Chapter 5

On-Line Computer Control – The z Transform

Analysis of Discrete-Time Systems

1. The sampling process

2. z-transform

3. Properties of z-transforms

4. Analysis of open-loop and closed-loop discrete time systems

5. Design of discrete-time controllers

Continuous signal and its discrete-time representation with different sampling rates

3 t (sec)

y*

6

9 12

3 t (sec)1

y

7 9 115 3 t (sec)1

y*

6

9 12

y y*Continuoussignal

Disontinuoussignal

t

t = nT t

y*

t

t = nT t

y*

t = nT t

y*

nT area Impulse

From the response of a real sampler to the response of an ideal impulse sample

(a) (b)

(c)

(a) (b)

(c)

T = 1 sec

T = 3 sec

The Sampling Process

1. At sampling times, strength of impulse is equal to value of input signal.2. Between sampling times, it is zero.

0n

*

nT)-(ty(nT)

...nT)-(ty(nT)....T)-(ty(T)(t)y(0)ty

ImpulseSamplery (t) y*

nT)]-(t[y(nT)(s)y0n

* L

nTs-

0n

* e y(nT)(s)y

or

Laplacing

The Hold Process :From Discrete to Continuous Time

Zero – Order Hold :

m* (t) m (t)

Continuousoutput

discreteimpulses

Hold Device

t

t

m* (t)

T

m (t)

1

Transfer Function :Response of an impulse input : (t)

Ts--Ts

e-1s

1

s

e-

s

1 H(s)

1)T(n t nTfor (t)m m(t) *

First Order Hold

tc2,3,4....en 1)T(n t nT

nT)-(tT

1)T]-m[(n -m(nT) m(nT) m(t)

Response to an impulse input

1

1

2

T 2T

-1

Transfer function:2-sT

s

e-1

T

sT1 H(S)

First Order versus Zero Order Hold

Comparison of reconstruction with zero-order and first-order holds, for slowly varying signals.

Comparison of reconstruction with zero-order and first-order holds, for rapidly changing signals.

0 2T 4T 6T 8T 10T t

m* (nT)

0 2T 4T 6T 8T 10T t

m (t)

0 2T 4T 6T 8T 10T t

m (t)

0 1T 3T 5T 7T t

m (t)

0 1T 3T 5T 7T t

m (t)

0 1T 3T 5T 7T t

m* (nT)

(a)

(a)

(b)

(b)

(c)

(c)

Z-Transforms

y(t) yz(t)

Remarks1. z-transform depends only on the discrete values y(0), y(ז),y(ז)..etc. If two

continuous functions have the same sampled values , then z-transform will be the same.

2. It is assumed that the summation exists and is finit.

3. We can also view t in the form Z[ y (s) ] = ŷ(z)

Sample

z)y(n (z)y )(

)(y

)()(

)()(

n-

0n

0

0

tyΖ

z

znysy

ezLet

enysy

z

n

nz

Ts

n

nTsz

Z-Transforms of Basic Functions

11

1

1111

1

321

z-

z

z

......zzz Z[u(t)]

1. Unit Step Function

onvergence for c ze

z-e

z

z-e

λ

ze , λλzeeZ

-aT

-aT

-aT

-aT

n

nn

n

anT-at

1

1

11

1

1

1

1

00

2. Exponential Function

Z-Transforms of Basic Functions - Continued

12zcoswTz

zcoswTzcoswt Z

12zcoswTz

zsinwTsinwt Z

2

2

2

1

13211

321

321

10)1(

2

320

z

aTz...(aT)z(aT)z)aT(zSz

(aT)zaTzSz

...(aT)z(aT)z)aT(zSatZ

3. Ramp Function

4. Trigonometric Functions

Z-Transforms of Basic Functions - Continued

5. Translation

( )kTs k

n

n 0

k

n k

Z f(t - kT) Z e f s f (z)z

Z f(t - kT) f(nT kT)z

let n - k,assume f( τ) 0 for 0

f( τ)z

- k

0

k

f( τ)z z

f (z) z

Z-transform for Numerical Derivative

y(z)

T

z1

T

yyZ

T

yydtdy

11nn

1nnt

1(z)zfT)-f(tz

z-1 is like a back shift operator

Properties of z-Transforms

y(t)t

nTyn

(z)y)z(z

As z

y(nT)....z-T)zT)-y(y(z--y(T)zy(T -)z)-y(y( n

n

nzy(nTn

nzy(nT n

zy(nT)z( y(z)f)-zoof: (

(z)y)-z( z

y(t) t

n--

-

limlimˆ111

lim1

212211)100lim

0

1)0

)0

)1ˆ1Pr

ˆ11

limlim

11

1

(z)fa(z)f a fafaZ

22112211

1. Linearity

2. Final Value Theorem

zf zz zy T ˆ

11

2ˆ

1

1

nT

)T(n-f(t)dt )T(n- ynT y

dtnTt

tft) y(

11

0

Numerical Integration in z-transform

Using Trapezoidal Rule

(z)fz(z)fz(z)y(z) y

T

Tn-ff(nT )T(ny

T ˆ1ˆ2

1ˆˆ

2

1)1

or solving

1.Partial fraction expansion

λ1, λ2,… λn are low-order polynomials in z-1 compute c1,c2,…cn.

Invert each part separately, we able

)(...

)()()(

)()(ˆ

112

21

1

11

1

z

c

z

c

z

c

zP

zQzy

n

n

11

311

1

1

11

1

1

12

11

11

1

21

1

2

31

21

1

21)(ˆ

2

1

1

2

1

31

311)31)(1(

34134)(ˆ

1

1

zzzy

z

zc

z

zc

z

c

z

c

zz

z

zz

z

zz

zzy

z

z

Inversion of z-transforms

y(nT) = -1/2 + 1/2 e11n

y:0,1,4,13,0,…

2.Inversion by Long-Division

1z-1+4z-2+13z-3

1-4z-1+3z-2 z-1

z-1-4z-2+3z-3

4z-2+3z-3

4z-2-16z-3+12z-4

13z-3-12z-4

y(0) = 0y(T) = 1y(2T) = 4y(3T) = 13

21

1

341)(ˆ

zz

zzy

From Tables of z-transforms

z-transforms of various functions

Function Lalpace transform z-transformin time domain

unit impluse 1

unit step 1/s

ramp: f(t) = at a/s2

f(t) = tn n!/sn+1

f(t) = e-at 1/s+a

f(t) =te-at 1/(s+a)2

21

1

1

10

21

1

1

)1(

1

1

1

1)1(

)1(

1

1

1

lim

ze

ze

ze

zea

z

aTz

z

aT

aT

aT

aTn

nn

a

z-transforms of various functions

Function Lalpace transform z-transformin time domain

f(t) = sinωt

f(t) = cosωt

f(t) = 1-e-at

f(t) = e-at sinωt

f(t) = e-at cosωt

221

1

221

1

21

21

1

cos21

cos1

cos21

sin

)1)(1(

)1(

cos21

cos1

cos21

sin

zeTez

Tez

zeTez

Tez

zez

ze

zTz

Tz

zTz

Tz

aTaT

aT

aTaT

aT

aT

aT

22

22

22

22

)(

)(

)(

as

as

as

ass

as

ss

Discrete-Time Response of systems

In computer control:

measurements are taken periodically and

control actions implemented periodically,

This results in a discrete input/discrete

output dynamic system.

Discrete System

encn

Example of Discrete Systems

Let

a discrete time approximation is

ckedt

dcn

nnn

nnnn

nnnn

kecT

cT

ckccT

cT

ckeT

cc

1

1

1

)1(

1

1

)1()(

)(ˆ

)(ˆ)(ˆ)(ˆ)1(

zTT

k

ze

zcor

zekzczT

zcT

Taking z-transform

Z-transform for a given continuous system with transfer function G(s) and a ZOH

1

1

1( ) ( ) ( )

1

Ts

Ts

eZ H s G s Z G s

s

G s G sZ Z e

s s

G s G sZ z Z

s s

G sz Z

s

Example: Pure Integrator with Hold

)1()1(]1[][]-[1

][][].1

[)(

1)(

1

1

21

11

21-

22

z

zTK

z

TzKz

s

KZz

s

eKZ

s

KZ

s

K

s

eZsHGZ

s

K

s

esHG

ppp

Tsppp

Ts

pTs

p

s

e ST1

c*(s) s

K p

y*(s)

Step response

Hence of

which impulse a ramp response

21

1

11

1

1

)1()1(

1

)1()(ˆ

1

1)(ˆ

1)(

z

zK

zz

zKzy

zzc

ssc

pp

Example： First order lag system

1

1

11 1

1( ) ( )

1

1[ ( )] [ ]

( 1)

[1 ] [ ]( 1)

1 1 [1 ] [ ]

1

1 1 [1 ][ ]

1 1

(

p

STp

pp

STp

p

p

p

pp

p T

p

KeH s G s

s s

KeZ HG s Z

s s s

Kz Z

s s

K z Zs s

K zz e z

K

1

1

1 )

1

p

p

T

T

e z

e z

Step Response for 1st order lag system

tKty

eKnTy

ze

K

z

K

zez

zeK(z)y

zzc

ssc

p

nTp

pp

p

p

p

p

p

to)(

]1[)(

)1()1(

)1)(1(

)1(ˆ

1

1)(ˆ

1)(

11

11

1

1

y(t)

time

＊

＊

＊

＊＊

＊ ＊ ＊＊ ＊＊ ＊

Note: Compare with discrete approximation to First-order system

From tables, for

Generalization

mm

kk

mnmn

nknknnn

zzcbzzcbzzcb

zzeazezazeazc

cbcb

cbeaeaeac

)(ˆ...)(ˆ)(ˆ

)(ˆ...)(ˆ)(ˆ)(ˆ

...

...

22

11

110

22

11110

)(...1

...

)(ˆ)(ˆ

22

11

22

110 zD

zbzbzb

zazazaa

ze

zcm

m

kk

or

D (z)=Transfer function relating e and c

Analogous to Laplace transfer

Discrete time input/output model

Remark：Note that D(z) is the z-transform of the response of the system to an impulse input 1)(ˆ ze

Z-transform of a continuous process with Sample and Hold

HoldH (s)

ProcessGp (s)

discrete input

c*(s)

continuousvariables

y (s) y*(s)

discreteoutput

we seek a relationship (Z-transfer function) between c and y.

Consider a impulse input c*(z)=1 c*(s)=1

)(

)]([

)(ˆ

)()(

)()()()(

zHG

sGsHZ

syZzy

sGsH

scsGsHsy

P

p

p

p

HGp(z) called the pulse transfer function (since it represents the z-transform of the pulse response of Gp (s) )

Then

Properties of pulse Transfer Function

1.

2.An impulse input is converted into a pulse input by the first order hold element . Hence HG(z) is the pulse response of G(s) sampled at z internals of T.

3.The pulse transfer function of two systems in series can be combined if there is a sample and hold in between.

)()()()( 2121 sGZsGZsGsGZ

)()(

)(1

1

2 zGzc

zc )(

)(2

)(2

3 zGzc

zc)(

)(

)(3

1

3 zGzc

zc

G1(z) G2(z)c1

Tc3

c2

Closed-Loop System)(ˆ ze

)(ˆ zc

)(2 sy

D (z) H (s) Gp (s)

set point

Hold Processdisturbance

ysp (z)

+

-

T

m (s)

T

y(2)y1(s)

sampled output)(ˆ zy

)(ˆ)(ˆ)()()(

)(ˆ)(ˆ)()(

)(ˆ)(ˆ)(

)()(

)()()()(ˆ

2

2

2

21

21

zyzyzyzDzHG

zyzezDzHG

zyzezHG

syZsyZ

sysyZsyZzy

spp

p

p

)()(1

)(ˆ

)()(1

)()()()(ˆ 2

zDzHG

zy

zDzHG

zyzDzHGzy

pp

spp

or

1. Roots of the Characteristic equation 1+HGp(z)D(z)=0 Determine stability of the closed-loop system2. Note similarity to continuous system.

Example: closed-loop response of a first-order system

1/

1/

1

)1()(

1)(

ze

zekzHG

S

ksG

p

p

T

T

pp

p

pp

)(ˆ11

)1(

1

)1(1

)(ˆ1

)1(

)(ˆ

)(

1

1

1/

1/

1/

1/

zyzkkbkk

zbkk

kze

zek

zykze

zek

zy

kzD

spcpcp

cp

cT

T

p

spcT

T

p

c

p

p

p

p

pTeb /

For proportional control

where

cp

pp

nT

cp

cp

cpcp

cp

kk

ekk

kknTy

zzkkbkk

zbkkzy

p

1

11

)(

)1(

1

)]1([1

)1()(ˆ

/

11

1

For a unit step change in set point

and

11

1)(ˆ

zzysp

The response is very similar to continuous control.

The steady state value of y(t) is

Hence the offset is

cp

pc

KK

KKy

1

cppc

pc

KKKK

KKoffset

1

1

11

Stability of Discrete Systems

0 zb.....zbzb1 -nn

-22

-11

1n

n1

2

21

1

1

nn

11

mm

110

zP1

c

zP1

c

zP1

c

zbzb1

zazaa

(z)e

(z)cD(z)

A system is consider to be stable if output remains bounded for boundedInputs. Consider a discrete system with transfer function

Where P1,P2,…,Pn are n roots of:

1) and between(-1 bounded always is termsecond The

eee

eee

jw)ln(PPkln

e|P|jβαPLet

eC(nT)f

zP-1

C(z)f

njw)nln(PnlnP

jw)ln(PlnP

K

jwKK

nlnPKk

1K

kK

KK

KK

K

termk heconsider t

D(z)C(z) and 1(z)eimput impulsean For th

bounded.remain willPplane,

complex on the circle-unit e within thlies P if:Hence

n as e and 0|P|ln then 1|P|

1e and 0|P|ln then 1|P|

n as 0e and 0|P|ln then 1|P|

IF

K

K

nlnPKK

|P|nlnKK

|P|nlnKK

K

K

K

plane"complex in the circleunit on theor

inside lies poles its all if stable is system discreteA "

Im

Unstable roots

real

Unit circle

STABLE

REGION

yoscillator:0y(t) then 0P1-

decay lexponentia:0y(t) then 1P0

tas y(t)1,|P| if

Py(nT)

Py(1)

....zPzP1zP-1

1(z)y

zP-1

1D(z)

withsystem a of response impule heConsider t

1

i

1

na1

1

2-21

1-11

1

11

poles of Location

Example: Stability of closed-loop

1][eKKe

KK- )eKK(1

KK-)bKK(1P

0)]zKKb(1-K[K1

PP

P

τ

T-

CPτ

T-

CPτ

T-

CP

CPCP1

1-CPCP

equation sticcharacteri Processorder -First of control lProporhona

Example: Stability of closed-loop - Continued

T of values highor K of values high by caused be may instabitythat note

where circle unti the cross will p

C

1

P

P

P

PP

PP

τ

T-

τ

T-

CP

CPτ

T-

τ

T-

CPτ

T-

τ

T-

CPτ

T-

e-1

e1KKor

-1KK 1 e sinceor

]e-[1KK-)e-1or(

]e-[1KK-e1

t

C 0 st 0

nD

n C n k n n-1 sk 0 I

n-1 C

1 de C(t) K [e(t) e(t)dt τ ] C

τ dt

τT C K [e (e ) (e -e )] C

τ T

C K [

1.Digital Apperoximation of classical controllers

Displacement form

Alternative form :

Velocity formn 1

Dn-1 k n-1 n-2 s

k 0 I

D D Dn C n C n-1 C n-2

I

-1 -2D D DC

I

τTe e (e -e )] C

τ T

τ 2τ τT ΔC K [1 ]e -K [1 ]e K e

τ T T T

τ 2τ τΔ τ(z) T D(z) K [(1 )-(1 )z z ]

τ T T Te(z)

Digital Feedback - Control

1. No initialization is necessary. [ Cs is not needed ] Bumpless transfer from manual / automatic

2. Automatic ‘reset-windup’ protection.3. Protection in case of computer failure

1. Since different modes are indistinguishable, on-line tuning methods will not work.

2. Difficult to put constraints on integral and / or derivative term.

1. Ziegler – Nichols

2. Cohen – Coon settings

3. Time - integral performance criteria

Disadvantages:

Tuning Digital Controllers:

Advantages of velocity Form

:responses Possible

controller Prototype Minimalor Deadbeat

1-

1-

1-

1-

1-sp

sp

z-1

z

HG(z)

1D(z)

:D(z)for solving

sampling oneafter 1 togoes ex.y(t) z-1

z)(y

want we

change-step z-1

1yLet

(z)yHG(z)D(z)1

HG(z)D(z)(z)y

have we

:changepoint set a following samplingfrist at theerror no

values)pledoutput(sam thesuch that designed is law-control This

2.

z

Y(t)

actual

time

ideal

Derivation of Deadbeat Controller- Continued

sp

1

sp 1 1

1

sp

-1

-1

-1

-1

HG(z)D(z)y (z) y (z)

1 HG(z)D(z)

1ˆy ;

1 1ˆ( ) HG(z)D(z)

y 1 1 HG(z)D(z)

z HG(z)D(z)

1-z 1

z 1

1-z HG(z)

zz y z

z z

y z z

z

D z

Deadbeat

Deadbeat

0~10 sec

Deadbeat control for (1/(s+1)3)

Sampling time: 2

321

321

1

1

2

23

1

1

01584.02915.0016.03233.0

002479.005495.0406.01

101584.03073.03233.0

002479.005495.0406.0

1)(

1

ZZZ

ZZZ

Z

Z

ZZ

ZZZ

Z

Z

ZHGZD

Ringing and Pole-placement

Ringing refers to excessive value movement caused by a widely oscillating controller output.

Caused by negative poles in D(z).

Hence avoid poles near -1.

Change controller design such that poles are on the side or near zero on negative side

SYS = TF(1,[1 3 3 1]) Transfer function: 1 --------------------- s^3 + 3 s^2 + 3 s + 1 >> sysd=c2d(SYS,2) Transfer function: 0.3233 z^2 + 0.3073 z + 0.01584 -------------------------------------- z^3 - 0.406 z^2 + 0.05495 z - 0.002479 Sampling time: 2 0.3233 0.6306 0.3231 0.0158

>> p1=[1 -1];p2=[0.3233 0.3073 0.01584]

p2 =

0.3233 0.3073 0.0158

>> c=conv(p1,p2)

c =

0.3233 -0.0160 -0.2915 -0.0158

Canceling the ringing pole at z=-0.8958

ans =

1.0000 -0.8958 -0.0547

>> p1=[1 0];p2=[1 -0.99];p3=[1 0.0547]; >> c=conv(p1,p2)

c =

1.0000 -0.9900 0

>> c=conv(c,p3)

c =

1.0000 -0.9353 -0.0542 0

Warning: Using a default value of 1 for maximum step size. The simulation step size will be limited to be less than this value.

>>

Smoothing the Control Action

p=[0.3233 -0.016 -0.2915 -0.01584]; r=roots(p) r = 1.0001 -0.8959 -0.0547 Delete the unstable pole z=-0.8959

p1=[1 -0.99]; p2=[1 0.0547]; c=conv(p1,p2)

c =

1.0000 -0.9353 -0.0542

Reconstruct the Control Loop

Zero-OrderHold

1

den(s)

Transfer FcnSubtract

Step

Scope1

Scope

num(z)

den(z)

DiscreteTransfer Fcn

0 5 10 15 20 25 30 35 40 45 500

0.2

0.4

0.6

0.8

1

1.2

1.4

The treatment of unstable poles sysd=c2d(SYS,1) Transfer function: 0.0803 z^2 + 0.1544 z + 0.01788 ----------------------------------- z^3 - 1.104 z^2 + 0.406 z - 0.04979

>> p2=[ 0.0803 0.1544 0.01788]

p2 =

0.0803 0.1544 0.0179

>> p1=[1 -1]

p1 =

1 -1

>> c=conv(p1,p2)

c =

0.0803 0.0741 -0.1365 -0.0179

The treatment of unstable poles

>> roots(c)

ans =

-1.7990 1.0000 -0.1238

>> p1=[1 0];p2=[1 -0.99];p3=[1 0.1238]; >> c=conv(p1,p2)

c =

1.0000 -0.9900 0

>> c=conv(c,p3)

c =

1.0000 -0.8662 -0.1226 0

0 5 10 15 20 25 30 35 40 45 500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

3.Dahlin’s Method

Require that the closed–loop system behave like a first-order system with dead-time.

response) (step S

1

1S

e y(S)

S -

)ze-)(1z-(1

)ze-(1z (z)y

1-T/-1-

-1-T/k-

1-sp z-1

1(z)y

1-k-T/-1-T/-

1--k-T/

)ze-(1-ze-1

)ze-(1

HG(z)

1 D(z)

1. Choose , such that D(z) is realizable2. Lot of algebra

Solving for D

we want

for

Dahlin’s Method

Dahlin’s Method

0~10sec0~50 sec

Dahlin’s Method for for (1/(s+1)3)

Sampling time: 2

4321

4321

11

1

2

23

11

1

01001.02001.03016.01884.03233.0

001567.003474.02566.06321.0

)1(1

)1(

01584.03073.03233.0

002479.005495.0406.0

)1(1

)1(

)(

1

ZZZZ

ZZZZ

ZeZe

Ze

ZZ

ZZZ

ZeZe

Ze

ZHGZD

k

TT

k

T

k

TT

k

T

Regulatory Control

Consider a process described by

yn = a1yn-1 + a2yn-2 + … + b1mn-1 + … + bk mn-k

In regulatory control, we want to keep y close to zero in presence of disturbances.

Ideally choose mn such that

yn ≡ ysp

ysp = a1yn + a2yn-2 + … + bkyn-k+1 + b1 mn + b2mn-1 + bkmn-k+1

Or mn = -1/b1 [ ysp – a1yn – a yn-1 - akyn-k+1 – b2mn-1 + … - bkmn-k+1 ]

Remark1. Problems can arise in practice if model parameters are not known

2. The above choice is equivalent to minimizing

3. If dead-time is present control will be unrealizable

N

0n

2spn )y-(y

N

1 P