lecture3_small

Sampling Sampling: period = T y(k) d

@

y(t)

T

Example (single pole signal) â&#x2C6;&#x2019;at e , tâ&#x2030;Ľ0 Consider, y(t) = 0 t<0 Laplace transform: Sampled signal:

with a > 0.

1 . s+a

k

y(k) = y(t)

= eâ&#x2C6;&#x2019;akT = eâ&#x2C6;&#x2019;aT .

y(s) =

t=kT

Z-transform,

y(z) =

z . z â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;aT

The s-plane pole is at s1 = â&#x2C6;&#x2019;a, and the corresponding z-plane pole is at z1 = eâ&#x2C6;&#x2019;aT .

Roy Smith: ECE 147b 3: 1

Sampling Example: (second order) Now consider a damped sinusoidal signal, Laplace transform: Sampled signal:

Z-transform:

y(s) =

β , (s + ι)2 + β 2

y(k) = eâ&#x2C6;&#x2019;ÎąkT sin(βkT ),

y(t) = eâ&#x2C6;&#x2019;Îąt sin(βt),

t â&#x2030;Ľ 0, with Îą > 0.

Poles: s1,2 = â&#x2C6;&#x2019;Îą Âą jβ. k â&#x2030;Ľ 0.

z â&#x2C6;&#x2019;1 eâ&#x2C6;&#x2019;ÎąT sin(βT ) y(z) = . â&#x2C6;&#x2019;1 1 â&#x2C6;&#x2019; z 2eâ&#x2C6;&#x2019;ÎąT cos(βT ) + z â&#x2C6;&#x2019;2 eâ&#x2C6;&#x2019;2ÎąT

Z domain poles given by: z 2 â&#x2C6;&#x2019; 2eâ&#x2C6;&#x2019;ÎąT cos(βT )z + eâ&#x2C6;&#x2019;2ÎąT = 0. q z1,2 = eâ&#x2C6;&#x2019;ÎąT cos(βT ) Âą e2ÎąT cos2 (βT ) â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;2ÎąT

p = eâ&#x2C6;&#x2019;ÎąT cos(βT ) Âą j 1 â&#x2C6;&#x2019; cos2 (βT )

Imaginary z-plane

eâ&#x2C6;&#x2019;ÎąT

βT Real

= eâ&#x2C6;&#x2019;ÎąT (cos(βT ) Âą j sin(βT )) = eâ&#x2C6;&#x2019;ÎąT eÂąjβT = e(â&#x2C6;&#x2019;ι¹jβ)T .

Roy Smith: ECE 147b 3: 2

Sampling General case: Sampling maps the s-domain poles to the z-domain via: zi = esi T .

Stable continuous-time signals (Re {si } < 0) map to stable discrete-time signals (|zi | < 1).

Pole locations under sampling: Imaginary

s-plane

Imaginary

z-plane

ω = π/T

1 Real

Real

ω = -π/T

Roy Smith: ECE 147b 3: 3

Sampling Sampled pole locations: (in detail) z-plane

Imaginary

N=4 N=3

N=5

ω n = 0.6π/T

ζ = 0.1 ωn = 0.3π/T

ω n = 0.7π/T ζ = 0.2

N=8 ζ = 0.3 ω n = 0.2π/T

ωn = 0.8π/T

N = 20

ωn = 0.9π/T

N=2

ζ = 0.9

Real

ω n = π/T

-1.0

-0.5

0

0.5

1.0

Changing the sampling frequency. (recall that zi = esi T . ) Decreasing T : decrease decay rate (r → 1) decrease oscillation frequency (θ → 0) poles track constant damping curves towards 1 This is simply because there are more samples taken in the same time period.

Roy Smith: ECE 147b 3: 4

Sampling Sample rate effects: zi = esi T , changing T changes the pole positions.

1.2

Continuous closed-loop system step response:

1.0

0.8

T1 = 0.25 seconds

G(s) =

T2 = 0.11 seconds

0.6

s2

ωn2 + 2ζωn s + ωn2

ζ = 0.6,

0.4

ωn = 5 rad./sec., s1,2 = −3 ± 4i.

0.2

0

0.5

1.0

1.5 Time [seconds]

z-plane

2.0

Imaginary

N=4 N=3

2.5

N=5

ω n = 0.6π/T

ζ = 0.1 ωn = 0.3π/T

ω n = 0.7π/T

ζ = 0.2

Discrete pole positions:

N=8 ζ = 0.3 ω n = 0.2π/T

ωn = 0.8π/T

N = 20

ωn = 0.9π/T

N=2

ζ = 0.9

Real

ω n = π/T

-1.0

-0.5

0

0.5

Period T1 : z1,2 = 0.255 ± 0.398i Period T2 : z1,2 = 0.650 ± 0.306i

1.0

Roy Smith: ECE 147b 3: 5

Aliasing Aliasing: What happens to signals of high frequencies (ω > π/T ) ?

As zi = esi T , sinusoids of frequencies from -π/T to π/T radians/second are mapped onto the unit disk by sampling.

Consider

y(t) = sin ω1 t,

which has Laplace transform:

ω1 . s2 + ω12

y(s) =

Poles are s1,2 = ±jω1 .

Imaginary z-plane

Sample at period T :

y(k) = sin ω1 kT , −jω1T

e jωaT = e

Z-transform:

y(z) =

z sin w1 T . z 2 − 2 cos ω1 T z + 1 ±jω1 T

Poles of y(z) are z1,2 = e

.

ωaT

Real 1

-1

−ω1T

Slow sampling, T > π/ω1 , implies that ω1 T > π.

The pole angle is greater than π.

Roy Smith: ECE 147b 3: 6

Aliasing Having w1 T > π, means that, e−jω1 T = ej(2π−ω1 T ) ,

and ejω1 T = e−j(2π−ω1 T ) .

Now, if (2π − ω1 T ) lies in the range 0 to π radians, the pole pattern is identical to that of a sinusoid of a lower frequency, ωa , where ωa T = 2π − ω1 T. 2π Equivalently, the apparent frequency is, ωa = − ω1 . (sampling freq: 2π/T rad/sec).

T

Example: 55 Hz Signal: y(t) = cos(2π55t) Sampling frequency: 1/T = 60 Hz Then y(k) = cos(2π55t)|t=kT = cos(2π5t)|t=kT Indistinguishable from a sampled 5 Hz signal!

Roy Smith: ECE 147b 3: 7

Aliasing Example: 55 Hz signal sampled at 60 Hz

1.5

Apparent (5 Hz) signal

Original (55 Hz) signal

1 0.5 0

0.1

0.2 time [seconds]

0.5 1 Sample points 1.5

Roy Smith: ECE 147b 3: 8

Aliasing Sampled signals The unit disk can only represent signals of frequency up to 1/2 the sampling frequency.

(Nyquist frequency).

Sampling operation maps signal poles via: zi = esi T .

Maps the horizontal strip from −jπ/T to jπ/T onto the whole z-plane. s-plane

Imaginary

π/T

z-plane

Imaginary

Mapping via sampling 1

-1 Real

Real

−π/T

And Re {s} < 0 in this strip maps to the inside of the unit disk.

Roy Smith: ECE 147b 3: 9

Aliasing Aliasing: (ambiguous mapping of higher frequency signals) Sampling also maps the next strip (from jπ/T to j3π/T ) onto the whole z-plane and adds it into the result.

Imaginary 3π/T Imaginary

Mapping via sampling π/T -1 Real

1 Real

−π/T s-plane

z-plane

Also true for all (infinite) 2π/T wide strips above and below the lowest frequency strip.

Roy Smith: ECE 147b 3: 10

Aliasing Consequences of aliasing: â&#x20AC;˘ Ambiguity. Our computer/controller cannot distinguish between frequencies inside the â&#x2C6;&#x2019;Ď&#x20AC;/T to Ď&#x20AC;/T range and those outside of it. â&#x20AC;&#x201C; Controller will respond incorrectly to an aliased signal (e.g. disturbance or error). â&#x20AC;&#x201C; An aliased signal cannot be reconstructed (signal processing).

Amelioration of the problem: y(k) e

@ @

F (s)

y(t)

T

â&#x20AC;˘ Anti-aliasing filter. Low pass, rejecting |Ď&#x2030;| > Ď&#x20AC;/T . â&#x20AC;&#x201C; High frequency signals no longer enter loop erroneously. â&#x20AC;&#x201C; High frequency disturbances/errors are â&#x20AC;&#x153;invisible.â&#x20AC;? â&#x20AC;&#x201C; Filter adds phase lag to the loop. (Potentially destabilizing!)

Roy Smith: ECE 147b 3: 11