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EPM 4056 Feedback Control Analysis and Design

Trim. 48- Trimester 2 2011/2012

Ching Seong Tan

Multimedia University



2

Content

Digital Control Systems - part 1

Sampled data systems.

The z-transform

Digital control concepts.



Continuous-Time Signals

• A signal that changes continuously in time:

• Defined at all times t (no gap)• Signal can take arbitrary values

• Example: temperature, position, velocity,…

0,),( t t t e

3



Continuous-Time LTI Systems

• A continuous-time system has input and output that are

both continuous-time signals

• Continuous-time linear time-invariant (LTI) system given

by differential equation:

• Model completely determined by the order n of the system, thecoefficients α and β

)(t e )(t y

dt

t dy

dt

t yd t e

dt

t de

dt

t ed t y

n

n

nn

n

n

)()()(

)()()(

101

4



Discrete-Time LTI Systems

• A discrete-time system has input and output that are both discrete-time signals

• Discrete-time linear time-invariant (LTI) system given by lineardifference equation

• Model completely determined by the order n of the system, and thecoefficients a and b

)(k e )(k y

)()1()()1()()( 0101 nk yak yank ebk ebk ebk y nnn

5



Example: Sampler and Zero-Order Hold

)(t e )(t e

)(s E Sampler and zero-order hold

)(s E

In general, original signal e(t) can not be fully reconstructed

6



Discrete-Time Signals

• A signal (sequence, or series) whose values are defined only atdiscrete times:

• Signal defined only at integer times k• Signal can still take arbitrary values

,2,1,0,1,0,1,),( k or k k e

7



Where are discrete-time signals from?

• Some come naturally

– Population of a species in different generations

– Annual growth percentage of GDP

– Results of a numerical algorithm in different rounds of iteration

• Some arise by sampling continuous-time signals at regular timeintervals, say, every T seconds

8



Digital Signals

• The process of representing an arbitrary continuous value y(kT) with

binary bits of finite word length is called quantization (A/D)• Quantized signal is discrete in both time and value, called a digital

signal

9



Sampling and Data Reconstruction

• In the control system, sampling and data reconstruction of signalsare needed to interface the digital computer with the physical world

10



Sampled-Data Control System

Digital controller is designed together with sampler anddata hold

• How to characterize the input-output relation from e(t) to u(t) based on the transfer function D(z) of the digital controller?

• Start from the simplest case: digital controller does nothing

)(t eSampler Data hold

)(kT e )(t e

)(t eSampler

Digitalcontroller

)(kT eData hold Plant

)(kT u )(t u )(t y

11



Mathematical Models of Physical Systems

• Control systems give desired output by controlling the input.Therefore control systems and mathematical modeling areinter-linked.

• Mathematical models of control systems are ordinary

differential equations

12



Transform methods

• Continuous System and Laplace Transform

– differential equations: describe continuous time systems

– Laplace Transform: analyze linear, time invariant, continuoustime systems (LTICS)

• Discrete Time Systems and z- Transforms

– difference equations: describe operation of discrete timesystems

– Z-transform: analyze linear, time invariant, discrete timesystems (LTIDS)

13



Laplace Transform

Differential

equations

Inputexcitation e(t)

Output

response r(t)

Time Domain Frequency Domain

Algebraic

equations

Inputexcitation E(s)

Output

response R(s)

Laplace Transform

Inverse Laplace Transform

14



Z-Transform

• Given a discrete-time signal

• Its (double-sided) z-transform is defined by

• Its (single-sided) z-transform is defined by

• z-transform F(z) is a function of the complex variable z,and is only well defined on a region of the z-plane (DOC)• We will always assume single-sided (causal) signals andsingle-sided z-transform, unless otherwise stated

,2,1,01,0,1,),( k or k k f

k

k k zk f k f Z zF ,1,0,1,,)()]([)(

0

,2,1,0,)()]([)(k

k k zk f k f Z zF

15



Z-Transform

• How to do Z-Transform?

• How to do inverse Z-Transform?

• How to infer properties of a signal from its Z-transform?

16



Z-Transform of a Signal

Z-1

e(0)

e(1)

e(2)

e(3)

e(4)

…

0k

k ze(k) E(z)

E(z)Z

e(0) · z 0

+e(1) · z -1

+e(2) · z -2

+e(3) · z -3

+e(4) · z -4

…

e(k)

Mapping from a discrete signal to a function of zMany Z-Transforms have this form:

m

j

j

j

n

i

i

i

zb

za

E(z)

0

0

Rational Function of z

Z-1

e(0)

e(1)

e(2)

e(3)

e(4)

…

0k

k ze(k) E(z)

E(z)Z

e(0) · z 0

+e(1) · z -1

+e(2) · z -2

+e(3) · z -3

+e(4) · z -4

…

e(k)

Mapping from a discrete signal to a function of zMany Z-Transforms have this form:

Rational Function of z

17



Z Transform of Unit Impulse Signal

e impulse (k) E impulse (z)Z

Z -1

e(0) = 1

e(1) = 0

e(2) = 0

e(3) = 0

e(4) = 0 …

1 · z 0

+0 · z -1

+0 · z -2

+0 · z -3

+0 · z -4

…

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

1(z) E impulse

18



Delayed Unit Impulse Signal

e delay (k) E delay (z)Z

Z -1

e(0) = 0

e(1) = 1

e(2) = 0

e(3) = 0

e(4) = 0 …

0 · z 0

+1 · z -1

+0 · z -2

+0 · z -3

+0 · z -4

…

1 z(z) E delay-1 0 1 2 3 4 5 6 7 8 9

0

0.5

1

19



Z-Transform of Unit Step Signal

e step (k) E step (z)Z

Z -1

e(0) = 1

e(1) = 1

e(2) = 1

e(3) = 1

e(4) = 1…

1 · z 0

+1 · z -1

+1 · z -2

+1 · z -3

+1 · z -4

…

... z z z(z) E step 3211-1 0 1 2 3 4 5 6 7 8 9

0

0.5

1

20



Z-Transform of Unit Step Signal

,n

,|a| 1

a

a

a

)a...aaa)((a...aa

n

nn

1

1

1

111

1

2

2

A little bit more math …

assuming

a

a

a

a

)a...aaa)((...aa

n

n

n

n

1

1

1

1lim

1

11lim1

1

22

11

11

1

321

z

z

-z... z z z(z) E step

21



Z-Transform of Exponential Signal

e exp (k) Eexp(z)Z

Z -1

e(0) = 1

e(1) = a

e(2) = a 2

e(3) = a 3

e(4) = a 4

…

1 · z 0

+a · z -1

+a 2 · z -2

+a 3 · z -3

+a 4

· z -4

…

1

33221

exp

1

1

1

--az

... za zaaz(z) E

-1 0 1 2 3 4 5 6 7 8 90

1

2

3

4

5

6

a=1.2

Rememberthis!

22



Z-Transform of sin/cos

ik θ ee(k)

11

1-iθ

z-e E(z)

2)cos

ik θ ik θ ee

(k θ e(k)

Time Domain Z-Transform

i

ee(k θ e(k)

ik θ ik θ

2)sin

-ik θ ee(k)

11

1--iθ

z-e E(z)

21

1

2121

1

1111

11

cos21

cos1

sincos1

cos1

2sincos1

1

sincos1

1

21

1

1

1

z zθ

zθ

) zθ () zθ (

zθ

)/ zθ i zθ zθ i zθ

(

)/ ze z-e

( E(z)iθ -iθ

21

1

cos21

sin

z zθ -

zθ E(z)

-

-

θ iθ eiθ sincos Euler Formula:

2cos

iθ iθ eeθ

2i

eesin

ii

23





Properties of z-Transform

)()1( )(

0

1

)(

limlim

lim

1zk

101

2121

1

0

0

22112211

0

z E zk e

E(z) )e(

E(z) z

z(z)E e(n)(k)e

(z)(z)E E (k)(k)*ee

dz

dE(z)-z ke(k)

)E(zε e(k)ε

e(k)z E(z)z n)u(k)e(k

E(z) z n)|n)u(k e(k

(z) E a(z) E a(k)ea(k)ea

e(k)zE(z)k e

z

k

n

-aak

n

k

k n

-n

n

k

-k

linearity

delayed

forwarded

convolution

initial value

final value

complex trans.

definition

25



Inverse Z-Transform

• Table Lookup – if the Z-Transform looks familiar,look it up in the Z-Transform table!

• Power series method (Long Division)

• Partial Fraction Expansion

e(k) E(z)Z

Z -1?

(k)u(k)ue(k) rampstep 23

(z) E (z) E

) z(

z

z E(z)

rampstep 23

1

2

1

321

1

1

Z -1?

26



Power Serious Method

• Sort both nominator and denominator withdescending order of z first

21

1

21

3

z z

z E(z)

• e(0)=3, e(1)=5, e(2)=7, e(3)=9 ,…, guess:e(k)=3estep(k)+2eramp(k)

27



Partial Fraction Expansion

• Many Z-transforms of interest can be expressedas division of polynomials of z

m

j

j

j

n

i

i

i

zb

za

E(z)

0

0

May be trickier:complex rootduplicate root

) p)...(z p)(z p(zb

zb... zb zbb

mm

m

m

21

2

210

m

j j

j

p z

cc E(z)

1

0

, p(k)ek

jd p j

1

exp

m

j

d pimpulse(k)e(k)ece(k)

j

1

exp0

where k>0

1

1

1

1

z p z

j

28



Example

86

141432

2

z z

z z E(z)

42

210

z

c

z

cc E(z)

042

01

2

1

1

0

k ,cc

k ,ce(k)

k k

(z- 2 )(z- 4 )

E 1(z)=c 0 Z-1 e1(k)=c0*eimpulse(k)

Z-1

Z-1 e 2(k)=c1*2k-1 , k>0

2

12

z

c(z) E

4

23

zc(z) E e 2(k)=c 2*4k-1 , k>0

c0? c1? c 2?

29



Get The Constants!

86

141432

2

z z

z z E(z)

42

210

z

c

z

cc E(z)

(z- 2 )(z- 4 )

, z

c

z

cc E(z)

42

210

, z ,c E(z) 0 3

86

14143lim

2

2

0

z z

z zc

z

,c z

)(zc)c(z)E(z)(z-K(z) 2

10

2

444

32

141434 4

2

2

z|

z

z zc)K(

30





Example – Complete Solution

386

14143limlim

2

2

0

z z

z z E(z)c

z z

4

14143

86

141432

2

2

2

2

z-

z z

z z

z z)(z(z) E

2

14143

86

141434

2

2

2

4

z-

z z z z

z z)(z(z) E

86

141432

2

z z

z z E(z)

42

210

z

c

z

cc E(z)

142

14214232

2

21

-

)( E c

324

14414434

2

42

-

)( E c

4

3

2

13

z z E(z)

0432

0311

k ,

k ,e(k)

k k

32



Solving Difference Equations

m)e(k b...)e(k bn) y(k a...) y(k a y(k) mn 11 11

E(z) zb... E(z) zbY(z) za...Y(z) zaY(z) mmnn

1111

E(z) za... za

zb... zbY(z)

n

n

m

m

1

1

1

1

1

... y(k)

Z

Z-1 Transfer Function

33





Solve it!

1

1

1

1

11

1

1

1

801

21

401

6080

21

40

60

8040

60

801401

60

401

60

z.

z.

z.

z.. z

.

. z

.

).)(z.(z

z.

) z.-)( z.-(

z. E(z)

z.

z.Y(z)

--

-

LTI: y(k)=0.4y(k-1)+0.6e(k-1)e (k)=0.8 k y (k)?

E(z) z.Y(z) z.Y(z)11

6040

Z

1801

1

z.

E(z)

Z

1180214060k-k

....- y(k)

Z-1 -1 0 1 2 3 4 5 6 7 8 9

0

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 90

0.2

0.4

0.6

0.8

1

35



Signal Characteristics from Z-Transform

• If E(z) is a rational function, and

• Then Y(z) is a rational function, too

• Poles are more important – determine keycharacteristics of y(k)

m)e(k b...)e(k bn) y(k a...) y(k a y(k) mn 11 11

m

j

j

n

i

i

) p(z

) z(z

D(z)

N(z)Y(z)

1

1

zeros

poles

36



Why are poles important?

m

j j

j

m

j

j

n

i

i

p z

cc

) p(z

) z(z

D(z)

N(z)Y(z)

1

0

1

1

m

j

k-

j jimpulse pc(k)ecY(k)1

1

0

Z-1

Z domain

Time domain

poles

components

37



Various pole values (1)

-1 0 1 2 3 4 5 6 7 8 90

0.5

1

1.5

2

2.5

-1 0 1 2 3 4 5 6 7 8 90

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 90

0.2

0.4

0.6

0.8

1

-1 0 1 2 3 4 5 6 7 8 9-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

p=1.1

p=1

p=0.9

p=-1.1

p=-1

p=-0.9

-1 0 1 2 3 4 5 6 7 8 9-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

38



Conclusion for Real Poles

• If and only if all poles’ absolute values are

smaller than 1, y(k) converges to 0

• The smaller the poles are, the faster thecorresponding component in y(k) converges

• A negative pole’s corresponding component is

oscillating, while a positive pole’s corresponding

component is monotonous

39



How fast does it converge?

• e(k)=a k , consider e(k)≈ 0 when the absolute value ofe(k) is smaller than or equal to 2% of e(0)’sabsolute value

|a|k

..|a|k

.|a|k

ln

4

9123020lnln

020

11360

4

70ln

4

70

.|.|k

.a

Remember

This!

0 2 4 6 8 10 120

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

y(k)=0.7k

y(11)=0.0198

40



Why do we need Z-Transform?

• A signal can be characterized with its Z-transform(poles, final value …)

• In an LTI system, Z-transform of Y(z) is themultiplication of Z-transform of E(z) and the

transfer function

• The LTI system can be characterized by thetransfer function, or the Z-transform of the unitimpulse response

41



Sampler and Zero-Order Hold

)(t e )(t e

)(s E Sampler and zero-order hold

)(s E

In general, original signal e(t) can not be fully reconstructed

42





Sampler and Zero-Order Hold

the sampler and zero-order hold can be broken up in two steps

In time domain

)(s E )(* s E )(s E

s

Ts

1

)(t e )(* t e )(t eT

T

T

44



Ideal Sampler

)(t e )(* t e

)(t T

)(t e )(* t eT

Representation of the idea sampler

Impulsemodulator

)()()(* t t et e T )(t e

45



First Perspectives of Sampler and Hold

In the time domain:

)(t e )(* t e )(t eT Zero-Order Hold

Sampler

In the frequency domain (a mixture of z- and Laplace transforms):

)(s E )( z E )(s E T

Zero-Order HoldSampler

0

)()(k

k zkT e z E is called the z-transform of E(s)

46



Second Perspectives of Sampler and Hold

In the time domain:

0

)()(*k

kTsekT es E

)(t e )(kT e )(t eT

Ideal Sampler

In the frequency domain:

)(s E )(* s E )(s E T

is called the star-transform of E(s)

T

Data hold

s

Ts

1

47



Relating Two Perspectives

48



Star-Transform

• Given E(s), find

• Method I

• Method II

• Method III find E(z) either from the table,

or by

then

49



Effect of Time Delay on Star-Transform

When is an integer multiple of the sampling period T, wehave the simple relation

nT t 0

50



Properties of E*(s)

• Property 1

E*(s) is periodic in s with period

Example

51



Properties of E*(s)

• Property 2

If E(s) has a pole at p , then E*(s) has poles at

Example

52



Data Reconstruction

Shannon’s Sampling Theorem

• the signal e(t) can be uniquely determined from its

sampling e*(t) with sampling period T if its Fouriertransform E(jω) contains no frequency componentgreater than π/T

• Or equivalently, one has to sample the signal twice as

fast as the fastest changing part in e(t) in order to beable to reconstruct it uniquely

53



Data Reconstruction in Frequency Domain

54



Zero-Order Hold

• A physically feasible (causal) data hold transfer functionis given by

with unit impulse response

Frequency response

55



First-Order Hold

• Given a sequence of sampled data e(kT ), k = 0, 1, . . .reconstruct the signal between sampling timesapproximately by first order Euler expansion

56



Fractional-Order Holds

• Given a sequence of sampled data reconstruct the signalbetween sampling times approximately as follows

57



58

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