1 Modern Control Theory (Digital Control) Lecture 2.

1

Modern Control Theory

(Digital Control)

Lecture 2

2

Outline Signal analysis and dynamic response

Discrete signals Discrete time – discrete signal plot z-Transform – poles and zeros in the z-plane

Correspondence with continuous signals Step response

Effect of additional zeros Effect of additional poles

s-Plane specifications z-Plane specifications Frequency response

3

Signal analysis – discrete signals Analysis

look at different characteristic signals z-transform, poles and zeros signals

unit pulse unit step exponential general sinusoid

4

Signal analysis – discrete signals The z transform

k

kk

k

zezE

keZzE

eeee

)(

))(()(

as transform-z thedefine we

........,.....,,

valuesdiskrete Given the

210

5

Signal analysis – discrete signals The Unit Pulse

k

kk

k

k

k

zzzE

Z

k

k

ke

1)(

0,0

0,1

)(

01

1

6

Signal analysis – discrete signals The unit Step

11

1

)(1)(

0,0)(1

0,1)(1

)(1)(

1

02

2

z

z

z

zzkzE

Z

kk

kk

kke

k

k

k

k

Zeros : z=0Poles : z=1

7

Signal analysis – discrete signals Exponential

rz

z

rz

rzzrzE

Z

k

krke

k

k

k

kk

k

1

0

1

03

3

1

1

)()(

0,0

0,)(

Zeros : z=0Poles : z=r

8

Signal analysis – discrete signals General Sinusoid

jj

k

kj

k

kjkk

jkk

jkjkkk

rez

z

zre

zrezerzE

Z

kerke

keerkkrke

1

0

1

05

5

4

1

1

)()(

)(1)(

)(12

1)(1)cos()(

(let us look at the terms, one by one, and use linearity)

9

Signal analysis – discrete signals

22

4

5

)cos(2

))cos((

))((

)()(

2

1

2

1)(

)(

rzrz

rzz

rezrez

rezzrezz

rez

z

rez

zzE

rez

zzE

jj

jj

jj

j

Zeros : z=0, z=r cosPoles : z=r exp(j) , z=r exp(-j)

Plots shown for

45,7.0 r

10


Transients r > 1, growing signal (unstable) r = 1, constant amplitude signal r < 1, decreasing signal (the closer r is to 0 the shorter the

settling time. In fact, we can compute settling time in terms of samples N.)

)(1)cos()(4 kkrke k

Conclusions General sinusoid

11

Signal analysis – discrete signals Samples per oscillation (cycle)

number of samples in a cycle is determined by or, N = samples/cycle depends on pole placements depend on

cycle

samples2

))(cos()cos(

N

Nkk

We have

dependence of

k=0

2

1

345

12

Signal analysis – discrete signals Samples per oscillation (cycle), cont.

cyclesamples845

360

have we45For

N

13


14

Signal analysis – discrete signalsPoleplacements

15

Correspondence with cont. signals Continuous signal

))(()(

)(1)cos()(

jbasjbas

assY

kbtety Lat

Discrete signal

))((

))cos(()(

)(1)cos()()(

jbTaTjbTaT

ZkaT

eezeez

bTrzzzY

kbTkeky

Poles: s = -a + jb,s = -a - jb

Poles: z = exp(-aT - jbT)z = exp(-aT + jbT)

sTez

Pole map

16

Correspondence with cont. signals

sTez

Pole map

10,0,

1,0,

1

:pole plane-s General

rrzs

rrzs

zjs

js

17

Correspondence with cont. signals Recall, poles in the s-plane

22

22

2))(()(

nn

n

dd

n

ssjsjssH

222222

222

1

part, Imag.

part, Real

frequency, natural Undamped

)sin(ratio, Damping

nnnnd

dn

n

n

18

Correspondence with cont. signals

sTez

Pole map

2

2

1,and

,where

1:pole plane-s General

nn

aT

j

nnd

ba

bTer

rezjbas

jjs

Fixed ,varying n

Fixed ,varying n

19

Correspondence with cont. signals Fixed n,

varying Fixed ,varying n

20

Correspondence with cont. signals Notice, in the vicinity of z = 1, the map of and n looks like the s-plane in the vicinity of s = 0.

21

Signal analysis – step response

Investigate effect of zeros fix z1 = p1, and explore effect of z2

a (delayed) second order sys is obtained = {0.5, 0.707} (by adjusting a1 and a2)

= {18°,45°,72°} (by adj. a1 and a2) a unit step U(z) = z/(z-1) is applied to the

system (pole, z=1, and zero, z=0)

22


Discrete stepresponsesfor = 18°

Overshootincreases withthe zero Z2

23


The zero has little infuence on the negative axis, large influence near +1

24


25


Investigate effect of extra pole fix z1 = z2 = -1, and explore effect of moving

singularity p1 (from -1 to 1) = 0.5 = {18°,45°,72°} a unit step is applied to the system

26


Mainly effect on rise time

Rise time expressedas number of samples.

The rise time increaseswith the pole

27

Signal analysis – step response Conclusions

Addition of a pole or a zero between -1 and 0 Only small effect

Addition of a zero between 0 and +1 Increasing overshoot when the zero is moving towards +1

Addition of a pole between 0 and +1 Increasing rise time when the pole is moving towards +1 (the

pole dominates)

28

s-Plane specifications Spec. on transients of dominant modes

dominant first order time constant (related to 3 dB bandwidth)

dominant second order rise time tr (related to natural frequency n ≈ 1.8/tr )

settling time ts (related to real part = 4.6 ts )

overshoot Mp, or damping ratio . Spec. on reference tracking

typically step or ramp input specification i.e. specifications on Kp and Kv , ess = r0 /Kv

ess is the steady state error for a ramp input of slope r0

29

s-Plane specifications

ExampleWe have system with dominant 2. order mode

Specifications:

)line vertical(

lines) angel(

43

21%

CCt

CCM

s

p

Notice, spec. on n not shown

30

z-Plane specifications Discrete system

similar specifications in addition, sample

time T2C

)(

)(

)(

5

43

21%

green

black

blue

Cer

CCt

CCM

T

s

p

Example (continued)Notice, sample time T must be chosen.

5Cr If fixed n

31

z-Plane specifications

Example (7.2 and 7.5)A system is given by

Specifications are1) Overshoot Mp less than 16%

2) Settling time ts (1%) less than 10 sec.

3) Chose sample time T such that)110(

1)(

sssG

)3(sec2.025.020

1

Hz25.0where20

T

fff bbs

32


)2(912.0,

46.010,6.4

)1(5.0%16)1(6.0

46.0

rerer

tt

MM

TT

ss

pp

1) Overshoot Mp less than 16%

2) Settling time ts (1%) less than 10 sec.

33


Cn

Also, we might have an additional specification on rise time tr

Damping, radius

r

n

Possibleregion

912.0

5.0

r

34

z-Plane specifications Steady-state errors

ZOH of plant transfer function, i.e. G(s) to G(z) Transfer function from R(z) to E(z), for investigating

the error.

)()()(1

1)( zR

zGzDzE

controllerD(z)

plantG(z)

R(z) U(z) Y(z)E(z)+

-

35

z-Plane specifications Now, if r(kT) is a step, then

zero. iserror theThus,

then,1at polea has If

1

1

)1()1(1

1

1)()(1

1)1(lim)(

),( valueFinal

1)()(1

1)(

1

DGzDG

KGD

z

z

zGzDze

e

z

z

zGzDzE

p

z

36

Frequency response Frequency response methods

Gain and phase can easily be plotted. Freq. response can be measured directly on a

physical plant. Nyquist's stability criterion can be applied. Error constants can be seen on gain plot. Corrections to gain a phase by additional poles and

zeros. Effect can easily be observed – in terms of cross over frequency, gain margin, phase margin.

Frequency response methods can also be applied for discrete systems (example).

37

Frequency responseDiscrete Bode Plot, Example (7.8)Plot the discrete frequency response corresponding to

)1(

1)(

sssG

Transform to z-domain by ZOH, with sample time T = 0.2, 1 and 2.Solution. Use Matlab c2d(sys,T).

Matlabsysc = tf([1],[1 1 0]);

sysd1 = c2d(sysc,0.2);sysd2 = c2d(sysc,1);sysd3 = c2d(sysc,2);

bode(sysc,'-',sysd1,'-.', sysd2,':', sysd3,'-',)

)135.0)(1(

523.0135.1)(

)368.0)(1(

718.0368.0)(

)8187.0)(1(

9355.00187.0)(

3

2

1

zz

zzG

zz

zzG

zz

zzG

38

Frequency response

Primary effect,Additional lag

Approx.phase lagT/2

Half samplefrequency

39

Frequency response

Approx.phase lagT/2

Accurate upto T = /2

40

Discrete Equivalents - Overview

controllerD(s)

plantG(s)

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

-

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

Translation todiscrete plantZero order hold (ZOH)

Lecture 3

1 Modern Control Theory (Digital Control) Lecture 2.

Documents

r slide

outline signal analysis

signals pole map slide

system slide

linearity slide

vicinity of z

splane slide

varying n slide