1

Analog and Digital Control

1Lecture 3

Kirsten Mølgaard Nielsen

Based on lecture notes by Jesper Sandberg Thomsen

2

OutlineDiscrete equivalents

Overview (emulation – finding a discrete equivalent to the continuous system)

Numerical integration Zero-Pole matching (mapping) Hold equivalents

Design by emulation Evaluation of design

Nyquist stability criterion Continuous case Discrete case

3

Discrete Equivalents - Overviewcontroller

D(s)plantG(s)

r(t) u(t) y(t)e(t)+

-

Translation to discrete controller (emulation)Numerical Integration• Forward rectangular rule• Backward rectangular rule• Trapeziod rule (Tustin’s method, bilinear transformation)• Bilinear with prewarpingZero-Pole MatchingHold Equivalents• Zero order hold (ZOH)• Triangle hold (FOH)

Translation todiscrete plantZero order hold (ZOH)

EmulationPurpose: Find a discrete transfer function which approximately has the same characteristics over the frequency range of interest.

Digital implementation: Control part constant between samples. Plant is not constant between samples.

4

Numerical Integration

Fundamental concept Represent H(s) as a differential equation. Derive an approximate difference equation.

We will use the following example Notice, by partial expansion of a transfer function this

example covers all real poles.

aeauuas

asH

sE

sU

)(

)(

)(

ExampleTransfer function Differential equation

5

Numerical Integration

kTTkTaeauTkTu

daeaudaeaukTu

continousdaeautu

aeauu

as

asH

sE

sU

TkT kT

TkT

t

,over of area)(

)(

)()()()(

Equation alDifferenti

)()(

)(

FunctionTransfer

0

0

6

Numerical Integration

Now, three simple ways to approximate the area.

Forward rectangle approx. by looking forward

from kT-T Backward rectangle

approx. by looking backward from kT

Trapezoid approx. by average

kT-T kT

kT-T kT

kT-T kT

7

Numerical Integration

Forward rectangular rule (Euler’s rule)(Approximation kT-T)

aTz

a

aTz

aT

zaT

aTz

zE

zUzH

zEaTzzUzaTzU

TkTaTeTkTuaT

TkTaeTkTauTTkTukTu

F

11)1(1)(

)()(

)()()1()(

unctionTransfer F

)()()1(

)()()()(

EquationDifference

1

11

11

11

1

111

8

Numerical IntegrationBackward rectangular rule (app kT)

aTzz

a

zaT

aTz

aT

aT

zaTzE

zUzH

zEaT

aTzU

zaTzU

kTeaT

aTTkTu

aTkTu

kTaTeTkTukTaTukTu

kTaekTauTTkTukTu

B

11)1()1()1(

11

1

)(

)()(

)(1

)()1(

1)(

unctionTransfer F

)(1

)(1

1)(

)()()()(

)()()()(

EquationDifference

2

22

22

222

222

9

Numerical IntegrationTrapezoid rule (Tustin’s Method, bilinear trans.)(app ½(old value + new value))

azz

T

a

aTzaT

zaT

zE

zUzH

kTeTkTeaT

aTTkTu

aT

aT

kTaekTauTkTaeTkTauT

TkTukTu

T

)1()1(22)2(

)1(

)(

)()(

unctionTransfer F

)()()2/(1

2/)(

)2/(1

)2/(1

)()()()(2

)()(

EquationDifference

3

3

3333

10

Numerical Integration

Comparison with H(s)

azz

T

azH

as

a

aTzz

azH

as

a

aTz

azH

as

a

sH

T

B

F

)1()1(2

)(rule Trapezoid

1)(rule Backward

1)(rule Forward

functionTransfer Method)(

11

Numerical IntegrationTransform s ↔ z

sT

sTz

z

z

Ts

Tsz

Tz

zs

TszT

zs

)2/(1

)2/(1or

)1(

)1(2rule Trapezoid

1

1or

1rule Backward

1or1

rule Forward

ionApproximatMethod

Comparison with respect to stability In the s-plane, s = j is the boundary between stability

and instability.

12

Numerical Integration - stability

jssT

sTz

jsTs

z

jsTsz

for )2/(1

)2/(1Trapezoid

for 1

1Backward

for 1Forward

(Forward)Region ofs-LHP

(Backward)Region ofs-LHP

(Trapezoid)Region ofs-LHP

13

Numerical IntegrationSome conclusions

Higher distortion with the forward and backward rule. The forward rectangular rule could cause a stable

continuous filter to be mapped into an unstable digital filter.

The trapezoid rule maps the stable region in the s-plane exactly into the stable region of the z-plane. Though, the j-axis is compressed into the unit circle with length 2.

The trapezoid rule appears superior. But, let us make some further investigations..... let us calculate the power |H(j|2 at, for example, = a – the change in frequency response.

14

Numerical Integration

2

1)(

)(

1)/(

1)(

)(

2

2222

22

jaH

a

aa

ajH

aj

ajH

What about HT(z) ? at which frequency is |HT(z)|2 = 0.5 ?

15

Numerical Integration

12/,)2

(2

Notice,

)2

arctan(2

or)2/tan(2

2

1

1)2/tan(21

1)(

en Power wh.with

1)2/tan(21

1

)2/tan(2

)(

1

11

22

1

22

1

11

aTaaT

T

aT

TT

Ta

TTa

zH

ezez

TjTa

aTjT

azH

T

TjTj

T

HT(z) as before but re-formulated (page 194)

I.e. high sampling frequency

16

Numerical IntegrationTrapezoid with pre-warping

maintain the power at some specified frequency

1. Choose a critical frequency

2. Replace by a´ such that new a = a’,

3. and insert in H(s). a’ = (2/T) tan( T/2)

4. Find the discrete filter H(z) based on the modified continuous transfer function H(s).

11

)2/tan(1

)()(

11

1

zz

Ts

sHzH

)1()(1 jHzH

17

Numerical Integration

1/

at Prewarping

2

1

1.0

with122

1

)(

filterorder 3th Example,

23

b

T

sss

sH

Plots for T = 0.1s ≈ 60 b

18

Numerical Integration

Plots for T = 1s ≈ 6 wb

Notice,prewarp

19

Numerical Integration

Plots for T = 2s ≈ 3 b

Notice,prewarp

20

Zero-Pole MatchingExact map between poles (z = esT) exists.

Basic idea, simply use same map for zeros.

Rules1. All poles are mapped by z = esT.

For example, s = -a maps to z = e-aT

2. All finite zeros are mapped by z = esT. For example, s = -b maps to z = e-bT

3. Basically, zeros at j= infinity maps to z = ej= -1 (representing the highest frequency)

4. Identical gain at some critical freq. (typically, s=0) H(s) at s=0 = H(z) at z=1

21

Zero-Pole Matching

z

zKzH

eK

as

a

ez

zK

zsH

ezas

as

aH(s)

aT

szaT

s

aT

1)(

2

11

14) (rule Gain

10)(3) (rule Zero

1) (rule Pole

01

Example (1)Compute the discrete equivalent by zero-pole matching

22

Zero-Pole Matching

s-plane z-plane

z

zKzH

as

aH(s)

1)(

match-zp

23

Zero-Pole Matching

)1()()1()(1

11)(

)(

)(1

1

kKekKekukuz

zK

z

zKzH

zE

zU

Including delay in computerNotice, u(k) depends on e(k). This comes from rule 3

Solution, first zero for s=infinity is mapped to z=infinity

(as in Matlab). This gives

)1()1()(1

1

)(

)(1

1

kKekukuz

zK

zK

zE

zU

24

Hold EquivalentsPurpose, design a discrete filter H(z), such that

input e(k) is samples of e(t) H(z) has an output u(k) u(k) approximates the output of the continuous filter

H(s) whose input is the continuous signal e(t).

Zero order hold (ZOH) (step invariant), approach below

HoldSampler H(s)e(t) e(k) u(k)

Samplereh(t) u(t)

25

Hold EquivalentsHold operation

gaina is)(

)()()(

11

)(

)(1)(1)(

ke

kesQsE

es

sQ

Ttttq

h

Ts

Q(s)

HoldSampler H(s)e(t) e(k) u(k)

Samplereh(t) u(t)

26

Hold Equivalents

ZOH transformation Notice, we can find H(z) by Take inverse Laplace to find a continuous signal. Sample this continuous signal to find a discrete signal.

kkette )()()(

)(11

)()()()()(

)()()()()()()(

11 kes

sHLksQsHLtu

ksQsHkesQsHsU

Ts

HoldSampler H(s)e(t) e(k) u(k)

Samplereh(t) u(t)

27

Hold Equivalents

)()(

)(1)(1)(11

)()( 11 ks

sHLTttke

ssHLtu Ts

s

sHLZ

s

sHZ

s

sHZzzHzUkuZ

tuku tkT

)()(where

)()1()()()(

)()(

1

1

Continuous output signal

Discrete output signal and transfer function H(z)

28

Hold EquivalentsTriangle hold equivalent

Extrapolate the samples to connect sample to sample in a straight line.

Impulse response for triangle hold

THold System

2

2

Ts

ee TsTs )(sH

)(ke )(tu

2

)()1()(

Result

s

sHZ

Tz

zzH

T

0.1

29

Hold Equivalents

Same third order filteras before

+ Zero-pole o ZOHx Triangle

T = 1(next T=2)

30

Hold Equivalents

Same filteras before

+ Zero-pole o ZOHx Triangle

T = 2

31

Hold EquivalentsConcluding remarks

Methods, we have looked at Numerical integration methods Zero-pole matching (mapping) Hold equivalents

All methods works fine for high sample frequency Matlab

32

Evaluation of Design

Discrete controller by emulation Can be calculated using one of several techniques.

Evaluation discrete equivalent (design) We need to analyze in the z-domain. Process expressed in z-domain (ZOH) Notice, other eval. might be necessary (simulation)

33

Evaluation of DesignExample, using Matlab

sysD_c = tf([10 1],[1 1]);T = 0.2; % around 30*wn. Also use, T = 1sysD_d = c2d(sysD_c,T,’matched’); % zero-pole matchedsysG_c = tf([1],[10 1 0]);sysG_d = c2d(sysG_c,T,’zoh’); % zero order holdsysCL_c = feedback(sysD_c*sysG_c,1);sysCL_d = feedback(sysD_d*sysG_d,1);step(sysCL_c); hold on;step(sysCL_d);

)110(

1)(

(antenna) sys. Cont.

sssG

1

)110()(

controller Cont.

s

ssD

34

Evaluation of Design

T = 0.2(s ≈ 30 n)

35

Evaluation of Design

T = 1(s ≈ 6 n)

The sample delay gives phase lag.Thus, less PM which gives higher overshoot

36

Nyquist Stability Criterion

Check closed loop stability with controller KD(s) and plant G(s)

)(

)()()(,

)()(1

)()(

)(

)(

sa

sbsGsD

sGsDK

sGsDK

sR

sY

Characteristic equation (CE)

0)(

)()(0)()(1

sa

sKbsasGsDK

Notice, number of poles in CE equals number of open-loop poles in D(s)G(S). (known)Notice, zeros in CE are closed-loop poles! (unknown)

37

Nyquist Stability CriterionApproach, check characteristic equation

Let us follow a contour encircling the RHP. Basic idea: Evaluate CE = 1+KD(s)G(s) at the

contour, and draw image. (We can just as well evaluate KD(s)G(s) and look at -1 )

)()( sGsDK

Notice, poles or zeros gives encirclements

38

Nyquist Stability Criterion

Criterion (continuous case)

Z = P + N

Z = number of unstable closed-loop poles. Thus, for stability we want Z=0. (unknown)

P = number of unstable open-loop poles of KDG (known).

N = total number of encirclements of the point -1 by evaluation of KDG (taken in same direction as s. Thus, N might be negative).

39

Nyquist Stability CriterionInterpretations and explanations

If the s-plane contour encircles a zero of 1+KDG in a certain direction, the image contour will encircle the origin in the same direction. (related to Z)

If the s-plane contour encircles a pole of 1+KDG in a certain direction, the image contour will encircle the origin in the opposite direction. (related to P)

The net number of same-direction encirclements, N, equals the difference N = Z- P.

Actually, we should investigate 1+KDG and encirclements around the origin, but easier to investigate KDG and encirclements around -1.

40

Nyquist Stability Criterion

Discrete case Same approach,

almost... Unstable region is

outside the unit circle.

Easier to evaluate using the stable region.

41

Nyquist Stability Criterion

Discrete case CE: 0 = 1+KD(z)G(z) Total number of poles = n Number of stable zeros of 1+KD(z)G(z) is n-Z Number of stable poles of 1+KD(z)G(z) is n-P

Criterion (discrete case) Z = (n-Z) – (n-P) or Z = P – N

42

Nyquist Stability Criterion

Discrete case Determin the number P of unstable poles of KDG Plot KD(z)G(z) for the unit circle. This is the counter-

clockwise path around the unit circle. Set N equal to the number of Counter-clockwise

encirclements of the point -1 on the plot Compute Z=P-N. The system is stable if and only if Z=0

43

Nyquist Stability CriterionExample (7.9)Evaluate the stability of the unity feedback discrete system with the plant G(s) with sampling time T=2 and ZOH. Use controller KD(z) = K.

)1(

1)(

sssG

Solution (matlab)sysC = tf([1],[1 1 0]);sysD = c2d(sysC,2);nyquist(sysD);

K=1

The system is openloop stable and there are no encirkelments of -1 => the system is closed loop stable for K=1