

2 − x /2σ dx = ∫xe 2

2πσ 2 0

σ2 2

or ∞

2 − x /2σ dx = ∫xe 2

σ 2 2πσ 2 2

=σ3

π 2

Therefore, E { x} =

{ }

σ3 π π =σ 2 σ 2 2 ∞

E x2 = ∫ 0

e − x /2σ dx = 2 2

∞

xe σ ∫

3 − x 2 /2σ 2

To use integration by parts ∫ udv = uv a − ∫ vdu , we let u = x 2 and v = e − x /2σ . b

Then dv = −

2 x − x2 /2σ 2 x − x2 /2σ 2 = − e dx e dx σ2 2σ 2

Therefore, we can write 1

∞

xe σ ∫

3 − x 2 /2σ 2

∞

(

dx = − ∫ x 2 d e − x /2σ 2

)

Now ∞

(

−∫ x d e 2

− x 2 /2σ 2

)

∞ 2 2 ⎪⎧ 2 − x2 /2σ 2 ∞ ⎪⎫ = − ⎨x e − ∫ e− x /2σ 2 xdx ⎬ 0 0 ⎩⎪ ⎭⎪

∞ ∞ 2 2 x − x2 /2σ 2 ⎪⎫ x ⎪⎧ 2 2 = − ⎨0 − 2σ ∫ 2 e dx ⎬ = 2σ ∫ 2 e − x /2σ dx = 2σ 2 σ σ 0 0 ⎩⎪ ⎭⎪ =1

{ }

Thus E x 2 = 2σ 2 . Now 2

⎛ π⎞ ⎛ π⎞ 2 Var ( x ) = E { x } − E { x} = 2σ − ⎜⎜ σ ⎟ = ⎜ 2 − ⎟σ 2 ⎟⎠ ⎝ 2⎠ ⎝ 2

b. Find the CDF Fx ( x ) . Plot it. Solution: x

Fx ( x ) = ∫ 0

e −t /2σ dt 2

To solve the integral, let us substitute z =

Fx ( x ) =

x 2 /2σ 2

∫ 0

t2 t . Then dz = 2 dt . 2 σ 2σ

x 2 /2σ 2

e− z e − z dz = −1 0

= 1 − e − x /2σ , 2

0≤ x<∞

Rayleigh CDF 1.2

0.8

0.6

0.4 2 σ =1

0.2

2 σ =2 2 σ =3

0.2

0.4

0.6

0.8

1.2

1.4

1.6

1.8

c. Find P { x ≤ 0.1× x RMS } , where x RMS = 2σ is the rms value of the random variable x. Solution: P { x ≤ 0.1× x RMS } = 1 − e −( 0.1) 2σ /2σ = 1 − e −0.01 = 1 − 0.99 = 0.01 2

The random variable x assumes values less than 0.1× xRMS with a probability of 1%. 6.19 Consider linear transformation y = A x + b of a vector x = ( x1 , x2 ,........, xn ) of T

continuous random variables. Assume that the n×n matrix A has rank n. a. Show that the PDF of y is given by f y ( y) =

( (

1 −1 fx A y − b A

))

Solution:

{

( )

}

( )

Let B = y ≤ y . Therefore, Fy y = ∫ f y y d y . Since the inverse of matrix A B

exists, we can define the inverse transformation T by x=A

−1

( y − b) ( )

T y

Since this is one-one transformation, we observe that y is in the set B

{

}

if and only if x is in the set T ( B ) = x A x + b ≤ y . Thus

( )

{

}

Fy y = P y in B = P { x in T ( B )} = ∫ f x ( x ) d x T ( B)

The n-dimensional elemental areas d x and d y are related by Jacobian as discussed in Section 6.8.1. That is dx1dx2 ...dxn = J ( y1 , y2 ,..., yn ) dy1dy2 ...dyn

where

⎛ ∂x1 ∂x1 ⎜ ∂y ∂y 2 ⎜ 1 ⎜ ∂x2 ∂x1 ⎜ J ( y1 , y2 ,..., yn ) = det ⎜ ∂y1 ∂y2 ⎜ ⎜ ⎜ ∂xn ∂xn ⎜ ∂y ∂y 2 ⎝ 1

∂x1 ⎞ ∂yn ⎟ ⎟ ∂x1 ⎟ ∂yn ⎟⎟ ⎟ ⎟ ∂xn1 ⎟ ∂yn ⎟⎠ x = A−1 ( y −b )

It is easy to prove that J ( y1 , y2 ,..., yn ) = det ( A−1 ) =

( )

Fy y = ∫ f x ( x ) d x = T ( B)

1 det ( A )

( (

))

1 −1 fx A y − b d y ∫ det ( A ) B

Therefore, f y ( y) =

( (

1 −1 fx A y − b A

))

b. If x = ( x1 , x2 ,........, xn ) is a Gaussian random vector with expected value T

m x and covariance matrix C x , the vector y is also Gaussian with expected value m y = Am x + b and covariance matrix C y = AC x A . T

Solution: ⎜ − ( A ( y −b ) − m x ) C x ( A ( y −b ) − m x ) ⎟ 1 1 −1 ⎝ 2 ⎠ f y ( y) = fx A y − b = e n /2 1/2 A ( 2π ) A C x

( (

⎛ 1

))

−1

⎞

Now

−1

( y − b ) − m = A ( y − b − Am ) = A ( y − m ) −1

−1

and

( A ( y − b ) − m ) = ( y − b − Am ) ( A ) = ( y − m ) ( A ) T

−1

−1 T

The argument of the exponential is therefore equal to

( y − b − Am ) ( A ) C A ( y − b − Am ) = ( y − m ) ( A ) C A ( y − m ) −1 T

−1 x

−1

−1 x

−1

( ) C A = ( AC A ) , we can express

where m y = Am x + b . Since A

−1

−1 x

−1

( y − m ) ( A ) C A ( y − m ) = ( y − m ) ( AC A ) ( y − m ) T

−1

−1 x

−1

Letting C y = AC x A and noting that C y = A C x A = A C x , the joint pdf T

of the Gaussian random vector y can be expressed as f y ( y) =

( 2π )

n /2

1/2

T −1 ⎛ 1 ⎞ ⎜ − y −m y C y y −m y ⎟ ⎝ 2 ⎠

(

)

(

)

6.20 Let y = x1 + x2 + ......... + xn be the sum of n independent random variables. a. Calculate the characteristic function Φ y ( jω ) . Solution:

{

}

Φ y (u ) = E {e jyu } = E e j ( x1 + x2 +.........+ xn )u = Φ x1 , x2 ,...., xn ( x1u, x2u,...., xn u ) If x1 , x2 ,........, xn are independent random variables we can write

{

Φ y (u ) = E e j ( x1 + x2 +.........+ xn )u n

}

{ }

= ∏ E e jxiu = Φ x1 (u )Φ x2 (u )..........Φ xn (u ) i =1

b. If x1 , x2 ,........, xn are independent Gaussian random variables, show that y is a Gaussian random variable. Solution:

For xi ~ N (mi , σ i2 ) , the characteristic function of the random variable y is given by n

i =1

Φ y (u ) = ∏ Φ x1 (u ) = ∏ e

1 jmi u − σ i2u 2 2

ju ( mi + m2 +...+ mn ) −

(

)

1 2 2 σ1 +σ 2 +.....+σ n2 u 2 2

1 jm y u − σ 2y u 2 2

where m y = m1 + m2 + .... + mn

σ y2 = σ 12 + σ 22 + ....... + σ n2 Thus y is N (m y , σ y2 ) random variable.

(

)

6.21 A random signal is defined as x (t ) = At + B where A is N 0, σ A2 random variable. a. Find the first-order PDF of x (t ) . Solution:

The PDF of x is given from (6.71) as f x ( x) =

f A ( A) 1 ⎛ x−B⎞ = fA ⎜ ⎟ g '( A) A= g −1 ( x ) t ⎝ t ⎠ 2

= t

2πσ A2

⎛ x−B ⎞ 2 −⎜ ⎟ /2σ A ⎝ t ⎠

b. Find the mean and correlation functions, mx (t ) and Rx (t1 , t2 ) . Solution: mx (t ) = E { At + B} = E{ A}t + B = B

Rx (t1 , t2 ) = E { x (t1 ) x (t2 )} = E {( At1 + B )( At2 + B )} = E { A2t1t2 + B 2 + At2 B + At1 B}

= t1t2 E { A2 } + t1 BE { A} + t2 BE { A} + B 2 = t1t2σ A2 + B 2 6.22

Let x (t ) be a zero-mean Gaussian random process with autocorrelation function − t −t

Rx (t1 , t2 ) = 2e 2 1 . Determine the joint PDF of x (t ) and x (t + τ ) .

Solution:

Let x1 = x (t ) and x2 = x (t + τ ) . The joint PDF of x1 and x2 is given from (6.169) as f x ( x) =

( 2π )

n /2

1/2

T −1 ⎛ 1 ⎞ ⎜ − ( x −m) C ( x −m) ⎟ ⎝ 2 ⎠

where ⎡ m ⎤ ⎡ E { x ( t1 )} ⎤ ⎥=0 m=⎢ 1⎥=⎢ ⎣ m2 ⎦ ⎢⎣ E { x ( t2 )}⎥⎦ and ⎡Var ( x1 ) COV ( x1 x2 ) ⎤ ⎡ 2 C=⎢ ⎥ = ⎢ −τ COV Var x x x ( ) ( ) ⎥ ⎣⎢ 2e 2 1 2 ⎦ ⎣⎢

2e − τ ⎤ ⎥ 2 ⎦⎥

Now −1

C =

(

2 1 − e −2 τ

(

C = 4 1 − e −2 τ

)

⎡1 ⎢ −τ ⎣⎢ −e

−τ −e ⎤ ⎥ 1 ⎦⎥

Also 1 1 1 T ( x − m x ) C ( x − m x ) = xT C x = 2 2 4 1 − e −2 τ

(

4 1− e

−2 τ

)

⎛1 ( x1 x2 ) ⎜⎜ − τ ⎝ −e

⎞ ⎛ x1 ⎞ ⎟⎜ ⎟ 1 ⎟⎠ ⎝ x2 ⎠

−e

−τ

( x + x − 2x x e ) 2 1

−τ

2 1

1 2

Substituting yields f x1 x2 ( x1 , x2 ) =

4π

( e

−1

4 1− e

(1 − e ) −2 τ

−2 τ

)

( x +x − 2 x x e ) 2 1

2 1

−τ

1 2

6.23

Let y (t ) = x (t ) cos(2π f c t + θ ) where x (t ) is a WSS random process and θ is statistically independent random variable uniformly distributed in the interval [0, 2π]. a. Find the autocorrelation function R y (t , t + τ ) . Is y (t ) WSS? Solution:

E { y (t )} = E { x (t ) cos ( 2π f c t + θ )} = E { x (t )} E {cos ( 2π f c t + θ )} = 0 R y (t , t + τ ) = E { y (t ) y (t + τ )} = E { x (t ) cos ( 2π f c t + θ ) x (t + τ ) cos[2π f c (t + τ ) + θ ]} E { x (t ) x (t + τ )}

E {cos(2π f cτ ) + cos ( 4π f c t + 2π f cτ + 2θ )} 2 R (t , t + τ ) ⎡cos(2π f cτ ) + E {cos ( 4π f c t + 2π f cτ + 2θ )}⎤ = x ⎣ ⎦ 2 =

Since x (t ) is WSS, Rx (t , t + τ ) = Rx (τ ) . Also E {cos [ 4π f c t + 2π f cτ + 2θ ]} = ∫

∞

−∞

fθ (θ ) cos [ 4π f c t + 2π f cτ + 2θ ] dθ

1 2π cos [ 4π f c t + 2π f cτ + 2θ ]dθ = 0 2π ∫0

Substituting R y (t , t + τ ) =

1 Rx (τ ) cos(2π f cτ ) 2

y (t ) is WSS process because E { y (t )} = constant and R y (t , t + τ ) is a function of

τ only . b. Show that spectral density of y (t ) is related to that of x (t ) by

⎧1 ⎫ G y ( f ) = ℑ{ R y (τ )} = ℑ ⎨ Rx (τ ) cos(2π f cτ ) ⎬ ⎩2 ⎭ 1 = ℑ{ Rx (τ )} ⊗ ℑ{cos(2π f cτ )} 2 1 = Gx ( f ) ⊗ [δ ( f − f c ) + δ ( f + f c ) ] 4 1 = [Gx ( f − f c ) + Gx ( f + f c ) ] 4

6.24 Let x (t ) = A cos(2π f c t ) + B sin(2π f c t ) be a random process where A and B are statistically independent N (0, σ 2 ) random variables. a. Find the mean and correlation functions, mx (t ) and Rx (t1 , t2 ) . Solution:

E { x (t )} = E { A cos 2π f c t + B sin 2π f ct} = E { A} cos 2π f c t + E { B} sin 2π f ct = 0 Rx (t1 , t2 ) = E {( A cos 2π f c t1 + B sin 2π f c t1 )( A cos 2π f c t2 + B sin 2π f c t2 )} = E { A2 } cos 2π f c t1 cos 2π f c t2 + E { B 2 } sin 2π f c t1 sin 2π f c t2

+ E { AB} cos 2π f c t1 sin 2π f c t2 + E { AB} cos 2π f c t2 sin 2π f c t1 0

Rx (t1 , t2 ) = σ 2 ( cos 2π f ct1 cos 2π f c t2 + sin 2π f ct1 sin 2π f c t2 ) = σ 2 cos ( 2π f cτ ) Cov x ( t1 , t2 ) = Rx ( t1 , t2 ) = σ 2 cos ( 2π f cτ )

where τ = t2 − t1 b. Is x (t ) WSS? Solution:

Yes. c. Determine the first-order PDF of x (t ) . Solution:

The random variables x (t ) is Gaussian because it is linear combination of Gaussian random variables A and B. Its mean and variance are given by E { x (t )} = 0 Var ( x (t ) ) = Cov x ( t , t ) = Rx ( t , t ) = σ 2

The first-order PDF of x (t ) is

f x ( x, t ) =

2πσ 2

e − x /2σ 2

6.25 Consider the random processes x (t ) = A cos ( 2π f c t + θ )

and y (t ) = B sin ( 2π f c t + θ )

where θ is a uniformly distributed random variable in the range [−π, π]. a. Are x (t ) and y (t ) each WSS? Solution:

It is easy to show that x (t ) and y (t ) each WSS as illustrated in Example 6.29. b. Find the crosscorrelation function Rxy (t1 , t2 ) . Solution:

Rxy (t1 , t2 ) = E { AB cos ( 2π f c t1 + θ ) sin ( 2π f c t2 + θ )} =

{

}

AB E sin ⎡⎣ 2π f c ( t2 − t1 ) ⎤⎦ + sin ⎡⎣ 2π f c ( t2 + t1 ) + 2θ ⎤⎦ 2

Now

{

}

E sin ⎡⎣ 2π f c ( t2 − t1 ) ⎤⎦ = sin ⎡⎣ 2π f c ( t2 − t1 ) ⎤⎦

{

}

E sin ⎡⎣ 2π f c ( t2 + t1 ) + 2θ ⎤⎦ = 0 Therefore, Rxy (t1 , t2 ) =

AB sin ⎡⎣ 2π f c ( t2 − t1 ) ⎤⎦ 2

b. Are the random processes jointly WSS? Solution:

Since x (t ) and y (t ) are each WSS and their cross-correlation is invariant under the shift of time origin, that is, Rxy (t , t + τ ) = Rxy (τ ) , they are jointly WSS. c. Are the random processes uncorrelated? Solution:

The random processes x (t ) and y (t ) are correlated because E { x (t ) y (t + τ )} =

AB sin ( 2π f cτ ) ≠ E { x (t )} E { y (t + τ )} = 0 for all τ 2 0

6.26 The input to an LTI system with the transfer function ⎧10e − j 2π fto , 0 ≤ f ≤ 4 ×103 H( f ) = ⎨ otherwise ⎩0,

⎛ f ⎞ . Let is a WSS process x (t ) with power spectral density Gx ( f ) = 10−8 Π ⎜ 5 ⎟ ⎝ 5 × 10 ⎠ y (t ) denote the output of the system. a. Find the output spectra density G y ( f ) . Solution:

G y ( f ) = H ( f ) Gx ( f ) 2

⎛ f ⎞ ⎛ f ⎞ = 100 × 10−8 Π ⎜ Π⎜ 3 ⎟ 5 ⎟ ⎝ 8 ×10 ⎠ ⎝ 5 ×10 ⎠ ⎛ f ⎞ = 10−6 Π ⎜ 3 ⎟ ⎝ 8 × 10 ⎠ ⎧10−6 , Gy ( f ) = ⎨ ⎩0,

0 ≤ f ≤ 4 × 103 otherwise

b. Determine the output autocorrelation function R y (τ ) . Solution:

Using the FT pair

ℑ → sinc(2Wt ) ←⎯

1 Π ( f / 2W ) , 2W

⎧ 8 ×10−3 ⎛ f ⎞ ⎫ Π⎜ = 8 × 10−3 × sinc(8 × 103τ ) R y (τ ) = ℑ−1 {G y ( f )} = ℑ−1 ⎨ 3 3 ⎟⎬ ⎝ 8 × 10 ⎠ ⎭ ⎩ 8 × 10

{

}

c. Find E y 2 (t ) . Solution:

{

}

E y 2 (t ) = R y (0) = 8 × 103 6.27

Let x (t ) be a white Gaussian noise process with power spectral density Gx ( f ) = N o / 2 . It is input to a first-order Butterworth filter with 3-dB cutoff frequency f3dB . The transfer function is given by H( f ) =

1 1 + j ( f / f 3dB )

a. Find the output spectra density G y ( f ) and autocorrelation function R y (τ ) . Solution:

G y ( f ) = H ( f ) Gx ( f ) 2

N N 1 1 = o = o 2 1 + j ( f / f 3dB ) 2 1 + ( f / f3dB )2 Using the FT pair e

−α t

ℑ ←⎯ →

2α , α + (2π f ) 2 2

⎧⎪ N ⎫⎪ 1 R y (τ ) = ℑ−1 {G y ( f )} = ℑ−1 ⎨ o 2⎬ ⎪⎩ 2 1 + ( f / f 3dB ) ⎪⎭ ⎧⎪ N π f ⎫⎪ 4π f3dB = ℑ−1 ⎨ o 3dB 2⎬ 2 2 4π 2 f3dB + ( 2π f ) ⎪⎭ ⎪⎩ Nπf = o 3dB e −2π f3dB τ 2

b. Determine the average output power. Solution:

{

}

E y 2 (t ) = R y (0) =

N oπ f3dB 2

c. Find the first-order PDF of y (t ) . Solution:

The random variables y (t ) is Gaussian with mean and variance are given by ∞

∞

E { y (t )} = ∫ E { x (t − u )} h(u )du = mx ∫ h(u )du = 0 −∞

Var ( y (t ) ) = Cov y ( t , t ) = R y ( t , t ) =

−∞

N oπ f3dB 2

The first-order PDF of y (t ) is f y ( y, t ) =

1 N oπ f3dB 2

e − x / Noπ f3dB 2

d. Calculate the cross spectral density Gxy ( f ) ? Solution:

Using (6.234), the cross-spectral density Gxy ( f ) is given by Gxy ( f ) = H ( f ) Gx ( f ) =

No 1 2 1 + j ( f / f3dB )

6.28 x (t ) is a WSS random signal with power spectral density Gx ( f ) . The signal passes through the system shown in Figure P6.3. a. Is y (t ) WSS? Solution:

The mean and autocorrelation function of output random process y(t) are

∞

−∞

m y (t ) = ∫ E { x (t − u )} h(u )du = mx ∫ h(u )du R y (τ ) = h(τ ) ⊗ h(−τ ) ⊗ Rx (τ ) Since the mean of y (t ) does not vary with the choice of t, and its autocorrelation function depends on the time difference τ only, the response of an LTI system to a wide-sense stationary random process is also wide-sense stationary. b. Find the output spectra density G y ( f ) . Solution:

The transfer function of the system is given by H ( f ) = 1 − e − j 2π fT The output spectra density G y ( f ) is related to the input spectra density by G y ( f ) = 1 − e − j 2π fT 6.29

Gx ( f ) = 2Gx ( f ) ⎡⎣1 − cos ( 2π fT ) ⎤⎦ = 4Gx ( f ) sin 2 (π fT )

Let x (t ) be WGN process with power spectral density N o / 2 . It is applied to an ideal LP filter of bandwidth B and transfer function

⎛ f ⎞ H( f ) = Π⎜ ⎟ ⎝ 2B ⎠ a. Find the output autocorrelation function R y (τ ) . Solution:

G y ( f ) = H ( f ) Gx ( f ) 2

No ⎛ f ⎞ Π⎜ ⎟ 2 ⎝ 2B ⎠

ℑ Using the FT pair sinc(2 Bt ) ←⎯ →

function R y (τ ) is given by

1 Π ( f / 2 B) , the output autocorrelation 2B

R y (τ ) = N o Bsinc(2 Bτ )

b. What is the maximum rate at which the filter output y (t ) can be sampled so that the resulting samples are independent? Solution:

R y (τ ) = 0 for τ =

n , n = ±1, ±2,....... 2B

1 . As a result, the 2B maximum rate at which samples of y (t ) are independent is 2B samples/second. The smallest value of τ for which R y (τ ) = 0 is

6.30 The autocorrelation function of a stationary process x (t ) is given by Rx (τ ) = e−πτ

a. Find the power spectral density Gx ( f ) . Solution:

{ }

Gx ( f ) = ℑ{ Rx (τ )} = ℑ e −πτ

ℑ From Table 2.2, e−πτ ←⎯ → e −π f 2

G x ( f ) = e −π f

{

}

b. Find the value of E x 2 (t ) . Solution:

E { x 2 (t )} = Rx (0) = 1 c. Determine the RMS and 3-dB bandwidth of the process. Solution:

The 3-dB bandwidth of the random process is given by solving Gx ( f3dB ) = 0.5Gx ( f 0 )

0.5 = e−π f

−π f 2 = log e ( 0.5 ) ⇒ f =

log e ( 2 )

The RMS bandwidth of the random process is ∞

2 −π f ∫ f e df 2

f RMS =

−∞ ∞

∫e

−π f 2

1/ 2π 1 . = 1 2π

−∞

6.31 Suppose that narrowband noise x (t ) is bandlimited to 2B Hz. Show that the crossspectral density of quadrature components xc (t ) and xs (t ) is given by ⎧⎪ j[Gx ( f − f c ) − Gx ( f + f c )], Gxc xs ( f ) = −Gxs xc ( f ) = ⎨ ⎪⎩0, Solution:

f ≤ fc otherwise

Rxc xs (τ ) = E { xc (t ) xs (t + τ )}

{

}

= E ⎡⎣ x (t ) cos ( 2π f c t ) + xˆ (t ) sin ( 2π f c t ) ⎤⎦ ⎡⎣ xˆ (t + τ ) cos ( 2π f c (t + τ ) ) − x (t + τ ) sin ( 2π f c (t + τ ) ) ⎤⎦ = E { x (t ) xˆ (t + τ ) cos ( 2π f c t ) cos ( 2π f c (t + τ ) )} − E { x (t ) x (t + τ ) cos ( 2π f c t ) sin ( 2π f c (t + τ ) )}

− E { xˆ (t ) x (t + τ ) sin ( 2π f c t ) sin ( 2π f c (t + τ ) )} + E { xˆ (t ) xˆ (t + τ ) sin ( 2π f c t ) cos ( 2π f c (t + τ ) )} Rxc xs (τ ) = Rxxˆ (τ ) cos ( 2π f c t ) cos ( 2π f c (t + τ ) ) − Rx (τ ) cos ( 2π f ct ) sin ( 2π f c (t + τ ) ) − Rxx ˆ (τ ) sin ( 2π f c t ) sin ( 2π f c (t + τ ) ) + Rxˆ (τ ) sin ( 2π f c t ) cos ( 2π f c (t + τ ) ) ˆ From Section 6.12, we apply Rxxˆ (τ ) = Rˆ x (τ ) , Rxx ˆ (τ ) = − Rx (τ ) , and Rxˆ (τ ) = Rx (τ ) to obtain

Rxc xs (τ ) = Rˆ x (τ ) cos ( 2π f c t ) cos ( 2π f c (t + τ ) ) − Rx (τ ) cos ( 2π f c t ) sin ( 2π f c (t + τ ) ) +Rˆ x (τ ) sin ( 2π f c t ) sin ( 2π f c (t + τ ) ) + Rx (τ ) sin ( 2π f c t ) cos ( 2π f c (t + τ ) ) = Rx (τ ) ⎡⎣ − cos ( 2π f ct ) sin ( 2π f c (t + τ ) ) + sin ( 2π f c t ) cos ( 2π f c (t + τ ) ) ⎤⎦ + Rˆ x (τ ) ⎡⎣cos ( 2π f c t ) cos ( 2π f c (t + τ ) ) − sin ( 2π f c t ) sin ( 2π f c (t + τ ) ) ⎤⎦ = − Rx (τ ) sin ( 2π f cτ ) + Rˆ x (τ ) cos ( 2π f cτ )

The spectral density Gxc xs ( f ) is now obtained by taking the FT and applying ℑ Rx (τ ) ←⎯ → Gx ( f ) ℑ Rˆ (τ ) ←⎯ → − j sgn ( f ) G ( f ) x

Gxc xs ( f ) = −Gx ( f ) ⊗

[δ ( f − f c ) − δ ( f + f c )] [δ ( f − f c ) + δ ( f + f c )] − j sgn ( f ) Gx ( f ) ⊗ 2j 2

⎡sgn ( f − f c ) Gx ( f − f c ) + sgn ( f + f c ) Gx ( f + f c ) ⎤⎦ [Gx ( f − f c ) − Gx ( f + f c )] − j⎣ 2 2 j j = ⎡⎣Gx ( f − f c ) − sgn ( f − f c ) Gx ( f − f c ) ⎤⎦ − ⎡⎣Gx ( f + f c ) + sgn ( f + f c ) Gx ( f + f c ) ⎤⎦ 2 2 j j = Gx ( f − f c ) ⎡⎣1 − sgn ( f − f c ) ⎤⎦ − Gx ( f + f c ) ⎡⎣1 + sgn ( f + f c ) ⎤⎦ 2 2 = j

This expression can be simplified to ⎪⎧ j[Gx ( f − f c ) − Gx ( f + f c )], Gxc xs ( f ) = −Gxs xc ( f ) = ⎨ ⎪⎩0,

f ≤ fc otherwise

6.32 Consider the block diagram of a receiver shown in Figure P6.4. Find the overall noise figure of the receiver. Figure P6.4 RF Receiver cascade Tantenna = 50o K TLNA = 50o K

Mixer Gain = 20dB Loss = 4.35 dB Post Amp

LNA Line Loss 0.2 dB

Line Loss 0.2 dB LO

Demodulator NF = 3 dB

TA = 50o K Gain = 35 dB

Solution:

The following table summarizes gains and noise factor values required to calculate the overall noise figure.

Line Loss LNA Line Loss LO

Post Amp Demodulator

Gain G (dB)

−0.2

−4.35

Gain G

0.955

100

0.955

0.3673

3162.3

Accumulated Gain G1G2 ......Gn −1

0.955

95.5

91.2

33.5

10.6×104

NF (dB)

0.2

4.35

1.047

2.728

13.63

501.12

Te = Te1 +

Te2 G1

Te3 G1G2

Te4 G1G2 G3

Te5 G1G2 G3G4

290

Te6 G1G2 G3G4 G5

50 13.63 501.12 50 290 + + + + 0.955 95.5 91.2 33.5 10.6 ×104 = 13.63 + 52.356 + 0.143 + 5.495 + 1.49 + 0.0027 = 73.12o K

= 13.63 +

This corresponds to FT = 1 +

Te 73.12 = 1+ = 1.252 290 To

Chapter 7 7.1

Consider the transmission of a 5 kHz message signal on a baseband communications channel with a power attenuation of 30 dB. The channel noise is AWGN with N o / 2 = 10−10 Watts/Hz. Calculate the minimum transmitted power PT to achieve a baseband SNRBB of at least 40 dB. Solution: SNRBB =

PR No B

PR 2 × 10 × 5 ×103 PR = 104 ×10 × 10−7 = 1× 10−2 W = 10 dBm

104 =

−10

PT = PR + Loss = 10 + 30 = 40 dBm = 10 W

7.2 The message signal s (t ) = 10 cos(1000π t ) + 5cos(2000π t ) modulates the carrier signal c(t ) = 10−3 cos(100 ×103 π t ) using DSB-SC AM scheme. White noise with power spectral density N o / 2 = 10−12 W/Hz is added during transmission. The received signal is coherently demodulated after ideal BP filtering.

a. Determine the receiver input CNR. Solution:

The receiver input CNR is given from (7.13) as CNRIN =

α 2 Ac2 s 2 2 No B

Now 102 52 125 s = + = = 62.5 2 2 2 2

Ac2 = 10−6 , α = 1, B = 1 kHz N o = 2 × 10−12 Substituting

CNRIN =

10−6 × 62.5 103 × 62.5 = = 15.625 ×103 = 41.93 dB −12 3 4 × 10 × 10 4

b. Determine the output SNR. Solution:

SNRDSB = CNRIN = 41.93 dB c. What degradation in postdetection SNR occurs if instead a 4th order Butterworth filter with 3-dB bandwidth of 1.5 kHz is used as a post-detection filter. (Hint: ratio of noise-equivalent bandwidth to 3-dB bandwidth BN / f3dB = 1.03 for a 4th order Butterworth filter). Solution:

The noise power at the postdetection filter output equals Pnc = 2 N o BN . The rolloff of the 4th order Butterworth filter at 1 kHz is obtained from (2.159) as 1/ 1 + (1/1.5 ) = 0.981 , i.e., about 2%. Therefore, 8

(α Ac ) s = 10−6 × 62.5 × 0.982 = 10 ×103 = 40 dB P SNRDSB = D = Pnc 2 N o BN 4 × 10−12 × 1.5 × 103 2

There is approximately 2 dB degradation in SNR using a 4th order Butterworth filter with 3-dB bandwidth of 1.5 kHz. 7.3 A message signal s (t ) with spectral density Gs ( f ) shown in Figure P7.1(a). It modulates the carrier signal c(t ) = 10−3 cos(2π fct ) using DSB-SC AM scheme. White noise with power spectral density N o / 2 = 10−12 W/Hz is added during transmission. The received signal is coherently demodulated. a. Assuming an ideal pre-detection BPF, determine the CNR at its output. Solution:

The ideal predetection filter has a bandwidth BT = 2 B for DSB-SC AM scheme. Therefore, the noise power, measured in the ideal predetection filter bandwidth is 2 N o B . The CNR, measured at the output of the predetection filter, is given by

CNRPD =

α 2 Ac2 s 2 Power in the input carrier signal = Input noise power in the predetection filter bandwidth 4No B

Figure P7.1

Gs ( f ) 1

(a) −5

f (kHz)

H PD ( f ) 10 kHz

(b)

f c − 7.5

f c + 7.5

f (kHz)

H LP ( f )

−5

−7.5

7.5

f (kHz)

From Figure P7.1 (a), we have ∞

s = ∫ Gs ( f )df = 5 2

−∞

Ac2 = 10−6 , α = 1, B = 5 kHz

N o = 2 × 10−12 Substituting 10−6 × 5 103 × 5 = = 125 = 21 dB CNRPD = 8 × 10−12 × 5 ×103 8× 5 b. Determine the output SNR assuming an ideal post-detection filter.

Solution:

SNRDSB = CNRIN =

PR = 21 + 3 = 24 dB No B

c. Modify your calculations (a) and (b) for a pre-detection filter with amplitude response shown in Figure P7.1(b). Solution:

The CNR, measured at the output of the predetection filter with amplitude response shown in Figure P7.1(b), is given by

α 2 Ac2 s 2 Power in the input carrier signal CNRPD = = Input noise power in the predetection filter bandwidth 4 N o BN where BN is noise-equivalent bandwidth of the predetection filter ∞

∫ H ( f ) df 2

BN = 0

H PD ( f ) max

2.5 ⎞ ⎛ = 2 ⎜ 5000 + 1000 × ⎟ = 11.667 kHz 3 ⎠ ⎝

Substituting 10−6 × 5 103 × 5 = = 53.57 = 17.3 dB CNRPD = 8 × 10−12 ×11.667 × 103 8 ×11.667

Although the spectral density of the in-phase noise nc (t ) extends up to 7.5 KHz, the demodulator output noise is limited by the ideal LPF to 5 kHz. Thus, the performance of the system is not negatively impacted by the non-ideal predetection filter. Therefore, SNRDSB =

PR = 24 dB No B

d. How is the performance in (c) modified if a non-ideal post-detection filter with amplitude responses in Figure P7.1(c) is used instead. Solution:

From Figure P7.1(c), the noise-equivalent bandwidth of the post-detection filter BN = 5.8335 kHz. Now

α Ac ) s 2 ( = = 2

SNRDSB

2 N o BN

10−6 × 5 = 214.28 = 23.3 dB 4 × 10−12 × 5.8335 ×103

7.4 The message signal s (t ) = 5cos ( 700π t ) + 2 cos ( 7000π t ) modulates the carrier signal c(t ) = cos(2π f ct ) using SSB-AM AM scheme. White noise with power spectral density N o / 2 = 10−12 W/Hz is added during transmission. The channel attenuates the transmitted signal by 60 dB. The received signal is coherently demodulated. a. Assuming an ideal pre-detection BPF, determine the CNR at its output. Solution:

The CNR at the ideal pre-detection BPF output is given by CNRPD =

α 2 Ac2 s 2 PR = 4 No B No B

Now 52 22 29 = 14.5 s = + = 2 2 2 2

Ac2 = 1, α = 10−3 , B = 3.5 kHz N o = 2 × 10−12 Substituting CNRPD =

10−6 × 14.5 103 × 14.5 = = 5.178 ×102 = 27.14 dB 8 × 10−12 × 3.5 ×103 8 × 3.5

b. Determine the output SNR. SNRSSB = CNRIN = CNRPD = 27.14 dB 7.5 A conventional AM system has a zero-mean Gaussian message signal whose spectrum is limited to B Hz. The peak value of the message signal is assumed to be

2.5σ m where σ m is the standard deviation of its amplitude variations. Determine the output SNR in terms of CNRIN . Assume ma = 0.875 . Solution:

SNRAM = η CNRIN where

η=

ma2 sn2 (1 + ma2 sn2 )

s (t ) can be expressed max s (t ) in terms the average and peak power of the message signal as

The power in the normalized message signal sn (t ) =

s2 s = Pmax 2 n

where Pmax is power corresponding to the peak value ( max s (t ) ) of the message

signal. For zero-mean Gaussian message signal, sn2 =

σ m2 1 = 2 ( 2.5σ m ) 6.25

Substituting

( 0.875) (1/ 6.25) = 0.109 η= = (1 + ma2 sn2 ) 1 + ( 0.875)2 (1/ 6.25) ma2 sn2

Therefore, SNRAM = 0.109 × CNRIN 7.6 The message signal s (t ) = 1.0 cos ( 600π t ) + 0.5cos ( 6000π t ) modulates the carrier signal c(t ) = 2cos(100 ×103 π t ) using conventional AM scheme. The modulator operates with a modulation index ma = 0.85 .We assume that the channel adds white Gaussian noise with power spectral density N o / 2 = 10−8 W/Hz during transmission. The received signal is demodulated using an envelope detector.

a. Determine the receiver input CNR. Solution:

The receiver input CNR from (7.42) is

(

α 2 Ac2 1 + ma2 sn2 PR PR CNRIN = = = 2 No B Pni N o B

)

Now ma = 0.85 sn2 =

1 ⎛ 12 0.52 ⎞ 1.25 s2 = 2⎜ + = 0.278 ⎟= Pmax 1.5 ⎝ 2 2 ⎠ 2.25 × 2

Ac2 = 4, α = 1, B = 3 kHz N o = 2 × 10−8 CNRIN =

(

4 1 + 0.852 × 0.278 −8

4 × 10 × 3 ×10

) = 40.03 ×10 = 46 dB 3

b. Determine the output SNR. Solution:

SNRAM = η CNRIN

0.852 × 0.278 η= = = 0.167 2 (1 + ma2 sn2 ) 1 + 0.85 × 0.278 ma2 sn2

Therefore, SNRAM = η CNRIN = 0.167 × 40.03 × 103 = 6.685 × 103 = 38.25 dB c. What degradation in postdetection SNR occurs if instead a fourth-order Butterworth filter with 3-dB bandwidth of 4.5 kHz is used as a postdetection filter.

Solution:

The noise-equivalent bandwidth BN of the fourth-order Butterworth filter equals its 3-dB bandwidth, i.e., 4.5 kHz. The roll-off of the filter at 3 kHz is about 2%, which is negligible. Therefore, SNRAM =

7.7

α 2 Ac2 ma2 sn2 2 N o BN

4 × (0.85) 2 × 0.278 = 4463 = 36.5 dB 4 × 10−8 × 4.5 × 103

A conventional AM system transmits speech signal with bandwidth 3.3kHz and amplitude PDF given by f s ( s ) = 2.5e −5 s . The peak value of the message signal is assumed to be 3.5σ m , where σ m is standard deviation of its amplitude variations. The channel introduces white Gaussian noise with power spectral density N o / 2 = 10−10 W/Hz is added during transmission. Assume ma = 0.875 . a. Find the carrier power required so that the output SNR exceeds 45 dB. Solution:

ma = 0.875 , sn2 =

SNRAM =

σ m2 = 0.0816 , N o = 2 × 10−10 , α = 1 , B = 3.3 kHz 2 ( 3.5σ m )

α 2 Ac2 ma2 sn2 2 No B

Ac2 × (0.875) 2 × 0.0816 = = 104.5 3 −10 4 × 10 × 3.3 × 10

or Ac2 =

104.5 × 10−7 × 4 × 3.3 = 0.668 (0.875) 2 × 0.0816

Carrier power Pc = Ac2 / 2 = 0.668 = 28.25 dBm b. Determine the threshold value of carrier power. Solution:

The threshold in a conventional AM system occurs when CNRPD ≈ 10 dB. That is,

10 = CNRPD =

(

Ac2 1 + ma2 sn2 4 No B

) = A × (1 + (0.875) × 0.0816) 2 c

8 × 10−10 × 3.3 ×103

Ac2 =

10 × 10−7 × 8 × 3.3 26.4 × 10−6 = = 24.9 × 10−6 2 1 + (0.875) × 0.0816 1.06

Threshold value of carrier power Pc = Ac2 / 2 = 12.45 × 10−6 = −19.04 dBm 7.8

Consider a message signal with bandwidth B = 5 kHz, and a normalized power sn2 = 0.25 . It is required to transmit this signal via a channel with that has a power attenuation of 30 dB. The channel noise is AWGN with N o / 2 = 10−10 W/Hz. It is desirable to have an SNR of at least 40 dB at the receiver output. a. Find the required carrier power if DSB-SC modulation is used. Solution:

The output SNR for DSB-SC scheme is given by

SNRDSB =

PR No B

Substituting

104 =

PR ⇒ PR = 0.01 W = 10 dBm 2 ×10 × 5 ×103 −10

Now 1 PR = α 2 Ac2 s 2 = α 2 s 2 Pc 2 where Pc = Ac2 / 2 is the carrier power. Since the channel produces an attenuation of 30 dB, the carrier power Pc is Pc =

PR 0.01 = −3 = 40 W = 46 dBm 10 × 0.25 10 × 0.25 −3

b. Find the carrier power required if conventional AM with modulation index of ma = 0.75 is used. Solution:

For the conventional AM scheme, the output SNR is given by

SNRAM = η CNRIN = η

η=

ma2 sn2 (1 + ma2 sn2 )

104 = 0.1233

PR , where No B

0.752 × 0.25 = 0.1233 1 + 0.752 × 0.25

PR ⇒ PR = 0.081 W = 19.09 dBm 2 ×10 × 5 × 103 −10

Now PR =

(1 + m s ) = α (1 + m s ) P 2

α 2 Ac2

2 2 a n

where Pc = Ac2 / 2 is the carrier power. Since the channel produces an attenuation of 30 dB, the required carrier power Pc in the conventional AM case is given by Pc =

(

α 1+ m s 2

2 2 a n

)

0.081 = 71 W = 48.54 dBm 10 × (1 + 0.752 × 0.25 ) −3

c. With the carrier power calculated in part (b), what is the maximum SNR that one can obtain for conventional AM (assuming of course that we do not overmodulate the message signal)? Solution:

Assuming ma = 1.0 yields

η=

ma2 sn2 (1 + ma2 sn2 )

1× 0.25 = 0.2 1 + 1× 0.25

Therefore,

(

)

α 2 1 + ma2 sn2 Pc PR SNRAM = η CNRIN = 0.2 × = 0.2 × No B No B = 0.2 ×

10−3 × (1 + 0.25 ) × 710 2 × 10

−10

× 5 ×10

= 177.5 × 103 = 52.49 dB

7.9 Consider an FM system with peak frequency deviation ∆f max = 60 kHz , message signal bandwidth B = 10 kHz ,and a normalized power sn2 = 0.25 . The channel introduces AWGN with N o / 2 = 10−9 W/Hz. Assuming received power level PR = 250 mW , determine a. Output SNR. Solution:

∆f max 60 = = 6 , sn2 = 0.25 , PR = 250 mW , N o = 2 × 10−9 10 B

Substituting SNRFM = 3D 2 sn2

PR 250 × 10−3 = 3 × 62 × 0.25 No B 2 ×10−9 × 10 × 103

= 27 ×125 × 102 = 337.5 × 103 = 55.28 dB b. Required transmit power PT for an output SNR = 45 dB when the channel introduces attenuation of 30 dB. Solution:

PR 2 ×10 ×10 ×103 2 ×10−0.5 2 × 10−0.5 = = 23.42 × 10−3 W = 13.7 dBm PR = 3 × 62 × 0.25 27

104.5 = 3 × 62 × 0.25

−9

PT = 13.7 + 30 = 43.7 dBm 7.10

A PM system with transmission bandwidth BT = 140 kHz is used to transmit a message signal bandwidth B = 10 kHz, and normalized power sn2 = 0.25 . The channel introduces AWGN with N o / 2 = 10−9 W/Hz. Assuming received power level PR = 250 mW , determine a. Output SNR. Solution:

140 = 2(∆φmax + 1)10 or ∆φmax = 7 − 1 = 6 SNRPM = ( ∆φmax ) sn2CNRIN 2

where PR 250 × 10−3 = = 125 × 102 CNRIN = 4 −9 N o B 2 × 10 ×10 SNRPM = 62 × 0.25 × 125 × 102 = 112.5 × 103 = 50.51 dB b. Required transmit power PT for output SNR = 45 dB when the channel introduces power attenuation of 30 dB. Solution: PR 2 × 10 × 10 × 103 2 ×10−0.5 2 PR = 2 = = 70.27 × 10−3 W = 18.47 dBm 6 × 0.25 9 10

104.5 = 62 × 0.25

−9

PT = 18.47 + 30 = 48.47 dBm c. Repeat (a) and (b) if the peak phase deviation ∆φmax is limited to π radians. Calculate the required transmission bandwidth BT . Solution:

SNRPM = ( ∆φmax ) sn2CNRIN = π 2 × 0.25 ×125 ×102 = 30.84 × 103 = 44.9 dB 2

104.5 = π 2 × 0.25

PR 2 ×10 ×10 × 103 −9

2 × 10−0.5 = 256.32 × 10−3 W = 24.087 dBm 2 π × 0.25 PT = 24.087 + 30 = 54.087 dBm PR =

BT = 2(π + 1)10 × 103 = 82.83 kHz

7.11 A message signal with bandwidth B = 15 kHz, and normalized power sn2 = 0.25 is transmitted using an FM system. Assuming the deviation ratio D = 3, find the output SNR for input CNR values (i) 11 dB, (ii) 25 (iii) 35 dB. In which case is the system operating below threshold? Solution:

( CNRIN )th ≈ 20(3 + 1) = 80 = 19 dB For input CNR values of 25 and 35 dB, the output SNR is given by SNRFM = 3D 2 sn2CNRIN In dB form

(

)

SNRFM = 10 log10 3D 2 sn2 + CNRIN = 10 log10 ( 6.75 ) + CNRIN = 8.29 + CNRIN For CNRIN = 25 dB , SNRFM = 8.29 + 25 = 33.29 For CNRIN = 35 dB , SNRFM = 8.29 + 35 = 43.29 For input CNR of 11 dB, the system is operating below threshold and the output SNR is given by 3 2 β CNRIN 2 SNRFM = CNRIN − 12 β 2(1+ β ) CNRIN e 1+

Substituting values for β and CNRIN , we obtain SNRFM = 7.4033 dB 7.12 An RC filter with time constant 25 µsec is used for de-emphasis in an FM system. Find the output SNR improvement in an FM system broadcasting signals with bandwidth 15 kHz. How much output SNR is improved if the signal bandwidth is increased to 53 kHz. Solution:

f1 =

1 2πτ

106 = 6.366 ×103 2π × 25

With deemphasis filtering, the output noise power is given by 2

⎛K ⎞ ⎡B B⎤ Pno = 2 ⎜ FD ⎟ N o f13 ⎢ − tan −1 ⎥ f1 ⎦ ⎣ f1 ⎝ Ac ⎠ For B = 15 kHz, 2

⎛K ⎞ 15 ⎤ ⎡ 15 Pno = 2 ⎜ FD ⎟ N o f13 ⎢ − tan −1 6.366 ⎥⎦ ⎣ 6.366 ⎝ Ac ⎠ 2

⎛K ⎞ ⎛K ⎞ = 2 ⎜ FD ⎟ N o f13 [ 2.356 − 1.17 ] = 2.37 ⎜ FD ⎟ N o f13 ⎝ Ac ⎠ ⎝ Ac ⎠ For B = 53 kHz, 2

⎛K ⎞ 53 ⎤ ⎡ 53 Pno = 2 ⎜ FD ⎟ N o f13 ⎢ − tan −1 6.366 ⎥⎦ ⎣ 6.366 ⎝ Ac ⎠ 2

⎛K ⎞ ⎛K ⎞ = 2 ⎜ FD ⎟ N o f13 [8.325 − 1.45] = 13.7 ⎜ FD ⎟ N o f13 ⎝ Ac ⎠ ⎝ Ac ⎠ Thus the output SNR is improved by 10 log10

13.7 = 7.62 dB if the signal 2.37

bandwidth is increased to 53 kHz. 7.13 Consider a message signal with bandwidth B = 5 kHz and a normalized power sn2 = 0.15 . It is required to transmit this signal via a channel that attenuates the transmitted signal by 30 dB. The channel noise is AWGN with spectral density N o / 2 = 10−10 W/Hz. It is desirable to have an SNR of at least 60 dB at the receiver output. a. Find the minimum required transmit power PT for a PM system with ∆φmax = 10. Solution:

SNRPM = ( ∆φmax ) sn2 2

PR No B

PR 2 ×10 × 5 × 103 1 1 = = 66.67 × 10−3 W = 18.24 dBm PR = 2 100 × 0.15 15

106 = 102 × 0.15

−10

PT = PR + channel attenuation (dB) = 18.24 + 30 = 48.24 dBm = 66.68 Watts b. Find the minimum required transmit power PT for an FM system with D = 10. Solution:

SNRFM = 3D 2 sn2

PR No B

PR 2 × 10 × 5 × 103 1 1 = = 22.22 ×10−3 W = 13.467 dBm PR = 3 ×100 × 0.15 45

106 = 3 × 102 × 0.15

−10

PT = PR + channel attenuation (dB) = 13.467 + 30 = 43.467 dBm = 22.22 Watts c. Find the minimum required transmit power PT for an FM system utilizing preemphasis and de-emphasis with D = 10 and f1 = 600 Hz . Solution: 2

⎛B⎞ P ( SNRFM ) DE = D ⎜ ⎟ sn2 R No B ⎝ f1 ⎠ 2

PR 2 ×10 × 5 × 103 1 1 PR = = = 0.96 × 10−3 W = −0.177 dBm 100 × 69.44 × 0.15 15 × 69.44

106 = 102 × 8.332 × 0.15

−10

PT = PR + channel attenuation (dB) = −0.177 + 30 = 29.82 dBm = 0.96 Watts 7.14 In a certain FM satellite communication system, the output SNR is found to be 40 dB with D = 6. The modulating signal has a bandwidth B = 10 kHz and a normalized power sn2 = 0.15 . The system with D = 6 is not in threshold, but the output SNR is required to be at least 50 dB. The increase of the output SNR can be accomplished by increasing either (i) D (that is, the transmission bandwidth) and/or (ii) the transmitted power.

a. If option (i) is selected, what are the maximum values of D and the corresponding transmission bandwidth that can be used without running into threshold? What is the corresponding output SNR? Solution:

For the output SNR = 40 dB and D = 6, the input CNRIN is given by CNRIN =

SNRFM 3D 2 sn2

104 = 617.3 (27.9 dB) 3 × 62 × 0.15

Next we determine the maximum value of D using the threshold condition (7.103).

( SNRFM )th = 105 = 60 D 2 ( D + 1) sn2 = 9 D 2 ( D + 1) Solving by trial and error or using MATLAB yields the maximum value of D ≈ 19.43. Transmission bandwidth BT corresponding to deviation ratio D ≈ 19.43 = 2( D + 1) B = 2(19.43 + 1)104 = 408.6 kHz The minimum input ( CNRIN )th required to operate above the threshold is

( CNRIN )th = 20( D + 1) = 408.6 = 26.113 dB For CNRIN = 27.9 dB , the system is not in the threshold and the output SNR is given by SNRFM = 3D 2 sn2CNRIN = 3 × 19.432 × 0.15 × 617.3 = 104.87 ×103 = 50.2 dB b. If option (ii) is chosen instead, calculate the minimum increase in transmitted power required to attain an output SNR of 50 dB subject to the constraint that the channel bandwidth is limited to 250 kHz? What is the corresponding value of D? Solution:

If the available channel bandwidth is limited to 250 kHz, the maximum value of D is given by

250 × 103 = 2( D + 1) × 104 or D=

250 × 103 − 1 = 11.5 2 × 104

CNRIN =

SNRFM 3D 2 sn2

105 = 1.68 × 103 (32.25 dB) 3 × 11.52 × 0.15

Increase in transmitted power required = 32.25 − 26.113 = 6.1375 dB

7.15 The baseband frequency spectrum for stereo FM broadcasting is depicted in Figure 5.41. a. Calculate the noise output power without pre-emphasis. Solution:

We refer to block diagram of the FM stereo receiver in Figure 5.42. L+R channel

The L+R channel occupies the frequency band [0 15] kHz. From (7.88), the power spectral density of the noise n1 (t ) at the FM demodulator output is ⎧⎛ K ⎞2 ⎪ FD f 2 N o , Gn1 = GnL+R = ⎨⎜⎝ α Ac ⎟⎠ ⎪ ⎩0,

0 ≤ f ≤ 15 kHz otherwise

The noise power at the FM demodulator output for the L+R channel is 2

⎛K ⎞ 2⎛ K ⎞ Pn1 = ⎜ FD ⎟ N o B 3 = ⎜ FD ⎟ N o × 2.25 × 1012 3 ⎝ α Ac ⎠ ⎝ α Ac ⎠ L-R channel

The L−R channel occupies the frequency band [23 53] kHz. From (7.88), the power spectral density of the noise at the FM demodulator output is

⎧⎛ K ⎞ 2 ⎪⎜ FD ⎟ f 2 N o , GnL−R = ⎨⎝ α Ac ⎠ ⎪ ⎩0,

23 ≤ f ≤ 53 kHz otherwise

The power spectral density of the noise n2 (t ) at the DSB-AM demodulator output is Gn2 = GnL−R ( f − 2 f P ) + GnL−R ( f + 2 f P ) 2 ⎧⎛ K ⎞ 2 ⎛ K FD ⎞ 2 2 2 2 FD ⎪ ( f − 2 f P ) + ( f + 2 f P ) No = ⎜ ⎟ f + 4 f P 2 No , − B ≤ f ≤ B = ⎨⎜⎝ α Ac ⎟⎠ α A ⎝ c⎠ ⎪ otherwise ⎩0,

(

)

where f P = 19 kHz is pilot tone frequency in stereo FM signal. The noise power at the output of the DSB-AM demodulator (L−R channel path) is given by 2

⎛K ⎞ B Pn2 = 2 N o ⎜ FD ⎟ ∫ f 2 + 4 f P2 df ⎝ α Ac ⎠ − B

(

)

B ⎤ ⎛K ⎞ ⎡B = 2 N o ⎜ FD ⎟ ⎢ ∫ f 2 df + 4 ∫ f P2 df ⎥ ⎝ α Ac ⎠ ⎣ − B −B ⎦ 2

⎛ K ⎞ ⎡ 2B3 ⎤ = 2 N o ⎜ FD ⎟ ⎢ + 8 f P2 B ⎥ ⎦ ⎝ α Ac ⎠ ⎣ 3 2

⎛K ⎞ = N o ⎜ FD ⎟ × 91.14 × 1012 ⎝ α Ac ⎠ b. Calculate the same for monophonic transmission (the baseband spectral range for mono FM signal is 15 kHz). Solution:

The noise output power at the FM demodulator output for a mono FM system is identical to that of the L+R channel. That is, 2

⎛K ⎞ 2⎛ K ⎞ PnFM −mono = ⎜ FD ⎟ N o B 3 = ⎜ FD ⎟ N o × 2.25 × 1012 3 ⎝ α Ac ⎠ ⎝ α Ac ⎠ c. How much noisier is stereo FM versus mono FM in dBs?

Solution:

The output of the upper branch after matrixing is given by L ( t ) + R ( t ) + L ( t ) − R ( t ) + n1 (t ) + n2 (t ) ≈ 2 L ( t ) + n2 (t )

Similarly, the output of the lower branch after matrixing is given by L ( t ) + R ( t ) − L ( t ) + R ( t ) + n1 (t ) − n2 (t ) ≈ 2 L ( t ) + n2 (t )

Therefore, the noise output power for both L+R and L−R channels after matrixing is given by 2

⎛K ⎞ Pno = Pn1 + Pn2 = ⎜ FD ⎟ N o × 93.39 × 1012 stereo ⎝ Ac ⎠

( )

Thus,

(P ) 10 log (P )

⎛ 93.39 ×1012 ⎞ = 10 log10 ⎜ ⎟ = 16.18 dB 2.25 × 1012 ⎠ ⎝ no mono

stereo

We conclude that stereo audio is 16.18 dB noisier than mono audio without deemphasis filtering. d. Repeat (a) through (c) taking 75 µsec pre-emphasis and de-emphasis into consideration. Show that stereo FM is 22 dB noisier than mono FM. Solution:

f1 =

1 2πτ

106 = 2.122 ×103 2π × 75

Mono audio with deemphasis

For mono audio case, the output noise power of the after deemphasis filtering is given by

⎛K ⎞ ⎛K ⎞ ⎡B B⎤ 15 ⎤ ⎡ 15 = 2 ⎜ FD ⎟ N o f13 ⎢ − tan −1 ⎥ = 2 ⎜ FD ⎟ N o f13 ⎢ − tan −1 Pno mono f1 ⎦ 2.122 ⎥⎦ ⎣ 2.122 ⎣ f1 ⎝ Ac ⎠ ⎝ Ac ⎠

( )

3 ⎛K ⎞ ⎛K ⎞ = 11.28 ⎜ FD ⎟ N o f13 = 11.28 ⎜ FD ⎟ N o 2.122 × 103 ⎝ Ac ⎠ ⎝ Ac ⎠

(

)

Stereo audio with deemphasis

For stereo audio case, the output noise power at the output of the L+R channel branch after deemphasis filtering is identical to that in the mono case. That is, 2

3 ⎛K ⎞ Pn1 = 11.28 ⎜ FD ⎟ N o 2.122 ×103 ⎝ Ac ⎠

(

)

For the L−R channel, the power spectral density of the noise n2 (t ) at the DSBAM modulator output after deemphasis filtering is given by 2 ⎧ 2 ⎛ K FD ⎞ 2 2 ⎪H (f) ⎜ ⎟ f + 4 f p 2 No , Gn2 = ⎨ DE ⎝ α Ac ⎠ ⎪ ⎩0,

(

)

− B ≤ f ≤ B kHz otherwise

The output noise power at the output of the L−R channel branch after deemphasis filtering is now obtained as

(

)

2 2 ⎛ K FD ⎞ B f + 4 f p Pn2 = 2 N o ⎜ df ⎟ ∫ 2 ⎝ α Ac ⎠ − B 1 + ( f / f1 ) 2

2 B ⎤ f p2 ⎛ K FD ⎞ ⎡ B f2 df df 4 = 2 No ⎜ + ⎥ ⎟ ⎢∫ 2 2 ∫ ⎥⎦ ⎝ α Ac ⎠ ⎢⎣ − B 1 + ( f / f1 ) − B 1 + ( f / f1 ) 2

⎛K ⎞ ⎡ ⎛B B⎞ B⎤ = 4 N o ⎜ FD ⎟ ⎢ f13 ⎜ − tan −1 ⎟ + 4 f p2 f1 tan −1 ⎥ f1 ⎠ f1 ⎦ ⎝ α Ac ⎠ ⎣ ⎝ f1 2

⎛K ⎞ ⎡ 15 ⎞ 3 ⎛ 15 2 −1 15 ⎤ × 109 = 4 N o ⎜ FD ⎟ ⎢( 2.122 ) ⎜ − tan −1 ⎟ + ( 38 ) × 2.122 × tan ⎥ 2.122 ⎠ 2.122 ⎦ ⎝ 2.122 ⎝ Ac ⎠ ⎣ 2

⎛K ⎞ = ⎜ FD ⎟ N o ×17.55 × 1012 ⎝ Ac ⎠

We observe that Pn2 Pn1 . Therefore, the noise output power for both L+R and L−R channels after matrixing is given by 2

⎛K ⎞ Pno = Pn1 + Pn2 ≈ Pn2 = ⎜ FD ⎟ N o × 17.55 × 1012 stereo ⎝ Ac ⎠

( )

Thus,

(P ) 10 log (P )

⎛ ⎞ 17.55 ×1012 ⎜ ⎟ = 22.1 dB = 10 log10 3 3 ⎜ ⎟ no mono ⎝ 11.28 × 2.122 ×10 ⎠

stereo

(

)

We conclude that stereo audio is 22.1 dB noisier than mono audio when deemphasis filtering is applied in both cases. 7.16 In FDM telephony system, 12 voice channels are frequency-division-multiplexed to form a group that occupies a frequency band from 60 to 108 kHz. Each 3.3-kHz voice signal is SSB-AM modulated and assigned a 4 kHz frequency slot to provide 0.7 kHz guardband between channels. The frequency-division-multiplexed group signal is then frequency modulated with a peak deviation ∆f max of 400 kHz. a. Determine the transmission bandwidth of the FM signal. Solution:

∆f max 400 ×103 = = 3.7 B 108 × 103

Transmission bandwidth BT = 2( D + 1) B = 2(3.7 + 1) × 108 × 103 = 1.0152 MHz . b. Determine the degradation in output SNR of the 12th channel compared with the first. Solution:

The 1st channel occupies the frequency band [60 64] kHz. From (7.88), the power spectral density of the noise at the FM demodulator output is ⎧⎛ K ⎞ 2 ⎪ FD f 2 N o , GnFM ( f ) = ⎨⎜⎝ α Ac ⎟⎠ ⎪ ⎩0,

f ≤ BT / 2 otherwise

The noise power at the output of the 1st channel USB-AM demodulator is given by 2

⎛K ⎞ P = 2 ∫ GnFM ( f )df = 2 ∫ ⎜ FD ⎟ f 2 N o df α Ac ⎠ 60 60 ⎝ 64

1 n

⎛ K FD ⎞ 2 × 109 ⎛ K FD ⎞ 3 3 12 = ×⎜ ⎟ N o ⎡⎣64 − 60 ⎤⎦ = ⎜ ⎟ N o × 30.76 × 10 3 ⎝ α Ac ⎠ ⎝ α Ac ⎠ The 12th channel occupies the frequency band [104 108] Hz. The noise power at the output of the 12th channel USB-AM demodulator is given by 2

⎛K ⎞ P = 2 ∫ GnFM ( f )df = 2 ∫ ⎜ FD ⎟ f 2 N o df α Ac ⎠ 104 104 ⎝ 108

108

12 n

⎛ K FD ⎞ 2 × 109 ⎛ K FD ⎞ 3 3 12 = ×⎜ ⎟ N o ⎡⎣108 − 104 ⎤⎦ = ⎜ ⎟ N o × 89.9 × 10 3 ⎝ α Ac ⎠ ⎝ α Ac ⎠ Assuming equal signal output powers for all channels in the FDM group, the degradation in output SNR of the 12th channel compared with the 1st is Pn12 89.9 10 log10 1 = 10 log10 = 4.657 dB Pn 30.76 To assure equal demodulator output SNRs, the voice channels in the multiplex at centered at higher frequencies are carried at increasingly higher levels (resulting in higher frequency deviation) to compensate for the parabolic noise spectral density of FM demodulator.

7.17 A radio receiver block diagram is shown in Figure P7.2. Each block is impedance matched to the source of 50. The radio is designed to receive a signal with a bandwidth of 20MHz centered about the BP filter frequency. a. Calculate the noise figure of the overall receiver. Solution:

The noise factor of a lossy two-port device is equal to its attenuation or loss. For example, NF1 = L1 = 3 dB . Using Frii’s

FT = F1 +

F2 − 1 F3 − 1 F4 − 1 + + + ... G1 G1G2 G1G2 G3

0.585 31.62 − 1 1.585 − 1 + + + ... 1.5 0.63 0.63 ×10 0.63 × 101.5 × 105 = 1.585 + 0.928 + 1.537 = 4.05 = 1.585 +

The noise figure NFT of the receiver is 6.07 dB b. Calculate the input signal to the receiver to achieve the SNR of at least 25 dB at the input of baseband stage. Specify the answer in dBm and in volts. Solution:

In order to achieve the SNR of at least SNRreq = 25 dB at the input of baseband stage, the SNR at the receiver input should be SNR1 = SNRreq + NFT = 25 + 6.07 = 31 The noise power at the receiver input from (6.295) is N in = kTB = 4 × 10−21 × 20 × 106 = 8 × 10−14 = −101 dBm Input signal power = −101 + 31 = −70 dBm (70.71 µV RMS)

1 2 BPF

GBPF = −2 dB G1 = 15 dB NF1 = 2 dB

GLPF = −3 dB 4 LPF

3 G2 = 50 dB NF2 = 15 dB

5 Detector

Baseband

Gdet = −4 dB NF2 = 10 dB

c. Calculate the signal and noise power at each stage of the receiver chain for input signal level determined in part (b). Solution:

The output noise power is computed using (6.324).

Stage (1) BP Filter (2) Low-noise Amplifier (3) Amplifier (4) LPF (5) Detector

Signal Power −72 dBm −57 −7 −10 −14

Noise Power −101 dBm −84 −32 −35 −39

d. The baseband stage can only work with signals as large as 500 mV. What is the maximum signal that can be received? Solution:

( 500 ×10 ) = 4 dBm Signal power corresponding to 500 mV maximum = −3

2 × 50

Since the total gain of the receiver is 56 dB, the maximum signal that can be received at the receiver input is 4 − 56 dBm = − 52 dBm.

Chapter 8 8.1

The signal x(t ) = 100 ⎡⎣1 + 0.707 cos (100π t ) ⎤⎦ cos ( 2000π t ) is sampled by an impulse train. a. Determine the minimum sampling rate. Solution: The bandwidth of x(t ) is 1050 Hz. Therefore, the minimum sampling rate is 2100 Hz.

b. What is the sampling rate if a guardband of 300 Hz is required. Solution:

The sampling rate to provide a guardband of 300 Hz is 2100 + 150 = 2.5 kHz. c. If the signal is sampled at f s = 1500 Hz , write an expression for the reconstructed signal xˆ (t ) . Display the spectra of original, sampled, and reconstructed signals in the same figure. Solution:

The signal x(t ) can be expressed as

x(t ) = 100 ⎡⎣1 + 0.707 cos (100π t ) ⎤⎦ cos ( 2000π t )

= 100 cos ( 2000π t ) + 70.7 cos (100π t ) cos ( 2000π t ) = 100 cos ( 2000π t ) + 35.35 ⎡⎣ cos ( 2100π t ) + cos (1900π t ) ⎤⎦

The recovered signal is xˆ (t ) obtained as

x [ n] = x ( nTs ) = 100 cos ( 4π n / 3) + 35.35 ⎡⎣cos ( 7π n / 5 ) + cos (19π n /15 ) ⎤⎦ = 100 cos ( 4π n / 3 − 2π n ) + 35.35 ⎡⎣cos ( 7π n / 5 − 2π n ) + cos (19π n /15 − 2π n ) ⎤⎦ = 100 cos ( 2π n / 3) + 35.35 ⎡⎣cos ( 3π n / 5) + cos (11π n /15) ⎤⎦ = 100 cos (1000π nTs ) + 35.35 ⎡⎣cos ( 900π nTs ) + cos (1100π nTs ) ⎤⎦ ⇒ 100 cos (1000π t ) + 35.35 ⎡⎣cos ( 900π t ) + cos (1100π t ) ⎤⎦ = xˆ (t )

The frequencies 1000, 2100, and 1900 Hz in the original signal are aliased by the aliased signals at frequencies 500, 450, and 550 Hz, respectively, in the recovered signal xˆ (t ) .

X(f) Xs (f)

1.2

Folding frequency window

Magnitude response

0.8

0.6

0.4

X̂ ( f )

0.2

-2.5

-2

-1.5

-1

-0.5 0 0.5 Frequency in kHz

1.5

2.5

8.2 Suppose the reconstruction filter output signal xˆ (t ) is cos ( 30π t + π / 6 ) . If the sampling rate fs = 200 Hz was used in obtaining it, determine two different input signals x1 (t ) and x2 (t ) that could have produced the above output signal. How many such input signals exist? Solution: xˆ (t ) = cos ( 30π t + π / 6 )

x1[n] = cos ( 30π nTs + π / 6 ) = cos ( 0.15π n + π / 6 ) = cos ( 0.15π n − 2π n + π / 6 ) = cos (1.85π n − π / 6 ) = cos (1.85π n 200Ts − π / 6 ) = cos ( 370π nTs − π / 6 ) → cos ( 370π t − π / 6 ) = x1 (t ) x2 [n] = cos ( 30π nTs + π / 6 ) = cos ( 0.15π n + π / 6 ) = cos ( 0.15π n + 2π n + π / 6 ) = cos ( 2.15π n + π / 6 ) = cos ( 2.15π n 200Ts + π / 6 ) = cos ( 430π nTs + π / 6 ) → cos ( 430π t + π / 6 ) = x2 (t )

Two different input signals could have produced the output signal xˆ (t ) = cos ( 30π t + π / 6 ) are x1 (t ) = cos ( 370π t − π / 6 ) x2 (t ) = cos ( 430π t + π / 6 )

An infinite number of signals, xA (t ) = cos ⎡⎣( 30π ± 2Af sπ ) t + π / 6 ⎤⎦ , A = ±1, ±2,... , exist that would produce the same output signal 8.3 A sinusoidal signal x(t ) = 5cos(10π t ) is naturally sampled using a 30% duty cycle square wave with period Ts . a. Determine the maximum value of the period Ts that allows the signal x(t ) to be reconstructed from the sampled signal. Solution:

Since the signal x(t ) is a sinusoidal waveform of frequency 5 Hz, the minimum sampling rate is 10 Hz, and the maximum value of the sampling period Ts = 0.1sec b. If the signal is sampled at twice the Nyquist rate, plot the spectrum X s ( f ) of the sampled signal. Solution:

For f s = 20 and τ / Ts = 0.3 , the spectrum X s ( f ) of the naturally sampled signal is given from (8.23) as ∞

∞

n =−∞

X s ( f ) = τ f s ∑ sinc ( nf sτ ) X ( f − nf s ) = 0.3 ∑ sinc ( 0.3n ) X ( f − 20n )

1 0.9 sinc

⎛ 0.3 f ⎞ ⎟ roll-off ⎝ fs ⎠

X(f) Xs (f)

( f τ ) = sinc ⎜

0.8

Magnitude response

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

-60

-40

-20

0 Frequency (Hz)

c. Specify the transfer function of the reconstruction filter to recover x(t ) from the sampled signal generated in (b). Sketch its frequency response. Solution:

The transfer function Hr (f) of the reconstruction filter is given by

⎧10 ⎪3, ⎪ ⎪ 4 f H r ( f ) = ⎨− + 10, ⎪ 3 ⎪0, ⎪ ⎩

f ≤5 5 ≤ f ≤ 7.5 otherwise

4 3.5

Magnitude response Hr(f)

3 2.5 2 1.5 1 0.5 0 -8

-6

-4

-2

0 2 Frequency (Hz)

d. Repeat (a) and (b) if the duty cycle was 10%. Solution:

The spectrum X s ( f ) of the naturally sampled signal for τ / Ts = 0.1 is given from (8.23) as ∞

∞

n =−∞

X s ( f ) = τ f s ∑ sinc ( nf sτ ) X ( f − nf s ) = 0.1 ∑ sinc ( 0.1n ) X ( f − 20n )

Magnitude response

⎛ 0.1 f ⎞ sinc ( f τ ) = sinc ⎜ ⎟ roll-off 0.25 ⎝ fs ⎠

X(f) Xs (f)

0.2

0.15

0.1

0.05

0 -200

-150

-100

-50

0 50 Frequency (Hz)

100

150

200

The transfer function Hr (f) of the reconstruction filter is given by ⎧10, ⎪ H r ( f ) = ⎨ −4 f + 30, ⎪0, ⎩

f ≤5 5 ≤ f ≤ 7.5 otherwise

11 10

Magnitude response Hr(f)

9 8 7 6 5 4 3 2 1 0 -8

-6

-4

-2

0 2 Frequency (Hz)

8.4 The signal with spectrum shown in Figure P8.1 is sampled by multiplying it with a 1 and 25% duty cycle. square wave with period Ts = 2.2 B Figure P8.1

X(f ) 1

− B −0.8B

f 0.8B B

a. Derive an expression for the spectrum X s ( f ) of the resultant sampled waveform and plot it. Solution:

The spectrum X s ( f ) of the naturally sampled signal for τ / Ts = 0.25 is given from (8.23) as ∞ ⎛ n⎞ X s ( f ) = 0.25 ∑ sinc ( 0.25n ) X ⎜ f − ⎟ Ts ⎠ n =−∞ ⎝

Natural Sampling 0.4 X(f) Xs (f)

0.35

sinc(f*tau) roll-off

Magnitude response

0.3 0.25 0.2 0.15 0.1 0.05 0

-8

-6

-4

-2

0 f/B (Hz)

b. If the samples are instead generated a by a sample and hold circuit, write an expression for the spectrum X s ( f ) of the sampled waveform and plot it. Solution:

The spectrum X s ( f ) of the sample and hold circuit output signal is given from (8.26) as ∞ ⎛ n⎞ X s ( f ) = 0.25sinc ( 0.25 fTs ) e − j 0.25π fTs ∑ X ⎜ f − ⎟ Ts ⎠ n =−∞ ⎝

Flat-top Sampling 0.4 X(f) Xs (f)

0.35

sinc(f*tau) roll-off

Magnitude response

0.3 0.25 0.2 0.15 0.1 0.05 0

-8

-6

-4

-2

0 f/B (Hz)

c. Specify the transfer function of the reconstruction filter in each case to recover the original signal without any distortion. The transfer function Hr (f) of the reconstruction filter for a naturally sampled signal is given by

⎧⎪4, Hr ( f ) = ⎨ ⎪⎩0,

f ≤B otherwise

The transfer function Hr (f) of the reconstruction filter for a flat-top sampled signal is given by

⎧⎪4sinc−1 ( 0.25 fTs ) , Hr ( f ) = ⎨ ⎪⎩0,

f ≤B otherwise

8.5 The samples of ±1 volt peak-to-peak analog signal with bandwidth 3.3 kHz are to be transmitted using a PCM system. a. Determine the minimum sampling rate if 700 Hz guard band is required. Solution:

Minimum sampling rate = 6.6 kHz with no guard band. Minimum sampling rate = 6.6 + 0.35 = 6.95 kHz with 700 Hz guard band. b. Calculate the number of quantization levels and bits/sample required, if the peak amplitude signal is to be represented within 0.1% accuracy. Solution:

Since the maximum quantization error allowed is 0.1% of the peak signal value, the peak quantization error e[n] = 0.001× 1 = 0.001 . The step size of a uniform quantizer is related to the maximum quantization error by e[n] max =

∆ 2

Therefore, ∆ = 0.001× 2 = 0.002

Number of quantization levels =

Full-scale range 2 = = 1000 ∆ 0.002

Assuming 210 = 1024 quantization levels are used, 10 bits/sample are required. c. What is the bit rate of PCM signal? Solution:

Serial bit rate of the PCM signal = 10 × 6.6 = 66 kbit/sec (no guard band) Serial bit rate of the PCM signal = 10 × 6.95 = 69.5 kbit/sec (750 Hz guard band)

8.6

For an m-bit ADC, show that peak SQNR is given by Peak SQNR = 6 m + 4.77 dB

where peak SQNR is defined as the ratio of peak signal power (i.e., at the crest of the signal waveform) to the average quantization noise power. Solution:

Assuming that the peak-to-peak signal equals the FS range of the quantizer, the peak signal-to-quantization noise ratio is given by Peak SQNR = 10 log10

⎛V2 ⎞ Peak signal power = 10 log10 ⎜ 2 ⎟ dB Average noise power ⎝ σe ⎠

For an m-bit quantizer with full-scale range of 2V, the noise variance σ e2 is given from (8.37) as

σ e2 = 2−2 m V 2 / 3 ⎛ ⎞ V2 Peak SQNR = 10 log10 ⎜ −2 m 2 ⎟ = 10 log10 3 × 22 m = 6m + 4.77 dB ⎝ 2 V /3⎠

(

)

8.7 Determine and illustrate the partition and quantization levels of a 4-bit, uniform midrise quantizer with ±1 volt FS range. For the sequence {0.2, −0.3, −0.7, 0.08, 0.25, 0.5, 0.8, 0.95}, a. Determine the quantized sequence. Also determine the mean square quantization error and SNR. Solution:

The quantization step size ∆ for a uniform quantizer with full-scale range of ±1V is given by ∆=

2 = 0.125 16

Figure illustrates the partition and quantization levels for 16-level midrise type quantizer.

Quantization Levels −0.9375 −0.6875 −0.4375 −0.1875

0.1875 0.4375 0.6875 0.9375

−0.8125 −0.5625 −0.3125 −0.0625 0.0625 0.3125 0.5625 0.8125 −1.0

−0.75

−0.25

−0.5

0.25

−0.875 0.625 −0.375 −0.125 0.125 0.375

0.5

0.75

0.625

1.0

0.875

Partition Levels For the sequence {0.2, -0.3, -0.7, 0.08, 0.25, 0.5, 0.8, 0.95}, the quantized sequence is given by

{0.1875, − 0.3125, − 0.6875, 0.0625, 0.1875, 0.4375, 0.8125, 0.9375} Mean square quantization error σ e2 =

1 2 ( x[n] − xˆ[n]) ∑ N n

2 2 2 2 1 ⎧⎪( 0.2 − 0.1875 ) + ( −0.3 + 0.3125 ) + ( −0.7 + 0.6875 ) + ( 0.08 − 0.0625 ) ⎫⎪ = ⎨ ⎬ 8 ⎪ + ( 0.25 − 0.1875 )2 + ( 0.5 − 0.4375 )2 + ( 0.8 − 0.8125 )2 + ( 0.95 − 0.9375 )2 ⎪ ⎩ ⎭ = 0.0011

Signal variance σ x2 =

(

)

2 1 x[n] − x[n] ∑ N n

Now the mean

x[n] =

1 {0.2 − 0.3 − 0.7 + 0.08+0.25 + 0.5 + 0.8 + 0.95} = 0.225 8

Substituting yields 2 2 2 2 1 ⎧⎪( 0.2 − 0.225 ) + ( −0.3 − 0.225 ) + ( −0.7 − 0.225 ) + ( 0.08 − 0.225 ) ⎫⎪ σ = ⎨ ⎬ 8 ⎪+ ( 0.25 − 0.225 )2 + ( 0.5 − 0.225 )2 + ( 0.8 − 0.225 )2 + ( 0.95 − 0.225 )2 ⎪ ⎩ ⎭ = 0.1482 2 x

⎛σ 2 ⎞ ⎛ 0.1482 ⎞ SQNR = 10 log10 ⎜ x2 ⎟ = 10 log10 ⎜ ⎟ = 21.3 dB ⎝ 0.0011 ⎠ ⎝σe ⎠

b. Determine 2's complement format binary stream corresponding to the quantized sequence. Solution:

The following table is generated using Two’s complement format to map quantization levels to codewords. Binary Word Quantization Level 15/16 0◊111 13/16 0◊110 11/16 0◊101 9/16 0◊100 7/16 0◊011 5/16 0◊010 3/16 0◊001 1/16 0◊000

Binary Word Quantization Level 1◊001 −15/16 1◊010 −13/16 1◊011 −11/16 1◊100 −9/16 1◊101 −7/16 1◊110 −5/16 1◊111 −3/16 1◊000 −1/16

The binary stream corresponding to the quantized sequence is

0001 1110 1011 0000 0001 0011 0110 0111 8.8 100 gigabyte hard disk is used to store PCM video signal. One volt peak-to peak NTSC video signal is sampled at 10.73864 MHz. The quantizing scheme is selected so that peak signal-to-quantization noise ratio is at least 45 dB. a. Calculate the number of quantization levels and bits/sample required. Solution:

From Problem 8.6, the peak SQNR is given by Peak SQNR = 6m + 4.77 dB 45 = 6m + 4.77 6m = 45 − 4.77 = 40.23 m≈7

We will choose m = 8 because an 8-bit ADC is more widely available. Number of quantization levels = 28 =256

b. What is the serial bit rate of the digital video signal? Solution:

Serial bit rate of the digital video signal = 10.73864 MHz × 8 = 85.91 Mb/s c. How many minutes of video program can the hard disk store? Solution:

Number of minutes of video program that can be stored on the hard disk =

Capacity of the disk 100 × 1000 × 8 Mbits = = 9312 seconds = 155 minutes Video ADC bit rate 85.91 Mb/s

8.9 A video signal x (t ) of bandwidth 4.2 MHz is to be digitized using an analog-todigital converter (ADC). a. Determine the Nyquist sampling rate. Solution:

Nyquist sampling rate = 2 × 4.2 = 8.4 MHz b. Suppose that the video signal is instead sampled at f s = 10.73864 MHz. Specify and draw the frequency response of a realizable LP filter that allows reconstruction of x (t ) from its samples. Solution:

We select the following specifications for the LP elliptic filter designed using the MATLAB code in Example 2.34. Passband frequency fp = 4.2 MHz Stopband frequency fs = 5.37 MHz Maximum passband ripple Rp = 0.1 dB Minimum stopband ripple Rs = 60 dB

Magnitude Response of Elliptic LP Filter 0

Magnitude Response(dB)

-10 -20 -30 -40 -50 -60 -70 -80 0

6 Frequency, Hz

c. A 10-bit ADC is used to digitize the video signal. What is the number of quantization levels? Assuming that the effective resolution of the ADC is 8.75 bits, calculate the ADC output SQNR. Solution:

Number of quantization levels = 210 = 1024 Peak SQNR = 6 × 8.75 + 4.77 = 57.27 dB

d. What is the serial bit rate of the digital video signal. Solution:

Serial bit rate of the digital video signal = 10.73864 MHz × 10 = 107.3864 Mb/s e. What practical implementation factors cause the SQNR loss in (c)? Solution:

Thermal noise Aperture-jitter noise − variation in sampling clock 8.10 Determine and illustrate the partition and quantization levels of a µ -law quantizer in the range of (−1,1) with 16 levels and µ = 9. For the same sequence as in Problem 8.7,

a. Determine the quantized sequence using the µ -law quantizer. Also determine the mean square quantization error and SNR. Solution:

The partition levels of a 16-level uniform quantizer with FS range of (−1,1) are {−1.0 −0.875 −0.75 −0.625 −0.5 −0.375 −0.25 −0.125 0 0.125 0.25 0.375 0.5 0.625 0.75 0.875 1.0} We now apply inverse µ -law mapping to obtain the partition levels for the nonuniform quantizer with µ = 9. x = F −1 ( y ) =

y ⎤ xmax ⎡ xmax − 1 sgn ( y ) 1 µ + ( ) ⎢ ⎥ µ ⎣ ⎦

The resulting partition levels for the nonuniform quantizer with µ = 9 are obtained as {−1.0 −0.7221 −0.5137 −0.3574 −0.2403 −0.1524 −0.0865 −0.0371 0 0.0371 0.0865 0.1524 0.2403 0.3574 0.5137 0.7221 1.0} The quantization levels for both uniform and non-uniform quantizer are {−0.9375 −0.8125 −0.6875 −0.5625 −0.4375 −0.3125 −0.1875 −0.0625 0.0625 0.1875 0.3125 0.4375 0.5625 0.6875 0.8125 0.9375}

1 0.9 0.8 0.7

Output

0.6 0.5 0.4 0.3 0.2 Non-uniform Quantization Mu-law Uniform Quantization

0.1 0

0.1

0.2

0.3

0.4 0.5 0.6 Normalized input

0.7

0.8

0.9

For the sequence {0.2, −0.3, −0.7, 0.08, 0.25, 0.5, 0.8, 0.95}, the µ-law quantized sequence is given by

{0.4375, − 0.5625, − 0.8125, 0.1875, 0.5625, 0.6875 , 0.9375, 0.9375} b. For this sequence, which method (uniform vs. µ-law) gives more accurate result? Mean square quantization error using µ -law quantizer σ e2 =

1 2 ( x[n] − xˆ[n]) ∑ N n

2 2 2 2 1 ⎪⎧( 0.2 − 0.4375 ) + ( −0.3 + 0.5625 ) + ( −0.7 + 0.8125 ) + ( 0.08 − 0.1875 ) ⎫⎪ = ⎨ ⎬ 8 ⎪ + ( 0.25 − 0.5625 )2 + ( 0.5 − 0.6875 )2 + ( 0.8 − 0.9375 )2 + ( 0.95 − 0.9375 )2 ⎪ ⎩ ⎭ = 0.0377

From problem 8.7 signal variance σ x2 = 0.1482 ⎛σ 2 ⎞ ⎛ 0.1482 ⎞ SQNR = 10 log10 ⎜ x2 ⎟ = 10 log10 ⎜ ⎟ = 5.945 dB ⎝ 0.0377 ⎠ ⎝σe ⎠

For this example, µ-law quantizer’s performance is worse than that of the uniform quantizer. This is because the given sequence is very short and more sample values are larger. The µ-law quantizer is designed to maximize SNR for voice signal with Laplacian type amplitude distribution. 8.11 Consider the performance of 8-bit, µ = 255 companded PCM system for a Gaussian message signal N 0, σ x2 with peak amplitude xmax .

(

)

a. Write the expression for the output SQNR achievable with this companded system. Solution:

Consider the compressor characteristic y = F ( x ) . The output of the compressor is uniformly quantized. Let yi and ∆ denote, respectively, the quantization level and the step size of the ith quantization interval for the compressed signal y. The corresponding input level and the partition interval are xi and ∆ i respectively.

y ymax

y = F ( x)

dF ( x) dx x = xi

∆ xi

xmin

xmax

∆i

ymin

Assuming that the number quantization levels M is large, ∆ i is small and xi is mid-point of the ith partition interval. Further, we assume that the pdf f x ( x) ≈ f x ( xi ) over the ith partition interval. Therefore, the quantization error ε i = xi − x has the mean-square value xi +

∆i 2

xi +

∆i 2

∆ i3 σ = ∫ ( xi − x ) f x ( x)dx ≈ f x ( xi ) ∫ ( xi − x ) dx = f x ( xi ) 12 ∆i ∆i 2

2 i

xi −

2 The variance of quantization noise σ nu is obtained by summing the noise over all M quantization intervals as

∆ 3i f x ( xi ) i =1 12 M

σ nu2 = ∑ σ i2 = ∑ i =1

We observe from the Figure that step size ∆ is related to ∆i by dF ( x) ∆ = dx x = xi ∆ i Substituting yields

σ nu2 =

f x ( xi )∆i ∆2 M ∑ 2 12 i =1 ⎛ dF ( x) ⎞ ⎜⎜ ⎟⎟ dx x x = i ⎠ ⎝

For M large, the summation can be approximated by an integral to obtain x

σ nu2 =

2 2 max max ymax f x ( x) ymax f x ( x) dx dx = 2 2 ∫ 2 ∫ 3M xmin ⎛ dF ( x) ⎞ 3M xmin ⎛ dF ( x) ⎞ 2 ⎜ ⎟ ⎜ ⎟ ⎝ dx ⎠ ⎝ dx ⎠

For µ-law type of compression characteristic,

dF ( x) is given by dx

( µ / xmax ) ymax dF ( x) = dx log e (1 + µ ) ⎡1 + µ ( x / xmax ) ⎤ ⎣ ⎦ The quantization noise added by a µ-law quantizer, σ µ2 , is now obtained as 2

2 ⎛ log e (1 + µ ) ⎞ xmax ⎡ ⎛ x ⎞⎤ ymax 2 σµ = 1 µ + ⎜ ⎟ ⎢ ∫ ⎢ ⎝⎜ xmax ⎠⎟ ⎥⎥ f x ( x)dx 3M 2 ⎜⎝ ymax ( µ / xmax ) ⎟⎠ xmin ⎣ ⎦ 2 2 x 2 ⎤ ⎛ x ⎞ ⎛ log e (1 + µ ) ⎞ max ⎡ xmax 2⎛ x ⎞ ⎢ ⎥ 1 2 µ µ = + + ⎜ ⎟ ⎜ ⎟ f x ( x)dx ⎜ ⎟ ∫ 3M 2 ⎝ µ ⎠ xmin ⎢⎣ ⎝ xmax ⎠ ⎝ xmax ⎠ ⎥⎦ 2 2 ⎤ 2 E{ x} ⎛ log e (1 + µ ) ⎞ ⎡ xmax 2 σx + µ 1 2 µ = + ⎢ ⎥ ⎜ ⎟ 2 xmax xmax 3M 2 ⎝ µ ⎠ ⎢⎣ ⎥⎦

Now

σ2 3M 2 SQNRµ = x2 = σ µ [ log e (1 + µ ) ]2 =

( µσ / x ) x

1 + 2µ

E{ x} xmax

3 × 22 m

max

+ µ2

σ x2 2 xmax

[log e (1 + µ )]

( µσ / x ) ( 2

max

E{ x} 2 +1 µσ x / xmax σ x

)

In dB form, the SQNR of the µ-law quantizer can be expressed as

SQNRµ = 6m + 4.77 − 20 log10 [ log e (1 + µ ) ] ⎡ ⎛ x ⎞2 ⎛ x ⎞ E{ x}⎤ ⎥ − 10 log10 ⎢1 + ⎜ max ⎟ + 2 ⎜ max ⎟ ⎢⎣ ⎝ µσ x ⎠ ⎝ µσ x ⎠ σ x ⎥⎦

(dB)

For a Gaussian message signal, E { x } / σ x = 0.798 . Therefore,

SQNRµ = 6m + 4.77 − 20 log10 [ log e (1 + µ )] ⎡ ⎛ x ⎞2 ⎛ x ⎞⎤ − 10log10 ⎢1 + ⎜ max ⎟ + 1.6 ⎜ max ⎟ ⎥ ⎢⎣ ⎝ µσ x ⎠ ⎝ µσ x ⎠ ⎥⎦

(dB)

b. Plot the SQNR performance as a function of the relative RMS input level ⎛σ ⎞ 20log ⎜ x ⎟ and compare it with that for the PCM system without ⎝ xmax ⎠ companding. Solution: SQNR comparison : Gaussian message signal 60

Signal-to-quantization noise ratio (dB)

0 u-law quantizer, u=255 Uniform quantizer -10 -60

-50

-40

-30

-20

-10

8.12 Suppose a message signal can be modeled as a LP stationary random process x (t ) with Laplacian first-order PDF f x ( x) =

1 − 2x e 2

The bandwidth of the random signal is 5 kHz, and we desire to transmit it using a PCM system. a. If sampling is done at the Nyquist rate and a uniform quantizer with 64 levels is utilized, what is the resulting SQNR? Solution:

A Laplacian PDF is given by 2x

f x ( x) =

− 1 e σx 2σ x

Therefore, in the present case, σ x2 = 1 . Now P {− K σ x < x ≤ K σ x } =

Kσ x

∫σ f ( x)dx x

−K

2 = 2σ x

Kσ x

∫e

−

σx

−

Kσ x

dx = − e σ x

= 1 − e− 2K 0

For K = 3, P {− Kσ x < x ≤ Kσ x } = 98.563% K = 4, P {− Kσ x < x ≤ Kσ x } = 99.65%

That is, for the signal amplitude has Laplacian distribution, only 0. 35% of the samples would lie outside the full-scale range of 8σ x . So if we choose

xmax ≈ 4σ x , the probability that an analog sample stays within the range

[− xmax , xmax ] is 0. 9965.

⎛x ⎞ SQNR = 6m + 4.77 − 20 log ⎜ max ⎟ = 6 × 6 + 4.77 − 20 log ( 4 ) ⎝ σx ⎠ = 28.77 dB

b. What is the bit rate of the PCM signal? Solution:

Serial bit rate = 10 × 6 = 60 kb/s

c. If we need to increase the SQNR by 12 dB, what is the resultant bit rate increase? Solution:

If the SQNR needs to be increased by by 12 dB, we require 2 additional bits/sample. Therefore, Serial bit rate = 10 × 8 = 80 kb/s Increase in the resultant bit rate =80 – 60 = 20 kb/s 8.13 Consider the µ -law compression characteristic. Assume µ = 255. If the FS range is ±1 volt and 256 quantization levels are used, what is the minimum and maximum effective separation between partition levels? If no compression is used, what is the interval between partition levels? Solution:

If no compression is used, the interval between levels is given by ∆=

2 = 7.8125 mV 256

The y-axis is uniformly quantized with step size ∆y =

1 1 = 128 128

in both +ve and −ve direction. The smallest segment on the x-axis corresponds to the segment [0 1/128] on the y-axis. The point on x-axis corresponding to y = 1/128 is given from (8.47) as x = F −1 ( y ) =

1 ⎡ 1/128 − 1⎤ = 1.73 × 10−4 1 + 255 ) ( ⎦ 255 ⎣

The smallest segment size on the x-axis is 0.173 mV. The largest segment on the x-axis corresponds to the last segment on the y-axis, that is, [127/128 1]. The point on x-axis corresponding to y = 127/128 is given from (8.47) as x = F −1 ( y ) =

1 ⎡ 127/128 − 1⎤ = 0.957 V 1 + 255 ) ( ⎦ 255 ⎣

The largest segment size on the x-axis is 1− 0.957 = 42.563 mV.

8.14

The spectrum of a BP signal x(t) is shown in Figure P8.3. Determine the smallest sampling frequency that allows perfect reconstruction of x(t) from its samples x[n] for each of the following cases: Figure P8.3

X(f )

− fH

f L = 60 , f H = 80

f L = 40 , f H = 50

f L = 55 , f H = 75

− fL

Sketch the spectrum of the sampled signal and frequency response of the ideal BP filter in each case. Solution (a):

f L = 60, f H = 80. BIF = 80 − 60 = 20 . nmax is given from (8.124) as ⎢ f ⎥ ⎢ 80 ⎥ nmax = ⎢ H ⎥ = ⎢ ⎥ = 4 ⎣ BIF ⎦ ⎣ 20 ⎦

Therefore, the smallest possible sampling frequency is obtained from (8.124) as fs =

2 f H 2 × 80 = = 40 Hz nmax 4

Xs( f )

−2 f s

−100

−60

− fs

−20

2 fs

100

Hr ( f )

, −40

−20

Solution (b):

f L = 40, f H = 50. BIF = 50 − 40 = 10 . nmax is given from (8.124) as ⎢ 50 ⎥ nmax = ⎢ ⎥ = 5 ⎣ 10 ⎦

Therefore, the smallest possible sampling frequency is obtained from (8.124) as fs =

2 f H 2 × 50 = = 20 Hz nmax 5 Xs( f )

−50

−2 f s

−30

− fs

−10

2 fs

Hr ( f )

, −30

−20

−10

Solution (c):

f L = 55, f H = 75. BIF = 75 − 55 = 20 . nmax is given from (8.124) as ⎢ 75 ⎥ nmax = ⎢ ⎥ = 3 ⎣ 20 ⎦

Therefore, the smallest possible sampling frequency is obtained from (8.124) as fs =

2 f H 2 × 75 = = 50 Hz nmax 3 Xs( f )

− fs −95

−75

−55

−45

−25

0 −5

95 2 f s

Hr ( f )

−50

−25

0 5

8.15 Consider an IF sampling base station receiver for CDMA cellular system operating at an IF of 70 MHz. The bandwidth of the IF channel is 1.25 MHz. a. Determine the smallest sampling frequency to avoid aliasing. What is the corresponding SQNR degradation? Solution:

nmax is given from (8.124) as ⎢ f ⎥ ⎢ 70.625 ⎥ nmax = ⎢ H ⎥ = ⎢ ⎥ = 56 ⎣ BIF ⎦ ⎣ 1.25 ⎦

Therefore, the smallest possible sampling frequency is obtained from (8.124) as

fs =

2 f H 2 × 70.625 = = 2.52 MHz nmax 56

Maximum SQNR degradation = 10log10 ( 56 ) = 17.48 dB

b. Determine the permissible sampling frequency range of that results in SQNR loss no larger than 7 dB. To keep SQNR degradation less than 7 dB, 10log10 ( n ) ≤ 7 . Therefore, n ≤ 5 . The smallest possible sampling frequency under this constraint is given from (8.124) as fs =

2 f H 2 × 70.625 = = 28.25 MHz 5 5

8.16 A linear DM system is tested with 1 V peak-to peak, 5-kHz sinusoidal test signal. The signal is sampled at 10 times the Nyquist rate. a. Determine the step size required to prevent slope overload. Solution:

A2π f m < ∆ / Ts 0.5 × 2π × 5 ×103 < ∆ ×100 ×103 5π ∆> = 0.157 V 100 b. Calculate the quantization noise power in the test signal bandwidth. Solution:

∆ 2 f m ( 0.157 ) × 5 ×10 NQ = = = 41.08 ×10−5 3 3 fs 3 ×100 ×10 2

c. Determine the demodulated SNR of the system. Solution:

( 0.5 ) / 2 = 3.04 × 10 2 = 24.83 dB A2 / 2 SNRDM = = NQ 41.08 × 10 −5 2

8.17 A linear DM system is designed to transmit speech signals with bandwidth 3.4 kHz. The signal is sampled at 10 times the Nyquist rate and a step size of 100 mV is used. a. The modulator is tested with a 1-kHz test signal. Determine the maximum amplitude of the test signal required to avoid slope overload. Solution:

Amax =

∆f s 0.1× 2 × 3.4 ×104 = = 1.08 V 2π f m 2π ×103

b. Determine the demodulated SNR of the system. Solution: 3 ∆ 2 f m ( 0.1) ×10 NQ = = = 49.02 ×10−6 3 fs 3 × 6.8 ×104 2

(1.08 ) / 2 = 11.9 × 103 = 40.75 dB A2 / 2 = SNRDM = max 49.02 × 10 −6 NQ 2

8.18 Consider the first-order linear predictor defined by

xˆ[n] = h1 x[n −1] where x[ n ] is a stationary random sequence with zero mean and h1 is coefficient of first-order predictor filter. Let us define the variance of prediction error as 2 σ d2 = E ( x[ n] − xˆ [ n]) .

{

}

a. Determine the optimal value of h1 that minimizes σ d2 and calculate the minimum value of σ d2 . Solution:

{

σ d2 = E ( x[ n] − xˆ [ n])

}

For first-order linear predictor xˆ [n] = h1 x[n − 1] . Substituting

{

σ d2 = E ( x[n] − h1 x[ n − 1])

} = E { x [n] − 2h x[n] x[n − 1] + h x [n − 1]} 2

2 1

= Rx ( 0 ) − 2h1 Rx (1) + h12 Rx ( 0 ) ⎛ R (1) ⎞ = Rx ( 0 ) ⎜ 1 + h12 − 2h1 x ⎟ ⎜ Rx ( 0 ) ⎟⎠ ⎝

Since x[n] is zero mean, Rx ( 0 ) = E { x 2 [n]} = σ x2 . Therefore, the variance of the prediction error is given by ⎛

σ d2 = σ x2 ⎜ 1 + h12 − 2h1 ⎝

Rx (1) ⎞ ⎟ σ x2 ⎠

To find h1 that minimizes σ d2 , we take the derivative with respect to h1 and solve

dσ d2 =0 dh1 R (1) R (1) 2h1 − 2 x 2 = 0 ⇒ h1opt = x 2 σx σx The minimum value σ d2 is ⎛ ⎛ R (1) ⎞ 2 ⎛ ⎛ R (1) ⎞ 2 ⎞ Rx (1) Rx (1) ⎞ x 2 ⎟ = σ x ⎜1 − ⎜ x 2 ⎟ ⎟ σ = σ ⎜1 + ⎜ 2 ⎟ − 2 2 2 ⎜ ⎜ σx ⎠ σx σx ⎟ σx ⎠ ⎟ ⎝ ⎝ ⎠ ⎝ ⎝ ⎠ 2 d

2 x

b. Determine the optimal prediction gain Gp . Solution: Gp =

σ x2 = σ d2

σ x2 ⎛ ⎛ R (1) ⎞ ⎞ σ ⎜1 − ⎜ x 2 ⎟ ⎟ ⎜ ⎝ σx ⎠ ⎟ ⎝ ⎠ 2

2 x

1 ⎛ ⎛ R (1) ⎞ 2 ⎞ ⎜1 − ⎜ x 2 ⎟ ⎟ ⎜ ⎝ σx ⎠ ⎟ ⎝ ⎠

8.19 Consider 2nd order sigma-delta modulator shown in Figure P8.3. Show that y[n] = x[n − 1] + ( e[n] − 2e[ n − 1] + e[n − 2])

(

Y ( z ) = z −1 X ( z ) + 1 − z −1

) E( z) 2

Figure P8.3

x[ n] + −

d1[n]

+ + +

Integrator 1

−

+ + +

d 2 [ n]

y[ n]

w[n]

−1

Integrator 2

z −1 1-bit DAC Solution:

Figure displays a linearized model of the second-order sigma-delta modulator where the 1-bit quantizer is replaced with a noise source e[n].

e[n] x[n]

−

+ + +

d1[n]

+ Integrator 1

−

+ +

d 2 [ n]

w[n]

z −1

y[n]

+ Integrator 2

z −1 1-bit DAC For the second loop, we have y[n] = w[n] + e[n]

(*)

d 2 [n] = d1[n] − y[n] + w[n]

Now w[n] = d 2 [n − 1] = d1[n − 1] − y[n − 1] + w[n − 1] = d1[n − 1] − e[n − 1] ,

Substituting into (*) yields y[n] = d1[n − 1] − e[n − 1] + e[n]

(**)

For the first loop, we have d1[n] = x[n] − y[n] + d1[n − 1]

Therefore, d1[n − 1] = x[n − 1] − y[n − 1] + d1[n − 2]

Substituting for y[n − 1] from (**), we obtain d1[n − 1] = x[n − 1] − d1[n − 2] + e[n − 2] − e[n − 1] + d1[n − 2] = x[n − 1] + e[n − 2] − e[n − 1]

Substituting into (**) yields y[n] = x[n − 1] + ( e[n] − 2e[ n − 1] + e[n − 2])

Taking the z-transform of both sides, we get

(

)

(

Y ( z ) = z −1 X ( z ) + 1 − 2 z −1 + z −2 E ( z ) = z −1 X ( z ) + 1 − z −1

) E( z) 2

a. Show that the magnitude-squared of the noise transfer function is given by

(

H e ( f ) = 1 − z −1 2

)

2 2 z = e jωTs

= 4 2 × sin 4 (π f / f s )

Solution:

The signal and noise transfer functions are obtained from (***) as H s ( z ) = z −1

(

H e ( z ) = 1 − z −1

)

The magnitude-squared of the noise transfer function is given by

(

H e ( f ) = 1 − z −1 2

)

2 2

2 −2 jπ fTs

= 4j e

z = e jωTs

(

= 1 − e j 2π fTs

⎛ e jπ fTs − e − jπ fTs ⎞ ⎜ ⎟ 2j ⎝ ⎠

)

2 2

= 4 ( sin (π fTs ) )

2 2

= 4 2 × sin 4 (π fTs )

b. The quantization noise power at the output is given by N out =

π 4∆2 60 K 5

Solution:

The PSD of output noise is given by

Gnout ( f ) = Ge ( f ) H e ( f )

where Ge ( f ) is a spectral density of the quantization noise.

Ge ( f ) =

σ e2 fs

∆2 12 f s

Therefore,

Gnout ( f ) =

4∆ 2 4 ⎛ π f ⎞ sin ⎜ ⎟ 3 fs ⎝ fs ⎠

The output quantization noise power in the signal band is given by B B ⎛π f ⎞ 4∆ 2 N out = ∫ Gnout ( f )df = sin 4 ⎜ ⎟ df ∫ 3 fs − B ⎝ fs ⎠ −B B ⎛π f ⎞ 8∆ 2 sin 4 ⎜ = ⎟ df ∫ 3 fs 0 ⎝ fs ⎠

For very large oversampling ratio K, sin 4 (π f / f s ) ≈ (π f / f s ) 4

B 8∆ 2 8∆ 2π 4 ⎛ B ⎞ ∆ 2π 4 4 N out = ( π f / f ) df = = ⎜ ⎟ s 3 f s ∫0 15 ⎝ f s ⎠ 60 K 5

c. The output SNR of the 2nd order sigma-delta system is given by ⎛ 15 ⎞ SNR = 10 log10 ⎜ 4 ⎟ + 10 log10 ( K 5 ) = −11.135 + 50 log10 K ⎝ 2π ⎠ Solution:

Assuming a sinusoidal input of peak amplitude ∆ / 2 , the signal power is given by 1⎛∆⎞ Pout = ⎜ ⎟ 2⎝ 2 ⎠

The output SNR of a first-order sigma-delta modulator is Pout 1 ⎛ ∆ ⎞ 60 K 5 15K 5 = ⎜ ⎟ = N out 2 ⎝ 2 ⎠ ∆ 2π 4 2π 4 2

SNR =

or in dB form ⎛ 15 ⎞ SNR = 10 log10 ⎜ 4 ⎟ + 10 log10 ( K 5 ) = −11.135 + 50 log10 K ⎝ 2π ⎠ d. Prove that for the 2nd order sigma-delta system every doubling of the oversampling ratio, the SNR improves by 15 dB and the resolution improves by 2 ½ bits. Solution:

Let the oversampling ratio K = 2 r . Substituting we obtain

SNR = −11.135 + 50log10 ( 2r ) = −11.135 + 15r For every doubling of the oversampling ratio K, that is, for every increment in r, the SNR improves by 15 dB, or equivalently, the resolution improves by 2.5 bits.

Chapter 9 9.1

What is the duration of a transmitted pulse or symbol for each of the following signals? (a) 16-PAM signal at 200kb/s_________________ (b) Manchester coded data at 10 Mb/s_______________________________ (c) Bipolar encoded data at 1.544 Mb/s________________________________ (d) 2B1Q encoded signal with a bit rate of 144 kb/s _____________________ What is the first-null bandwidth of signals in (a) and (d)? _________________ Solution: a. 16-PAM signal at 200kb/s Pulse or symbol duration = 20 µsec First-null bandwidth = 50 kHz b. Manchester coded data at 10 Mb/s Pulse or symbol duration = 100 nsec First-null bandwidth = 20 MHz c. Bipolar encoded data at 1.544 Mb/s Pulse or symbol duration = 0.647 µsec First-null bandwidth = 1.544 MHz d. 2B1Q encoded signal with a bit rate of 144 kb/s Pulse or symbol duration = 13.89 µsec First-null bandwidth = 72 kHz

9.2

An image is 1024 × 768 pixels with 3 bytes/pixel coding. a. How long does it take to transmit it over a 56 kbps modem link? Over a 1.5 Mbps ADSL link? Over a 10 Mbps cable modem link?

b. Compare the transmission times if JPEG compression is used prior to transmission. Assume that JPEG scheme achieves a compression ratio of 15:1.

Solution (a) and (b): An image with 1024 × 768 pixels and 3 bytes/pixel produces 2.36 Mbytes = 18.8744×106 bits. No compression Time required to transmit using 56 kb/s dial-up modem = 5.6174 min Time required to transmit using 1.5 Mb/s ADSL modem = 12.583 sec Time required to transmit over 10 Mb/s cable modem = 1.88744 sec Modem type

Transmit time No compression

Dial-up ADSL Cable modem

9.3

5.6174 min 12.583 sec 1.88744 sec

With JPEG compression (compression ratio =15) 22.47 sec 839 msec 125.83 msec

Consider a random data signal consisting of binary 1’s and 0’s occurring with equal probability. Calculate the power spectral density Gs ( f ) of unipolar RZ encoded signal assuming a 25% duty cycle. Solution: The FT of a rectangular pulse is obtained from Table 2.2 as ℑ Π (t / τ ) ←⎯ →τ sinc( f τ )

In the present case, V(f ) =

τ Tb

= 0.25 . Therefore,

Tb sinc ( fTb / 4 ) 4

Substituting into (9.22) along with (9.25) and (9.26), the spectral density for the unipolar RZ encoded signal is given by

∞ ⎛ Rb Tb T 1 A⎞ sinc ( fTb / 4 ) + Rb2 ∑ b sinc ( A / 4) δ ⎜ f − ⎟ 4 4 4 A=−∞ 4 Tb ⎠ ⎝ 2

Gs ( f ) =

⎛ Tb 1 ∞ A⎞ 2 sinc2 ( A / 4) δ ⎜ f − ⎟ = sinc ( fTb / 4 ) + ∑ 64 64 A =−∞ Tb ⎠ ⎝ a. Calculate the power spectral density Gx ( f ) of 25% duty cycle rectangular pulse train (Example 2.24) modulated by 101010… data pattern. Solution:

The periodic unipolar RZ signal, obtained by modulating the rectangular pulse train in Example 2.24 with 101010…data pattern, is shown in Figure.

x p (t )

…

T − b 8

−2Tb

…

Tb 8 0

−Tb

Tb To

2Tb

The PSD Gx ( f ) of a periodic signal x p (t ) is given from (2.173) as ∞

Gx ( f ) = ∑ Cn δ ( f − nf o ) 2

n =−∞

In the present case, To = 2Tb and τ = Tb / 4 . Therefore, ∞ nR ⎞ 2 ⎛ Gx ( f ) = ∑ Cn δ ⎜ f − b ⎟ 2 ⎠ ⎝ n =−∞

The exponential FS coefficients Cn of the periodic waveform are given by Cn =

⎛ nT / 4 ⎞ 1 Tb / 4 ⎛n⎞ sinc ⎜ b ⎟ = sinc ⎜ ⎟ 2Tb ⎝8⎠ ⎝ 2Tb ⎠ 8 ∞

nR ⎞ 1 ∞ nR ⎞ 1 ⎛n⎞ ⎛ ⎛n⎞ ⎛ sinc 2 ⎜ ⎟δ ⎜ f − b ⎟ Gx ( f ) = ∑ sinc ⎜ ⎟ δ ⎜ f − b ⎟ = ∑ 2 ⎠ 64 n =−∞ 2 ⎠ ⎝8⎠ ⎝ ⎝8⎠ ⎝ n =−∞ 8

b. Plot both on the same graph using the normalized frequency scale (i.e., f /Rb). What conclusions do you make? Solution:

Spectral density comparison: 25% duty cycle unipolar RZ signal vs rectangular pulse train 0.016 Unipolar RZ: Continuous portion Unipolar RZ: Discrete components 0.014 Rectangular pulse train

Spectral Density

0.012 0.01 0.008 0.006 0.004 0.002 0

9.4

3 4 5 Normalized Frequency f/Rb

Consider a random data signal consisting of binary 1’s and 0’s occurring with equal probability. Calculate the power spectral density Gs ( f ) of polar NRZencoded signal. Solution:

The spectral density of the binary polar NRZ signal is obtained from (9.33) as GpolarNRZ ( f ) =

⎛ f ⎞ 1 sinc2 ⎜ ⎟ Rb ⎝ Rb ⎠

a. Calculate the power spectral density Gs ( f ) of a deterministic data pattern 101010…..with polar NRZ encoding. Solution:

In the present case, To = 2Tb . The periodic polar NRZ signal x p (t ) ,

corresponding data pattern 101010….. , is shown in Figure.

x p (t )

…

−2Tb

… −Tb

−1

t 2Tb

The FS coefficients of the periodic polar NRZ signal x p (t ) are given by C0 = 0 −T /2 + Tb / 2 Tb ∞ ⎤ 1 1 ⎡ b − j 2π nfot − j 2π nf o t − j 2π nf o t − j 2π nf o t Cn = x t e dt e dt e dt e dt = − + − ( ) ⎢ ⎥ ∫ p ∫ ∫ To −∞ 2Tb ⎢⎣ −∫Tb ⎥⎦ −Tb / 2 Tb /2

Tb /2 Tb 1 ⎡ − jπ nt /Tb −Tb /2 − jπ nt /Tb Tb ⎤ +e − e e − jπ nt /Tb ⎢ ⎥ − /2 − Tb /2 T T b b ⎦ jπ n 2Tb j 2π n ⎣

1 1 ⎡ e jπ n / 2 − e jπ n + e − jπ n − e − jπ n /2 ⎦⎤ + ⎡e jπ n /2 − e − jπ n /2 ⎦⎤ ⎣ ⎣ j 2π n j 2π n

⎡⎣e jπ n /2 − e − jπ n / 2 ⎤⎦ = jπ n

sin (π n / 2 ) πn/ 2

⎛n⎞ = sinc ⎜ ⎟ ⎝2⎠

nR ⎞ ⎛n⎞ ⎛ Gx ( f ) = ∑ sinc 2 ⎜ ⎟δ ⎜ f − b ⎟ 2 ⎠ ⎝2⎠ ⎝ n≠0 b. Repeat (a) for a test pattern consisting of six binary 1’s followed by two binary 0’s. Solution:

In the present case, To = 8Tb . The periodic polar NRZ signal x p (t ) , corresponding to data pattern … 011111100111111001111110… , is shown in Figure.

x p (t )

…

−3Tb −8Tb

…

3Tb

-4Tb

4Tb

8Tb

−1

The FS coefficients of the periodic polar NRZ signal x p (t ) are given by

C0 =

3T −3Tb 4Tb ∞ ⎤ 4T 1 1 ⎡ b 1 = − − x ( t ) dt dt dt dt ⎥ = b = ⎢ ∫ p ∫ ∫ ∫ To −∞ 8Tb ⎢⎣ −3Tb ⎥⎦ 8Tb 2 −4Tb 3Tb

Cn =

3T −3Tb 4Tb ∞ ⎤ 1 1 ⎡ b − j 2π nfot − j 2π nf o t − j 2π nf o t − j 2π nfo t x t e dt e dt e dt e dt = − − ( ) ⎢ ⎥ p ∫ ∫ ∫ To −∞ 8Tb ⎣⎢ −3∫Tb −4Tb 3Tb ⎦⎥

4Tb 3Tb 1 ⎡ − jπ nt /4Tb −3Tb 1 + e − jπ nt /4Tb ⎤ − e e− jπ nt /4Tb ⎥ j 2π n −4Tb 3Tb ⎦ −3Tb j 2π n ⎣⎢ 1 1 ⎡e j 3π n / 4 − e jπ n + e − jπ n − e − j 3π n /4 ⎦⎤ + ⎡e j 3π n / 4 − e − j 3π n /4 ⎦⎤ = ⎣ ⎣ j 2π n j 2π n

2 3 sin ( 3π n / 4 ) ⎡⎣e j 3π n / 4 − e − j 3π n /4 ⎤⎦ = 2 3π n / 4 j 2π n

3 ⎛ 3n ⎞ = sinc ⎜ ⎟ 2 ⎝ 4 ⎠

Gx ( f ) =

nR ⎞ 1 9 ⎛ 3n ⎞ ⎛ + ∑ sinc 2 ⎜ ⎟δ ⎜ f − b ⎟ 4 4 n≠0 8 ⎠ ⎝ 4 ⎠ ⎝

c. Plot spectral densities on the same graph using the normalized frequency scale (i.e., f/Rb). Solution:

Spectral density comparison 2.5 Polar NRZ Square wave Six 1s and one 0 either side

Spectral Density

1.5

0.5

3 4 5 Normalized Frequency f/Rb

9.5 Optical fiber digital communication system uses unipolar Manchester pulses shown in Figure P9.1. a. Sketch the waveform for random data sequence 1011001. Solution: Unipolar Manchester signaling 1

1 0 2Tb

3Tb

4Tb

5Tb

6Tb

7Tb

b. Derive an expression for the spectral density Gs ( f ) assuming equiprobable binary data. How does this expression differ from (9.51) and why? Solution:

The spectral density of unipolar Manchester encoded signal is obtained by noting that x (t )um =

1 ⎡ x (t ) pm + 1⎤⎦ 2⎣

where

x(t )um = unipolar Manchester signal x(t ) pm = polar Manchester signal

The spectral density of unipolar Manchester encoded signal is Gum ( f ) =

G pm ( f ) 4

δ( f ) 4

where ⎛ fT ⎞ ⎛ π fTb ⎞ G pm ( f ) = Tbsinc 2 ⎜ b ⎟ sin 2 ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ Gum ( f ) =

Tb ⎡ δ( f )⎤ 2 ⎛ fTb ⎞ 2 ⎛ π fTb ⎞ ⎢sinc ⎜ ⎥ ⎟ sin ⎜ ⎟+ Tb ⎦ 4⎣ ⎝ 2 ⎠ ⎝ 2 ⎠

Note that the spectrum of unipolar Manchester encoded signal has DC component as expected. 9.6 We want to design a digital communications link to transmit data at 1.544 Mb/s. How much (first-null) bandwidth is required using the following signaling schemes: a. Bipolar RZ b. Polar RZ c. 4-ary PAM d. Manchester Solution:

a. 1.544 MHz b. 3.088 MHz c. 772 kHz d. 3.088 MHz

9.7 Binary data is to be transmitted using M-ary PAM over the ideal bandlimited channel with a bandwidth W = 4 kHz. We use RC pulses with roll-off factor α = 0.5 so that there is no ISI. If we want to achieve a bit rate of 56 kb/s, determine

a. the required size of signal set M. b. symbol rate and duration. Solution:

a. For M-ary signaling, Babs = 0.5(1 + α ) D . Substituting 4 × 103 = 0.5(1 + 0.5)D

Therefore, D =

4 × 103 16 ×103 = symbol/sec 0.75 3

Number of bits symbol k =

R 56 × 3 × 103 21 = = = 10.5 ≈ 11 bits/symbol D 16 × 103 2

Size of signal set M = 2k = 211 = 2048 56 × 103 b. Symbol rate D = = 5.09 × 103 symbols/sec 11

Symbol duration T =

11 = 196.43µ sec 56 × 103

9.8 We want to design a digital communications link to transmit data at 1.544 Mb/s. Compare the bandwidth required for following line encoding schemes using RC pulses with roll-off factor α = 0.5: a. Unipolar NRZ b. Bipolar RZ c. 4-ary PAM d. 16-ary PAM Solution:

a. For binary signaling, Babs = 0.5(1 + α ) R , where α = roll-off factor.

Using α = 0.5, the absolute bandwidth for NRZ signaling using RC pulses is

Babs = 0.5(1 + 0.5) × 1.544 = 0.75 × 1.544 = 1.158 MHz

b. Same as in (a). c. For M-ary signaling, Babs = 0.5(1 + α ) D

For 4-ary PAM, D = 772 ksymbols/second. Babs = 0.5(1 + α ) D = 0.75 × 772 = 579 kHz

d. For 16-ary PAM, D = 386 ksymbols/second. Babs = 0.5(1 + α ) D = 0.75 × 386 = 289.5 kHz

9.9 Consider a random data signal consisting of binary 1’s and 0’s occurring with equal probability. We assume that the digital communication system uses polar NRZ encoding and the basic pulse shape described by

⎧ ⎛ πt ⎞ ⎪cos ⎜ ⎟ , t < Tb / 2 v(t ) = ⎨ ⎝ Tb ⎠ ⎪ otherwise ⎩0,

a. Sketch the waveform for data sequence 1011001. Solution:

0.5

-0.5

-1

3 t/Tb

b. Find an expression for the power spectral density.

Solution:

The power spectral density Gs ( f ) of a binary polar NRZ signal is given from (9.33) as

Gs ( f ) =

A2 2 V( f ) Tb

Now

⎛ πt ⎞ ⎛ t ⎞ v(t ) = cos ⎜ ⎟ Π ⎜ ⎟ ⎝ Tb ⎠ ⎝ Tb ⎠ ⎛ t ⎞ ℑ Π ⎜ ⎟ ←⎯ → Tb sinc (Tb f ) ⎝ Tb ⎠ ⎛ πt ⎞ ℑ 1 ⎡ ⎛ ⎛ 1 ⎞ 1 ⎞⎤ → ⎢δ ⎜ f − cos ⎜ ⎟ ←⎯ ⎟+δ ⎜ f + ⎟⎥ 2⎣ ⎝ 2Tb ⎠ 2Tb ⎠ ⎦ ⎝ Tb ⎠ ⎝ Therefore, ⎧⎪ ⎛ t ⎞ ⎫⎪ ⎧⎪ ⎛ π t ⎞ ⎫⎪ ⎛ 1⎡ ⎛ 1 ⎞ 1 ⎞⎤ V ( f ) = ℑ ⎨Π ⎜ ⎟ ⎬ ⊗ ℑ ⎨cos ⎜ ⎟ ⎬ = Tb sinc (Tb f ) ⊗ ⎢δ ⎜ f − ⎟ +δ ⎜ f + ⎟⎥ 2⎣ ⎝ 2Tb ⎠ 2Tb ⎠ ⎦ ⎝ ⎩⎪ ⎝ Tb ⎠ ⎭⎪ ⎩⎪ ⎝ Tb ⎠ ⎭⎪ =

⎡ ⎛ ⎡ ⎛ Tb ⎡ 1 ⎞⎤ 1 ⎞⎤ ⎤ ⎢sinc ⎢Tb ⎜ f − ⎟ ⎥ + sinc ⎢Tb ⎜ f + ⎟⎥ ⎥ 2 ⎣⎢ 2Tb ⎠ ⎦ 2Tb ⎠ ⎦ ⎦⎥ ⎣ ⎝ ⎣ ⎝

Tb ⎡ 1 ⎞⎤ 1⎞ ⎛ ⎛ sinc ⎜ Tb f − ⎟ + sinc ⎜ Tb f + ⎟ ⎥ ⎢ 2 ⎠⎦ 2⎣ 2⎠ ⎝ ⎝

Substituting T ⎡ 1⎞ 1 ⎞⎤ ⎛ ⎛ Gs ( f ) = b ⎢sinc ⎜ Tb f − ⎟ + sinc ⎜ Tb f + ⎟ ⎥ 4⎣ 2⎠ 2 ⎠⎦ ⎝ ⎝

T ⎡ 1⎞ 1⎞ 1⎞ 1 ⎞⎤ ⎛ ⎛ ⎛ ⎛ = b ⎢sinc2 ⎜ Tb f − ⎟ + sinc 2 ⎜ Tb f + ⎟ + 2sinc ⎜ Tb f − ⎟ sinc ⎜ Tb f + ⎟ ⎥ 4⎣ 2⎠ 2⎠ 2⎠ 2 ⎠⎦ ⎝ ⎝ ⎝ ⎝

We observe that the spectral sidelobes decay very rapidly because of the smooth shape of the cosine pulse.

Spectral Density of Polar NRZ Signal 1.8 1.6

Spectral Density Gs (f)

1.4 1.2 1 0.8 0.6 0.4 0.2 0 -5

-4

-3

-2

-1 0 1 2 Normalized Frequency f/Rb

c. Compare the spectral efficiency with the system using rectangular pulses. Solution:

The first null bandwidth of the polar NRZ signal with cosine pulse shape is 1.5 Rb versus Rb for the rectangular pulse shape. However, the spectrum is much more compact thus avoiding significant adjacent channel interference. 9.10 Consider the unit impulse train in time with period T. ∞

δ p (t ) = ∑ δ (t − nT ) n =−∞

a. Show that the FS expansion for this signal is given by ∞

∑ δ (t − nT ) =

n =−∞

1 ∞ j 2π nt /T ∑e T n =−∞

Solution:

The FS expansion for this signal is given by ∞

δ p (t ) = ∑ Cn e j 2π nf t o

(*)

n =−∞

where fo =

1 T

T /2

Cn =

1 1 1 δ p (t )e − j 2π nfot dt = δ (t )e − j 2π nfot dt = ∫ ∫ TT T −T /2 T

Substituting into (*) yields

δ p (t ) =

1 ∞ j 2π nt /T ∑e T n =−∞

b. By taking FT of both sides, derive the identity ∞

∑ e− j 2π nfT =

n =−∞

1 ∞ ⎛ n⎞ δ⎜f − ⎟ ∑ T n =−∞ ⎝ T⎠

Solution:

By taking FT of both sides, we obtain ∞ ∞ ⎧ ∞ ⎫ ℑ ⎨ ∑ δ ( t − nT ) ⎬ = ∑ ℑ {δ ( t − nT )} = ∑ e − j 2π nfT n =−∞ ⎩ n =−∞ ⎭ n =−∞

1 ∞ ⎛ n⎞ ⎧1 ∞ ⎫ 1 ∞ ℑ ⎨ ∑ e j 2π nt /T ⎬ = ∑ ℑ e j 2π nt /T = ∑ δ ⎜ f − ⎟ T n=−∞ ⎝ T⎠ ⎩ T n =−∞ ⎭ T n =−∞

{

}

The identity is obtained by equating the left-hand sides as ∞

∑ e− j 2π nfT =

n =−∞

1 ∞ ⎛ n⎞ δ⎜f − ⎟ ∑ T n =−∞ ⎝ T⎠

9.11 Consider a baseband channel with bandwidth 64 kHz. a. If bipolar RZ signaling with RC pulses is used, show that the spectrum of 64 kb/s signal will fit into the channel when α =1. Find the absolute and 6-dB bandwidths. Solution:

For bipolar RZ signaling , Babs = 0.5(1 + α ) Rb . Substituting for Rb and α , we obtain Babs = 64 kHz = W = channel bandwidth

6-dB bandwidth of RC pulses = D / 2 = 32 kHz Alternatively, it can be calculated as 2

G RC ( f ) V(f) = = 1 0 − 0 .6 G R C (0 ) V (0 ) f = f − 6dB 2

⎡π T ⎛ 1⎧ 1 − α ⎞⎤ ⎫ −0.6 ⎨1 + cos ⎢ ⎜ f −6dB − ⎟ ⎥ ⎬ = 10 4⎩ 2T ⎠ ⎦ ⎭ ⎣α ⎝

⎡πT ⎛ 1⎧ 1 − α ⎞⎤ ⎫ −0.3 ⎨1 + cos ⎢ ⎜ f −6dB − ⎟ ⎥ ⎬ = 10 = 0.5 2⎩ 2T ⎠ ⎦ ⎭ ⎣α ⎝

⎡π T ⎛ 1 − α ⎞⎤ cos ⎢ ⎜ f −6dB − ⎟ =0 2T ⎠ ⎥⎦ ⎣α ⎝ πT ⎛ 1−α ⎞ π ⎜ f −6dB − ⎟= α ⎝ 2T ⎠ 2 2Tf −6dB − 1 + α = α 2Tf −6dB = 1 f −6dB =

D 1 = 2T 2

b. Repeat part (a) for the case of 144 kb/s quaternary PAM signaling. Choose appropriate value of α, if required. Solution:

For M-ary signaling, Babs = 0.5(1 + α ) D =

0.5(1 + α ) Rb . k

0.5(1 + α )144 × 103 2 128 1+ α = = 1.7778 72 α = 0.7778 64 × 103 =

9.12

A differential encoder is defined by the following Boolean expression d n = d n −1 ⊕ bn

where bn = current data bit into the differential encoder and dn = current differentially encoded bit. a. Assuming initial bit d 0 = 0 , calculate the differential encoder output for the input bit sequence 011110101. Solution:

Data bits bn Differentially encoded bits dn

0 1 1 1 1 0 1 0 1 0 1 1 1 1 1 0 0 1 1

b. Repeat (a) assuming initial bit d0 = 1 . Solution:

Data bits bn Differentially encoded bits dn

0 1 1 1 1 0 1 0 1 1 0 0 0 0 0 1 1 0 0

c. What conclusions do you make from (a) and (b) regarding the action of the encoder. Solution:

The effect of encoding in both cases is to toggle the differentially encoded bit an if bn is 0, and leave it unchanged if the incoming binary bit bn is 1. On the other hand, the differentially encoded bits in (b) are complement of those generated in (a). 9.13 Design the descrambler to recover the original data for the scrambler in Example 9.3. Verify that when the scrambled sequence cn is applied to the input of the descrambler, the output is the original sequence 10000000000. Solution:

The descrambler is shown in Figure.

cn-5

cn-4

cn-3

cn-2

cn-1 cn

Scrambled Data

User Data bn

We assume that the initial state of the shift register is set to all zeros. The descrambled data output is now be obtained from (9.63) as bn = cn ⊕ cn − 2 ⊕ cn −5

The calculation of descrambled bits is shown in the table below. cn

cn-1 Shift-register cn-2 contents cn-3 cn-4 cn-5 Data bits bn

0 0 0 0 0 1

1 0 0 0 0 0

0 1 0 0 0 0

1 0 1 0 0 0

0 1 0 1 0 0

1 0 1 0 1 0

1 1 0 1 0 0

1 1 1 0 1 0

0 1 1 1 0 0

1 0 1 1 1 0

1 1 0 1 1 0

0 1 1 0 1 0

Scrambled bits

9.14

A scrambler is shown in Figure P9.2. Design the descrambler to recover the original data. If a data sequence 100000001111 is applied to the input of the scrambler, determine the scrambled output sequence. Verify that the output is the original data sequence when the scrambled sequence cn is applied to the input of the descrambler. Solution:

The descrambler is shown in Figure.

cn-3

cn-2

cn-1

Scrambled data cn

User data bn

Descrambler

We assume that the initial state of the shift register is set to all zeros. The scrambled data output can now be obtained from (9.62) as cn = bn ⊕ cn − 2 ⊕ cn −3

Table below displays the calculation of scrambled bits.

Data bits

Shift-register contents Scrambled bits

cn-1 cn-2 cn-3 cn

0 0 0 1

1 0 0 0

0 1 0 1

1 0 1 1

1 1 0 1

1 1 1 0

0 1 1 0

0 0 1 1

1 0 0 1

1 1 0 0

0 1 1 1

1 0 1 0

We assume that the initial state of the shift register is set to all zeros. The descrambled data output can now be obtained from (9.63) as bn = cn ⊕ cn − 2 ⊕ cn −3

Table below displays the calculation of descrambled bits.

Scrambled bits

Shift-register contents Data bits

cn-1 cn-2 cn-3 bn

0 0 0 1

1 0 0 0

0 1 0 0

1 0 1 0

1 1 0 0

1 1 1 0

0 1 1 0

0 0 1 0

1 0 0 1

1 1 0 1

0 1 1 1

1 0 1 1

9.15 A 7-bit ADC samples video signal at 10.7386 MHz. a. Find the serial bit rate of the digital signal being generated. Solution:

Serial bit rate = 7 × 10.7386 = 75.17 Mb/s b. The digital signal is transmitted using RC pulses over a channel of bandwidth 12 MHz. If 16-ary PAM signaling is used, find the maximum roll-off factor α that can be used. Solution:

Symbol rate D =

75.17 × 106 = 18.8 × 106 symbols/sec 4

12 ×106 = 0.5(1 + α ) ×18.8 ×106 12 1+ α = = 1.2766 9.4 α = 0.2766

Chapter 10 10.1

Binary data is transmitted over an AWGN channel with a power spectral density N o / 2 = 10−9 W/Hz. Determine the signal amplitude A required to achieve a BER = 10 −6 , when the data rate is (a) 1 Mb/s, (b) 10 Mb/s, and (c) 100 Mb/s. Evaluate for unipolar NRZ and RZ line coding schemes. What is the channel first- null bandwidth required in each case? Solution: For unipolar NRZ signaling,

⎛ Eb ⎞ BER = Q ⎜⎜ ⎟⎟ using a MF receiver ⎝ No ⎠ N o / 2 = 10−9 ⇒ N o = 2 ×10−9 Using Table 6.1, we get x = 4.7535 for BER = 10-6.

∴

Eb 2 = ( 4.7535) = 22.6 ⇒ Eb = 4.519 ×10−8 No

Now Eb =

A 2Tb 2

(a) R = 1 Mbps, ∴Tb = 10−6 4.519 ×10 −8 × 2 A = ⇒ A = 0.3 V 10−6 2

(b) R = 10 Mbps, ∴Tb = 10−7 A2 =

4.519 ×10−8 × 2 ⇒ A = 0.95 V 10−7

⎛ Eb ⎞ A2Tb and BER = Q ⎜⎜ ⎟⎟ using a N 4 o ⎝ ⎠ MF receiver. From Table 6.1, we get x = 4.7535 for BER = 10-6. For unipolar RZ line coding scheme Eb =

∴

Eb 2 = ( 4.7535) = 22.6 ⇒ Eb = 4.519 ×10−8 No

Now (d) R = 1 Mbps, ∴Tb = 10−6 A2 =

4.519 × 10−8 × 4 ⇒ A = 0.425 V 10−6

(e) R = 10 Mbps, ∴Tb = 10−7 A2 =

4.519 ×10−8 × 4 ⇒ A = 1.34 V 10−7

(f) R = 100 Mbps, ∴Tb = 1×10−8 A2 =

4.519 ×10−8 × 4 ⇒ A = 4.25 V 10−8

The following table summarizes the results.

Bit Rate (Mbps)

Unipolar NRZ

Unipolar RZ

Amplitude (V) BW (MHz)

0.3

0.425

0.95

1.34

100

3.0

100

4.25

200

10.2 Consider a binary signaling system using sinusoidal pulses

⎛ πt ⎞ s1 (t ) = A sin ⎜ ⎟ , 0 ≤ t ≤ Tb ⎝ Tb ⎠ s 2 (t ) = 0

A correlation detector is used for the detection of transmitted symbols at the output of an AWGN channel with noise spectral density N o / 2 = 2.8 × 10−11 W/Hz. a. Determine an expression for the average probability of bit error assuming equiprobable binary data. Solution: The BER for a correlation or MF detector is given by ⎛ Eb ⎞ BERMF = Q ⎜ ⎜ N ⎟⎟ o ⎠ ⎝

where Eb = Average energy/bit =

1 1 E E1 + 0 = 1 2 2 2

⎛ πt ⎞ ⎛ 2π t ⎞ ⎤ A2Tb A2 Tb ⎡ E1 = ∫ A sin ⎜ ⎟ dt = ⎢1 − cos ⎜ ⎟ ⎥ dt = 0 2 ∫0 ⎢⎣ 2 ⎝ Tb ⎠ ⎝ Tb ⎠ ⎥⎦ Tb

N o / 2 = 2.8 ×10−11 ⇒ N o = 5.6 × 10−11 Substituting ⎛ A2T ⎞ b ⎟ BER = Q ⎜ ⎜ 4 No ⎟ ⎝ ⎠

b. Evaluate the bit error rate for A = 50 mV and the bit rate of 1 Mb/s. Solution: ⎛ ⎜ BER = Q ⎜ ⎜ ⎝

( 50 ×10 ) ×10 ⎞⎟ = Q ( 3.34 ) = 4.2 ×10 −3

4 × 5.6 × 10

−6

−11

⎟ ⎟ ⎠

−4

c. How much does the received pulse amplitude A have to increase for the same BER performance when the bit rate is doubled? Solution: To assure the same BER performance, the argument of Q function in the BER expression must remain the same when the bit rate is doubled. Therefore,

A2 × 0.5 ×10−6 11.16 × 4 × 5.6 ×10−5 2 = 3.34 ⇒ A = = 5 ×10−3 ⇒ A = 70.7 mV 4 × 5.6 ×10−11 0.5 10.3

A binary signaling system uses pulses s1 (t ) and s2 (t ), 0 ≤ t ≤ Tb to transmit binary data occurring with probabilities p and 1 − p , respectively . A matched filter detector is used for the detection of transmitted symbols at the output of an AWGN channel with noise spectral density N o / 2 W/Hz. a. Show that the optimum threshold value Vopt is given by

Vopt =

No s +s p log e + o1 o 2 2 1− p 2

where so1 and so 2 are given by (10.41) and (10.42), respectively. Solution: ⎛ s −V ⎞ ⎛V − s ⎞ BER = pQ ⎜ o1 T ⎟ + (1 − p )Q ⎜ T o 2 ⎟ ⎝ σo ⎠ ⎝ σo ⎠

d ( BER) p 1 1− p 1 − (V − s ) 2 /2σ 2 − (V − s )2 /2σ o2 = e opt o1 o − e opt o 2 =0 2 dVT V =V 2 2πσ o2 2 2 πσ o T opt 2 2 ⎛ p ⎞ (V − s ) ⎛ 1 − p ⎞ (Vopt − so 2 ) log e ⎜ ⎟ − opt 2 o1 = log e ⎜ − ⎟ 2σ o 2σ o2 ⎝2⎠ ⎝ 2 ⎠ ⎛ p⎞ ⎛ 1− p ⎞ 2 2σ o2 log e ⎜ ⎟ + 2Vopt so1 − so21 = 2σ o2 log e ⎜ ⎟ + 2Vopt so 2 − so 2 ⎝2⎠ ⎝ 2 ⎠

⎛ p ⎞ 2 2 2σ o2 log e ⎜ ⎟ + 2Vopt so1 − so1 = 2Vopt so 2 − so 2 ⎝ 1− p ⎠ Now solving for Vopt , we obtain

(

)

so22 − so21 ⎛ p ⎞ Vopt = + log ( so 2 − so1 ) e ⎜⎝ 1 − p ⎟⎠ 2 ( so 2 − so1 )

σ o2

⎛ p ⎞ so1 + so 2 + log e ⎜ 2 ( so 2 − so1 ) ⎝ 1 − p ⎟⎠

(*)

For MF detection

d 2 = so1 − so 2 = so 2 − so1

σ o2 =

N (s − s ) 2 No ∞ N H opt ( f ) df = o d 2 = o o 2 o1 ∫ 2 −∞ 2 2

Substituting into (*) yields Vopt =

No s +s p log e + o1 o 2 2 1− p 2

b. What is optimum threshold value when the binary signals are equiprobable? Solution: When the binary signals are equiprobable, p = 1 − p = Vopt =

so1 + so 2 2

1 . Substituting yields 2

c. Show that the average probability of error for the system is given by ⎛ so1 − so 2 N o ⎛ No s −s ⎞ p ⎞ p log e log e − + o1 o 2 ⎟ ⎜ ⎟ ⎜ 2 2 1− p ⎟ 2 1− p 2 ⎟ BER = pQ ⎜ + (1 − p )Q ⎜ ⎜ ⎟ ⎜ ⎟ N o ( so 2 − so1 ) N o ( so 2 − so1 ) ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠

Solution: Substituting Vopt into the BER expression yields

No so1 + so 2 ⎞ s +s p p ⎛ ⎛ No ⎞ + o1 o 2 − so 2 ⎟ log e ⎜ so1 − 2 log e 1 − p − ⎟ ⎜ 2 2 1− p 2 ⎟ + (1 − p )Q ⎜ ⎟ BER = pQ ⎜ σo σo ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ so1 − so 2 N o ⎛ No s −s ⎞ p ⎞ p − + o1 o 2 ⎟ log e log e ⎜ ⎟ ⎜ 2 2 1− p ⎟ 2 1− p 2 ⎟ = pQ ⎜ + (1 − p )Q ⎜ ⎜ ⎟ ⎜ ⎟ N o ( so 2 − so1 ) N o ( so 2 − so1 ) ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠

10.4 The binary signaling system in Problem 10.3 uses antipodal pulses s1 (t ) = AΠ ( t / Tb ) = − s2 (t ) . a. Show that the optimum threshold value Vopt is given by Vopt =

No p log e 2 1− p

Solution: For antipodal signaling, so1 = − so 2 . Substituting

σ o2

⎛ p ⎞ so1 + so 2 log e ⎜ + 2 ( so 2 − so1 ) ⎝ 1 − p ⎟⎠ N p = o log e 2 1− p

Vopt =

b. Calculate the value of optimum threshold Vopt for following a priori bit probabilities (i) p = 0.5,1 − p = 0.5 , (ii) p = 0.3,1 − p = 0.7 , p = 0.8,1 − p = 0.2 . Solution:

( p,1 − p ) ( 0.5, 0.5) ( 0.3,0.7 ) ( 0.8, 0.2 )

Vopt 0 −0.423N o 0.693 N o

c. Show that the average probability of error for the system is given by

No p ⎞ p ⎛ ⎛ No ⎞ ⎜ 2 Eb − 2 log e 1 − p ⎟ ⎜ 2 log e 1 − p + 2 Eb ⎟ ⎟ + (1 − p )Q ⎜ ⎟ BER = pQ ⎜ 2 N o Eb 2 N o Eb ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

Solution: By combining (10.43), (10.44), and (10.54), we obtain the following the following relationship for antipodal signaling . ∞

so1 − so 2 = d 2 = ∫ s1 (t ) − s2 (t ) dt = 4Eb . 2

−∞

Substituting No p ⎞ p ⎛ ⎛ No ⎞ ⎜ 2 Eb − 2 log e 1 − p ⎟ ⎜ 2 log e 1 − p + 2 Eb ⎟ ⎟ + (1 − p )Q ⎜ ⎟ BER = pQ ⎜ 2 N o Eb 2 N o Eb ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

When the binary signals are equiprobable

BER =

10.5

⎛ 2 Eb ⎞ 1 ⎛ 2 Eb ⎞ 1 ⎛ 2 Eb ⎞ Q⎜ + Q⎜ = Q⎜ ⎟ ⎟ ⎜ N ⎟⎟ 2 ⎜⎝ N o ⎟⎠ 2 ⎜⎝ N o ⎟⎠ o ⎠ ⎝

The binary signaling system in Problem 10.3 uses unipolar NRZ pulses s1 (t ) = AΠ ( t / Tb ) and s2 (t ) = 0 . a. Show that the optimum threshold value Vopt is given by

Vopt =

⎛ p ⎞ No log e ⎜ ⎟ + Eb 2 ⎝ 1− p ⎠

Solution:

Vopt =

σ o2

⎛ p ⎞ so1 + so 2 + log e ⎜ 2 ( so 2 − so1 ) ⎝ 1 − p ⎟⎠

For unipolar signaling

d 2 = so 2 − so1 = 2 Eb No 2 d = N o Eb 2 so1 + so 2 = Eb 2 A2Tb is average energy/bit. Substituting where Eb = 2 ⎛ p ⎞ N E Vopt = o b log e ⎜ ⎟ + Eb 2 Eb ⎝ 1− p ⎠

σ o2 =

⎛ p ⎞ No log e ⎜ ⎟ + Eb 2 ⎝ 1− p ⎠

b. Calculate the value of optimum threshold Vopt for following a priori bit probabilities (i) p = 0.5,1 − p = 0.5 , (ii) p = 0.3,1 − p = 0.7 , p = 0.8,1 − p = 0.2 . Solution:

( p,1 − p ) ( 0.5, 0.5) ( 0.3,0.7 ) ( 0.8, 0.2 )

Vopt Eb Eb − 0.423 N o Eb + 0.693N o

c. Write an expression for the average probability of error at the detector output. Solution: s −s ⎞ p ⎞ p ⎛ so1 − so 2 N o ⎛ No − + o1 o 2 ⎟ log e log e ⎜ ⎟ ⎜ 2 2 1− p 2 1− p 2 ⎟ + (1 − p )Q ⎜ ⎟ BER = pQ ⎜ σo σo ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ No p ⎞ p ⎛ ⎛ No ⎞ ⎜ Eb − 2 log e 1 − p ⎟ ⎜ 2 log e 1 − p + Eb ⎟ ⎟ + (1 − p )Q ⎜ ⎟ = pQ ⎜ N o Eb N o Eb ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

When the binary signals are equiprobable

BER =

10.6

⎛ Eb ⎞ 1 ⎛ Eb ⎞ 1 ⎛ Eb ⎞ Q⎜ + Q⎜ = Q⎜ ⎟ ⎟ ⎜ N ⎟⎟ 2 ⎜⎝ N o ⎟⎠ 2 ⎜⎝ N o ⎟⎠ o ⎠ ⎝

Consider the binary digital communication system using antipodal signaling. The received signal corresponding to binary “1” is given by r (t ) = s1 (t ) + n(t ) where s1 (t ) is shown in Figure P10.1 and n(t ) is AWGN with power spectral density N o / 2 Watts/Hz.

hopt (t )

(a)

s1 (t )

(b)

Tb / 2

−A

Tb / 2

−A

so1 (t )

(c)

(d)

A2Tb

φ1 (t ) − A Tb 0 −

Tb / 2

3Tb / 2

2Tb

A Tb

A2Tb 2

a. Sketch the impulse response of the filter matched to s1 (t ) . Solution: The impulse response of the filter matched to s1 (t ) is sketched in Figure (b). b. Sketch the output of the filter matched to the input s1 (t ) .

Solution:

The output of the filter matched to s1 (t ) is sketched in Figure (c). c. Draw the signal constellation. What is the d min for the signal set? Solution:

The signal constellation is sketched in Figure (d).

d min = 2 A Tb d. Write an expression for BER in terms of A and No. Solution:

The BER for antipodal signaling using a matched filter detector is given by ⎛ 2 Eb ⎞ BER = Q ⎜ ⎜ N ⎟⎟ o ⎠ ⎝

In the present case, Eb =

1 2 1 A Tb + A2Tb = A2Tb . Substituting 2 2

⎛ 2 A2T ⎞ b ⎟ BER = ⎜ ⎜ ⎟ N o ⎝ ⎠

10.7

A binary digital communication system uses orthogonal pulses shown in Figure P10.2. The channel noise is white Gaussian with a power spectral density N o / 2 W/Hz. a. Draw block diagram of the ML detector. Sketch impulse responses of matched filters. Solution:

In the matched-filter implementation of the ML detector, the impulse responses of matched filters are given by

h1 (t ) = φ1 (Tb − t ) h2 (t ) = φ2 (Tb − t ) where

φ1 ( t ) =

s1 (t ) A Tb

φ2 ( t ) =

s2 (t ) A Tb

The block diagram of ML detector is shown in Figure (a). The impulse responses of matched filters are displayed in Figure (b). T

r1 = ∫ r (t )φ1 (t )dt

(a)

h1 (t ) = φ1 (Tb − t )

Threshold Comparator

t = Tb

r (t ) = s1 (t ) or s2 (t )

+ −

n(t ) h2 (t ) = φ2 (Tb − t )

t = Tb T

r2 = ∫ r (t )φ2 (t )dt 0

(b)

h1 (t )

h2 (t )

1/ Tb

−1/ Tb

Tb 4

Tb 2

Tb / 2

−1/ Tb

b. Plot the output of the filter matched to s1 (t ) when the input is s1 (t ) . Repeat when the input is instead s2 (t ) . Solution:

The outputs of the matched filter with impulse response h1 (t ) to input pulses s1 (t ) and s2 (t ) are illustrated in Figure.

h1 (t ) * s1 (t ) / A Tb

1 0.8

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

0.2

0.4

0.6

0.8

h1 (t ) * s2 (t ) / A Tb

1 1.2 time (t/Tb)

1.4

1.6

1.8

-1

0.2

0.4

0.6

0.8

1 1.2 time (t/Tb)

1.4

1.6

1.8

1.6

1.8

c. Plot the output of the filter matched to s2 (t ) when the input is s2 (t ) . Repeat when the input is instead s1 (t ) . What do you conclude from (b) and (c)? Solution:

The outputs of the matched filter with impulse response h2 (t ) to input pulses s2 (t ) and s1 (t ) are illustrated in Figure.

h2 (t ) * s2 (t ) / A Tb

h2 (t ) * s1 (t ) / A Tb

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8

-1

0.2

0.4

0.6

0.8

1 1.2 time (t/Tb)

1.4

1.6

1.8

-1

0.2

0.4

0.6

0.8

1 1.2 time (t/Tb)

1.4

We observe that response of the filter matched to one signal pulse (say, s1 (t ) ) is zero at the sampling instant Tb when the input is other signal pulse (say s2 (t ) ).

d. Draw the signal constellation. What is the d min for the signal set? Solution:

φ2 (t )

(

s 2 = 0, Eb

)

d min s1 =

( E , 0)

φ1 (t )

The d min for the signal set is

d min = 2 Eb Substituting Eb = A2Tb , d min can be expressed as

d min = A 2Tb e. Write an expression for BER in terms of A and No. Solution:

Since the waveforms s1 (t ) and s2 (t ) are orthogonal, the BER is given by

⎛ A2T ⎞ ⎛ Eb ⎞ b BER = Q ⎜ = Q ⎜ ⎟ ⎜ N ⎟⎟ ⎜ ⎟ N o o ⎝ ⎠ ⎝ ⎠ 10.8 Consider the signal set displayed in Figure P10.3. a. Apply the Gram-Schmidt procedure to determine an orthonormal basis for the signal set. Solution:

We take s1 (t ) as the first basis function. T

E1 = ∫ s (t )dt = A ∫ dt = 2 1

T /2

A2T 2

Figure P10.3

s1 (t )

s2 (t )

T 2

s3 (t )

s4 (t )

0 T 2

−A

0 T 2

−A

The unit energy function φ1 (t ) is obtained by dividing s1 (t ) by E1 . That is, ⎧ 2 s1 (t ) , 0≤t ≤T ⎪ φ1 (t ) = ⎨ T A ⎪0, otherwise ⎩

Although, it is obvious that s2 (t ) is orthogonal to s1 (t ) , we will proceed to compute the function θ 2 (t ) using (10.88).

θ 2 (t ) = s2 (t ) − c21φ1 (t ) where

c21 = ( s2 (t ) • φ1 (t ) ) = ∫ s2 (t )φ1 (t )dt = ∫ 0dt = 0 T

This proves that s2 (t ) is orthogonal to s1 (t ) . Therefore,

θ 2 (t ) = s2 (t ) T

T /2

2 2 ∫ s2 (t )dt = A ∫ dt = A

Since θ2 = E2 =

T , the second orthonormal 2

function is obtained as

φ2 (t ) =

⎧ 2 s2 (t ) , 0≤t ≤T ⎪ =⎨ T A E2 ⎪ otherwise ⎩0,

θ 2 (t )

To obtain the next orthonormal function, the function θ 3 (t ) is computed as

θ3 (t ) = s3 (t ) − c31φ1 (t ) − c32φ2 (t ) where T

c31 = ( s3 (t ) • φ1 (t ) ) = ∫ s3 (t )φ1 (t )dt = 0

c32 = ( s3 (t ) • φ2 (t ) ) = ∫ s3 (t )φ2 (t )dt = − 0

1 2 2 AT T A2 dt = =A ∫ A T T /2 T 2 2 T /2

1 2 2 AT T A2 dt = − = −A ∫ A T 0 T 2 2

Substituting the values of c31 and c32 , we obtain

θ3 (t ) = s3 (t ) − A

T T φ1 (t ) + A φ2 (t ) = s3 (t ) − s1 (t ) + s2 (t ) = 0 2 2

This implies that s3 (t ) is a linear combination of φ1 (t ) and φ2 (t ) . That is, s3 (t ) = A

T T φ1 (t ) − A φ2 (t ) 2 2

Similarly, it is easy to conclude that s4 (t ) is a linear combination of φ1 (t ) and φ2 (t ) . That is, s4 (t ) = − A

T T φ1 (t ) + A φ2 (t ) 2 2

Figure (a) displays the orthonormal basis of the signal set in Figure P10.3. b. Plot the signal constellation corresponding to the signal set.

Solution:

The signal constellation for the signal set in Figure P10.3 is displayed in Figure (b).

(a)

φ1 (t )

φ2 (t )

2/T

T /2

φ2 (t )

2A T / 2

(b)

A T /2

s1 −A T / 2

2A T / 2

A T /2

−A T / 2

φ1 (t )

10.9 Consider the signal set si (t ) =

2 Es cos ( 2π f c t + ψ i ) , 0 ≤ t ≤ T T

i = 1,......,8

where 5π 3π 7π ⎫ ⎧ π π 3π , , ,π , , , ⎬ 4 2 4 ⎭ ⎩ 4 2 4

ψ i ∈ ⎨0,

a. What is the dimensionality of signal space? Solution:

Dimensionality of signal space = 2 16 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

b. Determine the basis vectors for the signal set. Solution:

By choosing the basis functions

φ1 (t ) =

2 cos ( 2π f c t ) T

φ2 (t ) =

2 sin ( 2π f c t ) , T

s i = ( si1 Es , si 2 Es )

si1 = cos(ψ i ) si 2 = sin(ψ i ) c. Draw the constellation diagram of the signal set.

s 2 = (0, Es )

φ2 (t )

s 3 = (− Es / 2, Es / 2)

s1 = ( Es / 2, Es / 2)

s 4 = (− Es , 0)

φ1 (t ) s 0 = ( Es , 0)

s 5 = (− Es / 2, − Es / 2)

s 7 = ( Es / 2, − Es / 2)

s 6 = (0, − Es )

10.10 Consider the 2 basis functions shown in Figure P10.4 (a). Sketch the waveforms corresponding to the points in the constellation shown in Figures P10.4 (b) and (c).

Figure P10.4

φ2 (t )

φ1 (t ) 1/ T

1/ T 0 T 4

3T 4

T 2

−1/ T

φ2 (t )

(a)

φ2 (t )

A − 2

s2 •

(c)

• s1

A − 2

33A A

φ1 (t )

−A

φ1 (t )

−A

(b)

−3A

Solution:

For constellation (b), the waveforms corresponding to signal points s1 and s 2 are obtained as s1 = ( A, A) ⇒ s1 (t ) = s11φ1 (t ) + s12φ2 (t ) = Aφ1 (t ) + Aφ2 (t ) s 2 = (− A / 2, − A / 2) ⇒ s2 (t ) = s21φ1 (t ) + s22φ2 (t ) = −

A A φ1 (t ) − φ2 (t ) 2 2

The waveforms are displayed in Figure.

s1 (t )

s2 (t )

2A / T A/ T 3T / 4

T /4

−2 A / T

t T

T /4

3T / 4

−A / T 18

For constellation (c), the waveforms corresponding to signal points s1 , s 2 , s 3 and s 4 are obtained as s1 = ( A, A) ⇒ s1 (t ) = s11φ1 (t ) + s12φ2 (t ) = Aφ1 (t ) + Aφ2 (t ) s 2 = (− A,3 A) ⇒ s2 (t ) = s21φ1 (t ) + s22φ2 (t ) = − Aφ1 (t ) + 3 Aφ2 (t ) s 3 = (− A, − A) ⇒ s2 (t ) = s31φ1 (t ) + s32φ2 (t ) = − Aφ1 (t ) − Aφ2 (t ) s 4 = ( A, −3 A) ⇒ s4 (t ) = s41φ1 (t ) + s42φ2 (t ) = Aφ1 (t ) − 3 Aφ2 (t )

The waveforms are displayed in Figure.

s1 (t )

s3 (t ) 2A / T

2A / T

3T / 4

0 T /4

3T / 4 T

−2 A / T

s2 (t )

s4 (t )

4A / T

2A / T 0

−2 A / T −4 A / T

T /4

T T /2

2A / T

−2 A / T

T /2

−4 A / T

10.11 Using the basis functions in Figure P10.5(a), sketch the waveforms corresponding the points of the constellation shown in Figure P10.5(b). Solution:

The signal waveforms corresponding to the constellation in Figure P10.5(b) are shown in Figure.

φ2 (t ) 2A

φ1 (t )

T s0 (t ) 2 A 0

T s1 (t ) 1 A 0

s 1(t)

s 0(t)

0.5

T s2 (t ) -1 A -2

T s3 (t ) -2 A

0.5

T s4 (t ) -2 A -4

T s5 (t ) -1 A

0.5

0.5 time (t/T)

s 5(t)

s 4(t)

-2 0

0.5

1 4

2 1

s 7(t)

s 6(t)

-4 0

T s6 (t ) A

0.5

s 3(t)

s 2(t)

T s7 (t ) 2 A 0

0 -1 0

0.5 time (t/T)

10.12 Consider the signal constellation shown in Figure P10.5(b). a. Derive the exact average probability of symbol error for the constellation. Solution:

For the constellation in Figure 10.5(b), d min = 2 A . The probability of symbol error can be exactly computed by noting that the conditional probability of a correct decision falls into one of two categories:

(i) Four corner points ( s 0 , s 3 , s 4 and s 7 ) P {correct decision s 0 sent} = P { n1 > − d min / 2 ∩ n2 > − d min / 2} = P { n1 > − d min / 2} P { n2 > − d min / 2} ⎛ ⎛ d ⎞⎞⎛ ⎛ d ⎞⎞ = ⎜ 1 − Q ⎜ min ⎟ ⎟ ⎜ 1 − Q ⎜ min ⎟ ⎟ ⎜ 2N ⎟ ⎟ ⎜ 2N ⎟ ⎟ ⎜ ⎜ o ⎠⎠⎝ o ⎠⎠ ⎝ ⎝ ⎝ 2

⎛ d ⎞ ⎡ ⎛ d ⎞⎤ ⎛ d ⎞ = 1 − 2Q ⎜ min ⎟ + ⎢Q ⎜ min ⎟ ⎥ ≈ 1 − 2Q ⎜ min ⎟ ⎜ 2 N ⎟ ⎢ ⎜ 2 N ⎟⎥ ⎜ 2N ⎟ o ⎠ o ⎠⎦ o ⎠ ⎝ ⎝ ⎣ ⎝

⎛ d ⎞ P {error s 0 sent} = 2Q ⎜ min ⎟ ⎜ 2N ⎟ o ⎠ ⎝

= P {error s 3 sent} = P {error s 4 sent} = P {error s 7 sent}

(ii) Four inner points ( s1 , s 2 , s 5 and s 6 ) P {correct decision s1 sent} = P {− d min / 2 < n1 < d min / 2 ∩ n2 > − d min / 2} = P {− d min / 2 < n1 < d min / 2} P {n2 > − d min / 2} ⎛ ⎛ d ⎞⎞⎛ ⎛ d ⎞⎞ = ⎜ 1 − 2Q ⎜ min ⎟ ⎟ ⎜ 1 − Q ⎜ min ⎟ ⎟ ⎜ 2N ⎟ ⎟ ⎜ ⎜ 2N ⎟ ⎟ ⎜ o ⎠⎠⎝ o ⎠⎠ ⎝ ⎝ ⎝ 2

⎡ ⎛ d ⎛ d ⎞ ⎞⎤ ⎛ d ⎞ = 1 − 3Q ⎜ min ⎟ + 2 ⎢Q ⎜ min ⎟ ⎥ ≈ 1 − 3Q ⎜ min ⎟ ⎜ 2N ⎟ ⎜ 2N ⎟ ⎢⎣ ⎜⎝ 2 N o ⎟⎠ ⎥⎦ o ⎠ o ⎠ ⎝ ⎝

⎛ d ⎞ P {error s1 sent} = 3Q ⎜ min ⎟ ⎜ 2N ⎟ o ⎠ ⎝

= P {error s 2 sent} = P {error s 5 sent} = P {error s 6 sent}

We now obtain the following expression for the average probability of symbol error Pe = =

⎛ d ⎞⎞ 4 ⎛ ⎛ d ⎞⎞ 1 M 4⎛ P {error s i sent} = ⎜ 2Q ⎜ min ⎟ ⎟ + ⎜ 3Q ⎜ min ⎟ ⎟ ∑ ⎜ 2N ⎟ ⎟ 8 ⎜ ⎜ 2N ⎟ ⎟ M i =1 8 ⎜⎝ o ⎠⎠ o ⎠⎠ ⎝ ⎝ ⎝ 5 ⎛ d min ⎞ 5 ⎛ 2 A2 ⎞ Q⎜ ⎟ ⎟ = Q⎜ 2 ⎜⎝ 2 N o ⎟⎠ 2 ⎜⎝ N o ⎟⎠

b. Compute the average probability of symbol error using the nearest neighbor estimate. Comment. Solution:

The number of minimum distance pairs for the constellation is M. Substituting K = M = 8 into (10.156) yields ⎛ 2 A2 ⎞ ⎞ ⎛ d ⎞ 8 ⎛ d Pe ≈ 2 Q ⎜ min ⎟ = 2Q ⎜ min ⎟ = 2Q ⎜ ⎟ ⎜ 2N ⎟ ⎜ No ⎟ 8 ⎜⎝ 2 N o ⎟⎠ o ⎝ ⎠ ⎝ ⎠ The average probability of symbol error estimate using the nearest neighbor approximation is very accurate for this constellation. 10.13 Consider the signal constellations shown in Figure P10.6. a. Determine the average energy of each signal constellation. Solution:

The average energy of a signal constellation Es is

Es =

1 M si 2 ∑ M i =1

Constellation (a) Es =

1 [ 2 Eb + 2Eb + 2Eb + 2 Eb ] = 2 Eb 4

Constellation (b) Es =

1 [0 + 4 Eb + 4 Eb + 8Eb ] = 4 Eb 4

Constellation (c) Es =

1 [ 2 Eb + 2Eb + 2Eb + 2 Eb ] = 2 Eb 4

b. Sketch decision regions in each case.

Solution:

φ2(t)

φ2(t) (− Eb , Eb )

( Eb , Eb )

s3 (0, 0)

(− Eb , − Eb )

φ1(t)

(2 Eb , 2 Eb )

(0, 2 Eb ) s2

( Eb , − Eb )

φ1(t)

(2 Eb , 0)

(b)

φ2(t)

(a)

s2 (0, 2 Eb )

D3 s3

(− 2 Eb , 0)

( 2 Eb , 0)

(0, − 2 Eb )

(c)

c. Write the average probability of symbol error for each constellation using the nearest neighbor bound. Solution:

For each constellation, the number of minimum distance pairs is 4. Also d min = 2 Eb in every case. Substituting K = 4 into (10.156) yields ⎞ ⎛ 2 Eb ⎞ ⎛ 2 Eb ⎞ 4 ⎛ d Pe ≈ 2 Q ⎜ min ⎟ = 2Q ⎜ ⎟ = 2Q ⎜⎜ ⎟⎟ ⎜ 2N ⎟ 4 ⎜⎝ 2 N o ⎟⎠ o ⎠ ⎝ No ⎠ ⎝

d. From this example, what conclusions can you make about the constellation rotation and translation on the error performance of the signal set. Solution:

Although average energy of the signal set in (b) is twice that in (a) and (c), d min remains the same. Consequently, twice the energy is required in (c) for the same error performance. Since the translation and rotation of a signal constellation does not change distances between signal pairs, the average probability of symbol error remains the same. The signal energy efficient constellations, however, are centerd about the centroid (origin in the present example) of the signal set. 10.14 Consider the 8-PAM signal with constellation in Figure P10.7. a. Draw the block diagram of the correlation detector. Solution:

r (t ) = s (t ) + n (t ) ×

∫

T 0

t =T

3-bit ADC

Bit sequence

φ1 (t ) ADC: Mid-rise Analog-to Digital converter

b. Sketch the transfer characteristic of the threshold comparator. Solution:

The transfer characteristic of the threshold comparator is displayed in Figure. c. Determine the estimated symbol sequence and corresponding bit sequence for the correlator output samples {+0.12, −0.201, +0.71, −1.55, −0.6,1.25} . Solution:

The estimated symbol sequence is {+0.25, −0.25, +0.75, −1.75, −0.75,1.25} . The corresponding bit sequence is {000,100, 001,110,101, 011} .

y[n] 010

1.75 011 1.25

001

0.75

0.25 −2.0

−1.5

−1.0

−0.5 100

101

111

000

ro[n] 0 0.5 −0.25

1.0

1.5

2.0

−0.75

−1.25

−1.75

110

10.15 Consider the 8-point signal constellation shown in Figure P10.8. a. Determine the symbol rate if the desired bit rate is 45 Mbps. Solution:

Since each symbol conveys 3 bits of information, the resultant symbol rate is D=

45 ×106 = 15 × 106 symbols/sec 3

b. Calculate the average energy of the signal set. Solution: 1 44 2 11 2 Es = ⎡⎣ 4 × ( 2 A2 ) + 4 × ( 9 A2 ) ⎤⎦ = A = A 8 8 2

c. What is the d min for the signal set? Solution: d min = 2 A =

8 Es 11

d. Use the nearest neighbor bound to estimate the probability of bit error. Can you improve the estimate by including additional terms? Solution:

Pe ≈

2 K ⎛ d min ⎞ Q⎜ ⎟ M ⎜⎝ 2 N o ⎟⎠

Here K = 4, M =8. Therefore,

Pe ≈

⎛ 4 Es ⎞ 8 Es ⎞ 2× 4 ⎛ Q ⎜⎜ ⎟⎟ = Q ⎜⎜ ⎟⎟ 8 ⎝ 11× 2 N o ⎠ ⎝ 11N o ⎠

Probability of bit error =

Pe 1 ⎛ 4 Es ⎞ 1 ⎛ 4 × 3Eb ⎞ ≈ Q⎜ ⎟ = Q⎜ ⎟ k 3 ⎜⎝ 11N o ⎟⎠ 3 ⎜⎝ 11N o ⎟⎠

If we include 8 signal pairs at distance 5, we get a better estimate

1 ⎛ 4 × 3Eb ⎞ 2 ⎛ 5 × 3Eb ⎞ Probability of bit error ≈ Q ⎜⎜ ⎟ + Q⎜ ⎟ 3 ⎝ 11N o ⎟⎠ 3 ⎜⎝ 11N o ⎟⎠ 10.16 Consider the 8-point signal constellation shown in Figure P10.9. a. Assign the bits to each point in the signal constellation using Gray coding. Solution:

See Figure

φ2 (t ) D5

D4 101

•

111 3A

011

•

001

φ1 (t ) D2

•

110

•

−3A

000

•

−3A

010

•

−A

100

−A

D7 A

b. Calculate the average energy of the signal set. Solution: 1 Es = ⎡⎣ 4 × ( 2 A2 ) + 4 × (18 A2 ) ⎤⎦ = 10 A2 8

c. Sketch ML decision regions Di for the signal set. Solution:

See Figure d. Use the nearest neighbor bound to estimate the probability of bit error. Solution:

The number of minimum distance pairs for the constellation is 4. Substituting K = 4 into (10.156) and applying (10.163) yields 2 × 4 ⎛ d min ⎞ 1 ⎛ 2 A ⎞ 1 ⎛ 2 A2 ⎞ BER ≈ Q⎜ ⎟ ⎟ = Q⎜ ⎟ = Q⎜ 3 × 8 ⎜⎝ 2 N o ⎟⎠ 3 ⎜⎝ 2 N o ⎟⎠ 3 ⎜⎝ N o ⎟⎠

Chapter 11 11.1 Digital data at 100 kbps is transmitted over an AWGN channel with a power spectral density N o / 2 = 10 −10 W/Hz. Calculate the average received power required to achieve a BER = 10-6 for the following modulation schemes: a. Coherent Binary ASK b. Coherent Binary PSK c. Coherent Binary FSK Solution: 10−6 = Q ( x) ⇒ x = 4.7534 a. For coherent BASK, BER = Q( Eb / N o ) . To achieve a BER = 10-6, we have Eb / N o = 4.7534 ⇒

Eb = 22.6 No

Eb = 22.6 × 2 ×10−10 For BASK, the required average power Pav is obtained as Pav =

Eb = Eb Rb = 22.6 × 2 × 10−10 × 105 = 452 µ W= − 3.45 dBm Tb

b. For coherent BPSK, BER = Q( 2 Eb / N o ) . To achieve a BER = 10-6, we have 2 Eb / N o = 4.7534 ⇒

Eb = 11.3 No

Eb = 11.3 × 2 ×10−10 For BPSK, the required average power Pav is obtained as Pav =

Eb = Eb Rb = 11.3 × 2 × 10−10 × 105 = 226 µ W= − 6.45 dBm Tb

c. For coherent BFSK, BER = Q( Eb / N o ) . To achieve a BER = 10-6, we have

Eb / N o = 4.7534 ⇒

Eb = 22.6 No

Eb = 22.6 × 2 ×10−10 For BFSK, the required average power Pav is obtained as Pav =

Eb = Eb Rb = 22.6 × 2 × 10−10 × 105 = 452 µ W= − 3.45 dBm Tb

11.2 A binary digital carrier system with average transmitted power 100 mW has to be designed for the worst case transmission loss of 60 dB, and N o = 10 −12 W/Hz. Find the maximum allowable bit rate for a BER = 10-5 using: (a) BPSK (b) DBPSK (c) noncoherent BFSK. Solution: Average transmitted power = 20 dBm Average received power Pav = 20 – 60 = − 40 dBm = 0.1 µW Pav 10−7 = Eb = PavTb = Rb Rb 10−5 = Q( x) ⇒ x = 4.265 BPSK BER = Q( 2 Eb / N o ) To achieve a BER = 10-5, we have 2 Eb / N o = 4.265 ⇒

Eb = 9.095 No

10−7 105 R = 9.095 ⇒ = = 11 kbps b 10−12 Rb 9.095 DBPSK

⎛ 0.5 ⎞ 5 BER = 0.5e− Eb / No ⇒ Eb / No = log e ⎜ ⎟ = log e ( 0.5 ×10 ) = 10.82 ⎝ BER ⎠

10−7 105 = 10.82 ⇒ R = = 9.242 kbps b 10−12 Rb 10.82 Noncoherent BFSK

⎛ 0.5 ⎞ 5 BER = 0.5e− Eb /2 No ⇒ Eb / 2 No = log e ⎜ ⎟ = log e ( 0.5 ×10 ) = 10.82 ⎝ BER ⎠ 10−7 105 10.82 = ⇒ R = = 4.62 kbps b 2 × 10−12 Rb 2 × 10.82 11.3

Binary data 01001011101 is to be transmitted using DBPSK. Assume that the carrier frequency fc = 2Rb and the pulse shape is unipolar NRZ. a. Sketch the differential encoder and modulator output waveforms

I(t)

Differential Bits Bit Sequence

Solution:

DBPSK Waveforms 1 0.5 0 0

t/Tb 1 0.5 0 0

5 t/Tb

1 0 -1 0

x(t)

t/Tb 2 0 -2 0

5 t/Tb

b. Show that the DBPSK demodulator reconstructs the original data in the absence of noise 3 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Solution: Differentially encoded bits d n Threshold-comparison sign Decoded differential bit dˆ

1 1 0 0 0 1 1 0 1 0 0 1 + − + + − + − − − + − 1 1 0 0 0 1 1 0 1 0 0 1

Regenerated data bits bˆn

0 1 0 0 1 0 1 1 1 0 1

11.4 Binary string 01001011101 is to be transmitted using QPSK signal. Assume that the carrier frequency fc = Rb and the pulse shape is unipolar NRZ. Sketch a. Sketch in-phase and quadrature baseband waveforms I(t) and Q(t) b. Sketch in-phase and quadrature carrier modulated waveforms I (t ) cos ( 2π f c t ) and Q (t ) sin ( 2π f c t ) c. Sketch the QPSK waveform.

Bit Sequence

Solution:

1 0.5 0 0

5 t/Tb

I(t)

1 0 -1

Q(t)

1 0

Q(t)sin(wc t)

I(t)cos(wc t)

-1

1 0 -1

x(t)

2 0 -2

11.5 Binary string 01001011101 is to be transmitted using OQPSK signal. Assume that the carrier frequency fc = Rb and the pulse shape is unipolar NRZ. a. Sketch in-phase and quadrature baseband waveforms I(t) and Q(t) b. Sketch in-phase and quadrature carrier modulated waveforms I (t ) cos ( 2π f c t ) and Q (t ) sin ( 2π f c t ) c. Sketch the OQPSK waveform

Bit Sequence

Solution:

1 0.5 0 0

5 t/Tb

I(t)

1 0 -1

Q(t)

1 0

Q(t)sin(wc t)

I(t)cos(wc t)

-1

1 0 -1

x(t)

2 0 -2

11.6

Binary string 01001011101 is to be transmitted using MSK signal. Assume that the carrier frequency fc = Rb and the pulse shape is unipolar NRZ. a. Sketch in-phase and quadrature baseband waveforms I(t) and Q(t) b. Sketch in-phase and quadrature carrier modulated waveforms I (t ) cos ( 2π f c t ) and Q (t ) sin ( 2π f c t ) c. Sketch the MSK waveform.

Bit Sequence

Solution:

1 0.5 0 0

5 t/Tb

I(t)

1 0 -1

Q(t)

1 0

Q(t)sin(wc t)

I(t)cos(wc t)

-1

1 0 -1

x(t)

1 0 -1

ψ(t) − ψ(0)

d. Plot the phase trellis for the bit sequence

180o 90o

−90o 0

0 0

−180o

t Tb

2Tb 3Tb 4Tb 5Tb 6Tb 7Tb 8Tb 9Tb 10Tb 11Tb

11.7 A channel with bandwidth of 4 kHz is available what data rates can be supported by the following modulation schemes. Assume RC pulses with a roll-off factor α = 0.5. Ignore noise. a. BPSK Solution: The absolute bandwidth is related to the bit rate (from Table 11.4) by BBPSK = (1 + α ) Rb = 4000 Hz Rb =

4000 = 2.666 kb/s 1.5

b. QPSK Solution: The absolute bandwidth is related to the bit rate (from Table 11.4) by BQPSK = (1 + α ) D = Rb =

(1 + α ) Rb = 4000 Hz 2

2 × 4000 = 5.333 kb/s 1.5

c. MSK

Solution: The absolute bandwidth is related to the bit rate (Table 11.4) by 3(1 + α ) Rb = 4000 Hz 4 4 × 4000 Rb = = 3.555 kb/s 3 × 1.5 BMSK =

d. 8-PSK Solution: The absolute bandwidth is related to the bit rate (from Table 11.4) by B8− PSK = (1 + α ) D = Rb =

(1 + α ) Rb = 4000 Hz 3

3 × 4000 = 8 kb/s 1.5

e. 64-QAM Solution: The absolute bandwidth is related to the bit rate (from Table 11.4) by B64 −QAM = (1 + α ) D = Rb =

(1 + α ) Rb = 4000 Hz 6

6 × 4000 = 16 kb/s 1.5

f. 256-QAM Solution: The absolute bandwidth is related to the bit rate (from Table 11.4) by B2564 −QAM = (1 + α ) D = Rb =

(1 + α ) Rb = 4000 Hz 8

8 × 4000 = 21.333 kb/s 1.5

11.8 Repeat Problem 11.7 assuming that the modulator uses unipolar NRZ pulse shaping. a. BPSK Solution: The null-null bandwidth is related to the bit rate (from Table 11.4) by BBPSK = 2 Rb = 4000 Hz Rb =

4000 = 2.0 kb/s 2

b. QPSK Solution: The null-null bandwidth is related to the bit rate (from Table 11.4) by BQPSK = 2 D = Rb = 4000 Hz Rb = 4 kb/s

c. MSK Solution: The null-null bandwidth is related to the bit rate (Table 11.4) by 3Rb = 4000 Hz 2 2 × 4000 Rb = = 2.666 kb/s 3 BMSK =

d. 8-PSK Solution: The null-null bandwidth is related to the bit rate (from Table 11.4) by B8− PSK = 2 D = Rb =

2 Rb = 4000 Hz 3

3 × 4000 = 6 kb/s 2

e. 64-QAM Solution: The null-null bandwidth is related to the bit rate (from Table 11.4) by B64−QAM = 2 D = Rb =

2 Rb = 4000 Hz 6

6 × 4000 = 12 kb/s 2

f. 256-QAM Solution: The null-null bandwidth is related to the bit rate (from Table 11.4) by B2564 −QAM = 2 D = Rb =

2 Rb = 4000 Hz 8

8 × 4000 = 16 kb/s 2

11.9 Consider a telephone line channel that is equalized to allow bandpass data transmission over a frequency range of 300 to 3300 Hz. The available channel bandwidth is 3000 Hz with the mid-channel frequency of 1800 Hz. Assume that the modulator uses RC pulse shaping. a. For QPSK transmission at 4,800 bps with fc = 1800 Hz, show that the spectrum of this signal will fit into the channel with roll-off factor α = 0.25. Find the absolute and 6-dB bandwidths of the QPSK signal. Solution: Rb = 4800 b/s, D = 2400 symbols/sec, α = 0.25 BQPSK = (1 + α ) D = 1.25 × 2400 = 3000 Hz

Thus spectrum of QPSK with RC pulses at 2400 symbols/second ( = 4800 bps) will fit into the allocated channel bandwidth. The 6-dB bandwidth of RC pulses for any α > 0 is D/2. Therefore, 6-dB bandwidth of QPSK signal using RC pulses is BQPSK (6-dB) = 2 × D / 2 = D = 2400 Hz

b. Repeat part (a) for the case of 6000 bps 8-PSK signaling. Choose the appropriate value of α. Solution: Rb = 6000 b/s, D = 2000 symbols/sec 3000 = B8 PSK = (1 + α )2000 ⇒ α = 0.5 Therefore, the spectrum of 8-PSK signaling at 2000 symbols/second ( = 6000 bps) using RC pulses ( α ≤ 0.5 ) will fit into the allocated channel bandwidth. The 6 dB bandwidth of 8-PSK signal is 2×D/2 = 2000 Hz. 11.10 A digital cellular system transmits data using QPSK over a BP channel with center frequency 1.93 GHz and bandwidth of 1.25 MHz. The modulator uses RC pulse shaping with a roll-off factor α = 0.5. Assume that the channel introduces AWGN of one-sided spectral density No = 10-9. a. Determine the maximum bit rate Rb of the system. Solution: Rb (1 + α ) 2 2 × 1.25 × 106 Rb = = 1.6667 × 106 1.5

1.25 × 106 =

b. Calculate the average power required at the receiver in order to guarantee a BER less than or equal to 10-6. Solution: BER = Q( 2 Eb / N o ) To achieve a BER = 10-6, we have 2 Eb / N o = 4.7534 ⇒

2 Eb = 22.6 No

Eb = 11.3 ⇒ Eb = 11.3 × 10−9 No Pav = Eb × Rb = 11.3 ×10−9 ×1.6667 × 106 = 18.83 mW = 12.75 dBm

11.11 Binary data is to be transmitted at 4 Mbps on a radio channel having 1.25 MHz bandwidth. a. Select the modulation method that minimizes signal energy, and calculate E b / N o in dB needed for a BER = 10-6. Solution: We need spectral efficiency of 4/1.25 = 3.2 bps/Hz. We consider M-PSK and M-QAM for achieving this. For both these schemes, spectral efficiency = k/2. Therefore we need k = 6.4. We choose k = 8 which corresponds to M = 256. We consider 256-PSK and 256-QAM. 256-PSK BERMPSK ≈

2 ⎡ 2 Eb k 2 ⎛ π ⎞ ⎤ 1 ⎡ 16 Eb ⎛ π ⎞⎤ Q⎢ sin ⎜ ⎟ ⎥ = Q ⎢ sin 2 ⎜ ⎟⎥ k ⎣⎢ N o ⎝ M ⎠ ⎦⎥ 4 ⎣⎢ N o ⎝ 256 ⎠ ⎦⎥

1 ⎛ 0.0024 Eb ⎞ = Q⎜ ⎟⎟ 4 ⎜⎝ No ⎠

To achieve a BER = 10-6, 0.0024 Eb E 19.936 = 4.465 ⇒ b = = 8.3 × 103 No N o 0.0024 Eb ( dB ) = 39.2 No

256-QAM BERMQAM ≈

⎛ 3 Eb ( log 2 256 ) ⎞ 15 ⎛ 0.094 E ⎞ 4( 256 − 1) b ⎟= Q⎜ Q⎜ ⎟ N o ⎟⎠ (log 2 256) 256 ⎜⎝ ( 256 − 1) N o ⎟⎠ 32 ⎜⎝

To achieve a BER = 10-6, 0.094 Eb E 21.27 = 4.612 ⇒ b = = 226 No N o 0.094 Eb ( dB ) = 23.54 No

b. Suppose that we are willing to relax the data rate requirement to limit our choice to 4-point signal constellations. What bit rate is achievable with a modulation scheme that ensures constant envelope modulated signal? What Eb / N o in dB is needed for a BER = 10-6. Solution: To ensure constant envelope signal, we choose MSK modulated scheme. The null-null bandwidth is related to the bit rate (Table 11.4) by 3Rb = 1.25 MHz 2 2 × 1.25 × 106 Rb = = 833.33 kb/s 3 BMSK =

Now ⎛ 2 Eb ⎞ BERMSK = Q ⎜ ⎜ N ⎟⎟ o ⎠ ⎝

To achieve a BER = 10-6, we need

2 Eb / N o = 4.7534 ⇒

This is equivalent to

Eb = 11.3 No

Eb = 10.53 dB . No

11.12 Consider the 8-ary PSK constellation shown in Figure 11.41(a). a. Sketch the modulated waveform corresponding to the sequence 100101111010. The carrier frequency is fc = 2D, and the pulse shape is unipolar NRZ. Solution:

Bit Sequence

1 0.5 0 0

6 t/Tb

I(t)

1 0 -1

Q(t)

1 0

Q(t)sin(wc t)

I(t)cos(wc t)

-1

1 0 -1

x(t)

2 0 -2

b. Draw the block diagram of the modulator and specify the appropriate symbol mapping table assuming the average signal energy Es = 5 Joules. Solution:

∞

I (t ) = ∑ anI v(t − nT ) n =−∞

anI

b2n even bits 00110

PSF

10 cos(2π fct) T

0100111101

2-bit SPC

−90o

Binary Data bn

+ 8-PSK signal + _ x(t)

10 sin(2π fct) T

odd bits 10111

anQ

b2n+1

PSF

× ∞

Q (t ) = ∑ anQ v(t − nT ) n =−∞

14 SM © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

c. Draw the block diagram of the demodulator.

I (t ) + noise

r ( t ) = x (t ) + n (t )

∫

cos ( 2π f c t )

t =T Compute

( r • s i ) for

BPF BPF

Binary Data bˆ

i =1,2,...., M 90o

− sin ( 2π f c t )

∫

Q (t ) + noise r2

and choose the largest

t =T

d. Specify the threshold comparator characteristic for in-phase and quadrature branches. Solution: Because it is not a rectangular constellation, the decision is made by computing the inner product ( r • si ) = r1si1 + r2 si 2 for each vector si in the signal constellation and declaring that s k was transmitted if i = k achieves the largest inner product value. 11.13 A QAM system operates on an AWGN channel with Es / N o = 28 dB and symbol rate D = 106. a. Select a square QAM constellation and specify a corresponding integer number of bits per symbol, k, for a modem with the highest data rate such that Pe < 10-6. Solution: ⎛ ⎛ M −1 ⎞ ⎛ 3 Es ⎞ 3 Es ⎞ Pe = 4 ⎜⎜ ⎟⎟ ≈ 4Q ⎜⎜ ⎟⎟ for large M ⎟⎟ Q ⎜⎜ − − 1 1 M N M N M o ⎠ o ⎠ ⎝ ⎠ ⎝ ⎝

For Pe < 10-6, we must have

3 Es > 5.0263 M − 1 No 3 ×102.8 M −1 < = 74.92 25.2638 M < 75.92 We choose M = 64 ⇒ k = 6 . b. Compute the bit rate for part (a). Solution: Bit rate = k × D = 6 × 106 = 6 Mbps c. Repeat part (a) if Pe < 10-9. Solution: For Pe < 10-9, we must have 3 Es > 6.2191 M − 1 No 3 × 102.8 M −1 < = 48.94 38.6773 M < 49.94 We choose M = 32 ⇒ k = 5 . d. Compute the bi rate for part (c). Solution: Bit rate = k × D = 5 × 106 = 5 Mbps 11.14 Consider the transmission of digital data using QAM over 6-MHz cable channel which introduces AWGN. We assume that the modem uses RC-shaped pulses with α = 0.5 . a. Determine the required Eb/No to achieve a bit error probability of 10-6 at 8 Mbps. Solution:

6 × 106 = (1 + α ) D = 1.5 D ⇒ 4 × 106 pulses or symbols/second Since k =

Rb 8 × 106 = = 2 , a 4-QAM constellation is used. The probability of D 4 × 106

bit error for M-QAM is given by

BER =

⎞ 4 ⎛ 1 ⎞ ⎛ 3log 2 M − 1 Q E / N ⎜ ⎟⎟ b o ⎟ log 2 M ⎜⎝ M ⎠ ⎜⎝ M − 1 ⎠

⎛ 2 Eb ⎞ For M = 4, BER = Q ⎜⎜ ⎟⎟ ⎝ No ⎠ ⎛ 2 Eb ⎞ 2 Eb = 4.7534 10−6 = Q ⎜⎜ ⎟⎟ ⇒ No ⎝ No ⎠ Eb E = 11.2975 or b = 10.53 dB No No b. Repeat Part (a) for a bit rate of 16 Mbps. Solution: Since k =

Rb 16 × 106 = = 4 , a 16-QAM constellation is used. 4 × 106 D

3 ⎛ 0.8Eb ⎞ For M = 16, BER = Q ⎜⎜ ⎟ 4 ⎝ N o ⎟⎠ 0.8Eb 3 ⎛ 0.8Eb ⎞ = 4.695 10−6 = Q ⎜⎜ ⎟⎟ ⇒ 4 ⎝ No ⎠ No Eb E = 27.5532 or b = 14.4 dB No No c. Repeat Part (a) for a bit rate of 32 Mbps. Solution:

Rb 32 × 106 Since k = = = 8 , a 256-QAM constellation is used. 4 × 106 D

For M = 256, BER =

10−6 =

15 ⎛ 24 Eb ⎞ Q⎜ ⎟ 32 ⎜⎝ 255 No ⎟⎠

15 ⎛ 24 Eb ⎞ 24 Eb = 4.6 Q ⎜⎜ ⎟⎟ ⇒ 32 ⎝ 255 N o ⎠ 255 N o

Eb E = 224.825 or b = 23.52 dB No No d. Comment on these results. Solution: k

Eb/No(dB)

10.53

14.4

23.52

It can be observed from the above table an increase on transmitted power of approximately 2 dB is required per additional bit per symbol. 11.15 Consider the 8-QAM signal constellation shown in Figure P11.1. a. Assign the bits to each point in the signal constellation using Gray coding. Solution:

φ2 (t ) 101

111 011

010 110

001 A

φ1 (t )

000 100

b. Sketch the modulated waveform corresponding to the sequence 100101111010. The carrier frequency is fc = 2D, and the pulse shape is unipolar NRZ.

Bit Sequence

Solution:

1 0.5 0 0

6 t/Tb

I(t)

2 0 -2

Q(t)

2 0

Q(t)sin(wc t)

I(t)cos(wc t)

-2

2 0 -2

x(t)

2 0 -2

c. Determine the symbol rate if the desired bit rate is 45 Mbps. Solution: Since each symbol conveys 3 bits of information, the resultant symbol rate is D=

45 ×106 = 15 ×106 symbols/sec 3

d. Specify the appropriate symbol mapping table and threshold comparator characteristic for in-phase and quadrature branches. Solution: For the given constellation, the normalization constant Co is calculated using (11.130) as

Co2 = M

M M

∑∑ ⎡⎣(a ) + (a ) ⎤⎦ m =1 n =1

I 2 m

Q 2 n

8 2 = 20 5

2 5

Co =

The symbol mapping tables for in-phase and quadrature branches are shown below:

The signal points in the constellation of Figure P11.1 can now be expressed using above symbol tables as

(

s = anI Co Es , anQ Co Es

)

where Es =

5 A2 2

Because it is not a rectangular constellation, the decision is made by computing the inner product ( r • si ) = r1si1 + r2 si 2 for each vector si in the signal constellation and declaring that s k was transmitted if i = k achieves the largest inner product value. 11.16 Consider the 8-QAM signal constellation shown in Figure P11.1. a. Use the nearest neighbor approximation to estimate the probability of bit error. Solution:

As illustrated in Figure P11.1, the minimum distance between two adjacent signal points is

dmin = A From Problem 11.15, we have 5 A2 Es = 2 Therefore, A=

2 Es 5

and

d min =

2 Es 5

Substituting into (10.156) and noting that the factor 2K/M = 1, we obtain the following estimate of the symbol error rate for the 8-QAM constellation:

⎛ 2 Es ⎞ ⎛ 3 Eb ⎞ Pe8-QAM = Q ⎜⎜ ⎟⎟ = Q ⎜⎜ ⎟⎟ ⎝ 5 2 No ⎠ ⎝ 5 No ⎠ b. Sketch the decision region boundaries for each point in the constellation. Solution:

φ2 (t ) 101

111 00 1

1 01

φ1 (t )

0 01

000

110

100

c. Compare the Eb/No required for the 8-QAM with that required for the 8-PSK to achieve the same symbol error probability. Solution:

The probability of symbol error for 8-PSK is given by ⎛ E ⎞ Pe8-PSK = 2Q ⎜⎜ 6 sin 2 (π / 8) b ⎟⎟ No ⎠ ⎝

whereas the probability of symbol error for 8-QAM is given by

⎛ 3 Eb ⎞ Pe8-QAM = Q ⎜⎜ ⎟⎟ ⎝ 5 No ⎠ Since the probability of symbol error is dominated by the argument of the Q function, the two signaling schemes will achieve the same probability of symbol error if ⎛E ⎞ = 6 sin 2 (π / 8) ⎜ b ⎟ ⎝ N o ⎠ M-PSK

3 ⎛ Eb ⎞ ⎜ ⎟ 5 ⎝ N o ⎠ M-QAM

( Eb / No )M-PSK 3 = = 0.683 ( Eb / No )M-QAM 5 × 6 × sin 2 (π / 8) d. Which of the above two signaling schemes is more vulnerable to phase errors? Comment. Solution:

Assuming that the magnitude of the signal points is detected correctly, then the detector for the 8-PSK signal will make an error if the phase error magnitude is greater than 22.5o. In the case of 8-QAM constellation an error will be made if the magnitude of the phase error exceeds 45o. Hence, the QAM constellation is more immune to phase errors. 11.17 Consider the 16-QAM signal constellation shown in Figure 11.50. a. Sketch the modulated waveform corresponding to the sequence 0111100110101101. The carrier frequency is fc = 2D, and the pulse shape is unipolar NRZ.

Bit Sequence

Solution:

1 0.5 0 0

8 t/Tb

I(t)

2 0 -2

Q(t)

2 0 -2

I(t)cos(wc t)

2 0 -2

Q(t)sin(wc t)

2 0 -2 0

x(t)

5 0 -5

b. Sketch the decision region boundaries for each point in the constellation. Solution:

φ2 (t ) Es 10

1011

•

1001

1110

Es 10

1010

•

1000

1100

−

Es 10

0001

•

−3

Es 10

0011

•

0000

•

0100

•

0110

0010

0101

0111

Es 10

•

Es 10

1101

•

−3

−

•

1111

•

φ1 (t )

Es 10

c. Specify the appropriate symbol mapping table and the threshold comparator characteristic for in-phase and quadrature branches. The symbol mapping tables for in-phase and quadrature branches are shown below:

Bits

Co anI

Bits

Co anI

0000

−1/ 10

1000

−1/ 10

0001

−3 / 10

1001

−1/ 10

0010

−1/ 10

1010

−3 / 10

0011

−3 / 10

1011

−3 / 10

0100 1/ 10

1100

1/ 10

0101 1/ 10

1101

3 / 10

0110

3 / 10

1110

1/ 10

0111

3 / 10

1111

3 / 10

Bits

Co anQ

Bits

Co anQ

0000

−1/ 10

1000

1/ 10

0001

−1/ 10

1001

3 / 10

0010

−3 / 10

1010

1/ 10

0011

−3 / 10

1011

3 / 10

0100

−1/ 10

1100

1/ 10

0101

−3 / 10

1101

1/ 10

0110

−1/ 10

1110

3 / 10

0111

−3 / 10

1111

3 / 10

Figure displays threshold comparator characteristic for in-phase and quadrature branches. y[n]

ro[n] −2A

−4A

−A

−3A

11.18 It is desired to transmit data at the rate of 500 kbps over a channel of bandwidth 200 kHz. Assume that the channel introduces AWGN of spectral density No = 10-9. The modulator uses RC pulse shaping with roll-off factor α = 0.5. a. Consider PSK, DPSK and QAM digital modulation schemes. Choose the constellation size M = 2n, where n is an integer. For QAM restrict to a square constellation. Solution: M-PSK 200 × 103 = (1 + 0.5) D =

1.5 × Rb 1.5 × 500 × 103 = k k

1.5 × 500 × 103 = 3.75 200 × 103 We choose k = 4. Therefore, M = 24 = 16

⇒k =

M-DPSK

1.5 × Rb 1.5 × 500 ×103 200 ×10 = (1 + 0.5) D = = k k 3 1.5 × 500 ×10 ⇒k = = 3.75 200 × 103 3

We choose k = 4. Therefore, M = 24 = 16 M-QAM

200 ×103 = (1 + 0.5) D =

1.25 × Rb 1.5 × 500 ×103 = k k

1.5 × 500 ×103 ⇒k = = 3.75 200 ×103 We choose k = 4. Therefore, M = 24 = 16 b. Select the most efficient modulation scheme on the basis of average power required to achieve a BER = 10-6 or better. Solution: 16-PSK

For 16-PSK, the BER is given by

1 ⎡ 8Eb ⎛ π ⎞⎤ BER16− PSK ≈ Q ⎢ sin ⎜ ⎟ ⎥ 2 ⎢⎣ N o ⎝ 16 ⎠ ⎥⎦ or 1 ⎡ 8Eb ⎛ π ⎞⎤ 2 ×10−6 ≈ Q ⎢ sin ⎜ ⎟ ⎥ 2 ⎣⎢ N o ⎝ 16 ⎠ ⎦⎥

From Table 6.1, Q(x) = 2 × 10-6 for x = 4.61. 4.612 =

8 Eb ⎛π ⎞ sin 2 ⎜ ⎟ No ⎝ 16 ⎠

Eb 21.252 = = 69.8 = 18.44 dB N o 8 × 0.038 Pav = Eb Rb = 2 × 10−9 × 69.8 × 500 × 103 = 69.8 mW = 18.44 dBm

16-DPSK

For 16-DPSK, the BER is given by

1 BER16− DPSK ≈ Q( 16sin 2 (π / 32) Eb / N o ) 2 or 1 ⎡ 16 Eb ⎛ π ⎞⎤ 2 ×10−6 ≈ Q ⎢ sin ⎜ ⎟ ⎥ 2 ⎣⎢ N o ⎝ 32 ⎠ ⎦⎥

From Table 6.1, Q(x) = 2 × 10-6 for x = 4.61. 4.612 =

16 Eb ⎛π ⎞ sin 2 ⎜ ⎟ No ⎝ 32 ⎠

Eb 21.252 = = 138.25 = 21.4 dB N o 16 × 0.0096 Pav = Eb Rb = 2 × 10−9 × 138.25 × 500 × 103 = 138.25 mW = 21.4 dBm

16-QAM

For 16-PSK, the BER is given by 3 ⎛ 0.8 Eb ⎞ BER16−QAM = Q ⎜ ⎟ 4 ⎜⎝ N o ⎟⎠ ⎛ 0.8 Eb ⎞ 4 ×10−6 = Q ⎜ ⎜ N ⎟⎟ 3 o ⎝ ⎠

From Table 6.1, Q(x) = 1.333 × 10-6 for x = 4.7. 4.7 2 =

0.64 × Eb No

Eb 22.09 = = 34.51 = 15.38 dB No 0.64 Pav = Eb Rb = 2 × 10 −9 × 34.51× 500 × 103 = 34.51 mW = 15.38 dBm

⎧69.8 mW (18.44 dBm ) ⎪ Pav = ⎨138.25 mW ( 21.4 dBm ) ⎪ ⎩34.51 mW (15.38 dBm )

16-PSK 16-DPSK 16-QAM

b. Discuss the relative advantages and disadvantages of the modulation schemes considered above for this application.

Solution:

16-QAM has significant SNR/bit advantage against 16-PSK and 16-DPSK schemes. Further, 16-QAM is attractive from the implementation point of view in silicon. 11.19 The CCITT V.29 16-QAM signal constellation shown in Figure P11.2. Assume that the pulse shape is unipolar NRZ and average energy per symbol is 2 Wattssec. a. Draw a block diagram of the modulator and specify the contents of symbol mapping tables. Solution:

∞

I (t ) = Co ∑ anI v(t − nT ) n =−∞

Co anI

b2n

PSF

even bits 00110

2Es cos(2π fct) T

0100111101

2-bit SPC Binary Data bn

−90o

+ 16-QAM signal + _ x(t)

2Es sin(2π fct) odd bits 10111 b2n+1

Co anQ

PSF

× ∞

Q (t ) = Co ∑ anQ v(t − nT ) n =−∞

For the given constellation (see Part (c) of the solution), the normalization constant Co is calculated using (11.130) as M 16 4 Co2 = M M = = 212 53 ∑∑ ⎡⎣(amI )2 + (anQ )2 ⎤⎦ m =1 n =1

Co =

4 = 0.275 53

The symbol mapping tables for in-phase and quadrature branches are shown below:

The signal points in the constellation of Figure P11.2 can now be expressed using above symbol tables as

(

s = anI Co Es , anQ Co Es

) 30

where

2 = Es =

4 2 A + 9 A2 + 25 A2 + 18 A2 ) ( 16

53 A2 ⇒ =2 4 A = 0.3885 b.

Sketch the modulated waveform corresponding to the sequence 0111100110101101. The carrier frequency is fc = 2D, and the pulse shape is unipolar NRZ.

Bit Sequence

Solution:

1 0.5 0 0

8 t/Tb

I(t)

5 0 -5

Q(t)

5 0

Q(t)sin(wc t)

I(t)cos(wc t)

-5

5 0 -5

x(t)

5 0 -5

c. Sketch the decision boundaries for the ML detector. Solution:

φ2 (t ) •

0100 5A 0101 3A

•

0011

0001

•

1101 A

−A

1010

1011

1111

•

1110•− A

1000

1100

0110

•

• •

•

0111

•

−3A

•

−5A

1001

φ1 (

0000

0010 −5A

•

d. Determine the estimated symbol sequences aˆ nI and aˆ nQ and corresponding bit sequences for the following outputs of symbol-rate samplers. Solution:

n In-phase sampler O/P Quadrature sampler O/P Co Es aˆ nI

1 0.93

2 −2

3 −1.97

4 −0.052

5 2.1

6 −0.95

−0.56

0.015

0.95

−1.95

−0.015

1.1655

−1.9425

1.9425

−1.1655

Co E s aˆ nQ

−0.3885

1.9425

1.1655

−1.9425

Bits

1000

1010

0011

0101

0000

1011

11.20 For MSK signaling, show that the carriers in (11.125) are orthogonal. Solution:

The MSK signal over the time interval 0 ≤ t ≤ Tb is

s1 (t ) =

⎡ ⎤ 2E b πt cos ⎢ 2π f c t + + ψ (0) ⎥ Tb 2Tb ⎣ ⎦

for binary 1 and s2 (t ) =

⎡ ⎤ 2E b πt cos ⎢ 2π f c t − + ψ (0) ⎥ Tb 2Tb ⎣ ⎦

for binary 0. For s1 (t ) and s2 (t ) to be orthogonal over 0 ≤ t ≤ Tb , it is required that Tb

⎡ ⎤ ⎡ ⎤ 2E b b πt πt s ( t ) s ( t ) dt cos ⎢ 2π f ct + = +ψ (0) ⎥ cos ⎢ 2π f c t − +ψ (0) ⎥ dt ∫0 1 2 ∫ Tb 0 2Tb 2Tb ⎣ ⎦ ⎣ ⎦ T Tb ⎛ π t ⎞ ⎫⎪ E b ⎧⎪ b = ⎨ ∫ cos [ 4π f ct + 2ψ (0)] dt + ∫ cos ⎜ ⎟dt ⎬ = 0 + 0 = 0 Tb ⎩⎪ 0 ⎝ Tb ⎠ ⎭⎪ 0

a. Sketch the constellation diagram and determine the decision regions. Solution:

Using (11.122), we can express the MSK waveform in terms of quadrature components as x(t ) =

2E b {cos [ψ (t )] cos ( 2π fct ) − sin [ψ (t )] sin ( 2π fct )} , Tb

0 ≤ t ≤ Tb

where

ψ (t ) = ψ (0) ±

πt 2Tb

Now ⎡ ⎛ πt ⎞ ⎛ πt ⎞ πt ⎤ cos [ψ (t )] = cos ⎢ψ (0) ± ⎟ ∓ sin [ψ (0)] sin ⎜ ⎟ ⎥ = cos [ψ (0)] cos ⎜ 2Tb ⎦ ⎣ ⎝ 2Tb ⎠ ⎝ 2Tb ⎠ 0 because ψ (0) = 0 or π

⎛ πt ⎞ = cos [ψ (0) ] cos ⎜ ⎟, ⎝ 2Tb ⎠

− Tb ≤ t ≤ Tb

The time interval has been extended to −Tb ≤ t ≤ Tb because cosine is an even

function. Similarly, it can be shown that sin [ψ (t )] can be expressed as

⎡ ⎛ πt ⎞ ⎛ πt ⎞ πt ⎤ sin [ψ (t ) ] = sin ⎢ψ (0) ± ⎟ ± cos [ψ (0) ] sin ⎜ ⎟ ⎥ = sin [ψ (0) ] cos ⎜ 2Tb ⎦ ⎣ ⎝ 2Tb ⎠ ⎝ 2Tb ⎠ ⎛ πt ⎞ = ± cos [ψ (0) ] sin ⎜ ⎟ ⎝ 2Tb ⎠

π⎤ ⎡ Now cos [ψ (0)] = cos ⎢ψ (Tb ) ∓ ⎥ = ± sin [ψ (Tb )] . Therefore, 2⎦ ⎣ ⎛ πt ⎞ sin [ψ (t ) ] = sin [ψ (Tb ) ] sin ⎜ 0 ≤ t ≤ 2Tb ⎟, 2 T b ⎝ ⎠ From discussion in Section 11.8, ψ (0) assumes values 0 or π. Similarly,

ψ (Tb ) assumes values ± •

π 2

. Thus four possibilities can arise:

ψ (0) = 0 and ψ (Tb ) =

1 •

ψ (0) = π and ψ (Tb ) =

−1 •

ψ (0) = π and ψ (Tb ) = − symbol 1

•

, corresponding to the transmission of symbol

ψ (0) = 0 and ψ (Tb ) = − symbol −1

, corresponding to the transmission of symbol

π 2

, corresponding to the transmission of

Thus the MSK signal can assume any one of four possible forms depending on values of ψ (0) and ψ (Tb ) . By choosing the basis functions

φ1 (t ) =

⎛ πt ⎞ 2 cos ⎜ ⎟ cos ( 2π f c t ) , Tb ⎝ 2Tb ⎠

− Tb ≤ t ≤ Tb

φ2 (t ) =

⎛ πt ⎞ 2 sin ⎜ ⎟ sin ( 2π f c t ) , Tb 2 T ⎝ b⎠

0 ≤ t ≤ 2Tb

we can express the MSK signal as x(t ) = x1φ1 (t ) + x2φ2 (t ),

0 ≤ t ≤ Tb

where Tb

x1 = ∫ x(t )φ1 (t )dt = Eb cos [ψ (0) ] ,

− Tb ≤ t ≤ Tb

−Tb

2Tb

x2 = ∫ x(t )φ2 (t )dt = − Eb sin [ψ (Tb )] ,

0 ≤ t ≤ 2Tb

The 2-dimensional signal constellation and decision regions of the MSK signal are shown in Figure. It is similar to that of QPSK in that both have four signal points. However, there is significant difference in their interpretation. In QPSK each signal point represents two information bits, whereas in MSK each signal point represents one information bit. Further, each decision region (i.e., quadrant) in QPSK corresponds to a unique two-bit symbol. In MSK, however, both decision regions D1 and D3 are for transmitted symbol −1 and D2 and D4 are for transmitted symbol 1.

φ2(t)

Signal point s 2

Signal point s1

Transmitted symbol 1

Transmitted symbol − 1

π⎞ ⎛ ⎜ψ (0) = π and ψ (Tb ) = − ⎟ 2⎠ ⎝ Signal point s 3

π⎞ ⎛ ⎜ψ (0) = 0 and ψ (Tb ) = − ⎟ 2⎠ ⎝ φ1(t) Signal point s 4

Transmitted symbol − 1

Transmitted symbol 1

π⎞ ⎛ ⎜ψ (0) = π and ψ (Tb ) = ⎟ 2⎠ ⎝

π⎞ ⎛ ⎜ψ (0) = 0 and ψ (Tb ) = ⎟ 2⎠ ⎝

b. Draw the block diagrams of the modulator and the demodulator.

MSK modulator

⎛ πt ⎞ 2 cos ⎜ ⎟ Tb ⎝ 2Tb ⎠ SM b2n

a2n

even bits 00110

Eb cos ( 2π f c t ) MSK signal

~ +

0100111101

x(t)

2-bit SPC

Binary Data bn

−90o

b2n+1

odd bits 10111

Eb sin ( 2π f c t ) a2 n+1

Delay Tb

⎛ πt ⎞ 2 sin ⎜ ⎟ Tb ⎝ 2Tb ⎠

MSK demodulator

I (t ) + noise r (t ) = x (t ) + n (t )

BPF BPF

∫

t =T

⎛ πt ⎞ 2 cos ⎜ ⎟ cos ( 2π f c t ) Tb ⎝ 2Tb ⎠ ⎛ πt ⎞ 2 sin ⎜ ⎟ sin ( 2π f c t ) Tb ⎝ 2Tb ⎠

∫

PSC & SM

Q (t ) + noise t =T

Binary Data bˆ

001011

aˆ 2 n+1

aˆ 2 n

Chapter 12 12.1 Let Pr ( f ) be the triangular spectrum given by

⎧ f ⎞ −4 ⎛ ⎪2.5 ×10 ⎜1 − ⎟, Pr ( f ) = ⎨ 4000 ⎝ ⎠ ⎪ ⎩0,

f < 4000 otherwise

a. For what symbol rate, pr ( t ) is a Nyquist pulse. Solution:

The folded spectrum Pr folded ( f ) is shown in Figure. If we choose 1 / 2T = 2000 , that is symbol rate D = 4000 symbols/second, then Pr ( f ) satisfies Nyquist criterion for ISI free transmission. Pr ( f ) 2.5 × 10−4

−4000

4000

Pr folded ( f ) 2.5 × 10−4

…

… −8000 −4000 1 − 2T 2T

0 1 4000 2T 2 T

8000

b. Determine pr ( t ) and verify the conclusion from (a). Solution:

The pulse pr ( t ) corresponding to the given Pr ( f ) is given from Table 2.2 and applying (2.81) as

pr ( t ) = sinc2 ( Dt ) c. At this symbol rate, what is the excess bandwidth of the pulse and roll-off factor? Solution:

Excess bandwidth = Babs − BNyquist = 4000 − 2000 = 2000

Roll-off factor α = 12.2

Excess bandwidth 2000 = =1 2000 BNyquist ( = 1/ 2T )

Consider an M-ary PAM system to transmit data at 50 kbps over an ideal bandlimited channel with bandwidth W = 7500 Hz. Assume polar RRC pulse signaling with a roll-off factor α = 0.5. Determine a. the required size of the signal set, M.

Solution:

7500 =

D (1 + α ) D (1 + 0.5) 15, 000 = ⇒D= = 10, 000 2 2 1.5

50, 000 = 5, 10, 000

M = 25 = 32

b. Eb / N o required to achieve Pe = 10−6 . Solution:

Pe =

⎞ 2( M − 1) ⎛ Eb 6 log 2 M ⎞ 6 Es 2( M − 1) ⎛ Q ⎜⎜ Q ⎜⎜ ⎟⎟ = ⎟⎟ 2 2 M M ⎝ N o ( M − 1) ⎠ ⎝ N o ( M − 1) ⎠

10−6 =

62 ⎛ Eb 30 ⎞ 62 ⎛ 0.0293Eb ⎞ Q ⎜⎜ ⎟ = Q⎜ ⎟⎟ 32 ⎝ N o (322 − 1) ⎟⎠ 32 ⎜⎝ No ⎠

⎛ 0.0293Eb ⎞ 32 ×10−6 = 0.516 ×10−6 = Q ⎜⎜ ⎟⎟ 62 N o ⎝ ⎠ From Table 6.1, we conclude that the argument of Q-function x = 4.88568

0.0293Eb E = 23.87 ⇒ b = 814.67 No No Eb (dB) = 29.11 No

12.3

Binary data is transmitted using polar RC pulses with spectrum displayed in Figure P12.1. Figure P12.1

Pr ( f ) 1 0.5 1.4 2.0 2.6

f (kHz)

a. Calculate the roll-off factor. Solution:

Excess bandwidth = Babs − BNyquist = 2600 − 2000 = 600 Roll-off factor α =

600 = 0.3 2000

b. What is the maximum signaling rate using above RC pulses? Solution:

2W 2 × 2600 = = 4000 symbols/sec (1 + α ) (1 + 0.3)

c. The channel distorts the signal pulses and the sample values of the matched filter output pulse are given by

{q [ k ]} = {0.0194, − 0.1452, 0.9677, 0.2032, 0.029}

List the bit sequence that will generate the worst-case ISI and the corresponding value of peak distortion. Solution:

The bit sequence is {−1, 1, 1, − 1, − 1} . The peak ISI distortion at the equalizer input is obtained applying (12.14) as ∞

D peak = ∑ ak q [ k ] k =−∞ k ≠0

= ( 0.0194 + 0.1452 + 0.2032 + 0.029 ) = 0.3968 d. How likely is such a sequence assuming that data bits are equiprobable and statistically independent? Solution: 5

1 ⎛1⎞ P ({−1, 1, 1, − 1, − 1} ) = ⎜ ⎟ = ⎝ 2 ⎠ 32

12.4 Consider a 4-PAM system to transmit data through a bandlimited channel at 8.4 kbps. The channel is characterized by the frequency response Hc ( f ) =

1 1 + j ( f / 4200)

The channel introduces AWGN noise with spectral density N o / 2 = 10−12 W/Hz. a. Write transfer functions of transmit and receive filters to deliver RC output pulse (α = 1) at the receiver. Solution:

Since 1/T =4200, the RC pulse spectrum for α = 1 is given from (9.66) as ⎧ 1 ⎧⎪ ⎛ π f ⎞ ⎫⎪ ⎪ ⎨1 + cos ⎜ ⎟ ⎬ , 0 ≤ f ≤ 4200 PRC ( f ) = ⎨ 8400 ⎩⎪ ⎝ 4200 ⎠ ⎭⎪ ⎪ elsewhere ⎩0,

Substituting into (12.46) and (12.47), the optimum transmit and receive filter characteristics are given by

⎧ 1 ⎛π f ⎞ 2 cos ⎜ ⎪ ⎟ 1 + ( f / 4200) , 0 ≤ f ≤ 4200 H T ( f ) = H R ( f ) = ⎨ 4200 ⎝ 8400 ⎠ ⎪ elsewhere ⎩0, b. Estimate the average energy/bit, Eb , required to achieve a BER = 10−6 . Solution:

The BER for M-ary PAM signaling over AWGN channel is given from and (10.177) as

BERM − PAM =

2(M − 1) ⎛ 6log 2 M Eb ⎞ Q⎜ ⎟ M log 2 M ⎜⎝ (M 2 − 1) No ⎟⎠

For 4-PAM, this simplifies to

3 ⎛ 4 Eb ⎞ BER4− PAM = Q ⎜⎜ ⎟ 4 ⎝ 5 No ⎟⎠ To achieve BER = 10−6, the argument x of Q-function is obtained from Table 6.1 as

E 4 Eb = 4.695 ⇒ b = 27.55 5 No No

Eb = 14.4 dB No In the process of channel equalization, noise enhancement occurs. The decrease in SNR/bit , Eb / N o , due to noise enhancement is given from (12.53) by ⎡ ⎛ W P ( f ) ⎞ −2 ⎤ Eb / N o penalty = 10 log10 ⎢⎜ ∫ RC df ⎟ ⎥ ⎢⎜⎝ −W H c ( f ) ⎟⎠ ⎥ ⎣ ⎦ W

PRC ( f ) df was calculated numerically using MATLAB. In the present Hc ( f ) −W

The integral ∫ W

PRC ( f ) df = 1.213 . Therefore, Hc ( f ) −W

case, ∫

Additional Eb / N o required = 10 log10 (1.2132 ) = 1.6773 dB

Eb / N o required to achieve BER of 10−6 = 14.4 + 1.6773 = 16.077 dB Eb required to achieve BER of 10−6 = 2 ×10−12 ×101.6077 = 0.081 nW = −70.9 dBm

12.5 The binary data 001011100 is applied to the input of a duobinary encoder. a. Construct the precoder output assuming initialization bits do = 1 . b. What is the corresponding received sequence? c. Derive the decoding rule at the receiver to reconstruct the data. Solution:

The following table shows the precoded sequence, the transmitted signal levels, the received sequence, and the decoded sequence. Data bits bk

Precoded sequence d k

Transmitted sequence ak

−A

Received sequence yk

−2A 0

Decoded

sequence bˆk

d. Assume that due to noise, the sampler output produced by the fourth bit is reduced to zero. Construct the decoded output sequence bˆk . How many errors are there in the decoded sequence? Solution: y4 = 0.The decoder will declare bˆ4 = 1 . Precoding means no error propagation.

12.6 The binary data 101100011 is applied to the input of a modified duobinary encoder.

a. Construct the precoder output assuming initialization bits ( 0, 0 ) . Solution:

The precoded sequence {dk } is generated by the operation d k = bk ⊕ d k −2

The sequence {d k } is then mapped to the sequence of polar transmitted signal levels

{ak } according to the relationship 0 → −A 1→ A

b. What is the corresponding received sequence? The noise-free samples at the output of the receive filter are given by y k = a k − a k −2

c. Derive the decoding rule at the receiver to reconstruct the data. ⎧±2 A, yk = ⎨ ⎩0,

with probability 1/2 (d k ≠ d k − 2 ) with probability 1/2 (d k = d k − 2 )

Since bk = 1 toggles d k − 2 bit, the decision rule in the absence of noise is given by ⎧1, bˆk = ⎨ ⎩0,

yk = ±2 A yk = 0

The following table shows the precoded sequence, the transmitted signal levels, the received sequence, and the decoded sequence. d. Assume that due to noise, the sampler output produced by the fourth bit is reduced to zero. Construct the decoded output sequence bˆk . How many errors are there in the decoded sequence? Solution: y4 = 0.The decoder will declare bˆ4 = 0 .

Data bits bk

Precoded sequence d k

Transmitted sequence ak

−A

Received sequence yk

−2A

−2A 2A

Decoded

sequence bˆk

12.7 Consider a channel with overall impulse response ⎧α k , q[k ] = ⎨ ⎩0,

α <1

k ≥ 0, otherwise

a. Calculate the peak ISI distortion at the equalizer input. Solution:

The peak ISI distortion is obtained by tha application of (12.14) as ∞

∞

k =−∞ k ≠0

k =1

k =0

D peak = ∑ q[k ] = ∑ α k = ∑ α k − 1 =

1 α −1 = 1−α 1−α

b. Calculate the coefficients of a linear equalizer that completely eliminates the ISI. Solution:

The z-transform of the sequence q[k ] is given by ∞

Q( z ) = ∑ α i z −i = i =0

1 1 − α z −1

The transfer function of the linear equalizer is obtained using (12.99) as H eq ( z ) =

1 = 1 − α z −1 Q( z)

The impulse response of the 2-tap equalizer is

heq [ k ] = {1, −α } c. Calculate the noise variance at the equalizer output. Solution:

The variance of noise at the equalizer output is given by

σ eq2 =

No N 2 No 2 2 N ci = c0 + c1 ) = o (1 + α 2 ) ( ∑ 2 i =− N 2 2

where N o / 2 is the equalizer input noise spectral density. d. Derive an expression for the bit error rate of the system. Solution:

In the present case, the equalizer is successful in completely eliminating the ISI. That is, it can be easily shown that peq [ 0] = 1 and peq [ k ] = 0, k ≠ 0 . Therefore, the BER is given from (12.172) as

⎛ 2 Eb ⎞ BER = Q ⎜ ⎜ η N ⎟⎟ ⎝ NE o ⎠ The noise enhancement factor of the equalizer is obtained by using (12.169) as N

η NE = ∑ ci2 = 1 + α 2 i =− N

Substituting ⎛ BER = Q ⎜ ⎜ ⎝

⎞ ⎟ 1 + α 2 No ⎟ ⎠

(

2 Eb

)

12.8 Consider a binary antipodal system with overall impulse response

{q [ k ]} = {1, 0.7}

The channel introduces AWGN noise with spectral density N o / 2 W/Hz. a. Design a 5-tap linear equalizer to eliminate ISI. Solution:

The z-transform of the sequence q[k ] is given by ∞

Q( z ) = ∑ q [ k ]z − k = 1 + 0.7 z −1 k =0

The transfer function of the linear equalizer is obtained using () as

H eq ( z ) =

1 1 = Q( z ) 1 + 0.7 z −1

(

= 1 − 0.7 z −1 + −0.7 z −1

) + ( −0.7 z ) + ( −0.7 z ) + ( −0.7 z ) + ( −0.7 z ) + ... + ... 2

−1

= 1 − 0.7 z −1 + 0.49 z −2 − 0.343z −3 + 0.24 z −4 − 0.168 z −5 + 0.1176 z −6 + ... + ... When the equalizer has 5 taps, its transfer function is given by H 5 ( z ) = 1 − 0.7 z −1 + 0.49 z −2 − 0.343 z −3 + 0.24 z −4 The impulse response of the 5-tap equalizer is

h5 [ k ] = {1, − 07, 0.49, − 0.343, 0.24} The z-transform of the equalizer output sequence is

(

)(

Peq ( z ) = Q( z ) H 5 ( z ) = 1 + 0.7 z −1 1 − 0.7 z −1 + 0.49 z −2 − 0.343z −3 + 0.24 z −4

)

= 1 + 0.168 z −5 Therefore,

peq [ k ] = {1, 0, 0, 0, 0, 0.168} b. Calculate the noise variance at the equalizer output Solution:

The variance of noise at the equalizer output is given by

No 4 2 No ⎡ 2 2 2 2 ci = σ = 1 + ( −0.7 ) + ( 0.49 ) + ( −0.343) + ( 0.24 ) ⎤ = 0.9527 N o ∑ ⎣ ⎦ 2 i =0 2 2 eq

c. Derive an expression for the bit error rate of the system. Solution:

The BER of the system can be expressed as 1 1 P {error ak = 1, ak −5 = 1} + P {error ak = 1, ak −5 = −1} 4 4 1 1 + P {error ak = −1, ak −5 = 1} P {error ak = −1, ak −5 = −1} 4 4

BER =

Now using (12.171), we can calculate the conditional error probabilities as follows: P {error ak = 1, ak −5 = 1} = P =P

{ E ( p [0] + p [5]) + n < 0} b

{

⎛ = Q⎜ ⎜ ⎝ P {error ak = 1, ak −5 = −1} = P =P

{

⎛ = Q⎜ ⎜ ⎝ P {error ak = −1, ak −5 = 1} = P

eq k

⎛ E 1.168 2 ⎞ ) ⎟ b( Eb 1.168 + n < 0 = Q ⎜ 2 ⎜ ⎟ σ eq ⎝ ⎠ 2 ⎛ 1.432 Eb ⎞ Eb (1.168 ) ⎞⎟ = Q⎜ ⎟ ⎜ 0.9527 N o ⎟ N o ⎟⎠ ⎝ ⎠

}

eq k

{ E ( p [0] − p [5]) + n < 0} b

eq k

⎛ E 0.832 2 ⎞ ) ⎟ b( Eb 0.832 + n < 0 = Q ⎜ 2 ⎜ ⎟ σ eq ⎝ ⎠ 2 ⎛ 0.7266 Eb ⎞ Eb ( 0.832 ) ⎞⎟ = Q⎜ ⎟⎟ ⎜ 0.9527 N o ⎟ No ⎝ ⎠ ⎠ eq k

}

{ E ( − p [0] + p [5]) + n < 0} b

eq k

⎛ E 0.832 2 ⎞ ) ⎟ b( = P − Eb 0.832 + n > 0 = Q ⎜ 2 ⎜ ⎟ σ eq ⎝ ⎠

{

eq k

}

⎛ E 0.832 2 ⎞ ) ⎟ = Q ⎛ 0.7266 Eb ⎞ b( = Q⎜ ⎜⎜ ⎟⎟ ⎜ 0.9527 N o ⎟ No ⎝ ⎠ ⎝ ⎠

P {error ak = −1, ak −5 = −1} = P

{ E ( − p [0] − p [5]) + n > 0} b

eq k

⎛ E 1.168 2 ⎞ ) ⎟ b( = P − Eb 1.168 + n > 0 = Q ⎜ 2 ⎜ ⎟ σ eq ⎝ ⎠

{

eq k

}

⎛ E 1.168 2 ⎞ ) ⎟ = Q ⎛ 1.432 Eb ⎞ b( = Q⎜ ⎜⎜ ⎟ ⎜ 0.9527 N o ⎟ N o ⎟⎠ ⎝ ⎝ ⎠

Therefore, BER =

⎛ 0.7266 Eb ⎞ ⎤ 1 ⎡ ⎛ 1.432 Eb ⎞ ⎢Q ⎜⎜ ⎟⎟ + Q ⎜⎜ ⎟⎟ ⎥ No ⎠ No 2 ⎣⎢ ⎝ ⎝ ⎠ ⎦⎥

d. Plot the BER curves of the system (i) without equalizer and (ii) with the equalizer designed in (a). Compare with AWGN performance with no ISI Solution:

Without equalizer, the BER of the system can be expressed as 1 1 P {error ak = 1, ak −1 = 1} + P {error ak = 1, ak −1 = −1} 4 4 1 1 + P {error ak = −1, ak −1 = 1} P {error ak = −1, ak −1 = −1} 4 4

BER =

Now using (12.171), we can calculate the conditional error probabilities as follows: P {error ak = 1, ak −1 = 1} = P

{ E ( q [0] + q [1]) + n < 0} b

⎛ E 1.7 2 ⎞ ) ⎟ b( = P Eb 1.7 + nk < 0 = Q ⎜ 2 ⎜ σ eq ⎟ ⎝ ⎠ ⎛ 2 Eb ⎞ = Q ⎜1.7 ⎟ ⎜ N o ⎟⎠ ⎝

{

}

P {error ak = 1, ak −1 = −1} = P

{ E ( q [0] − q [1]) + n < 0} b

⎛ E 0.3 2 ⎞ ) ⎟ b( = P Eb 0.3 + nk < 0 = Q ⎜ 2 ⎜ σ eq ⎟ ⎝ ⎠ ⎛ 2 Eb ⎞ = Q ⎜ 0.3 ⎟ ⎜ N o ⎟⎠ ⎝

{

P {error ak = −1, ak −1 = 1} = P

}

{ E ( −q [0] + q [1]) + n < 0} b

⎛ E 0.3 2 ⎞ ) ⎟ b( = P − Eb 0.3 + nk > 0 = Q ⎜ 2 ⎜ σ eq ⎟ ⎝ ⎠

{

}

⎛ 2 Eb ⎞ = Q ⎜ 0.3 ⎟ ⎜ N o ⎟⎠ ⎝ P {error ak = −1, ak −1 = −1} = P

{ E ( −q [0] − q [1]) + n > 0} b

⎛ E 1.7 2 ⎞ ) ⎟ b( = P − Eb 1.7 + nk > 0 = Q ⎜ 2 ⎜ σ eq ⎟ ⎝ ⎠ ⎛ 2 Eb ⎞ = Q ⎜1.7 ⎟ ⎜ N o ⎟⎠ ⎝

{

}

Therefore, BER =

⎛ 2 Eb ⎞ 2 Eb ⎞ ⎤ 1⎡ ⎛ ⎢Q ⎜⎜ 0.3 ⎟⎟ + Q ⎜⎜1.7 ⎟⎟ ⎥ 2 ⎢⎣ ⎝ No ⎠ N o ⎝ ⎠ ⎥⎦ Bit Error Rate vs. (Eb/No)dB

Equalized Unequalized No ISI

-1

-2

-3

BER

-4

-5

-6

-7

10 12 Eb/No [dB]

12.9

Consider a binary digital communication system with overall impulse response

{q [ k ]} = {0.0105 −0.0542 0.1570 0.9710 −0.1698 0.022, 0.0107} The channel introduces AWGN noise with spectral density N o / 2 = 0.05 W/Hz. a. Calculate the worst-case eye opening at the equalizer input. Solution:

The peak ISI distortion at the equalizer input is obtained by applying (12.14) and (12.86) as ∞

D peak = Eb ∑ q[n] n =−∞ n≠0

= Eb ( 0.0105 +0.0542+ 0.1570 + 0.17 + 0.022 + 0.0107 ) = 0.4244 Eb

The worst-case eye opening at the equalizer input is

( 0.971 − 0.4244 ) Eb = 0.5466 Eb b. Determine the coefficients of a 5-tap ZFE. Plot the equalized output waveform Solution:

⎛ 0.0746 ⎞ ⎜ ⎟ ⎜ −0.1355 ⎟ c opt = ⎜ 0.9772 ⎟ ⎜ ⎟ ⎜ 0.1716 ⎟ ⎜ 0.0094 ⎟ ⎝ ⎠

q[k]

0.5

-0.5

6 7 Samples k

peq[k]

0.5

-0.5

The equalizer output pulse sample values are given by

peq [k ] = {0.0008 − 0.0055

0.0294 0.0

0.0 1.0 0.0 0.0 0.0127

0.002

0.0001}

c. Calculate the noise enhancement and a worst-case estimate of BER at the equalizer output. Compare with AWGN performance with no ISI. Solution:

The noise enhancement factor of the equalizer is obtained by using (12.169) as 2

η NE = ∑ ci2 = 1.0085 i =−2

The worst-case BER estimate is given from (12.172) by ⎛ 2E ξ 2 ⎞ b ISI BERworst-case = Q ⎜ ⎟ ⎜ N o η NE ⎟ ⎝ ⎠

where

ξ ISI = peq [ 0] − Dpeak = 1 − 0.0504 = 0.9496 Substituting yields BERworst-case = 1.1731×10−5 for Eb / N o = 10 dB The BER for antipodal signaling over an AWGN for Eb / N o = 10 dB is given by

⎛ 2 Eb ⎞ BER = Q ⎜ =Q ⎜ N ⎟⎟ o ⎠ ⎝

( 20 ) = 0.3875 ×10

−5

It follows that in the present case the 5-tap ZFE is able to equalize the channel without significant noise enhancement. 12.10 Consider a binary digital communication system with overall impulse response

{q [ k ]} = {0.0101, −0.0891, 0.5752, 0.7423, 0.2714, 0.1762, 0.074 } The channel introduces AWGN noise with spectral density N o / 2 = 0.05 W/Hz.

a. Design a 7-tap MMSE equalizer for the channel. Plot the equalized output waveform Solution: ⎛ −0.0658 ⎞ ⎜ ⎟ ⎜ 0.1512 ⎟ ⎜ −0.2335 ⎟ ⎜ ⎟ c opt = ⎜ 0.5981 ⎟ ⎜ 0.5870 ⎟ ⎜ ⎟ ⎜ −0.6859 ⎟ ⎜ 0.2157 ⎟ ⎝ ⎠ 1

q[k]

0.5

-0.5

6 8 Samples k

peq[k]

0.5

-0.5

b. Calculate the worst-case eye opening at the equalizer output. Solution:

The equalizer output pulse sample values are given by

⎧ − 0.0007 0.0074 − 0.0537 0.0650 − 0.0873 0.1409 0.8033 0.1543⎫ peq [k ] = ⎨ ⎬ ⎩ − 0.1376 0.1217 − 0.0189 − 0.0127 0.0160 ⎭ Now Worst-case eye opening ξ ISI = peq [ 0] − D peak = 0.8033 − 0.8161 = −0.0128 . So the eye is closed under worst conditions. c. Calculate the noise enhancement at the equalizer output.

Solution:

The noise enhancement factor of the equalizer is obtained by using (12.169) as 3

η NE = ∑ ci2 = 1.3 i =−3

d. Plot DTFTs of the overall channel, equalizer, and combined responses.

|Q(ejw )|

Solution:

0 -10 -20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.6

0.7

0.8

0.9

0.6

0.7

0.8

0.9

|Heq(ejw )|

ω/π 0 -10 -20

0.1

0.2

0.3

0.4

0.5

|Peq(ejw )|

ω/π 0 -10 -20

0.1

0.2

0.3

0.4

0.5

ω/π

e. Repeat (a) through (d) for 31-tap MMSE equalizer. Comment about the performance improvement versus 7-tap equalizer. Solution:

The equalizer output pulse sample values are given by

⎧ 0.0 0.0 − 0.0001 0.0001 − 0.0002 0.0003 − 0.0005 0.0010 ⎫ ⎪ − 0.0017 0.0029 − 0.0051 0.0085 − 0.0131 0.0216 − 0.0365 0.0525 ⎪ ⎪⎪ ⎪⎪ peq [k ] = ⎨ − 0.0703 0.1166 0.8322 0.1166 − 0.0703 0.0525 − 0.0365 0.0216 ⎬ ⎪ − 0.0132 0.0085 − 0.0051 0.0029 − 0.0017 0.0011 − 0.0006 0.0004 ⎪ ⎪ ⎪ ⎪⎩ − 0.0004 0.0002 0.0002 0.0 0.0001 ⎪⎭ Worst-case eye opening ξ ISI = peq [ 0] − D peak = 0.8322 − 0.6628 = 0.1694 . So the eye is open worst conditions. 17 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

The noise enhancement factor of the equalizer is obtained by using (12.169) as 15

η NE = ∑ ci2 = 1.8570 i =−15

q[k]

0.5

-0.5

20 Samples k

0.6

0.7

0.8

0.9

0.6

0.7

0.8

0.9

0.6

0.7

0.8

0.9

peq[k]

0.5

|Q(ejw )|

-0.5

0 -10 -20

0.1

0.2

0.3

0.4

0.5

|Heq(ejw )|

ω/ π 0 -10 -20

0.1

0.2

0.3

0.4

0.5

|Peq(ejw )|

ω/ π 0 -10 -20

0.1

0.2

0.3

0.4

0.5

ω/ π

We observe that the channel has a high frequency null near the Nyquist bandedge. This makes it difficult to equalize even with a 31-tap linear equalizer. The equalizer

tries to provide gain at higher frequencies, but it is unsuccessful in compensating without significant amplification of the noise. 12.11 Consider a binary digital communication system with overall impulse response

{q [ k ]} = {0.8729, 0.4364, 0.2182} The channel introduces AWGN noise with spectral density N o / 2 = 0.025 W/Hz. a. Design a 5-tap MMSE equalizer for the channel. Plot the equalized output waveform Solution: ⎛ 0.0027 ⎞ ⎜ ⎟ ⎜ 0.0219 ⎟ c opt = ⎜ 1.0763 ⎟ ⎜ ⎟ ⎜ −0.5241⎟ ⎜ 0.0434 ⎟ ⎝ ⎠

The equalizer output pulse sample values are given by

peq [k ] = {0.0 0.0 0.0023 0.0203 0.9497 0.0169 0.0441 − 0.0954 0.0095}

q[k]

0.5

5 Samples k

peq[k]

0.5

-0.5

b. Calculate the worst-case eye opening at the equalizer output. Solution:

Worst-case eye opening ξ ISI = peq [ 0] − Dpeak = 0.9497 − 0.1886 = 0.7611 .

c. Calculate the noise enhancement at the equalizer output. Solution:

The noise enhancement factor of the equalizer is obtained by using (12.169) as 2

η NE = ∑ ci2 = 1.4354 i =−2

|Q(ejw )|

d. Plot DTFTs of the overall channel, equalizer, and combined responses.

0 -10 -20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.6

0.7

0.8

0.9

0.6

0.7

0.8

0.9

|Heq(ejw )|

ω/π 0 -10 -20

0.1

0.2

0.3

0.4

0.5

|Peq(ejw )|

ω/π 0 -10 -20

0.1

0.2

0.3

0.4

0.5

ω/

e. Repeat (a)-(d) with a 5-tap ZFE. Solution: ⎛0 ⎞ ⎜ ⎟ ⎜0 ⎟ c opt = ⎜1.1456 ⎟ ⎜ ⎟ ⎜ −0.5726 ⎟ ⎜ −0.0002 ⎟ ⎝ ⎠

The equalizer output pulse sample values are given by

peq [k ] = { 0.0 0.0 0.0 0.0 1.0 0.0 0.0 − 0.125 0.0}

q[k]

0.5

5 Samples k

peq[k]

0.5

-0.5

Worst-case eye opening ξ ISI = peq [ 0] − Dpeak = 1 − 0.1251 = 0.8749 . The noise enhancement factor of the equalizer is obtained by using (12.169) as 2

η NE = ∑ ci2 = 1.6402 i =−2

0 -10 -20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.6

0.7

0.8

0.9

0.6

0.7

0.8

0.9

ω/π 0 -10 -20

0.1

0.2

0.3

0.4

0.5

ω/π 0 -10 -20

0.1

0.2

0.3

0.4

0.5

ω/π

12.12 Show that the gradient vector of the mean-square error objective function

J (c) = c R yy c − 2c r ay + σ a2 T

is given by ∇ c J (c) = 2 ( R yy c − r ay ) . Solution:

The mean-square error objective function can be expanded as N

J (c) = ∑ ∑ ci c j R yy [i, j ] − 2 ∑ ci r ay [i ] + σ a2 i =− N j =− N

i =− N

We note that R yy [i, j ] = R yy [ j , i ] . Differentiating J (c) with respect to the equalizer coefficients ci , i = 0, ±1, ±2,....., ± N , we obtain N ∂J = 2 ∑ c j R yy [i, j ] − 2r ay [i], ∂ci j =− N

i = 0, ±1, ±2,....., ± N

This can be expressed in the matrix form as ⎛ ∂J ⎞ ⎜ ⎟ ⎜ ∂c− N ⎟ ⎜ # ⎟ ⎜ ⎟ ∂J ⎟ ⎜ ∇ c J (c ) = = 2 ( R yy c − r ay ) ⎜ ∂c0 ⎟ ⎜ ⎟ ⎜ # ⎟ ⎜ ∂J ⎟ ⎜⎜ ⎟⎟ ⎝ ∂cN ⎠

12.13 The overall channel model of a digital transmission system is described by the impulse response

{q [ k ]} = {0.743, 0.667} A DFE is used at the receiver to correct post-cursor ISI. a. Draw the block diagram of the equalizer? Solution:

Since there is only post-cursor ISI, no FFE is required. The FBE has only 1 tap. The feedback tap coefficient is given from (12.145) as

b1 =

peq [1] peq [0]

q[1] 0.667 = = 0.9 q[0] 0.743 0.9 T

vk Receive filter yk = zk output samples 1.34

Detected bits aˆ k

Threshold comparator

b. Write expressions for input to the equalizer yk and the threshold comparator zk assuming correct decisions are fed. Solution:

In the absence of FFE, the input to the equalizer yk is given by yk = ak q0 + ak −1q1 + nk

The input to the threshold comparator can be expressed using (12.141) as wk =

zk y q0 n ⎤ q ⎡ − vk = k − b1aˆ k −1 = ⎢ak + ak −1 1 + k ⎥ − b1aˆ k −1 peq [0] q[0] 0.743 0.743 ⎦ ⎣ 0.743

= [ ak + 0.9ak −1 + 1.34nk ] − 0.9aˆ k −1 Assuming that correct decision is fed, the feedback term cancels the post-cursor ISI term. That is,

wk = ak + 1.34nk = ± Eb + 1.34nk c. Calculate the noise amplification produced by the equalizer and probability of bit error. Solution:

The noise amplification η NE produced by the equalizer is given by 10 log10 (1.342 ) = 2.54 dB.

The probability of bit error is obtained from (12.172), by noting that ξ ISI = 1 , as

⎛ ⎛ 2 Eb ⎞ BER = Q ⎜ = Q⎜ ⎟ ⎜ η N ⎟ ⎜ ⎝ NE o ⎠ ⎝

⎞ ⎛ 1.114 Eb ⎞ ⎟ = Q⎜ ⎟ 2 N o ⎟⎠ (1.34 ) N o ⎟⎠ ⎜⎝ 2 Eb

12.14 The overall channel model of a digital transmission system is described by the impulse response

{q [ k ]} = {1, − 0.45, 0.3, −0.05} A DFE is used at the receiver to correct post-cursor ISI. a. What are the equalizer coefficients? Solution:

Since there is only post-cursor ISI, no FFE is required. The FBE has 3 taps. The feedback coefficients are given from (12.145) as b1 = q[1] = −0.45 b2 = q[2] = 0.3 b3 = q[3] = −0.05 b. Draw the block diagram of the equalizer. Solution:

Input

yk 1 peq [0]

Symbol decision

aˆk

−

vk TT

−0.45

0.3

−0.05 Feedback Equalizer (FBE)

c. For an input data bit ak = 1 , calculate the equalizer input assuming previous data bits have values ak −1 = 1, ak − 2 = −1, ak −3 = 1 . Assuming correct decisions are fed, calculate the threshold comparator input. Solution:

The input to the threshold comparator can be expressed using (12.141) as wk =

3 zk y − vk = k − vk = [ ak q0 + ak −1q1 + ak − 2 q2 + ak −3 q3 + nk ] − ∑ bn aˆ k − n peq [0] q[0] n =1

= [ ak − 0.45ak −1 + 0.3ak − 2 − 0.05ak − 3 + nk ] + 0.45aˆ k −1 − 0.3aˆ k − 2 + 0.05aˆ k −3

For an input data bit ak = 1 and previous data bits ak −1 = 1, ak − 2 = −1, ak −3 = 1 , the threshold comparator input samples are given by

wk = ⎡⎣1 − 0.45 (1) + 0.3 ( −1) − 0.05 (1) + nk ⎤⎦ + 0.45aˆk −1 − 0.3aˆk − 2 + 0.05aˆk −3

= 0.20 + nk + 0.45aˆk −1 − 0.3aˆk − 2 + 0.05aˆk −3 Assuming correct decisions are fed, the feedback term cancels the post-cursor ISI term. That is,

wk = 0.20 + nk + 0.45 (1) − 0.3 ( −1) + 0.05 (1) = 1 + nk d. Repeat (c) assuming that the previous decision is incorrect. Solution:

Assuming that the previous decision is incorrect, the threshold comparator input is given by

w k = 0.20 + nk + 0.45 ( −1) − 0.3 ( −1) + 0.05 (1) = 0.1 + nk e. Repeat (d) assuming that errors were made in both previous bits. Compare the noise margin in all 3 cases and comment. Solution:

Assuming that both previous decisions are incorrect, the threshold comparator input is given by

wk = 0.20 + nk + 0.45 ( −1) − 0.3 (1) + 0.05 (1) = −0.5 + nk

The noise margins are 1 and 0.1 V in cases (c) and (d) respectively. 12.15. Consider a binary antipodal system with overall channel impulse response

{q [ k ]} = {0.58, 0.813} a. Design a 1-tap DFE for the channel. What is the feedback coefficient value? Solution:

Since there is only post-cursor ISI, no FFE is required. The FBE has only 1 tap. The feedback coefficient is given from (12.145) as b1 =

peq [1] peq [0]

q[1] = 1.4 q[0] 1.4 TT

vk Receive filter yk = z k output samples 1 0.58

Detected bits aˆ k

Threshold comparator

b. Due to additive white Gaussian noise, the corrupted samples at the input to the equalizer for time instants k = 1,..., 6 are given by

{ y [ k ]} = {−0.116, 0.29, − 0.174, 1.508, 0, − 1.16} Assuming the correct previous decision value aˆ0 = −1 is fed back, calculate the detected values of bits corresponding the input sequence { y [ k ]} .

Solution:

The threshold comparator input is given by applying (12.145) as wk =

zk y y − v k = k − b1aˆ k −1 = k − 1.4aˆ k −1 peq [0] q0 0.58

Table below displays calculation of the detected values of bits. 0

1 2 −0.116 0.29 1.2 −0.9 −1 1 −1

k yk wk aˆk

3 4 5 −0.174 1.508 0.0 1.1 1.2 −1.4 1 1 −1

6 −1.16 −0.6 −1

c. Assuming instead that the wrong previous decision value aˆ0 = 1 is fed back, calculate the detected values of bits corresponding the input sequence y [ k ] . How many incorrect decisions are made assuming that decisions in (b) are correct. Solution:

Table below displays calculation of the detected values of bits. 0 1 2 3 −0.116 0.29 −0.174 1.9 −1.7 −1.6 1 −1 1 −1

k yk wk aˆk

4 5 1.508 0.0

6 −1.16

4.0

−1.4 −0.6

−1

Three incorrect decisions follow after wrong decision value aˆ0 = 1 is fed back. d. In both cases (b) and (c), calculate the mean-square error at the threshold comparator input assuming that decisions in (b) are correct. Solution:

The mean-square error at the threshold comparator input is defined as

{

J = E [ w k − ak ]

}

In case (b), the mean-square error at the threshold comparator input can be expressed as 1 6 1 2 2 2 2 2 2 2 ( wk − ak ) = ⎡⎣(1.2 − 1) + ( −0.9 + 1) + (1.1 − 1) + (1.2 − 1) + ( −1.4 + 1) + ( −0.6 + 1) ⎤⎦ ∑ 6 k =1 6 = 0.07

In case (c), the mean-square error at the threshold comparator input is given by

1 6 1 2 2 2 2 2 2 2 ( wk − ak ) = ⎡⎣( −1.6 − 1) + (1.9 + 1) + ( −1.7 − 1) + ( 4.0 − 1) + ( −1.4 + 1) + ( −0.6 + 1) ⎤⎦ ∑ 6 k =1 6 = 5.29

Chapter 13 13.1 Consider the synchronous time-division-multiplexing of 8 1-Mbps digital signals. The output frame is composed of 64 blocks of user data and each block is formed by interleaving one bit from each input line. Assume that the multiplexer adds 1 framing byte to each frame to facilitate demultiplexing operation at the receiver. a. Illustrate the TDM output frame format.

512 data bits I1 I 2 I 3 I 4 I 5 I 6 I 7 I8

Framing byte

…

Block # 1

I1 I 2 I 3 I 4 I 5 I 6 I 7 I8

Block # 64

b. Determine the input bit duration. Solution: Input bit duration = 1 µsec c. Determine the output bit duration. Solution: Number of bits from each data source per multiplexer output frame = 64 Total number of data bits per output frame = 64 × 8 = 512 bits Time to accumulate 512 data bits at the multiplexer input = 64 µsec Frame length = 512 + 8 = 520 bits Output bit duration = 64/520 = 8/65 µsec d. Determine the output frame rate. Solution: Output frame rate =

106 = 15, 625 frames/sec 64

e. Determine the output bit rate. Solution: Output bit rate =

1 106 = = 8.125Mbits/sec Output bit duration 8 / 65

13.2 Consider the synchronous time-division-multiplexing of sixteen 2 Mb/s digital signals. The output frame is composed of 256 blocks of user data and each block is formed by interleaving bytes from each input line. Assume that the frame synchronization circuitry adds 2 framing bytes to each frame. Illustrate the TDM output frame format. Determine a. Determine the input bit and byte durations Solution: Input bit duration = 0.5 µsec Input byte duration = 0.5×8 = 4 µsec b. Determine the output bit and byte durations Solution: Number of bytes from each data source in the multiplexer output frame = 256 Total number of data bytes per output frame = 256×16 = 4096 Frame length = 4096 + 2 = 4098 bytes = 32,784 bits Time to accumulate 4096 data bytes at the multiplexer input = 256 × 8/2×106 = 1,024 µsec Output bit duration =

1, 024 × 10−6 64 = µ sec 32, 784 2, 049

Output byte duration =

512 µsec 2, 049

c. Determine the output frame rate. Solution: Output frame duration =

4098 × 512 = 1024 µsec 2, 049

Output frame rate =

106 frames/sec 1024

d. Determine the output bit rate. Solution: Output bit rate =

1 2, 049 × 106 = = 32.01563 Mbits/sec Output bit duration 64

13.3 The E-1 (CEPT-1) carrier system multiplexes thirty voice channels. E-1 frame consists of 32 bytes. Byte 0 carries the frame alignment word 0011011. Bytes 1-15 and 17-31 carry PCM words corresponding to voice channels. Byte 16 carries the signaling information. The frame transmission rate is 8000 frames/second. a. Sketch E-1 frame format. Solution: The bit rate of E1signal is 2.048 Mbps. As depicted in Figure, the E1 frame is comprised of 32 bytes labeled as timeslots 0 to 31. We calculate this bit rate by multiplying the 32-byt E1 frame by 8000 frames per second. Subtracting time slots 0 and 16, we see that E1 lines offer 30 time slots to carry user data or a payload-carrying capacity of 1.920 Mbps.

Frame sync and alarms

Signaling

Timeslot 0

Timeslot 1

……

Timeslot 16

……

Timeslot 30

Timeslot 31

Speech or Data bits

b. What is the duration of each E-1 frame? Solution: Like all basic frames used in telecommunications, the duration of E-1 frame is 125 µseconds. c. What bit rate is assigned to each voice channel? Signaling channel? Solution: Bit rate of each voice channel (time slots 1−15 and 17−31) = 64 kbps Bit rate of signaling channel (time slot 16) = 64 kbps Bit rate of frame sync and user alarms channel (time slot 0) = 64 kbps d. Suppose 32 kb/s DPCM is used instead of PCM to digitize each voice channel. How many voice calls can be carried by E-1 carrier using this compressed format. Suggest the new format of E-1 frame. Solution: Time slots 1−15 and 17− 31now carry speech or data bits from 2 channels. This allows E-1 carrier to carry 60 voice calls. The digital streams for 2 voice channels can be bit- interleaved in each time slot. 13.4 Suppose a multiplexer combines 4 input streams each operating at a nominal rate of 1 Mb/s. The multiplexer frame format is shown in Figure P13.1. The frame is composed of a framing byte and 256 bit-multiplexed data bits. To facilitate stuffing synchronization, 4 flag and 4 stuffable bits are provided one for each input channel. The flag bit for each channel indicates whether the corresponding stuffable bit location contains a stuffing or user data bit. Figure P13.1

256 data bits I1 I 2 I 3 I 4 I1 I 2 I 3 I 4 …

Framing byte

… Flag Stuffable bits bits

a. Calculate the multiplexer output bit rate. Solution: Number bits in the frame = 8 + 256 + 4 + 4 = 272 Time to accumulate 256 data bits at the multiplexer input = 64/1×106 = 64 µsec Output bit duration =

Output bit rate =

64 × 10−6 8 = µ sec 272 34

34 Mb/s = 4.25 Mb/s 8

b. Determine the acceptable tolerance on input signal rates. Solution: The multiplexer provides either 64 or 65 bits for each stream per 272-bit frame (of duration 64 µsec). Thus, it allows each of the four input streams to transmit as low as 1 Mbps and as high as 1.015625 Mbps. c. What is the maximum stuffing rate per channel? Solution: The frame structure allows 1 bit to be stuffed for each input bit stream, i.e., 1 bit every 64 µsec per channel. Thus, the maximum stuffing rate per channel is 15.625 kbps. 13.5 The M12 multiplexer described in ITU-T Rec. G.473 combines four DS1 signals using stuffing synchronization. The multiplexer output frame format is shown in Figure 13.6. A tolerance of 50 ppm on nominal DS1 signal rates is allowed in the ITU-T standard. a. What is duration of a sub-frame? A frame? Solution: The bit rate of a DS2 signal is 6.312 Mbps. Each DS2 frame contains 49 × 6 × 4 = 1176 bits. Therefore, Frame duration =

1176 = 186.31 µ sec 6.312 × 106

Each DS2 sub-frame contains 49 × 6 = 294 bits. Therefore, 5 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Sub-frame duration =

294 = 46.578 µ sec 6.312 ×106

b. Determine the maximum and minimum input DS1signal rates. Solution: Tolerance of ± 50 ppm on the nominal DS1 rate implies that DS1 tributary rates into the multiplexer can vary 1.544Mbps × ±50 × 10−6= ± 77 bps. Therefore, DS1 rates at the M12 multiplexer input are expected to be in the range between 1.543,923 and 1.544,077 Mbps. As discussed in Example 13.3, the DS2 frame allows DS1 tributary rates between 1.540,429 and 1.545,796 Mbps. This wide range of rates allows DS2 signals the flexibility to transmit proprietary encoded DS1 signals as well as the commonly used 1.544Mbps ±50ppm.

DS1 nominal rate 1.540429 Mbps 1.543923 Mbps 1.544 Mbps 1.544077 Mbps 1.545796 Mbps

1.544 Mbps ±77 b/s

Range of acceptable DS1 rates multiplexed into a DS2

c. Determine the multiplexer synchronous rate per channel. Solution: The number of overhead bits per second in a DS2 signal is 5,367.35 frames/sec × 24 overhead bits/frame = 128.8164 kbps Therefore, the total number of DS1 bits allocated in a DS2 signal is 6.312 ×106 − 128.8164 kbps = 6.183184 Mbps

This capacity is divided equally among 4 DS1 signals. Therefore each DS1 signal may be input at a maximum rate of 6.183184×106 ÷ 4 = 1.545796 Mbps. It is also the multiplexer synchronous rate per channel. Therefore, the bit stuffing rate for a DS1input tributary operating at this rate is 0 bps. However, if the DS1input tributary is operating at the minimum rate of 1.540429 Mbps, the bit stuffing rate for it is 5.367 kbps. 13.6

The frame format for a M23 multiplexer which combines 7 DS2 signals to form the DS3 signal is displayed in Figure P13.2. In each DS3 frame one bit can be stuffed for each of the seven DS2 signals. Specifically, the state of the three C-bits in the ith subframe indicates whether or not bit stuffing occurs for the ith DS2 input during the multiplexing process.

Figure P13.2 M23 Multiplexer Frame Format DS21

DS22

DS23

DS24

DS25

DS26

DS27

……

DS27

Stuffing indicator bits

a. What is the number of bits in a DS3 frame? Solution:

The bit rate of a DS3 signal is 44.736 Mbps. Each DS3 frame contains 85 × 8 × 7 = 4,760 bits. b. How many frames/second are produced by the multiplexer? Solution: Number of frames produced by the multiplexer 44.736 Mbps ÷ 4,760 bits/frame = 9,398.32 frames/second c. Calculate the number of overhead bits/second added by the multiplexer. Solution: The DS3 frame is composed of seven subframes. Each subframe consists of eight blocks and each block contains 85 bits. The first bit in each block is a DS3 OH bit. Each DS3 frame contains 1 OH bit/block × 8 blocks/subframe × 7 subframes = 56 OH bits The number of overhead bits per second is given by 9,398.32 frames/sec × 56 overhead bits/frame = 526.30592 kbps d. How many bits are available for stuffing per second? Solution: Total number of DS3 bits

= 44.736 Mbps

Seven DS2 signals (7 × 6.312 Mb/s)

= −44.184 Mbps __________________ 552 kbps

DS3 OH bits

= −526.306 kbps __________________

Stuffing bits available

25.694 kbps

Therefore the bit stuffing rate for a DS2 signal operating at the nominal rate is 25.694 kbps ÷7 DS2 signals = 3.671 kbps e. Specify minimum and maximum allowable rates for DS2 input signals. Solution:

The maximum allowable DS2 rate is computed as follows: DS3 signal rate =

44.736 Mbps

DS3 OH bits

−526.306 kbps

Total DS2 bits =

44.209694 Mbps

The total number of DS2 bits is allocated evenly across the seven DS2 signals: 44.209694 Mbps ÷ 7 DS2 signals = 6.315671 Mbps Therefore, Maximum DS2 signal rate = 6.315671 Mbps. It is also the multiplexer synchronous rate per channel. The bit stuffing rate in the M23 multiplexer frame structure for a DS2 signal operating at this rate is 0 bps. The minimum allowable DS2 rate is computed by taking the maximum allowable DS2 rate and subtracting the maximum stuffing rate (i.e., the DS3 frame rate) as follows: Maximum DS2 rate = Maximum stuff rate = Minimum DS2 rate =

6.315671 Mbps −9.398 kbps 6.306272 Mbps

Therefore, Minimum DS2 signal rate = 6.306272 Mbps . The bit stuffing rate in the M23 multiplexer frame structure for a DS2 signal operating at this rate is 9.398 kbps. 13.7 The actual bit rates of 7 DS2 tributary signals input to a M23 multiplexer are given in the following table. Calculate the bit stuffing rate required for each of the DS2 signal for multiplexing them. Display the tributary and corresponding stuffing rates in a multiplexer block diagram. DS2 tributary 1 2 3 4 5 6 7

Bit rate (Mbps) 6.3075 6.306272 6.315671 6.310775 6.313225 6.31445 6.312

Solution: Stuff rate for DS2 tributary = Multiplexer synchronous rate – DS2 tributary rate

Stuff rate for DS2 tributary No. 1 = 6.315671Mbps− 6.3075 Mbps = 8.171 kbps Stuff rate for DS2 tributary No. 2 = 6.315671 Mbps− 6.306272 Mbps = 9.399 kbps Stuff rate for DS3 tributary No. 3 = 6.315671 Mbps− 6.315671 Mbps = 0 kbps Stuff rate for DS2 tributary No. 4 = 6.315671 Mbps− 6.310775 Mbps = 4,896 kbps Stuff rate for DS2 tributary No. 5 = 6.315671 Mbps− 6.313225 Mbps = 2.446 kbps Stuff rate for DS2 tributary No. 6 = 6.315671 Mbps− 6.31445 Mbps = 1.221 kbps Stuff rate for DS2 tributary No. 7 = 6.315671 Mbps− 6.312 Mbps = 3.671 kbps Figure displays seven DS2 tributary signals and corresponding stuff rates. DS2 tributary # 1 6.307500 Mbps

6.315671 Mb/s

Stuff rate = 8.171 kbps DS2 tributary # 2 6.306272 Mb/s

6.315671 Mb/s

Stuff rate = 9.399 kbps DS2 tributary # 3 6.315671 Mb/s Stuff rate = 0 bps DS2 tributary # 4 6.310775 Mb/s

6.315671 Mb/s

DS3 output = 44.736 Mb/s

6.315671 Mb/s

Stuff rate = 4.896 kbps DS2 tributary # 5 6.313,225 Mb/s Stuff rate = 2.446 kbps

6.315671 Mb/s

DS2 tributary # 6 6.314450 Mb/s Stuff rate = 1.221 kbps

6.315671 Mb/s

DS2 tributary # 7 6.312000 Mb/s

6.315671 Mb/s

Stuff rate = 3.671 kbps DS3 overhead = 526.306 kbps

13.8 Consider the M12 multiplexer with frame format illustrated in Figure 13.6. a. What is the probability of accepting false FAW in the search mode, assuming random data from tributaries, if a perfect match is required with 8 framing (F) bits of the framing pattern? Solution: The probability that random data matches the FAW is obtained from (13.58) as 8

1 ⎛1⎞ PFS = ⎜ ⎟ = ⎝ 2 ⎠ 256

b. What is the probability of rejecting true frame synchronization in the search mode for a link BER = 10-3? Solution: Probability of miss = 1 − PD Now

PD = {All bits received correctly} = (1 − p)8 where p = BER = 10−3 Probability of miss = 1 − (1 − 10−3 )8 = 7.972 ×10−3 13.9 The 8.448 Mb/s second-level multiplexer based on ITU-T Rec. G.742 combines four E1 (CEPT-1) signals at 2.048 Mb/s. A10-bit framing word is used at the beginning of each 848-bit frame. Frame alignment strategy requires the presence of the correct FAW without error. a. What is the probability of accepting a false FAW word in the in-frame mode assuming random binary data? Solution:

The probability that random data matches the FAW of length N is obtained from (13.58) as N

1 ⎛1⎞ ⎛1⎞ PFS = ⎜ ⎟ = ⎜ ⎟ = 1024 ⎝2⎠ ⎝2⎠

b. Assuming a random starting position in the search mode, calculate the average number of tests required for rejection of that position for false alignment? Solution:

Let the random variable y denote the number of frames examined before a location mismatches the framing pattern. The average number of frames E { y} , examined before a location mismatches the FAW, can be expressed as ∞

E { y} = ∑ yP { FAW mismatch detected on examining the ( y + 1) frame} y =0

= 0(1 − PFS ) + 1PFS (1 − PFS ) + 2 PFS2 (1 − PFS ) + 3PFS3 (1 − PFS ) + ......

{

} P(1 −(1P− P) )

= PFS (1 − PFS ) 1 + 2 PFS + 3PFS2 + 4 PFS3 + ... =

FS 2

PFS 1 1 ≈ = 1 − PFS 1024 − 1 1024

c. Calculate the average number of bits that elapse until the frame synchronizer locks onto the true FAW. Solution:

The length of a CEPT-2 frame n f is 106 bytes = 848 bits. On average, if the frame synchronizer starts at a random bit location in CEPT-2 frame, it will have to examine n f / 2 incorrect bit locations before locking onto the FAW. For each such incorrect bit location, on average, a FAW mismatch will be discovered after examining E { y} frames at which time the synchronizer will move one bit forward to the next bit. Since the synchronizer spends n f E { y} bit times in rejecting each incorrect bit location, the average number of bits elapsed until the nf nf n f E { y}) + frame synchronizer locks onto the framing bit is given by ( 2 2 1 , the average number of bits bits. Substituting n f = 848 and E { y} ≈ 1024 ⎛ 848 ⎞ + 1⎟ = 775 bits elapsed is given by 424 ⎜ ⎝ 1024 ⎠ 13.10 Consider a QPSK system operating at 1 Mb/s. Calculate the phase error variance if a quadrupling loop is used to recover the carrier. Assume that the received signal Eb / N o is 11 dB and the PLL noise bandwidth = 5 kHz. Solution:

Bit rate = 1 Mbps Loop filter noise bandwidth BN = 5 kHz Loop SNR ρ L =

Es = 11 + 3 dB ⇒ 25.12 No

The squaring loss is SL =

1 ⎡ 4.5 6 1.5 ⎤ + 2 + 3⎥ 1+ ⎢ ρ ρL ρL ⎦ L ⎣

1 1 = = 0.84 6 1.5 ⎤ 1.19 ⎡ 4.5 + + 1+ ⎢ ⎥ ⎣ 25.12 631 15,849 ⎦

The loop phase error variance is

BN T 5 × 2 ×10−3 = = 4.74 ×10−4 rad 2 σ = o S L ρ L 25.12 × 0.84 2 φˆ

13.11 A squaring loop is used for carrier recovery in a BPSK system. Calculate the normalized loop bandwidth BN Tb needed so that the output phase error variance after the prescaler is 0.04 rad2. Assume that the received signal Eb / N o is 11 dB. Solution

The output phase error variance due to AWGN is obtained from (13.23) as

σ φ2ˆ = o

BN Tb BN Tb = = 0.0834 BN Tb rad 2 S L ρ L 0.952 ×12.59

1 times that at the input. Since N the phase error variance at the prescaler output is required to be 0.04 rad2, the input phase error variance (i.e., inside the loop) should be less than 22 × 0.04 = 0.16 rad2. Therefore, we want to find BN Tb value such that 0.0834 BN Tb ≤ 0.16 . That is,

The phase (error) at the ÷ N prescaler output is

BN Tb ≤ 1.918 or BN ≤

1.918 Tb

Let us choose BN =

1.875 . Tb

a. Write an expression for the closed-loop transfer function of the two-pole PLL that meets the above phase error specification. Solution:

For the two-pole PLL, we note the following relationship between K, ζ, and ωn from Table 5.4: K=

ωn 2ζ

The noise bandwidth of the two-pole PLL is given From Table 5.4 as BN =

ωn 8ζ

We choose damping constant ζ = 1 to prevent an overshoot of the phase error. Substituting yields

ωn = 8ζ BN = 8 × 1×

1.875 15 = radians/sec Tb Tb

The closed-loop transfer function of the two-pole PLL is given by

H (s) =

ωn2 225 = 2 2 2 2 s + 2sζωn + ωn s Tb + 30Tb s + 225

From Table 5.4, the open loop gain K and time-constant τ 1 are given by K = 4 BN = 4 ×

τ1 =

1.875 7.5 = Tb Tb

Tb T 1 = = b 2 4Kζ 4 × 7.5 30

The loop filter is given by F (s) =

7.5 / Tb K = 1 + τ 1s 1 + s (Tb / 30 )

b. Sketch an implementation of the PLL loop filter.

τ 1 = RC =

υin

7.5 Tb

Tb 30

υout

13.12 A Costas loop is used for carrier recovery in a BPSK system with following parameters: Bit rate = 1 Mb/s Loop filter noise bandwidth BN = 50 kHz Loop SNR ρ L = 10 dB a. Calculate the squaring loss. Solution: SL =

1 = 0.5

ρL

1 1 = = 0.9524 0.5 1.05 1+ 10

b. Calculate the loop phase error variance. Solution:

σ φ2ˆ = o

BN T 50 ×103 ×10−6 = = 5.25 ×10−3 rad 2 0.9524 ×10 SL ρL

13.13 PLLs used in clock recovery circuits are characterized by very small bandwidth (ω3dBT << 1) and damping factor ζ ≥ 5. a. Show that the closed loop transfer function of the PLL in (13.35) can be approximated to that of a single-pole filter.

H ( s) ≈

2ζωn s + 2ζωn

Solution:

The phase transfer function of the resulting second-order PLL active lead-lag loop filter is given from Table 5.4 as ⎛

H (s) =

ωn2 + s ⎜ 2ζωn −

ωn2 ⎞

⎟ K ⎠ ⎝ s 2 + 2 sζωn + ωn2

For large K, this can be simplified to 2ζωn s + ωn2 H ( s) = 2 s + 2 sζωn + ωn2

This can be expressed as 2ζωn +

H (s) =

ωn2 s

s + 2ζωn +

ωn2 s

In the limit as s → ∞ , the phase transfer function asymptotically approaches the LP transfer function

H ( s) ≈

2ζωn s + 2ζωn

b. What is the 3-dB bandwidth of the loop ? Solution:

In ω-domain, we can express H(s) as

H ( jω ) ≈

2ζωn 1 1 = = jω + 2ζωn ⎛ ω ⎞ ⎛ ω ⎞ j⎜ ⎟ +1 1+ j ⎜ ⎟ ⎝ 2ζωn ⎠ ⎝ B−3dB ⎠

where B−3dB = 2ζωn radians/sec is the 3-dB bandwidth of the loop. 13.14 Consider length-13 Neuman-Hofman sequence [1, 1, 1, 1, 1, 1, -1, -1, 1, 1, -1, 1, 1] . a. Compute and sketch its autocorrelation function. Solution:

The autocorrelation function Rc (k ) of the FAW is given by Rc (k )

1 N − k −1 ∑ ci ci + k N i =0

Autocorrelation of length-13 Neuman-Hofman sequence: 1, 1, 1, 1, 1, 1, -1, -1, 1, 1, -1, 1, -1 1 0.9 0.8 0.7

Rc (k)

0.6 0.5 0.4 0.3 0.2 0.1 0 -10

-5

0 k

b. What is the magnitude of its maximum side-lobe values? Solution:

The autocorrelation function side-lobes for this sequence are no larger than 0.1538.

Rc (k )

−12 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0

−0.0769 0.0 −0.0769 0.0 0.0769 0.0 0.0769 0.0 0.0769 0.1538 0.0769 0.1538 1.0

12 11 10 9 8 7 6 5 4 3 2 1

−0.0769 0.0 −0.0769 0.0 0.0769 0.0 0.0769 0.0 0.0769 0.1538 0.0769 0.1538

c. Comment on its correlation properties versus that of length-13 Barker sequence. Solution:

1. This sequence satisfies Barker property Rc (k ) ≤ 1, k ≠ 0 2. The maximum sidelobe magnitude is

2 1 versus for Barker sequences. N N

13.15 Consider length-15 PN sequence [-1, 1, 1, 1, -1, -1, -1, -1, 1, -1, 1, -1, -1, 1, 1]. a. Compute and sketch its autocorrelation function. Solution: Autocorrelation of length-15 PN sequence: -1, 1, 1, 1, -1, -1, -1, -1, 1, -1, 1, -1, -1, 1, 1

0.8

Rc (k)

0.6

0.4

0.2

-0.2 -10

-5

0 k

Rc (k )

−14 −13 −12 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 0

−0.0667 0.0 0.2 0.133 −0.2 −0.133 −0.2 0.133 0.0667 −0.133 −0.2 −0.2677 −0.0667 0.0 1.0

14 13 12 11 10 9 8 7 6 5 4 3 2 1

−0.0667 0.0 0.2 0.133 −0.2 −0.133 −0.2 0.133 0.0667 −0.133 −0.2 −0.2677 −0.0667 0.0

b. What is the magnitude of its maximum side-lobe values? Solution:

The autocorrelation function side-lobes for this sequence are no larger than 0.2677. c. Comment on its correlation properties versus that of length-13 Barker sequence. Solution:

1. This sequence satisfies Barker property Rc (k ) ≤ 1, k ≠ 0 2. The maximum sidelobe magnitude is

4 1 versus for Barker sequences. N N

Chapter 14

14.1 A discrete memoryless source x emits symbols from the alphabet set A = {a, b, c, d , e} with probabilities {0.5, 0.3, 0.1, 0.05, 0.05} respectively. a. Determine the information content of the source in bits/symbol. Solution: H ( x ) = − ∑ p x ( xi ) log 2 p x ( xi ) xi ∈A

= −0.5log 2 ( 0.5 ) − 0.3log 2 ( 0.3) − 0.1log 2 ( 0.1) − 0.05log 2 ( 0.05 ) − 0.05log 2 ( 0.05 ) = 1.7855 bits/symbol b. What is the maximum information content of the source? Specify the corresponding source PMF. Solution: The entropy of the information source is maximum when all symbols in the alphabet are equiprobable. That is, the probabilities of symbols in the alphabet set A = {a, b, c, d , e} are {0.2, 0.2, 0.2, 0.2, 0.2} . The maximum value of source entropy is then log 2 5 = 2.3219 bits/symbol . c. Determine the number of bits per symbol required to transmit the source with simple binary coding. Solution: Since the number of symbols in the alphabet is 5, k = 3 bits are required with simple binary coding to satisfy 2k ≥ 5. 14.2 A discrete memoryless source emits symbols every 125 µsec from the alphabet set A = {−1, −3 / 4, −1/ 2, −1/ 4,1/ 4,1/ 2,3 / 4,1} with respective probabilities

{1/16,1/16,1/ 8,1/ 4,1/ 4,1/ 8,1/16,1/16} . a. Determine the entropy of the source. Solution:

H ( x ) = − ∑ p x ( xi ) log 2 p x ( xi ) xi ∈A

= −0.0625log 2 ( 0.0625 ) − 0.0625log 2 ( 0.0625 ) − 0.125log 2 ( 0.125 ) − 0.25log 2 ( 0.25 ) − 0.25log 2 ( 0.25 ) − 0.125log 2 ( 0.125 ) − 0.0625log 2 ( 0.0625 ) − 0.0625log 2 ( 0.0625 ) = 2.75 bits/symbol

b. What is the minimum bit rate required to transmit the source? Solution: Minimum bit rate required to transmit the source =

2.75 = 22 kbps 125 × 10−6

14.3 A discrete memoryless source emits symbols from the alphabet set A = {0,1/ 8,1/ 4,1 / 2,3 / 4, 7 / 8,15 / 16,1} with respective probabilities

{1/16, 1/ 8, 1/ 8, 1/ 4, 5 / 32, 1/ 8, 3 / 32, 1/16} . a. Determine the entropy of the source. Solution:

H ( x ) = − ∑ p x ( xi ) log 2 p x ( xi ) xi ∈A

1 1 1 1 5 log 2 (1/16 ) − log 2 (1/ 8 ) − log 2 (1/ 8 ) − log 2 (1/ 4 ) − log 2 ( 5 / 32 ) 16 8 8 4 32 1 3 1 − log 2 (1/ 8 ) − log 2 ( 3 / 32 ) − log 2 (1/16 ) 8 32 16 = 2.8636 bits/symbol =−

b. Assuming that the source is quantized according to the following quantization table Input symbol 0, 1/8, 1/4 1/2, 3/4 7/8,15/16, 1

Quantizer output symbol 1/8 1/2 15/16

Find the entropy of the quantized source. Solution:

The following table summarizes probabilities of the quantizer output symbols: Input symbol

Quantizer output symbol

0, 1/8, 1/4 1/2, 3/4 7/8,15/16, 1

1/8 1/2 15/16

Probability of quantizer output symbol 10/32 13/32 9/32

The entropy of the quantized source is given by H ( x ) = − ∑ p x ( xi ) log 2 p x ( xi ) xi ∈AQ

10 13 9 log 2 (10 / 32 ) − log 2 (13 / 32 ) − log 2 ( 9 / 32 ) 32 32 32 = 1.5671 bits/symbol =−

14.4 Consider a discrete-time, continuous amplitude source x characterized by the Laplacian PDF 2x

− 1 e σx 2σ x

f x ( x) =

a. Calculate the differential entropy of x. Solution:

The differential entropy of x is obtained using (14.20) as ∞

∞

−∞

h ( x ) = − ∫ f x ( x) log 2 f x ( x)dx = − ∫ f x ( x)

log e f x ( x) dx log e 2

2x ⎤ ⎡ 1 − σx ⎥ log e ⎢ e ∞ − 2x 2 σ ⎢ ⎥ x 1 σx ⎣ ⎦ dx e =− ∫ log e 2 2σ x −∞ ∞ − ⎡ ⎛ 1 ⎞ 2 x⎤ 1 e σ x ⎢log e ⎜ dx =− − ⎟ ∫ ⎜ 2σ ⎟ σ ⎥ ( log e 2 ) 2σ x −∞ ⎢⎣ ⎥ x x ⎠ ⎝ ⎦ 2x

⎛ 1 ⎞ log e ⎜ ⎟ 2x 2x ∞ 2σ x ⎠ ∞ − σ x 2 x − σx 1 ⎝ e dx + =− e dx ( log e 2 ) 2σ x −∞∫ ( log e 2 ) 2σ x −∞∫ σ x

The expression can be further simplified as follows h( x) =

( 2σ ) 1

log e

∞

∫e

( log e 2 )

−

σx

2σ x −∞

∞

− 2 1 σx dx + x e dx ∫ σ x ( log e 2 ) 2σ x −∞

σx

= log 2

( 2σ ) ×1 + σ ( log2 2) σ 2 x

= log 2

log e = log ( 2σ e ) ( 2σ ) + log 2 e

b. Assume that the source x is uniformly quantized with a quantization step size ∆. Let p xQ ( i∆ ) be the probability that x is quantized to i∆, where i = 0, ±1, ±2,... . Solution:

If the source x with Laplacian PDF is uniformly quantized with a quantization step size ∆, the PMF of the quantized source xQ is given by 2x ⎧( i + 0.5) ∆ 1 − σx ⎪ ∫ e dx, i = 0, ±1, ±2,... p xQ ( i∆ ) = ⎨( i −0.5 2 σ ∆ x ) ⎪ otherwise ⎩0,

For i = 0, ∆/2

∆/2

− − − 1 2 σx σx σx px ( 0 ) = e dx = e dx = − e ∫ ∫ σx 0 2σ x −∆ /2

where α =

∆ /2 −

= 1− e 2 0

2∆

σx

For i ≥ 1, ( i + 0.5) ∆

− − 1 σx σx = − p x ( i∆ ) = e dx e ∫ 2σ x (i −0.5)∆ −

2i∆

σx

( i + 0.5) ∆

1⎡ − = ⎢e 2⎢ ⎣ − ( i − 0.5) ∆

2 ( i − 0.5) ∆

σx

−e

−

2 ( i + 0.5) ∆

σx

⎤ ⎥ ⎥ ⎦

2∆ ⎞ 2i∆ ⎛ + 2∆ − − ⎛ 2∆ ⎞ − iα ⎛α ⎞ ⎜ e 2σ x − e 2σ x ⎟ = e σ x sinh ⎜ ⎟⎟ = e sinh ⎜ ⎟ ⎜ ⎜ ⎟ ⎝2⎠ ⎝ 2σ x ⎠ ⎝ ⎠

Because f x ( x) is even function (symmetric), we can show that

p x ( i ∆ ) = p x ( −i ∆ ) = e

−iα

⎛α ⎞ sinh ⎜ ⎟ ⎝2⎠

c. Calculate the entropy of the quantized source xQ . Solution:

The entropy of the quantized source xQ can now be calculated as H ( xQ ) = − ∑ p x ( xi ) log 2 p x ( xi ) xi ∈AQ

∞

= − p x (0) log 2 p xQ (0) − 2∑ p x ( xi ) log 2 p x ( xi ) i =1

where A Q = {i∆, i = 0, ±1, ±2,...} . Now ∞

−∑ p x ( xi ) log 2 p x ( xi ) = − i =1

1 ∞ ∑ px ( xi ) log e px ( xi ) log e 2 i =1

=−

1 ⎫ ⎛α ⎞⎧ ∞ sinh ⎜ ⎟ ⎨∑ e − iα ⎡⎣ −iα + log e ( sinh(α / 2) ) ⎤⎦ ⎬ log e 2 ⎝ 2 ⎠ ⎩ i =1 ⎭

=−

∞ 1 ⎫ ⎛α ⎞⎧ ∞ sinh ⎜ ⎟ ⎨α ∑ −ie −iα + log e ( sinh(α / 2) ) ∑ e − iα ⎬ log e 2 ⎝ 2 ⎠ ⎩ i =1 i =1 ⎭

From Appendix A, we have ∞

∑ e−iα = i =1

e −α 1 = α −α 1− e e −1

Taking derivative of both sides ∞

−∑ ie − iα = i =1

d ⎛ 1 ⎞ eα = − ⎜ ⎟ 2 dα ⎝ eα − 1 ⎠ ( eα − 1)

Substituting ⎧ ⎫ ∞ 1 ⎪ −α eα ⎛α ⎞⎪ log sinh( α / 2) −∑ p x ( xi ) log 2 p x ( xi ) = − sinh ⎜ ⎟ ⎨ + ( ) ⎬ 2 2 eα − 1 ⎪ ⎝ 2 ⎠ ⎪ log e 2 ( eα − 1) i =1 ⎩ ⎭

⎧ ⎫ α α − ⎞ − ⎞ ⎛ ⎛ 1 ⎪ −eα ⎛α ⎞⎪ α H ( xQ ) = − ⎜1 − e 2 ⎟ log 2 ⎜1 − e 2 ⎟ − 2sinh ⎜ ⎟ ⎨ log sinh( α / 2) + ) α ⎬ 2( 2 e − 1⎪ ⎝ 2 ⎠ ⎪ log e 2 ( eα − 1) ⎝ ⎠ ⎝ ⎠ ⎩ ⎭

14.5

For a continuous random variable x with x ≤ a , the entropy is maximum if x is uniformly distributed over [−a, a] . Show that the maximum entropy is given by log 2 2a . Solution:

We want to solve the following optimization problem: Max h ( x ) f x ( x)

subject to the constraint ∞

g1 ( x) = ∫ f x ( x)dx − 1 = 0 −∞

x ≤a We use the calculus of variation to solve this constrained optimization problem1. In the present case, L ( x, f x ( x), λ ) = h [ f x ( x) ] + λ g1 [ f x ( x)]

The optimum f x ( x) and Lagrange multiplier λ satisfy

Maximize:

h [ y ( x) ] = ∫ F ( x, y, y ') dx a b

Subject to:

g1 [ y ( x) ] = ∫ G ( x, y, y ') dx a

To solve this problem, the Lagrangian function L ( x, y, y ', λ ) is formed as shown below:

L ( x, y, y ', λ ) = F ( x, y, y ') + λG ( x, y, y ' ) To find the optimum y(x), the following unconstrained Euler equation is solved along with the constraint equation.

d ⎛ ∂L ⎞ ∂L =0 ⎜ ⎟− dx ⎝ ∂y ' ⎠ ∂y

∂L =0 ∂y That is, dh ( x ) dg ( x ) +λ 1 =0 df x ( x) df x ( x)

Now a

h ( x ) = − ∫ f x ( x) log 2 f x ( x)dx = − −a

1 f x ( x) log e f x ( x)dx log e 2 −∫a

Differentiating it with respect to f x ( x ) , we obtain dh ( x ) 1 =− [1 + log e f x ( x)] log e 2 df x ( x)

=−

log e e log 2 f x ( x) − log e 2 log e 2

= − log 2 e − log 2 f x ( x)

dg1 ( x ) =1 df x ( x)

Substituting, we obtain − log 2 e − log 2 f x ( x) + λ = 0 or log 2 f x ( x) = − log 2 e + λ We note that optimum f x ( x ) is independent of x. Hence, the random variable x ≤ a must be uniformly distributed in the range [−a, a] . That is,

⎧1 ⎪ , f x ( x ) = ⎨ 2a ⎪⎩0,

x ≤a otherwise

The maximum differential entropy of the random variable x ≤ a is given by 1 1 ⎛ 1 ⎞ h( x) = − ∫ log 2 ⎜ ⎟ dx = log 2 ( 2a ) ∫ dx = log 2 ( 2a ) 2a 2a ⎝ 2a ⎠ −a −a a

14.6

For a continuous random variable x with x 2 = σ x2 , the Gaussian PDF maximizes the entropy. Show that

1 h ( x ) ≤ log 2 ( 2π eσ x2 ) bits 2 where the equality is achieved if x ~ N ( 0, σ x2 ) . Solution:

We want to solve the following optimization problem: Max h ( x ) f x ( x)

subject to the constraints ∞

g1 ( x) = ∫ f x ( x)dx − 1 = 0 −∞ ∞

g 2 ( x) = ∫ x 2 f x ( x)dx − σ x2 = 0 −∞

As discussed in Problem 14.5, the optimum f x ( x ) and multipliers λ1 , λ2 satisfy dh ( x ) dg ( x ) dg ( x ) + λ1 1 + λ2 2 =0 df x ( x) df x ( x) df x ( x)

(*)

Now dh ( x ) 1 =− [1 + log e f x ( x)] df x ( x) log e 2 dg1 ( x ) =1 df x ( x) dg 2 ( x ) = x2 df x ( x) Substituting into (*), we obtain the following condition for optimal solution.

−1 − log e f x ( x) + λ1 log e 2 + ( λ2 log e 2 ) x 2 = 0 or log e f x ( x) = ( λ1 log e 2 − 1) + ( λ2 log e 2 ) x 2 Solving for the optimal f x ( x) yields

f x ( x ) = e( 1

λ log e 2 −1) + ( λ2 log e 2 ) x 2

(**)

We can now determine λ1 and λ2 by substituting the expression for f x ( x) into the constraint equations as follows: ∞

1 = ∫ e( 1

λ log e 2 −1) + ( λ2 log e 2 ) x 2

−∞

= 2e( 1

λ log e 2 −1)

∞

∫e

( λ2 loge 2 ) x 2

= 2e( 1

λ log e 2 −1)

e( 1

λ log e 2 −1)

⎛1 ⎞ π ⎜⎜ ⎟⎟ , ⎝ 2 −λ2 log e 2 ⎠

λ2 < 0

−λ2 log e 2

Also, ∞

σ = ∫ x 2e( λ log 2−1)+( λ log 2) x dx 2

2 x

−∞

= e( 1

λ log e 2 −1)

∞

∫xe 2

( λ2 loge 2 ) x 2

−∞

= =−

−λ2 log e 2

∞

∫xe 2

( λ2 loge 2 ) x 2

−∞

1 2λ2 log e 2

λ2 log e 2 = −

1 2σ x2

Substituting the value of λ2 yields

e( 1

λ log e 2 −1)

−λ2 log e 2

1 2πσ

2 x

1 2πσ x2

and

e( 2

λ log e 2 ) x 2

= e − x /σ x 2

Substituting λ1 and λ2 into (**), the optimal f x ( x) is obtained as

f x ( x ) = e( 1

λ log e 2 −1)

14.7

e( 2

λ log e 2 ) x 2

e − x /σ x 2

2πσ x2

A Gaussian random signal with zero mean, variance σ 2 and bandwidth B Hz is sampled at the Nyquist rate. The sampled signal is then applied to a quantizer with the following quantization table:

Quantizer Input (−∞, −4σ) (−4σ, −3σ) (−3σ, −2σ) (−2σ, −σ) (−σ, σ) (σ, 2σ) (2σ, 3σ) (3σ, 4σ) (4σ, ∞)

Quantizer output y y0 y1 y2 y3 y4 y5 y6 y7 y8

a. Determine the entropy of the quantizer output. Solution:

We calculate probabilities of the quantizer outputs by noting that P {k1σ ≤ x ≤ k2σ } = Fx (k2σ ) − Fx (k1σ ) ⎛ −k σ ⎞ ⎛ −k σ ⎞ = Q ⎜ 2 ⎟ − Q ⎜ 1 ⎟ = Q ( −k2 ) − Q ( − k1 ) ⎝ σ ⎠ ⎝ σ ⎠ = 1 − Q ( k2 ) − 1 + Q ( k1 ) = Q ( k1 ) − Q ( k2 )

Quantizer output y

Probability p y ( y j )

y0 y1 y2 y3 y4 y5 y6 y7 y8

3.1671×10−5 0.0013 0.0214 0.1359 0.6827 0.1359 0.0214 0.0013 3.1671×10−5

The entropy of the quantized source y is given by H ( y ) = − ∑ p y ( y j ) log 2 p y ( y j ) y j ∈A y

= −2 × 0.0013log 2 ( 0.0013) − 2 × 0.0214 log 2 ( 0.0214 ) − 2 × 0.1359 log 2 ( 0.1359 ) − 0.6827 log 2 ( 0.6827 ) = 1.4209 bits/symbol b. Calculate the minimum bit rate required to transmit the quantized source. Solution:

Minimum bit rate = 2B×1.4209 =2.8418 B bits/sec 14.8

A discrete memoryless source has the alphabet A x = {a, b, c, d } with respective probabilities {1/2, 1/4, 1/8, 1/8}. a. Determine the number of typical sequences of length 8. Solution:

H ( x ) = − ∑ p x ( xi ) log 2 p x ( xi ) xi ∈AQ

= −0.5log 2 ( 0.5 ) − 0.25log 2 ( 0.25 ) − 0.125log 2 ( 0.125 ) − 0.125log 2 ( 0.125 ) = 14 / 8 bits/symbol The number of typical sequences of length 8 is 2nH ( x ) = 28×14/8 = 214 . b. Determine the number of nontypical sequences of length 8. Solution:

The number of atypical sequences is 48 − 214 = 216 − 214 = 3 × 214 . c. Write two examples each of typical and nontypical sequences of length 8. Solution:

Examples of typical sequences include those with a appearing 8×1/2 = 4 times, b appearing 8×1/4 = 2 times, etc.. Two such sequences are {a,d,b,b,a,a,c,a} and {a,a,a,a,c,d,b,b}. Examples of nontypical sequences of length 8: {d,d,b,c,c,a,b,d} and {c,c,c,c,c,b,c,c}. 14.9

A discrete memoryless source emits symbols every 1 ms from the alphabet A x = { x1 , x2 ,..., x8 } with respective probabilities

{0.35, 0.2, 0.15, 0.12, 0.08, 0.05, 0.03, 0.02} . a. Determine the entropy of the source. Solution:

H ( x ) = − ∑ p x ( xi ) log 2 p x ( xi ) xi ∈AQ

= −0.35log 2 ( 0.35 ) − 0.2 log 2 ( 0.2 ) − 0.15log 2 ( 0.15 ) − 0.12 log 2 ( 0.12 ) − 0.08log 2 ( 0.08 ) − 0.05log 2 ( 0.05 ) − 0.03log 2 ( 0.03) − 0.02 log 2 ( 0.02 ) = 2.5443 bits/symbol b. Use the Huffman coding procedure to assign binary codewords to all symbols. Solution:

A Huffman tree for this source is developed in Figure. The Huffman code is displayed in Table.

0.35

0.62 0

x1 x4 x5

x1 x2 x3 x4 x5 x6 x7 x8

0.2

0.38

x2 x5 x6 x7 x8

x2 0.15

0.27 0

x4 x5 1

0.12

0.08

0.18 0

x5 x6 x7 x8

0.05

0.03

0.05 0

x7 x8

0.1 0

x6 x7 x8

1 0 1

0.02

Symbol Codeword x1

010

011

110

1110

11110

11111

c. What is the average length of the Huffman code and how does it compare with the entropy of the source? Solution:

The average number of bits/symbol for this code is given by using (14.97) as

l = ∑ p x (ai )l (ai ) = 2 × 0.35 + 2 × 0.2 + 3 × 0.15 + 3 × 0.12 ai ∈A x

+ 3 × 0.08 + 4 × 0.05 + 5 × 0.03 + 5 × 0.02 = 2.6 bits/symbol

d. Calculate the minimum bit rate required to transmit the source. Solution:

Minimum bit rate required to transmit the source =

2.5443 = 2.5443 kbps 1× 10−3

14.10 A discrete memoryless source emits symbols from the alphabet A x = {a, b, c, d } with respective probabilities {0.2, 0.5, 0.15, 0.15} . a. Design a Huffman code to assign binary codewords to all symbols. Solution:

A Huffman tree for this source is developed in Figure. The Huffman code is displayed in Table. 0.5

1.0 0

abcd

1 0.2 0

0.3

0.15

c 0.15

acd

cd 1

Symbol

Codeword

b a c d

0 10 110 111

b. Now design a Huffman code for encoding pairs of symbols at a time. Solution:

If we now encode pairs of source symbols, we will have the alphabet A x2 = {bb, ba, ab, bc, bd , cb, db, aa, ac, ca, ad , da, cc, cd , dc, dd } with probabilities

{0.25, 0.1, 0.1, 0.075, 0.075, 0.075, 0.075, 0.04, 0.03, 0.03, 0.03, 0.03, 0.0225, 0.0225, 0.0225, 0.0225}

A Huffman tree for this source is developed in Figure. The Huffman code is displayed in Table.

0.25

0.56 0

bb 0.1

bbdbaccaaddabacd

baab

0.1

0.44

ab 0

bc cc

bcaadcdd

0.075

0.31 dbaccaaddabacd

0.15 bdcb

0.075 1

0.135

0.075

0 dbacca

db 0.04

0.04

1 0

0.24

0.03

0.06 0

dbaccaaddacccd

acca

0.03 ca

baabdbaccaaddacccd 0.16

0.075

bbdbaccaaddabacd baabdbaccaaddacccd

0.2

0 1

0.03

0.06 0

0.105

addacccd

adda

0.03 1 da

0.12

0.0225 cc

0.045

cccd

0.0225 cd

0.085 1 aadcdd

0.0225 dc 0.0225

0.045 dcdd

Symbol pair

Probability

Codeword

bb ba ab bc bd cb db aa ac ca ad da cc cd dc dd

0.25 0.1 0.1 0.075 0.075 0.075 0.075 0.04 0.03 0.03 0.03 0.03 0.0225 0.0225 0.0225 0.0225

00 100 101 0100 0110 0111 1100 01010 11010 11011 11100 11101 11110 11111 010110 010111

c. Compare the performance of the two codes. How do they compare with the entropy of the source? Solution:

H ( x ) = − ∑ p x ( xi ) log 2 p x ( xi ) xi ∈A x

= −0.5log 2 ( 0.5 ) − 0.2 log 2 ( 0.2 ) − 0.15log 2 ( 0.15 ) − 0.15log 2 ( 0.15) = 1.7855 bits/symbol The average number of bits/symbol for the code in (a) is given by using (14.97) as l = ∑ p x (ai )l (ai ) = 1× 0.5 + 2 × 0.2 + 3 × 0.3 ai ∈A x

= 0.5 + 0.4 + 0.9 = 1.8 bits/symbol

The average number of bits/symbol pair for the code in (b) is given by using (14.101) as

l2 = ∑ p x ( x)l ( x) = 2 × 0.25 + 6 × 0.1 + 16 × 0.075 + 5 × 0.04 + 20 × 0.03 x

+ 10 × 0.0225 + 12 × 0.0225 = 0.5 + 0.6 + 1.2 + 0.2 + 0.6 + 0.225 + 0.27 = 3.595 bits/symbol pair Therefore, the average code length = 1.7975 bits/symbol. Coding method

Bits/symbol

Entropy Huffman coding symbols of alphabet Huffman coding pair of symbols

1.7855 1.8 1.7975

The entropy of this code is also 1.7855 bits/symbol. The average number of bits/symbol for the Huffman code is 1.8 bits. This code very closely approaches the entropy of the source. By encoding pair of symbols, we achieve a slight improvement but it is not worth the complexity. 14.11 Let a binary channel be described by the transition probabilities matrix ⎛ 2 / 3 1/ 3 ⎞ P=⎜ ⎟ ⎝ 1/ 8 7 / 8 ⎠ 3 1 and P ( x2 ) = . 4 4 a. Determine the output symbol probabilities.

Assume the source probabilities are P ( x1 ) =

Solution: The joint PMF p xy ( xi , y j ) of random variables x and y can be computed using the

transition probabilities matrix P and source probabilities as follows:

1/2

1/4

1/32

7/32

From the joint probability matrix, we can compute marginal PMF of the channel output random variable.

p y = (17 / 32 15 / 32 )

b. Calculate H ( x ) , H ( x y ) , and mutual information I ( x; y ) . Solution:

H ( x ) = − ∑ p x ( xi ) log 2 p x ( xi ) xi ∈A x

= −0.75log 2 ( 0.75 ) − 0.25log 2 ( 0.25 ) = 0.8113 bits/symbol

H ( y ) = − ∑ p y ( y j ) log 2 p y ( y j ) y j ∈A y

= − (17 / 32 ) log 2 (17 / 32 ) − (15 / 32 ) log 2 (15 / 32 ) = 0.9972 bits/symbol H ( x, y ) = − ∑ ∑ p xy ( xi , y j ) log 2 p xy ( xi , y j ) xi ∈A x y j ∈A y

= − (1/ 2 ) log 2 (1/ 2 ) − (1/ 4 ) log 2 (1/ 4 ) − (1/ 32 ) log 2 (1/ 32 ) − ( 7 / 32 ) log 2 ( 7 / 32 ) = 1.6359 bits/symbol Now H ( x y ) = H ( x , y ) − H ( y ) = 1.6359 − 0.9972 = 0.6387 I ( x; y ) = H ( x ) − H ( x y ) = 0.8113 − 0.6387 = 0.1726 14.12 Consider the channel illustrated in Figure P14.1. Figure P14.1 1-p

1-p

a. Write the transition probabilities matrix P of the channel. Solution:

0 ⎞ ⎛1 − p p ⎜ ⎟ 1− p p ⎟ P = ⎜0 ⎜p 0 1 − p ⎟⎠ ⎝

b. Calculate the capacity of the channel. For what input distribution is this capacity achieved? Solution:

Since the channel is symmetric, its capacity is given by C = log 2 N − H ( p,1 − p, 0 ) = log 2 3 − H ( p,1 − p ) = 1.585 − H ( p ) bits/channel use

The capacity is achieved for the input PMF p x = (1/ 3 1/ 3 1/ 3) . 14.13 A channel is described by the transition probabilities matrix ⎛3/ 8 P=⎜ ⎝ 1/ 8

1/8 3/8

3/8 1/8

1/8 ⎞ ⎟ 3/8 ⎠

Find the capacity of the channel. For what input distribution is this capacity achieved? Solution:

Since the channel is symmetric, its capacity is given by C = log 2 N − H ( 3 / 8,1/ 8,3 / 8,1/ 8 ) = log 2 4 − 1.8113 = 0.1887 bits/channel use .

The capacity is achieved for the input PMF p x = (1/ 2 1/ 2 ) 14.14 Two binary symmetric channels are connected in cascade as shown in Figure P14.2. a. Determine the transition probabilities matrix for the cascaded channel. Solution:

The transition probabilities matrices of channel 1 and 2 are given by

⎛ 0.8 0.2 ⎞ ⎛ 0.8 0.2 ⎞ P1 = ⎜ ⎟ , P2 = ⎜ ⎟ ⎝ 0.2 0.8 ⎠ ⎝ 0.2 0.8 ⎠ Using (14.39), we can write p y = p x P1 p z = p y P 2 = p x P1 P 2 ⎛ 0.8 0.2 ⎞ ⎛ 0.8 0.2 ⎞ P1 = ⎜ ⎟ , P2 = ⎜ ⎟ ⎝ 0.2 0.8 ⎠ ⎝ 0.2 0.8 ⎠ The transition probabilities matrix of the cascaded channel is given by ⎛ 0.68 0.32 ⎞ P = P1 P 2 = ⎜ ⎟ ⎝ 0.32 0.68 ⎠ Figure P14.2

0.8

0.2

0.8

0.2

z2 0.8

Channel 1

Channel 2

b. Calculate P ( z1 ) and P ( z2 ) when P ( x1 ) = P ( x2 ) = 0.5 . Solution:

p z = p x P1 P 2 = ( 0.5

⎛ 0.68 0.32 ⎞ 0.5 ) ⎜ ⎟ = ( 0.5 ⎝ 0.32 0.68 ⎠

0.5 )

Therefore, P ( z1 ) = P ( z2 ) = 0.5 c. Determine the capacity of each channel. Solution:

Since both channels are BSC, the capacity is given by C1 = C2 = log 2 (2) − H ( 0.8, 02 ) = 1 − 0.7219 = 0.2781 bits/channel use d. Determine the capacity of the cascaded channel. Solution:

Since the cascaded channel is BSC, its capacity is given by

C = log 2 (2) − H ( 0.68, 0.32 ) = 1 − 0.9044 = 0.0956 bits/channel use 14.15 The Z-channel has input alphabet {0, 1}, and output alphabet {0, 1}, like the BSC. However, the Z-channel is asymmetrical. The 0 symbol is always transmitted correctly, but the 1 symbol is received incorrectly (as 0) with probability ε. a. Sketch the channel diagram showing all transition probabilities. Solution:

P(0|0) = 1 |1 P(0

Input x 1

ε )=

P(1|1) = 1− ε

0 Output y 1

0 ⎞ ⎛1 P=⎜ ⎟ ⎝ε 1− ε ⎠

b. Find the capacity of the channel and the input channel PMF that achieves it. Solution:

Let P { x = 0} = po and P { x = 1} = 1 − po . p y = p x P = ( po

0 ⎞ ⎛1 1 − po ) ⎜ ⎟ = ( po + ε (1 − po ) ⎝ε 1− ε ⎠

(1 − po )(1 − ε ) ) 21

H ( y ) = − ∑ p y ( y j ) log 2 p y ( y j ) y j ∈A y

= − ⎡⎣ po + ε (1 − po ) ⎤⎦ log 2 ( po + ε (1 − po ) ) − (1 − po )(1 − ε ) log 2 ⎡⎣(1 − po )(1 − ε ) ⎤⎦

= H ( (1 − po )(1 − ε ) )

Now H ( y x = 0) = 0 H ( y x = 1) = − ( ε ) log 2 ( ε ) − (1 − ε ) log 2 (1 − ε ) = H ( ε ) H ( y x ) = (1 − po ) H ( ε ) The channel capacity is given by

{

}

{

}

C = Max H ( y ) − H ( y x ) = Max H ( (1 − po )(1 − ε ) ) − (1 − po ) H ( ε ) px

Taking the derivative of H ( (1 − po ) (1 − ε ) ) − (1 − po )H ( ε )

(*)

with respect to po and setting the result to zero, we obtain d H ( (1 − po )(1 − ε ) ) dpo

+ H (ε ) = 0

(**)

Now d H ( (1 − po )(1 − ε ) ) dpo

{

}

d − ⎡ po + ε (1 − po ) ⎤⎦ log 2 ( po + ε (1 − po ) ) − (1 − po )(1 − ε ) log 2 ⎡⎣(1 − po )(1 − ε ) ⎤⎦ dpo ⎣

=−

(1 − ε ) ⎡1 + log

{

log e 2 ⎣

( p + ε (1 − p ) )⎤⎦ − ⎡⎣1 + log ( (1 − p )(1 − ε ) )⎤⎦} o

⎡ p + ε (1 − po ) ⎤ ⎡ (1 − po )(1 − ε ) ⎤ = − (1 − ε ) log 2 ⎢ o ⎥ = (1 − ε ) log 2 ⎢ ⎥ ⎣ (1 − po )(1 − ε ) ⎦ ⎣ po + ε (1 − po ) ⎦

Substituting into equation (**), we obtain

⎡ (1 − po )(1 − ε ) ⎤ ⎥ + H (ε ) = 0 ⎣ po + ε (1 − po ) ⎦

(1 − ε ) log 2 ⎢ or

⎡ (1 − po )(1 − ε ) ⎤ H (ε ) log 2 ⎢ ⎥=− (1 − ε ) ⎣ po + ε (1 − po ) ⎦ or H (ε )

(1 − po )(1 − ε ) = 2− (1−ε ) po + ε (1 − po )

for 0 ≤ ε < 1 . Solving for po , we obtain po =

1 − ε (1 + K ( ε ) )

(1 − ε ) (1 + K (ε ) )

−

H (ε )

, K ( ε ) = 2 (1−ε )

⎛ 1 − ε (1 + K ( ε ) ) The optimal PMF is ⎜ ⎜ (1 − ε ) (1 + K ( ε ) ) ⎝

K (ε )

⎞ ⎟. (1 − ε ) (1 + K (ε ) ) ⎟⎠

Substituting into equation (*), the channel capacity is given by ⎛ K (ε ) ⎞ K (ε ) C = H ( (1 − po )(1 − ε ) ) = H ⎜⎜ H (ε ) ⎟⎟ − ⎝ 1 + K ( ε ) ⎠ (1 − ε ) (1 + K ( ε ) ) c. Discuss the special cases of (b) when ε =0 and ε =1. Solution:

For ε = 0 , K ( 0 ) = 1. C = H (1/ 2 ) = 1 bit/channel use . The answer is obvious because the channel is exactly the noiseless channel as discussed in Example 14.11. For ε = 1 , the capacity is 0 because it is not possible to determine from the output whether a 0 or 1 was transmitted. d. Evaluate the channel input probabilities that achieve the capacity for the case ε = 1/2. Solution:

For ε = 0.5 , K ( 0.5 ) = 0.25. Substituting, we obtain

2 C = H ( 0.2 ) − H ( 0.5 ) = 0.7219 − 0.4 = 0.3219 bit/channel use 5 14.16 Encode the following string using the LZW algorithm abracadabra Assume that the LZW implementation initializes the dictionary with values 0 − 255, where the first 128 entries are ASCII characters. Solution:

The encoder dictionary is initialized with values 0 − 255, where the first 128 entries are ASCII characters. Prefix w Nil a b r a c a d a ab r ra

Input a b r a c a d a b r a -

Dictionary

Output

Pointer

New Phrase

Phrase

Pointer

256 257 258 259 260 261 262 263 -

ab br ra ac ca ad da abr -

a b r a c a d ab ra

97 98 114 97 99 97 100 256 258

14.17 Encode the following string using the LZW algorithm This_love_ This_love _is_ a_strange_love_ A_faded_ kind_of_day_love _This_love Solution:

The encoder dictionary is initialized with values 0 − 255, where the first 128 entries are ASCII characters. Prefix w nil T h

Input T h i

Dictionary

Output

Pointer

New Phrase

Phrase

Pointer

256 257

Th hi

T h

84 104

i s _ l o v e _ T Th i is _ _l o ov e e_ i is is_ a _ s t r a n g e e_ l lo v ve _ A _ f a d e d _ k

s _ l o v e _ T h i s _ l o v e _ i s _ a _ s t r a n g e _ l o v e _ A _ f a d e d _ k i

258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292

is s_ _l lo ov ve e_ _T Thi is_ _lo ove e_i is_a a_ _s st tr ra an ng ge e_l lov ve_ _A A_ _f fa ad de ed d_ _k ki

i s _ l o v e _ Th is _l ov e_ is_ a _ s t r a n g e_ lo ve _ A _ f a d e d _ k

105 115 45 108 111 118 101 45 256 258 260 262 264

267 97 45 115 116 114 97 110 103 264 261 263 45 45 102 97 100 101 100 45 107 25

i n d d_ o f _ d a y _ _l _lo v ve ve_ T Th Thi s s_ l lo lov e

n d _ o f _ d a y _ l o v e _ T h i s _ l o v e -

293 294 295 296 297 298 299 300 301 302 303 304 305 306 -

in nd d_o of f_ _d da ay y_ _lov ve_T This s_l love -

i n d_ o f _ d a y _lo ve_ Thi s_ lov e

105 110 290 111 102 45 100 97 121 268 282

266 259

281 101

14.18 Decode the following LZW-encoded string

1213346176889 The source encoder alphabet is binary. List the complete dictionary. Solution:

The decoder dictionary is initialized with codewords 1 and 2 for binary characters 0 and 1, respectively.

Prefix w (Previous O/P)

NIL 0 1 0 01 01 10 010 0 011 010 100 100

Dictionary Input pointer

1 2 1 3 3 4 6 1 7 6 8 8 9

Output phrase

0 1 0 01 01 10 010 0 011 010 100 100 0100

Pointer

Phrase

1 2

0 1

3 4 5 6 7 8 9

01 10 00 010 011 100 0100

10 11 12 13

0110 0101 1001 1000

14.19 Consider a CATV channel with a bandwidth 6 MHz. Assume that the channel can be modeled as an AWGN channel. a. What is the capacity of the channel for a SNR of 35 dB. Solution:

C = W log 2 (1 + SNR ) bits/second =6 ×106 × log 2 ( 3.16328 ×103 )

= 69.76 Mbps b. Calculate the minimum SNR required for a data rate of 43 Mb/s? Solution:

SNR = 2C /W − 1 = 243/6 − 1 = 142.675 SNR = 21.54 dB c. Compare the bit rate calculated in (a) with that achievable using 256-QAM. The modulator uses RRC pulses with roll-off factor α = 0.5.

Solution:

Using Table 11.5, the bit rate achievable using 256-QAM is given by 6 ×106 = Rb (1 + α ) / log 2 M = Rb (1 + 0.5) / 8 Rb =

6 × 8 ×106 = 32 ×106 bits/second 1.5

14.20 Calculate the maximum information rate that can be transmitted over an AWGN channel of bandwidth 200 kHz. Assume that the received power level is 1 mW and the one-sided noise spectral density is 10-9 Watts/Hz. Solution: ⎛ P ⎞ C = W log 2 ⎜1 + s ⎟ ⎝ N oW ⎠ ⎛ ⎞ 10−3 = 200 ×103 × log 2 ⎜1 + −9 . 3 ⎟ ⎝ 10 × 200 ×10 ⎠ = 200 × 103 × log 2 ( 6 ) = 517 kbps

a. How much the capacity is increased by doubling the received power? Solution:

⎛ ⎞ 2 × 10−3 C = 200 ×103 × log 2 ⎜1 + −9 3 ⎟ ⎝ 10 × 200 ×10 ⎠ = 200 × 103 × log 2 (11) = 691.8863 kbps b. How much the capacity is increased by doubling the bandwidth instead? Solution:

⎛ ⎞ 10−3 C = 400 ×10 × log 2 ⎜1 + −9 3 ⎟ ⎝ 10 × 400 × 10 ⎠ = 400 × 103 × log 2 ( 3.5 ) = 722.942 kbps 3

c. Comment on the power-bandwidth trade-off in (a) and (b). Solution:

The increase in the channel capacity as a function of SNR is logarithmic. The effect of bandwidth on the capacity is more complicated because of two competing effects. On the one hand, the capacity increases linearly as a function of bandwidth. On the other hand, however, a higher bandwidth means higher input noise at the receiver which, in turn, decreases the SNR. In general, the increase in bandwidth provides more favorable tradeoff versus power. 14.21. Find the channel capacity of the DMC shown in Figure P14.3: Figure P14.3

+ n

The output y of the channel is given by y = x+n

where P {n = 0} = P {n = ε } = 1/ 2 and x is a binary random variable with alphabet A x = {−1,1} . Assume that x and n are statistically independent. Calculate the capacity of channel for values of ε = 0, ±1, and 0.5. Solution:

ε = 0 . In this case, y = x , and the channel is identical to the noise-free channel discussed in Example 14.11. Hence the capacity is 1 bit per channel/use.

ε = 1 In this case, the channel has two possible outputs corresponding to each of the two inputs, as illustrated in Figure. Since the channel has nonoverlapping outputs, the input x can be determined without error by knowing the output. Hence, H ( x y ) = 0. Therefore,

{

}

C = Max H ( x ) − H ( x y ) = Max { H ( x )} = 1 for px

p x = (1/ 2 1/ 2 ) .Thus, the capacity of this channel is also 1 bit

channel/use.

1/2

−1

1/2

Input

1/2

Output

1 1/2

ε = −1 In this case, the channel has two possible outputs corresponding to each of the two inputs, as illustrated in Figure. Knowing y, we know the x which was sent, and hence H ( x y ) = 0. Therefore,

C = Max { I ( x; y )} = Max { H ( x )} = 1 . The capacity of this channel is px

also 1 bit channel/use.

1/2

−1

1/2

Input

1/2 1 1/2

−1

−2 Output

ε = 0.5 In this case, the channel has two possible outputs corresponding to each of the two inputs, as illustrated in Figure. Knowing y , we know the x which was sent, and hence H ( x y ) = 0. Therefore,

C = Max { I ( x; y )} = Max { H ( x )} = 1 . The capacity of this channel is px

also 1 bit channel/use. −1

1/2

Input

1/2 1

−1

−0.5 1

1/2 1.5

14.22 The DCT coefficients corresponding to an 8 × 8 block of pixels is given below: ⎡1260 − 1 − 12 − 5 2 − 2 − 3 1 ⎤ ⎢ −23 − 17 − 6 − 3 − 3 0 0 − 1 ⎥⎥ ⎢ ⎢ −11 9 − 2 2 0 − 1 − 1 0 ⎥ ⎢ ⎥ 0 1 1 0 0 0⎥ ⎢ −7 −2 ⎢ −1 −1 1 2 0 −1 1 1 ⎥ ⎢ ⎥ 0 2 0 −1 1 1 −1⎥ ⎢ 2 ⎢ −1 0 0 −1 0 2 1 −1 ⎥ ⎢ ⎥ 2 1 − 1 0 ⎦⎥ ⎣⎢ − 3 2 − 4 − 2 a. Using JPEG luminance quantization table, find the resulting block of quantized DCT coefficients. Solution:

The quantized DCT coefficients are obtained using (14.109) and table in Figure 14.34 as ⎛ 79 0 ⎜ ⎜ −2 − 1 ⎜ −1 1 ⎜ −1 0 G =⎜ ⎜ 0 0 ⎜ ⎜ 0 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 ⎟⎠

b. Find the one-dimensional sequence that results after zig-zag scanning. Solution:

The one-dimensional sequence that results after zigzag scanning quantized DCT AC coefficients in (14.110) is 0 –2 −1 –1 –1 0 0 1 − 1 EOB 14.23 Consider a black-and- white HDTV picture with the frame rate of 60 Hz. Each frame is made up of 1,080 scan lines × 1,920 pixels/line. We assume that pixels

can occupy one of 16 brightness levels with equal probability and independent of others. a. Calculate the information content of the black-and- white HDTV image. What is minimum bit rate required to transmit the signal? Solution:

Since each pixel can assume one of 16 brightness levels with equal probability and independent of others, the entropy (information content) of the image is 4 bits/ pixel. . Total number of pixels in a frame = 1080× 1920 = 2,073,600 Since each pixel can be represented by 4 bits, the minimum bit rate required to transmit the signal 2,073,600 × 4 × 60 = 497.664 Mbps. b. Repeat (a) for color HDTV if the pixels can additionally occupy one of 256 colors with equal probability and independent of others. Solution:

Since each pixel can assume one of 16 brightness levels and one of 256 colors levels with equal probability and independent of others, it can be represented by 4 + 8 = 12 bits. Minimum bit rate required to transmit the signal 2,073,600 × 12 × 60 = 1.492992 Gbps.

Chapter 15 15.1

Suppose a (3,1) repetition code is used on a BSC with bit error rate p. a. Assuming that the receiver makes decision by taking majority vote of the three bits, find the probability of decoded bit error. Solution: The (3,1) repetition code has d min = 3 . It can correct t = 1 error. The probability of decoded bit error is upper-bounded by

d min n ⎛ n ⎞ j n− j ∑ ⎜ ⎟ p (1 − p) n j =t +1 ⎝ j ⎠

Pb ≤

3 3 ⎛3 ⎞ j 3− j ∑ ⎜ ⎟ p (1 − p) 3 j =2 ⎝ j ⎠

= p 3 + 3 p 2 (1 − p) b. How many errors this code can correct and detect? Solution:

The (3,1) repetition code can correct t = 1 error and detect s ≤ d min − 1 = 2 errors. c. Plot error performance of the (3,1) code as a function of p. Solution: 0

Comparison of Probability of decoded bit error for (3,1)and (5,1) repitition codes

(3,1)repitition code (5,1) repitition code

-2

Probability of decoded bit error vs BER

-4

-6

-8

-10

-12

-14

-5

-4

-3

-2

-1

BER p

d. Repeat (a) through (c) for (5,1) repetition code and compare its performance with the (3,1) code. Solution:

The (5,1) repetition code has d min = 5 . Therefore, it can correct t = 2 errors and detect 4 errors. The probability of decoded bit error is upper-bounded by

5 5 ⎛5 ⎞ j Pb ≤ ∑ ⎜ ⎟ p (1 − p)3− j 5 j =3 ⎝ j ⎠ = p 5 + 5 p 4 (1 − p) 4 + 10 p 3 (1 − p) 2 Figure displays performance comparison with the (3,1) code. 15.2

Consider the (4,1) repetition code a. Write generator and parity-check matrices for the code. Solution:

In this case, k =1, n =4. We have 3 parity bits that are the same as the message bit. Since G = ( I k # P ) for a systematic binary linear (n, k) code, the identity matrix I 1 = 1 and the parity matrix P is (1 1 1) of dimension 1×3. Therefore,

G = ( I 1 # P ) = (1 # 1 1 1) The parity-check matrix is now obtained as

⎛1 # 1 0 0 ⎞ ⎜ ⎟ H = P # I 3 = ⎜1 # 0 1 0 ⎟ ⎜ ⎟ ⎝1 # 0 0 1⎠

(

)

(a) What is minimum distance d min of the code. How many errors can this code correct and detect? Solution:

The (4,1) repetition code has d min = 4 . Therefore, it can correct 1 error and detect 3 errors.

15.3

Consider the systematic binary linear (6, 3) code with generator matrix

⎛1 0 0 # 1 1 0 ⎞ ⎜ ⎟ G = ⎜ 0 1 0 # 0 1 1⎟ ⎜ ⎟ ⎝ 0 0 1 # 1 0 1⎠ a. Determine the parity-check matrix H of the code. Solution:

We observe that for a systematic binary linear (n, k) code G = ( I k # P ) and

(

)

H = P # I n −k . Therefore, T

⎛1 0 1 # 1 ⎜ H = ⎜1 1 0 # 0 ⎜ ⎝0 1 1 # 0

0 0⎞ ⎟ 1 0⎟ ⎟ 0 1⎠

b. What is minimum distance d min of the code. How many errors this code can correct and detect? Solution:

The codewords and their Hamming weights are tabulated below: Message vector 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

Codeword 0 0 0 0 0 0 0 0 1 1 0 1 0 1 0 0 1 1 0 1 1 1 1 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 1 0 1 1 1 1 0 0 0

Hamming weight 0 3 3 4 3 4 4 3

From the Table, d min = 3. Therefore, the code can correct all single errors and detect all double errors. c. Using the standard array, show that the code is able to correct one error pattern of Hamming weight 2, in addition to all error patterns of Hamming weight 1.

Solution:

The standard array for the code is shown in the table. As an example, we will show that the error pattern (100001) can be corrected. Suppose that the code word x = (011110) is transmitted and y = (111111) is received. We calculate the syndrome of y as (111). This syndrome corresponds to the correctable error pattern e = (100001) from the table. Thus y + e = (111111) + (100001) = (011110) is identified as the transmitted codeword. We will see in parts (d) and (e) that the code correct all single errors but not all double errors. Standard array for (6, 3) code

Coset leaders 000000

Codewords

Syndrome

001101 010011

011110

100110

101011

110101

111000

000

000001

001100 010010

011111

100111

101010

110100

111001

001

000010

001111 010001

011100

100100

101001

110111

111010

010

000100

0010101 010111

011010

100010

101111

110001

111100

100

001000

000101 011011

010110

101110

100011

111101

110000

101

010000

011101 000011

001110

110110

111011

100101

101000

011

100000

101101 110011

111110

000110

001011

010101

011000

110

100001

101100 110010

111111

000111

001010

010100

011001

111

d. Use the standard array to find the most likely codeword, given that the noisy received vector is (010101). Solution:

We calculate the syndrome of y as (110). This syndrome corresponds to the correctable error pattern e = (100000) from the table. Thus y + e = (010101)+ (100000) = (110101) is identified as the transmitted codeword. . d. Now suppose (001101) is transmitted and (111101) is received. What is the decoded word? Comment on your result. Solution:

(001101) is transmitted and (111101) is received. We calculate the syndrome of y as (101). This syndrome corresponds to the error pattern e = (001000) from the table. The decoded word is identified as (111101) + (001000) = (110101). This is an incorrect decoding since the error pattern caused by the channel (110000) is not a correctable error pattern. This code corrects single errors in any position and one error pattern of double errors. Thus, as noted earlier, it is a single error correcting code. 15.4

` The parity bits of (7, 3) linear block code are given by b0 = m1 + m2 b1 = m2 + m3 b2 = m1 + m2 + m3 b3 = m1 + m3 a. Find the generator and parity check matrices G and H for this code. Solution:

The generator matrix of the (7, 3) linear block code is given by ⎛1 0 0 # p00 p01 p02 p03 ⎞ ⎜ ⎟ G = ( I 3 # P ) = ⎜ 0 1 0 # p10 p11 p12 p13 ⎟ ⎜ ⎟ ⎝ 0 0 1 # p20 p21 p22 p23 ⎠

The parity bits are linear combinations of the message bits and are determined by the application of (15.12) as

b j = p0 j m1 + p1 j m2 + p2 j m3 ,

j = 0,1, 2,3

Therefore, ⎛1 0 0 # 1 0 1 1 ⎞ ⎜ ⎟ G = ⎜0 1 0 # 1 1 1 0⎟ ⎜ ⎟ ⎝0 0 1 # 0 1 1 1 ⎠

(

H = P # I n−k T

)

⎛1 1 0 # 1 ⎜ ⎜0 1 1 # 0 =⎜ 1 1 1 # 0 ⎜ ⎜1 0 1 # 0 ⎝

0 0 0⎞ ⎟ 1 0 0⎟ 0 1 0⎟ ⎟ 0 0 1⎟⎠ 5

b. What is d min for the code? Solution:

dmin = 4 c. List all the codewords for this code. Solution:

The codewords and their Hamming weights are tabulated below: Message vector 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

Codeword 0 0 0 0 0 0 0 0 0 1 0 1 1 1 0 1 0 1 1 1 0 0 1 1 1 0 0 1 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 0 1 0 1 1 1 1 0 0 1 0

Hamming weight 0 4 4 4 4 4 4 4

d. Construct the standard array and determine the correctable error patterns and their syndromes. Solution:

The standard array for (7, 3) code is shown in the Table. The correctible error patterns and their syndromes are displayed in the first and last columns, respectively.

Coset leaders 0000000 0010111

0101110 0111001 1001011 1011100 1100101 1110010 0000

0000001 0010110

0101111 0111000 1001010 1011101 1100100 1110011 0001

0000010 0010101

0101100 0111011 1001001 1011110 1100111 1110000 0010

0000100 0010011

0101010 0111101 1001111 1011000 1100001 1110110 0100

0001000 0011111

0111110 0101001 1011011 1010100 1101101 1111010 1000

0010000 0000111

0111110 0101001 1011011 1001100 1110101 1100010 0111

0100000 0110111

0001110 0011001 1101011 1111100 1000101 1010010 1110

1000000 1010111

1101110 1111001 0001011 0011100 0100101 0110010 1011

1000001 1010110

1101111 1111000 0001010 0011101 0100100 0110011 1010

1100000 1110111

1001110 1011001 0101011 0111100 0000101 0010010 0101

1010000 1000111

1111110 1101001 0011011 0001100 0110101 0100010 1100

1001000 1011111

1100110 1110001 0000011 0010100 0101101 0111010 0011

1000100 1010011

1101010 1111101 0001111 0011000 0100001 0110110 1111

1000010 1010101

1101100 1111011 0001001 0011110 0100111 0110000 1001

15.5

Codewords

Syndrome

Consider the systematic binary linear code with parity-check matrix

⎛1 0 # 1 0 0 ⎞ ⎜ ⎟ H = ⎜1 1 # 0 1 0 ⎟ ⎜ 0 1 # 0 0 1⎟ ⎝ ⎠ a. Determine the generator matrix G of the code. Solution:

⎛1 0 # 1 1 0 ⎞ G=⎜ ⎟ ⎝ 0 1 # 0 1 1⎠

b. Write all the codewords of the code and determine the minimum distance dmin of the code. How many errors can this code correct and detect? Solution:

Message vector 0 0 0 1 1 0 1 1

Codeword

Hamming weight

0 0 0 0 0 0 1 0 1 1 1 0 1 1 0 1 1 1 0 1

0 3 3 4

From the Table, d min = 3. Therefore, the code can correct all single errors and detect all double errors. c. Using the standard array, show that the code is able to correct two error patterns of Hamming weight equal to 2, in addition to all error patterns of Hamming weight 1. Solution:

The standard array for (5, 2) code is shown in the Table. The code is able to correct two error patterns (11000) and (10001) of Hamming weight equal to 2, in addition to all error patterns of Hamming weight 1. Standard array for (5, 2) code

Coset leaders 00000 01011

Codewords

Syndrome

10110

11101

000

00001

01010

10111

11100

001

00010

01001

10100

11111

010

00100

01111

10010

11001

100

01000

00011

11110

10101

011

10000

11011

00110

01101

110

11000

10011

01110

00101

101

10001

11010

00111

01100

111

d. What is the asymptotic coding gain with respect to uncoded BPSK modulation? Solution:

For the hard-decision ML decoding, the asymptotic coding gain is given from (15.90) as

Ghd = 10 log10

Rc d min 0.4 × 3 = 10 log10 = −2.21 dB 2 2

For the soft-decision ML decoding, the asymptotic coding gain is given from (15.95)) as Gsd = 10 log10 Rc d min = 10 log10 ( 0.4 × 3) = 0.792 dB

15.6 Consider the systematic binary linear code with parity-check matrix

⎛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⎟ ⎜ H= ⎜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 ⎠ a. Determine the generator matrix G of the code. Solution:

⎛1 0 0 0 0 0 0 0 0 0 0 # 0 0 1 1 ⎞ ⎜ ⎟ ⎜0 1 0 0 0 0 0 0 0 0 0 # 0 1 0 1⎟ ⎜0 0 1 0 0 0 0 0 0 0 0 # 0 1 1 0⎟ ⎜ ⎟ ⎜0 0 0 1 0 0 0 0 0 0 0 # 0 1 1 1⎟ ⎜0 0 0 0 1 0 0 0 0 0 0 # 1 0 0 1⎟ ⎜ ⎟ G = ⎜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 0 0 0 1 # 1 1 1 1⎠ b. Determine the minimum distance d min of the code. How many errors can this code correct and detect? 9 © 2012 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

Solution:

All Hamming codes have d min = 3. Therefore, it can correct 1 error and detect 2 errors. c. What is the codewords corresponding to the message vector m = (1 0 1 1 1 0 0 0 1 1 1)? Solution:

The codewords corresponding to the message vector m = (1 0 1 1 1 0 0 0 1 1 1) is x = (1 0 1 1 1 0 0 0 1 1 1 0 1 1 1) d. If y = (1 1 1 0 1 1 0 0 0 0 1 1 0 1 0) is received at the receiver, what is the most likely code word x transmitted? Solution:

The syndrome table for (15,11) code is shown below: Error vector 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0 0 0 0 0 0

Syndrome 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 1 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 1 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 1 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

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 syndrome of the received vector y is obtained as s = yH = ( 0 1 1 0 ) T

From table, the most likely error vector is ( 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 ) . Therefore,

xˆ = y + e j = (1 1 1 0 1 1 0 0 0 0 1 1 0 1 0 ) + ( 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 ) = (1 1 0 0 1 1 0 0 0 0 1 1 0 1 0 ) e.

If y = (0 1 1 0 1 0 1 0 1 1 1 0 1 1 0) is received at the receiver, what is the most likely message m transmitted?

Solution:

The syndrome of the received vector y is obtained as s = (1 0 1 1) From table, the most likely error vector is ( 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 ) . Therefore,

xˆ = y + e j = ( 0 1 1 0 1 0 1 0 1 1 1 0 1 1 0 ) + ( 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 ) = ( 0 1 1 0 1 0 0 0 1 1 1 0 1 1 0) 15.7

Prove that an (n,k) block code can correct t errors if its code rate Rc statisfies the following inequality

⎡ t ⎛ n ⎞⎤ 1 Rc ≤ 1 − log 2 ⎢ ∑ ⎜ ⎟ ⎥ n ⎣ i =0 ⎝ i ⎠ ⎦ Solution:

An (n,k) block code that can correct t errors satisfies the Hamming bound (15.34). That is, t ⎛n⎞ 2n − k ≥ ∑ ⎜ ⎟ i =0 ⎝ i ⎠

Taking logarithm of both sides, we obtain

⎡ t ⎛ n ⎞⎤ n − k ≥ log 2 ⎢ ∑ ⎜ ⎟ ⎥ ⎣ i =0 ⎝ i ⎠ ⎦ or ⎡ t ⎛ n ⎞⎤ k ≤ n − log 2 ⎢ ∑ ⎜ ⎟ ⎥ ⎣ i =0 ⎝ i ⎠ ⎦ Rc =

⎡ t ⎛ n ⎞⎤ k 1 ≤ 1 − log 2 ⎢ ∑ ⎜ ⎟ ⎥ n n ⎣ i =0 ⎝ i ⎠ ⎦

a. Show that a (63, 39) block code capable of correcting 4 errors can exist. What is the redundancy of this code? Solution:

Using the Hamming bound (15.34), we obtain 4 ⎛ 63 ⎞ ⎛ 63 ⎞ ⎛ 63 ⎞ ⎛ 63 ⎞ ⎛ 63 ⎞ ⎛ 63 ⎞ 263−39 ≥ ∑ ⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ i =0 ⎝ i ⎠ ⎝ 0 ⎠ ⎝1 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝ 4 ⎠ ?

16,777,216 ≥ 1 + 63 + 1,953 + 39,711 + 595,665 = 637,393

Yes. A (63, 39) block code capable of correcting 4 errors can exist.

Redundancy = 15.8

n−k k 39 = 1 − = 1 − ≈ 38.1% n n 63

Consider the generator matrix of the (7, 4, 3) Hamming code given in Example 15.4. The code is extended to an (8, 4) code by adding an overall parity check bit to (7, 4, 3) code. a. Find the parity check matrix H for this code. Solution:

The G and H matrices of the (7, 4) Hamming code are given by ⎛1 0 ⎜ ⎜0 1 G=⎜ 0 0 ⎜ ⎜0 0 ⎝

0 0 # 0 0 # 1 0 # 0 1 #

1 0 1⎞ ⎟ 1 1 1⎟ , 1 1 0⎟ ⎟ 0 1 1 ⎟⎠

⎛1 1 1 ⎜ H = ⎜0 1 1 ⎜1 1 0 ⎝

0 1

# 1 0 # 0 1

0⎞ ⎟ 0⎟ 1 ⎟⎠

The G and H matrices of the (8, 4) Hamming code are given by ⎛1 0 ⎜ ⎜0 1 G =⎜ 0 0 ⎜ ⎜0 0 ⎝

0 0 # 0 0 # 1 0 # 0 1 #

1 0 1 1⎞ ⎟ 1 1 1 0⎟ , 1 1 0 1⎟ ⎟ 0 1 1 1 ⎟⎠

⎛1 1 1 ⎜ 0 1 1 H =⎜ ⎜1 1 0 ⎜ ⎝1 0 1

1 1

# #

0 0

1 0

0 0 ⎞ ⎟ 0 1⎟ 1 0⎟ ⎟ 0 1⎠

b. What is d min for the code? Solution:

The codewords of (8, 4) extended Hamming code are tabulated below: Information Bits Code word

Weight

0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

0 4 4 4 4 4 4 4 4 4 4 4 4 4 4 8

00000000 00010111 00101101 00111010 01001110 01011001 01100011 01110100 10001011 10011100 10100110 10110001 11000101 11010010 11101000 11111111 From the table, dmin = 4 .

c. Does the extra check bit increase the error-correcting and error-detecting capabilities of the code? Solution:

The (8, 4) extended Hamming code has dmin = 4 . Therefore, it can correct 1 error and detect 3 errors. The (8, 4) extended Hamming code doesn’t improve the error-correcting capability versus (7, 4) Hamming code. However, it can detect triple errors. 15.9

Let g ( D) = D8 + D7 + D6 + D4 + 1 be the generator polynomial of (15, 7) cyclic code. a. Verify that g(D) can be a generator polynomial for the code. Solution:

To verify that g ( D) = D8 + D7 + D6 + D4 + 1 is a generator polynomial for (15, 7) cyclic code, it should divide D15 + 1 without any remainder.

D7 + D6 + D4 +1 D8 + D7 + D6 + D4 +1) D15 + 1 D15 + D14 + D13 + D11 + D7 D14 + D13 + D11 + D7 +1 D14 + D13 + D12 + D10 + D6 D12 + D11 + D10 + D7 + D6 +1 D12 + D11 + D10 + D8 + D4 D8 + D7 + D6 + D4 +1 D8 + D7 + D6 + D4 +1 0 b. Determine the code polynomial (code word) in systematical form for the message m( D) = D5 + D4 + D2 + 1 . Solution:

For this code, n − k = 8. Therefore,

D8m( D) = D13 + D12 + D10 + D8 D5 + D3 + 1 D8 + D 7 + D 6 + D 4 + 1 ) D13 + D12 + D10 + D 8 D13 + D12 + D11 + D9 + D5 D11 + D10 + D9 + D8 + D5 D11 + D10 + D9 + D 7 + D 3 D8 + D 7 + D 5 + D 3 D8 + D 7 + D 6 + D 4 + 1 b ( D ) = D 6 + D 5 + D 4 + D 3 + 1 ↔ b = ( 01111001)

y( D) = D8m( D) + b ( D ) = D13 + D12 + D10 + D8 + D6 + D5 + D4 + D3 + 1 ↔ y = ( 0110101# 01111001) c. Is the polynomial y( D) = D14 + D11 + D5 + D2 + 1 a code word?

Solution:

D 6 + D5 + D 2 + 1 D8 + D 7 + D 6 + D 4 + 1 ) D14 + D11 + D 5 + D 2 + 1 D14 + D13 + D12 + D10 + D 6 D13 + D12 + D11 + D10 + D 6 + D 5 + D 2 + 1 D13 + D12 + D11 + D9 + D5 D10 + D 9 + D 6 + D 2 + 1 D10 + D 9 + D8 + D 6 + D 2 D8 + 1 D8 + D 7 + D 6 + D 4 + 1 s ( D ) = D 7 + D 6 + D 4 ↔ s = (11010000 )

Since the syndrome is not zero, y ( D) is not a valid code polynomial. 15.10 Show that for an (n,k) block code, the probability of an undetected error Pu (e) over a BSC with crossover probability p is given by n

Pu (e) = ∑ Ai p i (1 − p ) n −i i =1

where A1 , A1 ,......., An represents the weight distribution of the code. Solution:

The probability of undetected error Pu (e) is given from (15.30) as Pu (e) = P {Error pattern is a nonzero codeword x}

∑

p wH ( x ) (1 − p )

n − wH ( x )

2k −1 nonzero codewords x

By rearranging terms in the summation on the basis of Hamming weight of the error patterns, we can write the above equation as n

Pu (e) = ∑ Ai p i (1 − p ) n −i i =1

where Ai is the number of code words with the Hamming weight i. For an (n,k,dmin) code, A1 to Admin −1 are zero. That is, n

Pu ( e) = ∑ Ai p i (1 − p ) n − i i = d min

a. Suppose (7,4) Hamming code is used for error detection over a BSC with crossover probability p = 10-3. Compute the probability of an undetected error for this code. Solution:

dmin = 3 for the (7,4) Hamming code. The weight distribution of the code from Table 15.6 is A3 = 7, A4 = 7, and A7 = 1 . Therefore, Pu (e) = 7 p 3 (1 − p ) + 7 p 4 (1 − p ) + p 7 4

If p = 10−3, the probability of an undetected error is Pu (e) = 7 p 3 (1 − p ) + 7 p 4 (1 − p ) + p 7 4

= 7 × 10 −9 × 0.996 + 7 × 10 −12 × 0.997 + 10 −21 = 6.97 × 10 −9

15.11 Consider the m = 4 Hamming code. a. What is the block length n and dmin for the code? Solution:

The key parameters of the m = 4 Hamming code are Block length:

n = 2m – 1 = 15

Number of data bits: k = 2m – m – 1 = 11 Number of parity bits: n – k = 4 Minimum distance: dmin = 3 b. Write the parity check matrix H for this code. Solution:

Recall that columns of the (n − k) × n parity-check matrix for a Hamming code consists of all nonzero binary sequences of length n − k.

(

H = P #I 4 T

)

⎡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 ⎦

c. Give the set of equations for computing the parity bits in terms of the message bits. Solution:

The set of equations for computing the parity bits are obtained using (15.11) as ⎛0 0 11 ⎞ ⎜ ⎟ ⎜0 1 0 1⎟ ⎜0 1 1 0⎟ ⎜ ⎟ ⎜0 1 11 ⎟ ⎜1 0 0 1 ⎟ ⎜ ⎟ b = ( b0 , b1 ,..., b3 ) = mP = ( m0 m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 ) ⎜1 0 1 0 ⎟ ⎜1 0 1 1 ⎟ ⎜ ⎟ ⎜1 1 0 0 ⎟ ⎜1 1 0 1 ⎟ ⎜ ⎟ ⎜1 1 1 0 ⎟ ⎜ ⎟ ⎝1 1 1 1 ⎠ b0 = m4 + m5 + m6 + m7 + m8 + m9 + m10 b1 = m1 + m2 + m3 + m7 + m8 + m9 + m10 b2 = m0 + m2 + m3 + m5 + m6 + m9 + m10 b3 = m0 + m1 + m3 + m4 + m6 + m8 + m10

d. Write an estimate of the probability of decoded bit error if the channel BER p = 10-4? Repeat for p = 10-6. Solution:

An estimate of the probability of decoded bit error is obtained by using (15.60) as

Pb ≈

(

)

d min d min Admin 2 p(1 − p ) n

From Table 15.6, Admin = A3 = 35 . Substituting yields

( (

) )

3 3 × 35 × 2 10−4 (1 − 10−4 ) = 5.6 ×10−5 15 3 3 Pb ≈ × 35 × 2 10−6 (1 − 10−6 ) = 5.6 ×10−8 15

Pb ≈

for p = 10−4 for p = 10−6

15.12 Show that the coding gain for soft-decision decoded ( n, k , dmin ) block code is given by

⎡ k log e 2 ⎤ Gc = 10 log10 ⎢ Rc d min − ⎥ dB Eb / N o ⎦ ⎣ where Eb / No = SNR/bit of the link. Solution:

For a coded system, the logarithm of the probability of code word error is given from (15.92) as ⎛E ⎞ log e Pcw (e) ≈ k log e 2 − Rc d min ⎜ b ⎟ ⎝ N o ⎠coded For an uncoded system, the logarithm of probability of error for a word or packet of k bits is given from (15.84) by ⎛k⎞ ⎛ E ⎞ log e Pw (e) ≈ log e ⎜ ⎟ − ⎜ b ⎟ ⎝ 2 ⎠ ⎝ N o ⎠ uncoded To achieve the same probability of error, the right-hand sides of above equations must be equal. That is, ⎛E ⎞ ⎛k⎞ ⎛ E ⎞ log e ⎜ ⎟ − ⎜ b ⎟ = k log e 2 − Rc d min ⎜ b ⎟ ⎝ 2 ⎠ ⎝ N o ⎠ uncoded ⎝ N o ⎠coded Dividing both sides by ( Eb / No )coded yields

⎛k⎞ log e ⎜ ⎟ ⎝ 2 ⎠ − ( Eb / N o )uncoded = k log e 2 − R d ( Eb / N o )coded ( Eb / N o )coded ( Eb / N o )coded c min

( Eb / N o )uncoded ( Eb / N o )coded

⎛k⎞ log e ⎜ ⎟ k log e 2 ⎝ 2 ⎠ ≈ R d − k log e 2 = Rc d min − + ( Eb / N o )coded ( Eb / N o )coded c min ( Eb / N o )coded

Taking logarithm of both sides, we obtain ⎡ ( Eb / N o ) uncoded ⎤ ⎡ k log e ( 2 ) ⎤ G = 10 log10 ⎢ ⎥ = 10 log10 ⎢ Rc d min − ⎥ ( Eb / N o )coded ⎦⎥ ⎣⎢ ( Eb / N o )coded ⎦⎥ ⎣⎢ a. Calculate the coding gain for (127, 36, 31) BCH code at ( Eb / No )coded = 8 dB. Solution: 36 × log e ( 2 ) ⎤ ⎡ G = 10 log10 ⎢8.7874 − ⎥ = 6.836 dB 6.3 ⎣ ⎦

15.13 Consider the performance of extended Golay code (24,12,8) over AWGN channel using binary antipodal signaling. a. Plot the probability of codeword error for hard-decision decoding for Eb/No in the range 4-14 dB. Solution:

The upper bound on the probability of codeword error for hard-decision decoding is given from (15.49) as n ⎛n⎞ Pcw (e) ≤ ∑ ⎜ ⎟ p j (1 − p ) n − j j = t +1 ⎝ j ⎠

where p for binary antipodal signaling over an AWGN channel is given by

⎛ 2 Ecb ⎞ ⎛ R 2E ⎞ = Q⎜ c b ⎟ p = Q⎜ ⎟ ⎜ N ⎟ ⎜ No ⎟⎠ o ⎠ ⎝ ⎝ The probability of decoded bit error is now given by

d min Pcw (e) n

Pb ≈

Error peformance of Golay Code - Soft vs Hard decision decoding

-1

Uncoded (24,12) Golay code - soft decision decoding (24,12) Golay code- hard decision decoding

-2

Probability of Decoded Bit Error

-3

-4

-5

-6

-7

-8

9 10 Eb/No in dB

b. Repeat (a) for the soft-decision decoding. Solution:

The upper bound on the probability of codeword error for soft-decision decoding is given from (15.77) as n n ⎛ 2 Ecb d ⎞ Pcw (e) = ∑ Ad P2 (d ) ≤ ∑ Ad Q ⎜⎜ ⎟⎟ d = dmin d = d min ⎝ No ⎠

c. What is the difference in coding gain for two cases. Solution:

The coding gains at probability of decoded bit error Pb = 10−8 are Hard-decision decoding:

G = 2.4 dB

Soft-decision decoding:

G = 4.8 dB

Soft-decision decoding provides a coding gain advantage of 2.4 dB over hard decision decoding in the present case. 15.14 Consider the error-detecting capability of an (n,k) CRC code. Prove the following: a. If the generator polynomial g ( D) has more than one term, it can detect single errors. Solution:

If the code polynomial x(D) is transmitted over a noisy channel, the received polynomial y(D) is given by

y ( D ) = x ( D ) + e( D ) If y(D) is divisible by g ( D) , it is a valid codeword polynomial. This implies that the received vector is either error-free or has been corrupted by an undetectable error pattern. Since x(D) is constructed as a multiple of the generator polynomial g ( D) , the error pattern is undetectable if the corresponding error polynomial e( D) is also a multiple of g ( D) ..So if we want a polynomial code to be able to detect a particular error pattern, then the corresponding error polynomial should not be divisible by g ( D) . For a single error, the error polynomial is of the form e( D) = Di . Now if g ( D) has at least two nonzero terms, it is easy to see that g ( D) can’t divide e( D) without remainder. Hence all single errors are detectable. b. If the generator polynomial g ( D) of the code is a primitive polynomial of degree n− k, it will detect all double errors as long as the total codeword length doesn’t exceed 2 n − k − 1 . Solution:

An error polynomial corresponding to double errors will have the form e( D) = Di + D j = Di 1 + D j −i for some i and j > i . We know from discussion

(

)

i in (a) that if g ( D) has more than 1 term, it can’t divide D .Thus e( D) will be divisible by g ( D) if g ( D) divides 1 + D j −i . This implies that 1 + D m

(

)

( 1 ≤ m ≤ n ) should not be divisible by g ( D) in order that all double errors can be detected. Now if g ( D) is a primitive polynomial of degree n− k, it cannot divide 1 + D m for all m < 2n − k − 1 . Thus if g ( D) is selected to be a primitive polynomial of degree n− k, then it will detect all double errors as long as the codeword length does not exceed 2 n − k − 1 .

c. If the generator polynomial g ( D) of the code has (1 + D ) as a factor, it will be able to detect all odd number of errors. Solution:

Suppose all codewords have an even number of 1s, then all odd numbers of errors can be detected. If we evaluate the codeword polynomial x( D ) at D = 1, we then obtain the sum of binary coefficients of x( D ) . If x(1) = 0 for all codeword polynomials, then (1 + D ) must be a factor of x( D ) and hence g ( D) must contain (1 + D ) as a factor. The generator polynomial g ( D) is, therefore, frequently selected with (1 + D ) as a factor so that it can detect all odd number of errors. 15.15 Let g1 ( D) = 1 + D and let g 2 ( D) = 1 + D + D3 . Consider the message vector (1 0 0 1 0 1). a. If g1 ( D) is used as the generator polynomial, find the codeword. Solution:

n − k =1 m ( D ) = D5 +D 2 + 1 Dm ( D ) = D 6 +D3 + D D5 + D 4 + D3 + 1

D + 1 ) D 6 +D 3 + D D 6 + D5 D5 + D3 + D D5 + D4 D 4 + D3 + D D 4 + D3

D D +1 b ( D ) = 1 ↔ b = (1)

y ( D) = Dm( D) + b ( D ) = D6 +D3 + D + 1 ↔ y = (100101#1) b. If g2 ( D) is used as the generator polynomial, find the codeword. Solution:

n−k =3 m ( D ) = D5 +D 2 + 1 D3m ( D ) = D8 +D5 + D3 D5 + D3 + D D 3 + D + 1 ) D8 +D 5 + D3 D8 + D 6 + D 5 D 6 + D3 D6 + D 4 + D3 D4 D4 + D2 + D b ( D ) = D 2 + D ↔ b = (110 )

y( D) = D3m( D) + b ( D ) = D8 +D5 + D3 + D2 + D ↔ y = (100101#110) c. Can g2 ( D) detect single errors? double errors? triple errors? Solution:

Single errors can be detected since g2 ( D) has more than one term. Double errors cannot be detected even though g2 ( D) is primitive because the codeword length exceeds 2 n − k − 1 = 7. Triple errors cannot be detected since (D + 1) is not a factor of g2 ( D) . d. Find the codeword corresponding the above message vector if g ( D) = g1 ( D) g2 ( D) is the generator polynomial. Comment on the errordetecting capabilities of g ( D) .

Solution:

g ( D ) = g1 ( D ) g 2 ( D ) = (1 + D ) (1 + D + D3 ) = 1 + D + D3 + D + D 2 + D 4 = 1 + D 2 + D3 + D 4 n−k = 4 m ( D ) = D5 +D 2 + 1 D 4 m ( D ) = D9 +D 6 + D 4 D5 + D4 + D + 1 D 4 + D 3 + D 2 + 1 ) D 9 +D 6 + D 4 D 9 + D8 + D 7 + D5 D8 + D 7 + D 6 + D 5 + D 4 D8 + D 7 + D 6 + D 4 D5 D5 + D 4 + D3 + D D 4 + D3 + D D 4 + D3 + D 2 + 1 b ( D ) = D 2 + D + 1 ↔ b = ( 0111)

y( D) = D4 m( D) + b ( D ) = D9 +D6 + D4 + D2 + D + 1 ↔ y = (100101# 0111) The new code can detect all single and all odd errors. It cannot detect double errors because the codeword length exceeds 23 − 1 = 7. (Degree of primitive polynomial factor g 2 ( D) = 3). 15.16 A rate Rc = ½ convolution encoder is displayed in Figure P15.1. Figure P15.1 xi1 mi

mi-1

xi2 24

a. Find the generators polynomials for this code. Solution:

The generators polynomials for this code are g1 ( D) = 1 g1 ( D) = 1 + D b. Draw the state transition diagram. Solution:

Current state 0 1

Input

Next state 0 1 0 1

0 1 0 1

Output 00 11 01 10

0 input 1 input 0/00

S0 0

1/11

0/01

S1 1

1/10 c. Is the code systematic? Prove by an example. Solution:

The code is systematic since g1 ( D) is of degree 0. For an input 011010, the output of the encoder is 00 11 10 01 11 01. Note that the input sequence appears in the output. c. Draw the decoding trellis to depth 8, and decode the received sequence 11 10 10 01 00 11 10 11.

11 ×

01 × 11 ×

01 ×11

× × 1 × 00× 01 00 01 × 00 1 00 0 0 11 × 11 11 ×11 × 01 × × × 10 10 10 10 ×

The decoded sequence is 11100110. d. Draw the modified state transition diagram and determine the corresponding free distance dfree. Solution:

S0start 0

1/11

0/01

D 2 KJ

1/10

DKJ

S0end 0

0 input 1 input

S1 = D 2 KJS0start + DKJS1 S0end = DKS1

T ( D, K , J ) =

S0end D3 K 2 J = S0start 1 − DKJ

= D 3 K 2 J (1 + DKJ + D 2 K 2 J 2 + ...)

d free = 3 from D3 K 2 J term.

15.17 A rate Rc = 1/3 convolution code is described by the generator polynomials g1 ( D ) = 1 + D 2 g2 ( D) = 1 + D + D 2

g3 ( D) = 1 + D a. Draw the shift-register convolution encoder for this code. Solution: xi1 = mi ⊕ mi−2

+ Information Bits mi

Encoder Output Bits mi-1

mi-2

xi2 = mi ⊕ mi−1 ⊕ mi−2 xi3 = mi ⊕ mi−1

Encoder output x11 x12 x13 x12 x22 x23 x31 x32 x33 x14 x42 x43 x51 x52 x53 .......

b. Draw the state transition diagram. What is the constraint length of this code? Solution:

Current state 00 01 10 11

Input 0 1 0 1 0 1 0 1

Next state 00 10 00 10 01 11 01 11

Output 000 111 110 001 011 100 101 010

0/000

1/111

0/110

1/001

0 input 1 input

0/011

1/100

0/101

1/010

Constraint length of the code = 3 c. Draw the trellis diagram and calculate the encoder output sequence corresponding to input sequence 0 1 1 0 1 0. Solution: 0/000

S0 (00)

1/111

S1 (01)

1/001

S2 (10)

S3 (11) Depth

0/011

0/101

1/100

1 input 0 input

Input bit sequence:

0 1

Encoder output sequence:

000 111 100 101 001 011

d. If the input information rate is 1 Mb/s, what is the bit rate of the encoded sequence? Solution:

Since the encoder generates 3 output bits per input bit, the bit rate of the encoded sequence is 3 Mb/s. The actual bit rate of the encoded sequence is slightly higher due to the overhead of tail bits.

15.18 A rate Rc = 1/3 convolutional code is defined by the generator polynomials g1 ( D ) = 1 + D

g2 ( D) = 1 + D + D 2 g3 ( D) = D 2

a. Draw the state transition diagram Solution:

Current state 00 01 10 11

Input

Next state 00 10 00 10 01 11 01 11

0 1 0 1 0 1 0 1

Output 000 110 011 101 110 000 101 011

0/000 1/110 0 input 1 input

S0 00

1/101

S2 10

1/000

0/011

S1 01

0/110

0/101

1/011

b. Determine its transfer function. Solution:

0 input 1 input

D 2 KJ S

start 0

1/101

1/110

D 2 KJ

D2 K 1/000

0/110

0/011

D2 K

S0end 00

0/101

D2 K

1/011

D 2 KJ S 2 = D 2 KJS0start + D 2 KJS1 S1 = D 2 KS2 + D 2 KS3 S3 = JKS2 + D 2 KJS3 S0end = D 2 KS1 Substituting for S2 , we obtain

S1 = D2 K ( D2 KJS0start + D2 KJS1 ) + D2 KS3 = D4 K 2 J ( S0start + S1 ) + D2 KS3 Substituting for S2 , we obtain S3 = D 2 K 2 J 2 ( S0start + S1 ) + D 2 KJS3 or S3 =

D 2 K 2 J 2 ( S0start + S1 ) 1 − D 2 KJ

S1 = D K J ( S 4

start 0

+ S1 ) + D K 2

D 2 K 2 J 2 ( S0start + S1 ) 1 − D 2 KJ

or S1 ⎣⎡(1 − D 4 K 2 J )(1 − D 2 KJ ) − D 4 K 3 J 2 ⎦⎤ = S0start ⎣⎡ D 4 K 2 J (1 − D 2 KJ ) + D 4 K 3 J 2 ⎦⎤ or S0end ⎡(1 − D 4 K 2 J )(1 − D 2 KJ ) − D 4 K 3 J 2 ⎤ = S0start ⎡ D 4 K 2 J (1 − D 2 KJ ) + D 4 K 3 J 2 ⎤ 2 ⎦ ⎣ ⎦ D K⎣

The transfer function T ( D, K , J ) is given by S0end D K ⎡⎣ D K J (1 − D KJ ) + D K J ⎤⎦ T ( D, K , J ) = start = S0 (1 − D 4 K 2 J )(1 − D 2 KJ ) − D 4 K 3 J 2 2

D 6 K 3 J (1 − D 2 KJ + KJ )

(1 − D KJ ) − D K J (1 − D KJ ) − D K J D K J (1 − D KJ + KJ ) = (1 − D KJ ) − D K J (1 − D KJ + KJ ) 2

c. Calculate dfree for the code. Solution:

T ( D, K , J ) = D6 K 3 J (1 + KJ ) + D8 K 5 J 3 + ... There are two paths with Hamming weight 6. Thus, dfree = 6. d. Assuming hard-decision decoding and BPSK modulation, plot the probability of decoded bit error for Eb / N o = 4 −11 dB. Compare it with the performance achieved using soft-decision decoding. Solution:

The probability of decoded bit error for hard- and soft-decision decoding, computed using (15.126) and (15.129) respectively, are displayed.

e. What is the coding gain advantage of soft- versus hard-decision decoding at Pb = 10 −6 . Solution:

The coding gain advantage of soft- versus hard-decision decoding at Pb = 10 −6 is approximately 2.6 dB. 15.19 Consider (2, 1, 3) convolutional code with generator polynomials g1 = (1 0 1) and g2 = (1 1 1). Use soft-decision Viterbi decoding to find the estimate of the information sequence m̂ when the received sequence is r = [(−1,−2) (2,−1) (−1,3 ) (1,−2) (−3,2) (1,2)] Solution: The transition diagram of the convolutional coder is shown in Figure.

0 input 1 input 0/00 1/11

S0 00

1/00

S2 10

1/10

0/11

S1 01

0/01

0/10

1/01 Figure (a) displays the results of applying the VA at depth 1. We calculate the soft-decoding branch metrics for paths originating at state S0 at depth 0 and ending at depth 1 states using (15.111). To calculate branch metrics, we calculate the correlation between the first received symbol (−1, −2) and symbols on the trellis branches leading to depth 1 states. We use the polar mapping sAj,i = 1 − 2 xAj,i for the symbols on trellis branches in computation of branch metrics.

m 1 2 d1 (0, 0) = ∑ r1,m s0,0 = r1,0 s0,0 + r1,0 s0,0 = −1 × 1 + − 2 × 1 = − 3 m =1 n

m 1 2 d1 (0, 2) = ∑ r1, m s0,2 = r1,0 s0,0 + r1,0 s0,2 = −1 × −1 + − 2 × − 1 = 3 m =1

These branch metrics are labeled next to state nodes at depth 1. The decoder stores these branch metrics and proceeds to depth 2. Figure (b) displays the results of applying the VA at depth 2. We see that there are now four paths, one for each state of the encoder. The computation of metrics for branches leading to depth 2 states follows: n

m 2 d 2 (0, 0) = ∑ r2,m s0,0 = r2,0 s10,0 + r2,0 s0,0 = 2 × 1 + −1 × 1 = 1 m =1 n

m 1 2 d 2 (0, 2) = ∑ r2,m s0,2 = r2,0 s0,0 + r2,0 s0,2 = 2 × −1 + − 1 × − 1 = − 1 m =1 n

m 1 2 d 2 (2,1) = ∑ r2,m s2,1 = r2,0 s0,0 + r2,0 s0,0 = 2 × 1 + −1 × − 1 = 3 m =1 n

m 2 d 2 (2,3) = ∑ r2,m s2,2 = r2,0 s10,0 + r2,0 s0,2 = 2 × −1 + − 1 × 1 = − 3 m =1

The accumulated partial metric for each of the four paths at depth 2 is now determined by adding these branch metrics to metrics noted above the states at depth 1. These partial metrics are now noted above state nodes at depth 2. Figure(a) 1

Depth Received sequence

S0 (00)

S2 (10)

2,−1

−1,3

1,−2

−3,2

1,2

-3

S1 (01)

−1,−2

S3 (11)

Figure(b) 1

Depth

S0 (00)

−1,−2 -3

-2

S1 (01)

1,−2

−3,2

1,2

-4

S2 (10)

−1,3

2,−1

Received sequence

S3 (11)

Figure (c) displays the results of applying the VA at depth 3. There are now two paths leading to each state at depth 3. The decoder computes partial metrics for both paths entering each state by adding the corresponding branch metric to the partial metric of the connecting survivor at depth 2. The decoder discards the path segment with the smaller accumulated partial metric. This leaves one survivor path passing through each of the depth 3 state nodes. The partial metrics for survivor paths are now noted above state nodes at depth 3. In Figure (d), the surviving paths up to depth 3 are shown by solid lines and the paths not yet decoded by the decoder are shown as dashed lines. The decoding process is repeated for depths 5 and 6 as illustrated in Figures (e) and (g). At depth 6 a single path with the largest accumulated metric is selected by the algorithm, and decoding is complete. The decoded message sequence from Figures (g) is 101000. Figure(c) 1

Depth

S3 (11)

-2

-4

S 2 (10)

1,−2

S1 (01)

-3

−1,3

2,−1

S0 (00)

−1,−2

3 10

Received sequence

−3,2

1,2

Figure(d) 1

Depth Received sequence

−1,−2

−1,3

2,−1

11 01

S3 (11)

1,2

× ×

−3,2

S2 (10)

1,−2 4

S0 (00) S1 (01)

Figure(e) 1

Depth Received sequence

−1,−2

−1,3

2,−1

S1 (01)

S 2 (10)

1,−2 4

S0 (00)

−3,2

1,2 12

× 10

S3 (11)

Figure (f) 1

Depth Received sequence

−1,−2

−1,3

2,−1

S2 (10)

S3 (11)

1,2 12

11 8

−3,2

S1 (01)

1,−2 4

S0 (00)

× ×

10 5

Figure (g) 1

Depth Received sequence

−1,−2

−1,3

2,−1

1,−2

−3,2

S0 (00)

1,2 00

S1 (01)

S 2 (10)

S3 (11) 15.20 Consider a TCM scheme shown in Figure P15.2.

Information bits

Figure P15.2

8-PSK Constellation Mapping

+ Convolutional Encoder (2,1,3)

a. Draw the trellis for this scheme, and assign subsets to the transitions according to the heuristic rules of Ungerboeck. Solution:

The trellis for the TCM scheme is shown in Figure.

00 / 000 10 /100

11/111

00 / 000 10 /100

01/ 011

00 01 00

011 00/ /111 10

/ 10/ 000 100

/1 1

/ 01

0 0

1 /00 0 0 1 /10 10 0 10 Input bits 1 10 0 1/0 Output bits 11/ 10 01/001 01/001 110 11 11/101 11 11/101

……

01/001 11 11/101 11

Next the three output (code) symbols on trellis branches are assigned signal phases according to the heuristic rules of Ungerboeck. This process is called set partitioning and is displayed in Figure. Note that the partitioning does not result in the natural mapping of the output bits to the constellation signal points. For example, the output symbol triplet (111) is assigned the constellation point 6 with corresponding phase shift of 270o. 8 − PSK ∆ 0 = 2sin(π / 8) = 0.765

∆1 = 2

∆2 = 2

000(0)

100(4)

011(2)

111(6)

001(1)

101(5)

010(3)

110(7)

The mapping of the output bits to the signal constellation points or phases is illustrated in the Figure.

011(2) 010(3)

001(1)

100(4)

000(0)

101(5)

110(7) 111(6)

The mapping of signal points in the constellation to the trellis branches is displayed in Figure.

0 4

0 00 4

1 5

2 6

01 0 1 10 5 7

0 4

10 3 11

1 5

b. What is the minimum distance error event through the trellis relative to the path generated by the all-zero bit sequence? Solution:

Figure displays two candidate paths, one generated by the all-zero bit sequence and the other erroneous, that diverges from the all-zero path and merges later. If they go to different states in the first transition, they will need at least three symbol intervals to rejoin. The total Euclidean distance between those two paths is d 2 = ( d leaving same node ) + ( d entering different nodes ) + ( d entering same node ) 2

> ∆12 +∆ 02 + ∆12 = 2 + 0.586 + 2 = 4.586

All other error events correspond to paths that are separated by even greater Euclidean distances.

Erroneous non-parallel path at minimum Euclidean distance

0 4

0 00 4

All-zero path

1 5

2 6

01 0 1 10 5 7

0 4

10 3 11

1 5

The minimum distance error event through the trellis for this TCM scheme corresponds to parallel arms (e.g., paths labeled 0 and 4). They rejoin in one symbol interval and distance between them is 2 (i.e., d2 = 4). c. Assuming that your answer to part (b) is the minimum distance error event for the trellis, what is dfree of the code? Solution:

The minimum distance error event correspond to parallel transitions . i.e., the Euclidean distance d = 2. Therefore, d free = 2 . d. Assuming that parallel transitions dominate the error probability, what is the coding gain of this trellis code relative to uncoded 8-PSK, given that dmin for the 8-PSK is .765? Solution:

G = 10 log10

2 (TCM 8 − PSK ) d free 4 = 10 log10 = 8.347 dB 2 2 d min (uncoded 8 − PSK) (.765)