3. Systems and Transfer function

Discrete-time system revision• Discrete-time system• A/D and D/A converters• Sampling frequency and sampling theorem• Nyquist frequency• Aliasings• Z-transform & inverse Z-transform• The output of a D/A converter

3.1 Zero-order-hold (ZOH)

A Zero-order hold in a system

x(t) x(kT) h(t)Zero-orderHold

Sampler

3.1 Zero-order-hold (ZOH)

How does a signal change its form in a discrete-time system?

The input signal x(t) is sampled at discrete instants and the sampled signal is passed through the zero-order-hold (ZOH). The ZOH circuit smoothes the sampled signal to produce the signal h(t), which is a constant from the last sampled value until the next sample is available. That is

TtforkTxtkTh 0 ),()(

3.1 Zero-order-hold (ZOH)

Transfer function of Zero-order-holdThe figure below shows a combination of a

sampler and a zero-order hold.

x(t) x(kT) h(t)Zero-orderHold

Sampler

3.1 Zero-order-hold (ZOH)

Assume that the signal x(t) is zero for t<0, then the output h(t)is related to x(t) as follows:

0k

1)T)](k-1(t-kT)-x(kT)[1(t

)]2(1)(1)[()](1)(1)[0()( TtTtTxTttxth

h(t)

t

3.1 Zero-order-hold (ZOH)

As

The Laplace transform of the above equation becomes

000

)1(

)1(

00

0k

11

][1)T)](k-1(t-kT)-1(t[

1)T)]](k-1(t-kT)-x(kT)[1(t[)]([

k

kTsTs

k

kTsTs

k

TskkTs

Tsk

k

kTs

k

x(kT)es

ee

s

ex(kT)

s

eex(kT)

s

e

s

ex(kT)x(kT)L

LthL

kTs

kTskTs

ekTtLtL

s

ekTtLesFkTtfL

)( ,1)]([

)](1[ ,)()]([

3.1 Zero-order-hold (ZOH)

As

Therefore

Finally, we obtain the transfer function of a ZOH as

)(11

)]([)(0

sXs

ex(kT)e

s

ethLsH

Ts

k

kTsTs

00

)()()()(k

kTs

k

ekTxkTtkTxLsX

s

e

sX

sHsGsG

Ts

h

1

)(

)()()( 0

3.1 Zero-order-hold (ZOH)

There are also first-order-hold and high-order-hold although they are not used in control system.

3.1 Zero-order-hold (ZOH)

A zero-order-hold creates one sampling interval delay in input signal.

3.1 Zero-order-hold (ZOH)

First-order-hold

T

Ts

s

esG

Ts

h

11)(

2

1

3.1 Zero-order-hold (ZOH)

First-order-hold and high-order-hold does not bring us much advantages except in some special cases. Therefore, in a control system, usually a ZOH is employed. The device to implement a ZOH is a D/A converter.

If not told, always suppose there is a ZOH in a digital control system.

3.2 Plants with ZOH

Given a discrete-time system, the transfer function of a combination of a ZOH and the plant can be written as GHP(z) in Z-domain. HP, here, means the ZOH and the Plant.

ZOH GP(s)

GHP(z)

3.2 Plants with ZOH

The continuous time transfer function GHP(s)=G0(s)GP(s)

The discrete time transfer function

sT1

11

ez ,)(

)1()(

)()1(

)()(

)()()()1()()(

s

sGZzzG

s

sGZz

s

sGZz

s

sGZ

s

sGeZ

s

sGZ

s

sGeZsGZzG

PHP

PPP

PsTPPsTHPHP

3.2 Plants with ZOH

Example 1: Given a ZOH and a plant

Determine their Z-domain transfer function.1

1)(

s

sGP

1

1

11

111

111

11

11

1

)1(

)1)(1(

11)1(

1

1

1

1)1(

1

11)1(

1

11)1(

)1(

1)1(

)()1()(

ze

ze

zez

zzez

zezz

sZ

sZz

ssZz

ssZz

s

sGZzzG

T

T

T

T

T

PHP

3.2 Plants with ZOH

Example 2: Given a ZOH and a plant

Determine their z-domain transfer function.)1(

1)(

sssGP

)1)(1(

)1()1(

1

1

1

1

)1()1(

1

111)1(

1

111)1(

)1(

1)1(

)()1()(

11

21

1121

11

21

21

211

zez

zTeezeT

zezz

Tzz

sZ

sZ

sZz

sssZz

ssZz

s

sGZzzG

T

TTT

T

PHP

3.2 Plants with ZOH

Exercise 1: Given a ZOH and a plant

Determine their z-domain transfer function.

Answer:

))(()(

bsas

absGP

111

1

111

1)(

ze

a

ze

b

z

ba

ba

zzG

aTaTHP

Assignment 1

You are required to implement a digital PID controller which will enable a control object with a transfer function of

where K=0.2, n=10 rad/s, and =0.3.to track a) a unit step signal, and b) a unit ramp

signal.1) Simulate this control object and find the responses using

Matlab or other packages/computer languages.

22

2

2)(

nn

n

ss

KsG

Assignment 1

2) Choose a suitable sample period for a control loop for G(s) and explain your choice.

3)* Derive the discrete-time system transfer function GHP(z) from G(s).

4) Design a digital PID controller for the discrete-time system, and optimize its parameters with respect to the performance criterion below using steepest descent minimization process .

5) Simulate the resulting closed-loop system and find the responses. Swapping the input signals a) and b), discuss the resulting responses.

M

kkekS

0

3.3 Represent a system in difference equation

For we have

Let A=1-e-T and B=e-T, then the transfer function can be rewritten as

1

1

1

)1()(

ze

zezG

T

T

HP

)1()1()(

)1()1()(

)()()()()1)((

)(

)(

11

)1()(

111

1

1

1

1

kBykAxky

kAxkByky

zzAXzzBYzYzAXBzzY

zX

zY

Bz

Az

ze

zezG

T

T

HP

1

1)(

s

sGP

3.3 Represent a system in difference equation

Simulate the above system1) Parameters and input: A=1-e-T, B=e-T , x(k)=1

2) initial condition: x(k-1)=0, y(k)=y(k-1)=0, k=0

3) Simulation

While k<100 do

y(k)=Ax(k-1)+By(k-1); Calculate output

x(k-1)=x(k); y(k-1)=y(k); x(k)=1; k=k+1; Update data

print k, x(k), y(k); Display step, input & output

End

3.3 Represent a system in difference equation

Let T=1, we have A=0.6321 and B=0.3679

For a unit step input, the response is

y(k)=0.6321x(k-1)+0.3679y(k-1)

k= 0 1 2 3 4 5

x(k) 1 1 1 1 1 1

y(k) 0 0.6321 1 1 1 1

3.3 Represent a system in difference equation

Time (sec.)

Am

plitu

de

Step Response

0 1 2 3 4 5 6

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Assignment 1

1)* Simulate this control object and find the responses using Matlab or other packages/computer languages.

Hints: Method 1

22

2

2)(

nn

n

ss

KsG

)(2

)]()([

)(2

)()()(

)(

)(

2)(

22

211

22

2

22

2

sXss

KLsXsGLy(t)

sXss

KsXsGsY

sX

sY

ss

KsG

nn

n

nn

n

nn

n

Assignment 1

Hints: Method 2

2,1,0

))(())(()]([

))(())(()(

))(()1)((

12)(

)(

)(

)()]([)( ,

)(

)(

2)(

22

11

22

110

11

22

11

22

110

22

110

22

11

22

11

22

110

22

2

22

2

k

zazazYzbzbbzXLzYLy(kT)

zazazYzbzbbzXzY

zbzbbzXzazazY

zaza

zbzbb

ss

KZ

zX

zY

zX

zYsGZzG

sX

sY

ss

KsG

nn

n

nn

n

3.4 System stability

We can rewrite the difference equation as

If A=1 and =0.9, for an impulse input we have

k 0 1 2 3 4 ...x(k) 1 0 0 0 0 ...

y(k) 0 1 0.9 0.81 0.729 …It decreases exponentially, a stable system.

ABkykxA

kyA

BkxAkBykAxky

)),1()1((

))1()1(()1()1()(

))1(9.0)1()( kykxky

3.4 System stability

If K=1 and =1.2, we have

k 0 1 2 3 4 ...

x(k) 1 0 0 0 0 ...

y(k) 0 1 1.2 1.44 1.728 2.074…

It increases exponentially, an unstable system.

))1(2.1)1()( kykxky

3.4 System stability

If K=1 and = -0.8, we have

k 0 1 2 3 4 ...

x(k) 1 0 0 0 0 ...

y(k) 0 1 -0.8 0.64 -0.512 …

It decays exponentially, and alternates in sign, a gradual stable system.

))1(8.0)1()( kykxky

3.4 System stability

It is clear that the value of determines the system stability. Why is so important?

First, let A=1, we have

From the transfer function, we can see that z= is a pole of the system. The pole of the system will determine the nature of the response.

zz

z

zX

zYzG

zzYzzXzYkykxky

1

1)(

)()(

)()()()1()1()(

1

1

11

3.4 System stability

For continuous system, we have stable, critical stable and unstable areas in s domain.

Stable area Unstable area

Critical stable area

3.4 System stability

What is the stable area, critical stable area and unstable area for a discrete system in Z domain ?

Stable area: unit circle

Critical stable: on the unit circle

Unstable area: outside of the unit circle

3.4 System stability

As

For the critical stable area in s domain s=j,

As is from 0 to , then the angle will be greater than 2. That is the critical area forms a unit circle in Z domain.

sT1 ez ,)(

)1()(

s

sGZzzG P

HP

1)(sin)(cos

sincosez

22

sT je j

3.4 System stability

If we choose a point from the stable area at S domain, eg s=- a + j, we have

Let eg s=- + j

The stable area in Z domain is within a unit circle around the origin.

0)(sin)(cos

)sin(cosez

22

sT

ee

jee j

aa

aja

ee

jee

22

sT

)(sin)(cos

)sin(cosez

3.4 System stability

Exercise 2: Prove that the unstable area in Z domain is the area outside the unit circle.

Hint: Follow the above procedures.

3.4 System stability

Z domain responses

0

1

3.5 Closed-loop transfer function

Computer controlled system

Gc(z) ZOH GP(s)R(z)E(z)

M(z)

GHP(z)

Computer system

C(z)

Plant

3.5 Closed-loop transfer function

Let’s find out the closed-loop transfer function

)()(1

)()(

)(

)()(

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

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

)()()(

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

zGzG

zGzG

zR

zCzT

zGzGzRzGzGzCzC

zGzGzCzRzC

zCzRzE

zGzGzEzGzMzC

HPC

HPC

HPCHPC

HPC

HPCHP

3.5 Closed-loop transfer function

C(z): output; E(z): error

R(z): input; M(z): controller output

GC(z): controller

GP(z)/G(z): plant transfer function

GHP(z): transfer function of plant + ZOH

T(z): closed-loop transfer function

GC(z)GHP(z): open-loop transfer function

1+ GC(z)GHP(z)=0: characteristic equation

3.6 System block diagram

G(s)

H(s)

-

+

R(s)

C(s) C(z)

)(1

)(

)(

)(

zGH

zG

zR

zC

G(s)

H(s)

-

+

R(s)

C(z)

)()(1

)(

)(

)(

zHzG

zG

zR

zC

3.6 System block diagram

The difference between G(z)H(z) and GH(z)

G(z)H(z)=Z[G(s)]Z[H(s)]

GH(z)=Z[G(s)H(s)]

Usually, G(z)H(z) GH(z)

G(z)H(z) means they are connected through a sampler. Whereas GH(z) they are connected directly.

3.6 System block diagram

Example: Find the closed-loop transfer function for the system below.

Solution: The open-loop is G1(z)G2H(z).

The forward path is G1(z)G2(z).

G1(s)

H(s)

-

+

R(s)

C(z)G2(s)

3.6 System block diagram

)()(1

)()()(

)(

)()(

))()(1/()()()()()()()(

))()(1/()()(

)()()()()( );()()()()(

21

21

212121

21

2121

zHGzG

zGzGzR

zR

zCzT

zHGzGzRzGzGzGzGzEzC

zHGzGzRzE

zHGzGzEzEzRzHGzGzEzRzE

G1(s)

H(s)

-

+

R(s)

C(z)G2(s)

3.6 System block diagram

*Exercise 3: Find the output for the closed-loop system below.

G(s)

H(s)-

+

R(s)

C(s) C(z)

)(1

)()(

zGH

zGRzC

3.6 System block diagram

*Exercise 4: Find the output for the closed-loop system below.

G1(s)

H(s)

-

+

R(s)

C(z)G2(s)

)(1

)()()(

21

21

zHGG

zGzRGzC

Reading

Study book

• Module 3: Systems and transfer functions (Please try the problems on page 3.46-47)

Textbook

• Chapter 3 : Z-plane analysis of discrete-time control system (pages 74-83 & 104-114).

Tutorial

Exercise 1: Given a ZOH and a plant

Determine their z-domain transfer function.))((

)(bsas

absGP

111

1

1111

1

13211

11

111

)1(

1

1

1

1

1

1)1(

111)1(

)()(1)1()1(

))(()1(

)()1()(

ze

a

ze

b

z

ba

ba

z

zeab

a

zeba

b

zz

bsZ

ab

a

asZ

ba

b

sZz

bs

aba

as

bab

sZz

bs

k

as

k

s

kZz

bsass

abZz

s

sGZzzG

aTaT

bTaT

PHP

Tutorial

You are required to implement a digital PID controller which will enable a control object with a transfer function of

where K=0.2, n=10 rad/s, and =0.3.to track a) a unit step signal, and b) a unit ramp

signal.1) Simulate this control object and find the responses using

Matlab or other packages/computer languages.

22

2

2)(

nn

n

ss

KsG

Tutorial

2) Choose a suitable sample period for a control loop for G(s) and explain your choice.

3) Derive the discrete-time system transfer function GHP(z) from G(s).

4) Design a digital PID controller for the discrete-time system, and optimize its parameters with respect to the performance criterion below using steepest descent minimization process .

5) Simulate the resulting closed-loop system and find the responses. Swapping the input signals a) and b), discuss the resulting responses.

M

kkekS

0

Tutorial

2) Choose a suitable sample period for a control loop for G(s) and explain your choice.

a. Sampling theoremb. Input signalc. Bandwidth of a systemd. Bold plotse. Applying sampling theoremf. Sampling frequency