ELG4157: Digital Control Systems

amaranthgymnophoriaElectronics - Devices

Nov 15, 2013 (3 years and 6 months ago)

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ELG4157: Digital Control Systems


Discrete Equivalents

Z
-
Transform

Stability Criteria

Steady State Error

Design of Digital Control Systems




1

Advantages and Disadvantages


Improved sensitivity.


Use digital components.


Control algorithms easily
modified.


Many systems inherently
are digital.


Develop complex math
algorithms.


Lose information during
conversions due to
technical problems.


Most signals continuous
in nature.


Digitization


The

difference

between

the

continuous

and

digital

systems

is

that

the

digital

system

operates

on

samples

of

the

sensed

plant

rather

than

the

continuous

signal

and

that

the

control

provided

by

the

digital

controller

D(s)

must

be

generated

by

algebraic

equations
.


In

this

regard,

we

will

consider

the

action

of

the

analog
-
to
-
digital

(A/D)

converter

on

the

signal
.

This

device

samples

a

physical

signal,

mostly

voltage,

and

convert

it

to

binary

number

that

usually

consists

of

10

to

16

bits
.


Conversion

from

the

analog

signal

y
(
t
)

to

the

samples

y
(
kt
),

occurs

repeatedly

at

instants

of

time

T

seconds

apart
.



A

system

having

both

discrete

and

continuous

signals

is

called

sampled

data

system
.


The

sample

rate

required

depends

on

the

closed
-
loop

bandwidth

of

the

system
.

Generally,

sample

rates

should

be

about

20

times

the

bandwidth

or

faster

in

order

to

assure

that

the

digital

controller

will

match

the

performance

of

the

continuous

controller
.

3

4

Digital Control System


ADC

Micro

Processor

DAC

Correction

Element

Process

Clock

Measurement

+

-

A

D

D

A

A: Analog

D: Digital

5

Continuous Controller and Digital Control


G
c
(
s
)

Plant

R
(
t
)

y(
t
)

Continuous Controller

+

-

A/D

Digital

Controller

D/A and

Hold

Plant

D/A

+

-

r
(
t
)

Digital Controller

y
(
t
)

r
(
kT
)

p
(
t
)

m
(
t
)

m
(
kT
)

6

Applications of Automatic Computer

Controlled Systems


Most

control

systems

today

use

digital

computers

(usually

microprocessors)

to

implement

the

controllers)
.

Some

applications

are
:


Machine

Tools


Metal

Working

Processes


Chemical

Processes


Aircraft

Control


Automobile

Traffic

Control


Automobile

Air
-
Fuel

Ratio


Digital

Control

Improves

Sensitivity

to

Signal

Noise
.


7

Digital Control System


Analog electronics can integrate and differentiate signals. In order
for a digital computer to accomplish these tasks, the differential
equations describing compensation must be approximated by
reducing them to algebraic equations involving addition, division,
and multiplication.


A digital computer may serve as a compensator or controller in a
feedback control system. Since the computer receives data only at
specific intervals, it is necessary to develop a method for describing
and analyzing the performance of computer control systems.


The computer system uses data sampled at prescribed intervals,
resulting in a series of signals. These time series, called sampled
data, can be transformed to the
s
-
domain, and then to the z
-
domain
by the relation
z

=
e
zt
.


Assume that all numbers that enter or leave the computer has the
same fixed period
T
, called the sampling period.


A sampler is basically a switch that closes every
T

seconds for one
instant of time.

8

r
(
t
)

r
*(
t
)

Continuous

Sampled

Sampler

r
(
T
)

r(2
T
)

r
(3
T
)

r
(
kT
)

r
(4
T
)

0

T

2T

3
T

T

2T

3
T

4
T

4
T

Zero
-
order

Hold

G
o
(
s
)

P
(
t
)



s
e
e
s
s
s
G
sT
sT






1
1
1
)
(
0

D/A

A/D

Computer

Proces
s

Measure

r(
t
)

c
(
t
)

e
(
t
)



e*
(
t
)

u
*(
t
)

u
(
t
)


Sampling analysis


Expression of the sampling signal

Modeling of Digital Computer

)
(
)
(
)
(
)
(
)
(
)
(
)
(
*
0
0
kT
t
kT
x
kT
t
t
x
t
t
x
t
x
k
k
T
















10


Analog to Digital Conversion: Sampling


An input signal is converted from continuous
-
varying
physical value (e.g. pressure in air, or frequency or
wavelength of light), by some electro
-
mechanical device
into a continuously varying electrical signal. This signal has
a range of amplitude, and a range of frequencies that can
present. This continuously varying electrical signal may
then be converted to a sequence of digital values, called
samples, by some analog to digital conversion circuit.


There are two factors which determine the accuracy with which the
digital sequence of values captures the original continuous signal: the
maximum rate at which we sample, and the number of bits used in
each sample. This latter value is known as the quantization level

11

Zero
-
Order Hold


The

Zero
-
Order

Hold

block

samples

and

holds

its

input

for

the

specified

sample

period
.



The

block

accepts

one

input

and

generates

one

output,

both

of

which

can

be

scalar

or

vector
.

If

the

input

is

a

vector,

all

elements

of

the

vector

are

held

for

the

same

sample

period
.


This

device

provides

a

mechanism

for

discretizing

one

or

more

signals

in

time,

or

resampling

the

signal

at

a

different

rate
.



The

sample

rate

of

the

Zero
-
Order

Hold

must

be

set

to

that

of

the

slower

block
.

For

slow
-
to
-
fast

transitions,

use

the

unit

delay

block
.


12

The
z
-
Transform

The z
-
Transform is used to take discrete time domain signals into a complex
-
variable frequency domain. It plays a similar role to the one the Laplace
transform does in the continuous time domain. The z
-
transform opens up new
ways of solving problems and designing discrete domain applications. The z
-
transform converts a discrete time domain signal, which is a sequence of real

numbers
, into a complex frequency domain representation.




























0
0
0
0
)
(
)
(
)}
(
{
1
)
(
)
(
)}
(
*
{
)}
(
{
)
(
)}
(
*
{
have

we
s,
transform
Laplace

the
Using
0,



signal

a
For
)
(

)
(
)
(
*
k
k
k
k
sT
k
ksT
k
z
kT
f
z
F
t
f
Z
z
z
z
U
z
kT
r
t
r
Z
t
r
Z
e
z
e
kT
r
t
r
t
kT
t
kT
r
t
r

13

Transfer Function of Open
-
Loop System

Zero
-
order

Hold
G
o
(
s
)

Process

r
(
t
)

T
=1

r*
(
t
)



3678
.
0
3678
.
1
2644
.
0
3678
.
0
)
(
)
1
1
1
1
(
1
)
(

:
fraction

partial

into

Expanding
)
1
(
1
)
(
)
(
)
(
)
(
*
)
(
)
1
(
1
)
(

;
)
1
(
)
(
2
2
2





















z
z
z
z
G
s
s
s
e
s
G
s
s
e
s
G
s
G
s
G
s
R
s
Y
s
s
s
G
s
e
s
G
st
st
p
o
p
st
o
14























n
i
T
a
i
n
n
n
i
e
z
z
K
z
X
Then
a
s
K
a
s
K
a
s
K
a
s
a
s
a
s
A(s)
X(s)
If
1
2
2
1
1
2
1
)
(

:
)
(
)
)(
(

:
Example:

T
T
e
z
z
e
z
z
z
z
s
s
s
Z
s
s
s
s
Z
2
5
15
1
10
2
5
1
15
10
)
2
)(
1
(
)
4
(
5




























Z
-
Transform

Z
-
transform method: Partial
-
fraction expansion approaches

Inverse Z
-
transform method: Partial
-
fraction expansion approaches

























n
i
kT
a
i
T
a
T
a
T
a
T
a
T
a
i
n
e
K
kT
X
then
e
s
z
K
e
z
z
K
e
s
e
z
e
z
A(z)
X(z)
If
1
2
1
)
(

:
)
(
)
)(
(

:
2
1
2
1
Example:

kT
T
T
T
e
e
z
z
z
z
Z
e
z
z
e
z
Z
kT
x
2
2
1
2
2
1
1
1
)
)(
1
(
)
1
(
)
(






























16

Closed
-
Loop Feedback Sampled
-
Data Systems

G
(z)

r
(
t
)

R
(
z
)

E
(
z
)

Y
(
z
)

Y
(
z
)

)
(
)
(
1
)
(
)
(
)
(
1
)
(
)
(
)
(
)
(
z
D
z
G
z
D
z
G
z
G
z
G
z
T
z
R
z
Y





G
(z)

R
(
z
)

E
(
z
)

Y
(
z
)

Y
(
z
)

D
(z)

17

Now Let us Continue with the Closed
-
Loop System for the
Same Problem

5
4
3
2
1
2
3
2
2
2
147
.
1
4
.
1
4
.
1
3678
.
0
)
(
6322
.
0
6322
.
1
2
2644
.
0
3678
.
0
)
6322
.
0
)(
1
(
)
2644
.
0
3678
.
0
(
)
(
1
)
(

:
input

step
unit

a
an

Assume
6322
.
0
2644
.
0
3678
.
0
)
(
1
)
(
)
(
)
(




























z
z
z
z
z
z
Y
z
z
z
z
z
z
z
z
z
z
z
Y
z
z
z
R
z
z
z
z
G
z
G
z
R
z
Y
Stability


The

difference

between

the

stability

of

the

continuous

system

and

digital

system

is

the

effect

of

sampling

rate

on

the

transient

response
.


Changes

in

sampling

rate

not

only

change

the

nature

of

the

response

from

overdamped

to

underdamped,

but

also

can

turn

the

system

to

an

unstable
.


Stability

of

a

digital

system

can

be

discussed

from

two

perspectives
:


z
-
plane


s
-
plane

18

19

Stability Analysis in the
z
-
Plane


A linear
continuous

feedback control system is stable if all poles of the
closed
-
loop transfer function
T
(
s
) lie in the left half of the
s
-
plane.


In the left
-
hand
s
-
plane,




0; therefore, the related magnitude of
z

varies between 0 and 1. Accordingly the imaginary axis of the
s
-
plane
corresponds to the unit circle in the
z
-
plane, and the inside of the
unit
circle

corresponds to the left half of the
s
-
plane.


A sampled system is stable if all the poles of the closed
-
loop transfer
function
T
(
z
) lie within the unit circle of the
z
-
plane.

T
z
e
z
e
e
z
T
T
j
sT










)
(
1



Re

Im

z
-
plane

Stable zone

The graphic expression of the stability

condition for the sampling control systems

The stability criterion

In the characteristic equation 1+
GH
(
z
)=0, substitute
z
with

1
1



s
s
z
——

Bilinear transformation


We can analyze the stability of the sampling control systems the same as we did
in chapter 3 (Routh criterion in the
s
-
plane) .
































































)
(


)


(
1
0
1


0



)
1
(
2
)
1
(
1
1
1
1
1
1
1

:

,


,



:
2
2
2
2
2
2
2
2
2
2
z-plane
le of the
unit circ
inside the
e
the s-plan
of
ft half
for the le
y
x
y
x
y
x
y
j
y
x
y
x
jy
x
jy
x
jy
x
jy
x
z
z
j
s
then
jy
x
z
j
w
suppose
Proof





The Stability Analysis

Unstable zone

Critical stability

0
368
.
0
368
.
1
632
.
0
1
)
(
1
2






z
z
Kz
z
G
Determine
K

for the stable system

Solution:

0
)
632
.
0
736
.
2
(
264
.
1
632
0


0
368
.
0
368
.
1
632
.
0
1
2









K
s
Ks
.
z
z
Kz
K
K
K
.
n
h criterio
f the Rout
In terms o
632
.
0
736
.
2
264
.
1
632
.
0
736
.
2
632
0

:


We have: 0 <
K
< 4.33

1
1



s
s
z
Make


The Stability Analysis

22

Example: Stability of a closed
-
loop system


G
p
(s)

r
(
t
)

Y
(t)

G
o
(
s
)

gain.

of

values
all
for

stable
is

continuous

the
gain where

increased
for

unstable

is

system

sampled
order
-
Second
39
.
2




0

:
for

stable

is

system

This
unstable)
(

)
295
.
1
115
.
1
(

)
295
.
1
115
.
1
(
012
.
3
310
.
2
2
10


When
circle,
unit

e
within th
lie

roots

the
because

stable

is

system

The
0
)
6182
.
0
5
.
0
)(
6182
.
0
5
.
0
(
6322
.
0
2

;
1
0
)
1
(
2

:
0
G(z)]
[1
equation

the
of

roots

the
are

(z)
function t
transfer
loop
-
losed

the
of

poles

The
)
1
(
2
)
(
3678
.
0
3678
.
1
2
)
2644
.
0
3678
.
0
(
)
(
;
)
1
(
)
(






































K
j
z
j
z
z
z
K
j
z
j
z
z
z
K
Kb
Kaz
a
z
a
z
a
z
a
z
b
az
K
z
z
z
K
z
G
s
s
K
s
p
G
Example


23

The Steady State Error Analysis

)
(
1
)
(
)
(
1
)
(
)
(
)
(
)
(
)
(
)
(
z
G
z
R
z
G
z
G
z
R
z
R
z
c
z
R
z
E







G(s)

r

c



e





















*
2
*
*
1
1
1
1

)
(
1
)
(
)
1
(
lim
)
(
)
1
(
lim
a
v
p
z
z
ss
K
T
K
T
K
z
G
z
R
z
z
E
z
e
)
(
)
1
(
lim

;
)
1
(
)
1
(
)
(

)
(
)
(
)
1
(
lim

;
)
1
(
)
(

)
(
)
(
lim


;
1
)
(
)
(
1
)
(
2
1
*
3
2
2
1
*
2
1
*
z
G
z
K
z
z
z
T
z
R
t
t
r
z
G
z
K
z
Tz
z
R
t
t
r
z
G
K
z
z
z
R
t
t
r
z
a
z
v
z
p





















)
(
)
1
(
lim
1
z
E
z
e
z
ss



)
5
(
)
(

1




s
s
K
s
G
s
T
2) If
r
(
t
) = 1+
t,

determine
e
ss
=


1) Determine
K

for the stable system.

Solution










































5
25
5
5
)
1
(

)
5
(
)
1
(

)
5
(
1
)
(
2
2
s
K
s
K
s
K
Z
e
s
s
K
Z
e
s
s
K
s
e
Z
z
G
Ts
Ts
Ts
1)

)
0067
.
0
)(
1
(
2135
.
0
2067
.
2
5
25
1
5
)
1
(
5
)
1
(
2
1
5
2
1



























z
z
z
z
K
e
z
Kz
z
Kz
z
KTz
z
T
T
r



G

(s)

c

Z.O.H

e

Example

0
)
4202
.
2
067
.
10
(
)
1.573
9.993
(
0.9932


1
1
0
)
2135
.
0
0335
.
0
(
)
0335
.
5
2067
.
2
(
)
5
(


0
)
0067
.
0
)(
1
(
2135
.
0
2067
.
2
5
1
)
(
1


:
system

the
of
equation

stic
charecteri

The
2
2























K
w
K
w
s
s
z
K
z
K
z
-K
z
z
z
z
K
z
G
16
.
4
0



K
2)

K
K
T
K
T
K
e
K
z
z
z
K
z
G
z
K
z
z
z
z
K
z
G
K
v
p
ss
z
z
v
z
z
p
5
2
.
0
0

1
1
2
.
0
)
0067
.
0
(
2135
.
0
2067
.
2
5
lim
)
(
)
1
(
lim
)
0067
.
0
)(
1
(
2135
.
0
2067
.
2
5
lim
)
(
lim
1
T
*
*
2
1
1
*
2
1
1
*





























Steady State Error and System Type


1) For unity feedback in figure below,

2)

30

Design of Digital Control Systems

The

Procedure
:



Start

with

continuous

system
.


Add

sampled
-
data

system

elements
.


Chose

sample

period,

usually

small

but

not

too

small
.

Use

sampling

period

T

=

1

/

10

f
B
,

where

f
B

=


B

/

2


and


B

is

the

bandwidth

of

the

closed
-
loop

system
.


Practical

limit

for

sampling

frequency
:

20

˂


s

/


B

˃
40


Digitize

control

law
.


Check

performance

using

discrete

model

or

SIMULINK
.

31

32

Start with a Continuous Design

D
(
s
) may be given as an existing design or by using root
locus or bode design.

G
(z)

r
(
t
)

R
(
z
)

E
(
z
)

Y
(
z
)

Y
(
z
)

D
(z)

33

Add Samples Necessary for Digital Control


Transform
D
(
s
) to
D
(z): We will obtain a discrete system
with a similar behavior to the continuous one.



Include D/A converter, usually a zero
-
order
-
device.



Include A/D converter modeled as an ideal sampler.



And an antialiasing filter, a low pass filter, unity gain filter
with a sharp cutoff frequency.



Chose a sample frequency
based
on the closed
-
loop
bandwidth

B
of
the continuous system.

34

Closed
-
Loop System with Digital Computer Compensation





















b
a
K
B
A
C
e
B
e
A
z
D
s
G
Z
B
z
A
z
C
z
D
b
s
a
s
K
s
G
z
z
z
z
D
z
G
z
z
z
D
K
r
z
z
G
r
z
z
k
z
D
z
z
z
G
T
s
s
s
Gp
z
D
z
E
z
U
z
D
z
G
z
D
z
G
z
T
z
R
z
Y
bT
aT
c
c




































1
1

;
;
);
(
)}
(
{
;
)
(

;
)
(
240
.
0
1
7189
.
0
5
.
0
)
(
)
(

;
240
.
0
7189
.
0
359
.
1
)
(
.

and


parameters

two
the
have

and

3678
.
0
at
)
(

of

pole

cancer the

We
)
(
)
3678
.
0
(
)
(
select

we
If

;
3678
.
0
1
0.7189
z
0.3678
)
(
1;

when
)
1
(
1
)
(
plant

a

and

hold
order
-
zero

a
with
system
order

second

he
Consider t
)
(
)
(
)
(

is
computer

the
of
function
tranfer
The
)
(
)
(
1
)
(
)
(
)
(
)
(
)
(
35

Compensation Networks (10.3; page
747)

The compensation network,
G
c
(
s
) is cascaded with the unalterable process
G
(
s
) in order to provide a suitable loop transfer function
G
c
(
s
)
G
(
s
)
H
(
s
).

G
(s)

R
(s)

G
c
(
s
)

+

-

H
(
s
)

Y(s)

Compensation



network

lead
-
phase

a

called

is
network

the
p,
z
When
r
compensato
order
First
)
(
)
(
)
(
)
(
)
(
)
(
1
1












p
s
z
s
K
s
G
p
s
z
s
K
s
G
c
N
j
i
M
i
i
c
j




-
z

-
p

36

Closed
-
Loop System with Digital Computer Compensation

There are two methods of compensator design:

(1)
G
c
(
s
)
-
to
-
D
(
z
) conversion method, and

(2) Root locus z
-
plane method.



The
G
c
(
s
)
-
to
-
D
(
z
) Conversion Method





0

when
1
1

;
;
transform)
-
(z

)
(
)}
(
{
)
Controller

(Digital

)
(

r)
Compensato
Order
-
(First

)
(















s
b
a
K
B
A
C
e
B
e
A
z
D
s
G
Z
B
z
A
z
C
z
D
b
s
a
s
K
s
G
bT
aT
c
c
The Frequency Response

The frequency response of a system is defined as the
steady
-
state response of the system to a sinusoidal input
signal.

The sinusoid is a unique input signal, and the resulting
output signal for a linear system, as well as signals
throughout the system, is sinusoidal in the steady
-
state; it
differs form the input waveform only in amplitude and
phase.


37

Phase
-
Lead Compensator Using Frequency Response


A

first
-
order

phase
-
lead

compensator

can

be

designed

using

the

frequency

response
.

A

lead

compensator

in

frequency

response

form

is

given

by







In

frequency

response

design,

the

phase
-
lead

compensator

adds

positive

phase

to

the

system

over

the

frequency

range
.

A

bode

plot

of

a

phase
-
lead

compensator

looks

like

the

following



G
c
s
(
)
1



s





1

s





p
1

z
1


m
z
p

sin

m



1


1

Phase
-
Lead Compensator Using Frequency Response


Additional positive phase increases the phase margin and
thus increases the stability of the system. This type of
compensator is designed by determining alfa from the
amount of phase needed to satisfy the phase margin
requirements.


Another effect of the lead compensator can be seen in the
magnitude plot. The lead compensator increases the gain of
the system at high frequencies (the amount of this gain is
equal to alfa. This can increase the crossover frequency,
which will help to decrease the rise time and settling time of
the system.


Phase
-
Lag Compensator Using Root Locus


A first
-
order lag compensator can be designed using the root locus. A lag
compensator in root locus form is given by






where the magnitude of z is greater than the magnitude of p. A phase
-
lag
compensator tends to shift the root locus to the right, which is undesirable. For this
reason, the pole and zero of a lag compensator must be placed close together
(usually near the origin) so they do not appreciably change the transient response
or stability characteristics of the system.



When a lag compensator is added to a system, the value of this intersection will be
a smaller negative number than it was before. The net number of zeros and poles
will be the same (one zero and one pole are added), but the added pole is a
smaller negative number than the added zero. Thus, the result of a lag
compensator is that the asymptotes' intersection is moved closer to the right half
plane, and the entire root locus will be shifted to the right.

G
c
s
(
)
s
z

(
)
s
p

(
)
Lag or Phase
-
Lag Compensator using Frequency Response


A first
-
order phase
-
lag compensator can be designed using the frequency
response. A lag compensator in frequency response form is given by






The phase
-
lag compensator looks similar to a phase
-
lead compensator, except
that a is now less than 1. The main difference is that the lag compensator adds
negative phase to the system over the specified frequency range, while a lead
compensator adds positive phase over the specified frequency. A bode plot of a
phase
-
lag compensator looks like the following


G
c
s
(
)
1



s





1

s





42

Example: Design to meet a Phase Margin
Specification

Based on Chapter 10 (
Dorf
): Example 13.7



differ!

would
)
(

of
t
coefficien

the
the
period,

sampling

for the

lue
another va
select

we
If
)
73
.
0
(
)
95
.
0
(
85
.
4
)
(
4.85;

and

0.73,
e

,
95
.
0
have

We
second.

0.001


Set

).
(
by

realized

be

to
is

)
(
r
compensato

the
Now
5.6.


Then

rad/s.

125
When
1
)
(

yield

order to
in

select

We
)
312
(
)
50
(
)
(
;
312

and

;
50

;

10.18).

(Eq

6.25

is

ratio

zero
-
pole

required

that the
find

we
10.4,
on

Based
10.24).

(Eq

2

is
margin

phase

that the
find

we
),
(

of

diagram

Bode

the
Using
10.10).

(Fig

rad/s

125
frequeny
crossover

a
with
45

of
margin

phase

a

achieve

that we
so

)
(
design

attempt to

will
We
.
)
1
25
.
0
(
1740
)
(
0.312
-
05
.
0
2
1
o
c
o
z
D
z
z
z
D
C
B
e
A
T
z
D
s
G
K
j
ω
GG
K
s
s
K
s
G
b
a
ab
s
G
s
G
s
s
s
G
c
c
c
c
c
p
c
p





























43

The Root Locus of Digital Control Systems

D
(
z
)

Zero

Order

hold

KG
p
(
s
)

R
(
s
)

+

-

Y
(
s
)

o
k
o
z
D
z
KG
z
D
z
KG
z
D
z
KG
z
D
z
KG
z
D
z
KG
z
D
z
KG
z
R
z
Y
360
180
)
(
)
(

and

1
)
(
)
(
or

0
)
(
)
(
1

4.
axis.

real

horizontal

the
respect to
with
l
symmetrica

is

locus
root

The

3.
zeros.

and

poles

of
number

odd
an

of
left

the
to
axis

real

the
of
section

a
on

lies

locus
root

The

2.
zeros.

the
to
progresses

and

poles

at the

starts

locus
root

The

1.
K varies.

as

system

sampled

the
of
equation

stic
characteri

for the

locus
root

Plot the
equation)

istic
(Character

0
)
(
)
(
1

;
)
(
)
(
1
)
(
)
(
)
(
)
(










44

Re {
z
}

Im {
z
}

2 poles at

z

= 1

0

-
1

One zero

At z =
-
1

-
3

-
2

Root locus

1
;
3
;
0
)
(
)
(
)
1
(
)
1
(
for

solve

and

Let
0
)
1
(
)
1
(
1
)
(
1
2
1
2
2
























d
dF
F
K
K
z
z
z
K
z
KG
Unit circle

K increasing

Unstable

System
Order

Second

a

of

Locus
Root
45

Design of a Digital Controller

plane.
-
z

on the

circle
unit

in the
point with

desired

a
at
roots
complex

of
set

a

give

will
system

d
compensate

the
of

locus

that the
so

b)
-
(z
Select
plane.
-
z

the
of

axis

real

positive

on the

lies
that
G(z)
at

pole

one

cancel

to
a)
-
(z

Use
)
(
)
(
)
(

controller

a
select

will
we

method,

locus
root

a

utilizing

response

specified

a

achieve

order to
In
b
z
a
z
z
D



46

Example: Design of a digital compensator

0.8.
K
for

stable

is

system

the
Thus

0.8.


at

circle
unit

on the

is

locus
root

The
-2.56.
z

as
point
entry

obtain the

we
),
(
for
equation

the
Using
)
2
.
0
)(
1
(
)
1
(
)
(
)
(

have

we

0.2,



and

1


select

we
If
)
(
)
1
(
)
)(
1
(
)
(
)
(
)
(
Select

system.

unstable

have

we
1,
)
(
With

13.8.

Example
in

described

as

is

)
(
when
system

stable

a
in
result

that will
D(z)
r
compensato

a
design

us
Let
2


















K
F
z
z
z
k
z
D
z
KG
b
a
b
z
z
a
z
z
K
z
D
z
KG
b
z
a
z
z
D
z
D
s
G
p

47

origin.

at the

is
equation

stic
characteri

the
of
root

the
1,
K
When

plane.
-

the
of

axis

real

on the

lie

would
locus
root

Then the
)
1
(
)
98
.
0
)(
1
(
)
1
(
)
(
)
(
that
so

0.98
-



and

1



selecting
by

locus
root

the
improve

would
we
,
inadequate

were
e
performanc

system

the
If









z
z
K
z
z
z
K
z
D
z
KG
b
a
48


+1

0.2

-
1

Unit circle

K
=0.8

Im{
z
}

Re{
z
}

K

increasing

Entry point at

z

=
-
2.56

Root locus

49

P13.10 Dorf

0.55.

is
overshoot

The

.
6843
.
0
4641
.
0

are

poles

the
119.5,
n
Whe
(f)
5661
.
0
092
.
1
1982
.
0
2759
.
0
)
(

75;

When
(e)
75.



find

we
0.3,

of
overshoot

maximum

and

1
for

13.19

Figure

Using
(d)
239.

is


of

value
maximum

method,

locus
root

Using
(c)
0
3679
.
0
368
.
1
0026
.
0
0037
.
0
1

is
equation

stic
characteri

system

loop
-
closed

The

(b)
3679
.
0
368
.
1
0026
.
0
0037
.
0
)
(
)
(
function
transfer
The

(a)
)
(
;
1
.
0
;
)
10
(
1
)
(
2
2
2
j
z
K
z
z
z
z
T
K
K
T/
τ
K
z
z
z
K
z
z
z
K
z
D
z
G
K
z
D
T
s
s
s
G
p























50

P13.11 Dorf

150
;
999
.
0
;
993
.
0
1
1
;
;
999
.
0
993
.
0
150
)
(

:
0.01)
(

method

)
(

to
)
(

Use
(d)
3
.
155
;
99
.
0
;
9324
.
0
1
1
;
;
99
.
0
9324
.
0
3
.
155
)
(

:
0.1)
(

method

)
(

to
)
(

Use
(b)
.
01
.
0



and

30%

are

input)

ramp

a
(for
error

tracking
state
-
safety

and
overshoot

system

d
compensate

The
150.

and

0.1,

0.7,


select
may

we
Plot,

Bode

using
By
)
(

(a)
01
.
0
07
.
0
01
.
0
007
.
0


















































C
e
B
e
A
b
a
K
B
A
C
e
B
e
A
z
z
B
z
A
z
C
z
D
T
z
D
s
G
C
e
B
e
A
b
a
K
B
A
C
e
B
e
A
z
z
B
z
A
z
C
z
D
T
z
D
s
G
e
PO
K
b
a
b
s
a
s
K
s
G
bT
aT
c
bT
aT
c
ss
c