Design of Compensators Using Root Locus Diagram

7/28/2019 Design of Compensators Using Root Locus Diagram

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AUTOMATIC CONTROL

THIRD POWER

Prepared by

Dr. Helmy El-Zoghby

2013

3/9/2013 2

Design of compensators

using root locus diagram


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A compensator (filter or controller) is a circuit whichadded to the control system to improve the system

performance.

Different types of controllers1. Two Position Controller2. Proportional Controller

3. Integral Controller4. Differential Controller

5. PI Controller6. PD Controller7. PID-Controller

8.Lead compensator9.Lag compensator

10.Lag-lead compensator11.Intelligent compensators (fuzzy , neural , genetic , .)

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Benefits of compensators

Lead compensator (as PD-controller) is used to:

- improve transient response

- improve stability

Lag compensator ( as PI-controller) is used to:- reduce or eliminate steady-state error

Lag- Lead compensator ( as PID-controller)

- used to improve all issues.


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Note:

For a closed loop system:

By placing the closed-loop poles at the desired location

-if the new closed-loop poles lies on the old root locus , then k

is tuned or using P-controller (amplifier).

-if the new closed-loop poles not lies on the old root locus ,then reshape the root locus by adding poles/zeros

(compensator) to system.

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Effect of adding poles on control system response


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Effect of adding zeros on control system response

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Lag and lead compensators concept

)(s

)z(sK)(G ccc

cps

+

+=

The controller transfer function


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Lead compensator

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Lag compensator


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Electronic circuit of lead or lag compensator

)(s

)z(s

)1

(

)1

(

.)(

)()(G c

22

11

23

14c

cc

i

o

ps

s

s

ss K

CR

CRCRCR

vv

+

+=

+

+

==

For lead R1C1> R2C2

For lag R1C1< R2C2

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Lead compensator design example

Radar tracking system


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draw the uncompensated system root locus and find the damping ratio and

natural frequency at k=4

design a suitable compensator using three different methods to rise the natural

frequency to 4 rad/sec at the same damping ratio

draw the compensated root locus

draw the designed controller electronic circuit

check the steady state error after adding the controller

draw the transient response before and after compensation.

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solution

For the uncompensated system

5.0)cos(

60,2then

4ppK

0

n

22

==

==

==

and


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030210180

)90120(0180

polesallofzerosallof180

==

++=

+= o

Angle of deficiency is:

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1-Design of compensator using minimum steady-

state error method


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Selection of the controller pole and zero

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The compensated system T.F is :

4.68*4/ppK c21c == czp


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Steady-state error before and after compensation at

unit ramp input

%202.01

02.5)2(

4

)4.5(

)9.2(68.4.lim

)()()(limk

kv

0

0

new

v

===

=++

+=

=

=

=

newnew

ss

s

sc

e

sss

ss

sHsGssG

%505.01

2)2(

4.lim)()(limk

kv

00v

===

=+

====

ess

ss ssssHssG

Before compensation

After compensation

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Note : If the input is unit step :

%01

)2(

4lim

k1

k

p

0p

==

=+

=

+

=

e ss

sss

%01

)2(4

)4.5()9.2(68.4lim

k1

k

p

0newp

==

=++

+=

+

=

new

new

ss

s

e

ssss

Before compensation

After compensation


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Root locus before and after compensation

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Compensator electronic circuit

)1

(

)1

(

.)4.5(

)9.2(68.4)(G

22

11

23

14c

CR

CRCRCR

s

s

s

ss

+

+

=+

+=

=

===

ktake

uftake

R

CCCR CR

10

10,68.4

3

21

23

14

=== kkkthen RRR 5.18,5.34,8.46 214


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2-Design of compensator using zero-pole cancellation

method

)(s

)z(s)(G cc

c

c ps K

+

+=

Since the Angle of deficiency =300

and

-Select the controller zero equal to the nearest pole

of the system to the desired point to cancel thispole

-Find the controller pole from angle condition

-Find kc from magnitude condition

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Take Zc=2 and from angle condition Pc=4

From magnitude condition

44/4*44/ppK c1c ===

)4(s

)2(s4

)(s

)z(s)(G cc

+

+=

+

+=

c

c ps K

%505.01

,2)2(

4.limk

kv0v

====+

== esssss

s

%2525.01

,4)2(

4

)4(

)2(4.limk

kv0

new

v ====++

+=

= new

new

sss essss

s

xxx o

p

30

jw

Scale 1:2

-4 -1 0-2

Steady-state error before design

The controller T.F become :

Steady-state error after design


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Design of compensator using PD-controller

)()(. zkkk

kkk cdd

p

ddpsssFT +=+=+=

RR

k

k

p

dRC

1

2=

=

PD-controller block diagramPD-controller electronic circuit

PD-controller transfer function

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Since the Angle of deficiency =300

-the angle of controller zero must equal to 300

-draw a line of 300 with real axis from point (p)

-Find the controller zero from angle condition

-Find kd

from magnitude condition

)()(. zkkk

kkk cdd

p

ddpsssFT +=+=+=


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xxo

p

30

jw

Scale 1:2

Zc=7.8-1 0-2

Compensated and

uncompensated root locus

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==

==

==

====

kthenktakeR RR

R

6.41

6.4then

k60Rthen10ufCtake

C*R0.66.3*3.5/4*4z4/ppK

21

1

2

p

c21d

k

From angle condition

8.7==kk

zd

p

c

From magnitude conditionPD-controller electronic circuit


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%505.01

,2)2(

4.limk

kv0v

====+

== esssss

s

%9.10109.01

,2.9)2(

4).6.06.4.(limk

kv0

new

v ====+

+== new

new

sss essss

Steady-state error before design

The controller T.F become :

Steady-state error after design

ssFT kk dp 6.06.4. +=+=

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Using zero-pole cancellation method try to solve this problem

and simulate by matlab

A-If it required to design a lead

controller to get overshoot=2%

and peak time equal 1 sec

Check these values by calculations

Lead angle =96 degree , wd=3.14 ,

wn=5, zeta=.779 ,beta= 38.

0 1 2 3 4 5 6 7 8 9 1 00

0.2

0.4

0.6

0.8

1

1.2

1.4

Time

res

p

o

n

c

e


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B-at 1 sec peak time and 10% overshoot

Check these values by calculationszeta=.57 , wd=3.14 , wn=3.89 , lead angle

=35, beta=55

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Time

response

32

Proportional-Integral controller (lag compensator)

Time output equation is :

+= dttetetp kk ip )()()(

where Kp is the proportional constant and Ki is the integral constant.

The block diagram of the PI controller

The PI - controller electronic

circuit.


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Design of lag compensator (PI-controller)

The PI-controller transfer function is :

( )pz

kkk

k

kkkk

c

c

c

p

i

p

ipi

p

s

s

s

s

s

s

sFT

+

+=

+

=

+

=+=.

The pole of the controller at s=0

k

kp

is =The zero of the controller at

pczc

x

p

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PI-controller design steps

The zero of the controller at s=-ki/kp is obtained from angle condition

For the following open loop transfer function design a controller to satisfy 0.8

damping ratio , natural frequency of 5 rad/sec , and elimination of steady-state

error

The controller gain constant kp is obtained from magnitude condition

Example

The pole of the controller at s=0

)7)(3(

1)()(

++=

sssHsG


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011

n37)8.0(cos)(cos,5 ====

The steady-state error is zero , PI controller is used

( )pz

kkk

k

kkkk

c

c

c

p

i

p

ipi

p

s

s

s

s

s

s

sFT

+

+=

+

=

+

=+=.

27)45108(0180

polesallofzerosallof180

=++=

+= o

Angle of deficiency is:

Lag compensator is required

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From angle condition

From magnitude condition

50*5.2

then

20*1/ppK

ki

c21c

==

===

k

K

p

cpzp

( )s

s

s

s

skk

kGp

i

pc

5.220)(

+=

+

=

5.2==kkZ p

ic

The controller transfer function is


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PI-controller electronic circuit

kthen

k

take

R

R

RRk p

20

1

20

2

1

1

2

=

=

==

kthen

ufC

take

R

RCk i

2

10

501

=

=

==

3/9/2013 38

XX XO

p

Pc=0Zc=-2.5-3-7

27

45 108

Scale 1:1

Re

Jw

37

Wn=5


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Thank You&

Any Questions?

Date post:	03-Apr-2018
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Design of Compensators Using Root Locus Diagram

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