Feedback Control of Computing Systems M2: Signals and Z-Transforms

© 2004 HellersteinFeedback Control of Computing Systems

Feedback Control of Computing SystemsM2: Signals and Z-Transforms

Joseph L. HellersteinIBM Thomas J Watson Research Center, [email protected]

September 21, 2004

2 © 2004 HellersteinFeedback Control of Computing Systems: M2 - Signals

Motivating Example

( )r k ( )y k( )e kController

NotesServer

NotesSensor

( )u k

( 1) ( ) ( 1)Iu k u k K e k ( 1) (0.43) ( ) (0.47) ( )w k w k u k ( 1) 0.8 ( ) 0.72 ( ) 0.66 ( 1)y k y k w k w k

( )w k

( ) ( ) ( )e k r k y k

The problemWant to find y(k) in terms of KI so can design control system that is stable, accurate,

settles quickly, and has small overshoot.But this is difficult to do with ARX models.

The SolutionUse a different representation

This module focuses on signals: time varying data


M2:Lecture


Agenda

Z-Transform representation of signals Z-Transforms of common signals Properties of z-transforms of signals Poles Effect of poles Final value theorem

Reference: “Feedback Control of Computer Systems”, Chapter 3.


Z-Transform of a Signal

0 1 2 3 40

1

2

3

4

5

6

Time domain representationu(0)=1u(1)=3u(2)=2u(3)=5u(4)=6

u(k)

z domain representation1z0 +3z-1 +2z-2 +5z-3 +6z-4

0

If { ( )} (0), (1),... is a signal, then its -Transform is( ) ( ) k

k

u k u u zU z u k z

0

1

2

1: 0 (current time)

: 1 (one time unit in the future)

: 2 (two time units in the future)

z k

z k

z k

z is time shift; z-1 is time delay

k


Write the Z-Transforms or Plots for the Following Finite Signals

0 1 2 3 4 5-2

-1

0

1

2

3

4

5

k

u(k

) 4321 4542)( zzzzzU

(Drop exponents >0.)

0 1 2 3 4-1

0

1

2

3

4

5

k

v(k

)

32154)()( zzzzzUzV


0 1 2 3

1

time (k)

Impulse 0,0)(;1)0( kkyyy(k)

1 ...001)( 210

zzzzY

Common Signals: Impulse

5)3(5 :Example 3 uz


0 1 2 3

1Step

time (k)

y(k)

0 1 2

1

( ) 1 1 1 ...1

1

1

Y z z z z

zz

z

0,1)( kky

Common Signals: Step


Properties of z-Transforms of Signals

0 1 2

0 1 2

1 0 1

0 1

1 2 3

Signals: ( ) (0) (1) (2) ...

( ) (0) (1) (2) ...

Shift: ( ) (0) (1) (2) ...

(1) (2) ...

Delay: ( ) / (0) (1) (2) ..

U z u z u z u z

V z v z v z v z

zU z u z u z u z

u z u z

U z z u z u z u z

0 1 2

0 1 2 0 1 2

.

Scaling: ( ) (0) (1) (2) ... -Transform of { ( )}

Sum of signals: (0) (1) (2) ... (0) (1) (2) ...

aU z au z au z au zz au k

u z u z u z v z v z v z

0 1 2 ( (0) (0)) ( (1) (1)) ( (2) (2)) ...

( ) ( )u v z u v z u v z

U z V z


Infinite Length Signals

0 5 10 15 200

2

4

6

8

10

15

...555)( 21

zz

zzzU


Construct a Z-Transform for the Following Infinite Length Signal

0 5 10 15 200

2

4

6

8

10

( )Y z

0 5 10 15 200

2

4

6

8

10

5 ( ), ( ) is the unit stepW z W z

0 5 10 15 200

2

4

6

8

10103z

0 5 10 15 200

2

4

6

8

10

115 ( )z W z

)(53)(5)( 1110 zWzzzWzY


Common Signals: Geometric

azz

zaazzY

...1)( 22

kaky )( :Geometric

0 5 10 15 200

0.2

0.4

0.6

0.8

1

a=0.8

8.0

...64.08.01)( 21

zz

zzzY


Poles of a Z-TransformDefinition: Values of z for which the denominator is 0

Easy to find the poles of a geometric:az

zzV

)(

5 2( )1 0.8

z zY zz z

Quick exercises: What are the poles of the following Z-Transforms?

Easy if sum of geometrics

Harder if expanded polynomial

Poles determine key behaviors of signals

Pole is a.

06.05.0

3)(

2

zz

zzV


Effect of Pole on the Signal

-5

0

5a=0.4

-5

0

5a=0.9

-5

0

5a=1.2

0 5 10-5

0

5a=-0.4

0 5 10-5

0

5a=-0.9

0 5 10-5

0

5a=-1.2

azzaky k

)(

What happens when|a| is larger?|a|>1?a<0?

Larger |a|Slower convergence

|a|>1Does not converge

a<0Oscillates

,...),,1(...1

Why?

222 aazaazaz

z


Final Value Theorem

Provides an easy way to determine the steady state value of a signal

Limit as k becomes large

)()1(lim)( 1 zVzv z

(if V(z) has all of its poles inside the unit circle)

circleunit theinside is if ,0)1(lim

01)1(lim

11

)1(lim

1

1

1

aaz

zz

zz

zz

z

z

z

Final value of the unit step is 1.

Final value of the impulse is 0.


Applying the Final Value Theorem

)3.0)(4.1(2.0)(

:signal for the valuefinal theisWhat

42.5)|()1()3.0)(1(

38.0)(

1

zzzzV

zWzzzzzW

z


Describe the following Signals

Components of descriptionDoes the signal convergeFast or slow convergenceOscillations, if any 5 2( )

1 0.8z zY z

z z

7 3 10( )0.6 1 0.2zU z

z z z

Step of 5 Geometric at .8

5 (2)(0.8)k

13 (7)( 0.6) (10)(.02) , 0 7, 0

k k kk

0 2 4 6 8 10-10

-5

0

5

10

15

20

3)()()1(lim 1 uzUzz

(1) 3 (7)( .6) 10 17.2u Delayed geometric

Delayed step


M2:Labs


Plot The Signals

4.03.13.12)( .3

)1(410

12)( 2.

/)34(2)( 1.

2

2

42

53

zzzzzY

zzz

zzzY

zzzzY


Find the Final Values

1272)(

)2.02.1(16

09.3)(

2

2

zzzzY

zzz

zzzY


Describe the Following Signals

Which converges more quickly? Which of the following oscillate? Do any of the following fail to converge?

24.02.1

32.02.12

2

2

zzz

zzz

Feedback Control of Computing Systems M2: Signals and Z-Transforms

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