ele 3202 cat 1-solutions [compatibility mode].pdf

7/28/2019 ELE 3202 Cat 1-Solutions [Compatibility Mode].pdf

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EE 3202: CAT 1

Questions and Answers

13 March 2010

1


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Question 1

(a) Use block diagrams to show why a closedloop control system is preferred to an open

one. [6]

(b) Use a block diagram to determine the overalltransfer function of a negative feedback

control system. [4]

(c) Analyse the stability of the system in (b)

above using the complex plane and hence

show how at least one root of the

characteristic equation can make the system

unstable. [10]2


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Solution 1

a) Block diagram open loop control system

From the block diagram, the output (b) is not

compared with the input (a) so as to knowwhen the desired/final value of (b) has been

attained/reached.

Control

system

Input signal (a) Output signal (b)

[1

[13


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Solution 1 Contd

Block diagram closed loop control system

The output signal C is compared with the input

signal R, so the control signal E is used by the

control system. Hence control capability

which is not the case with open loop

Controlsystem

feedback

-

+

Controlsignal

Inputsignal

Feedback

Signal

Output signalR KG

H

EC

F

[2

[1

[1

4


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Solution 1 Contd

(b) Block diagram:

Controlsystem

feedback

-

+

Controlsignal

Inputsignal

Feedback

Signal

Output signalR KG

H

EC

F

HCF

EKGCFRE

*

*

KGH

KG

R

C

orHG

KG

R

C

1

;*1

[1

[1[1

5

1:[Workin


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Solution 1 Contd

c) All the analysis of a control system is based on

the characteristic equation, and from the

transfer function in (b) above, x-tic eqn is given

by:

The solution of this equation gives the values of

S.

For the system to be stable, all values of S

must have negative real parts.

1+GH(S)=0 [1

[1

[1

[1

6


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Solution 1 Contd

This implies that all values of S lie on the

LHS of the s-plane.

Therefore the system will be unstable if atleast one of the roots lie on the RHS of the s-

plane, thus having a positive real part.

Hence on the complex plane below, the

system will be unstable because S6 lies on the

RHS of that plane.

[1

[1

[27


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Solution 1 Contd

complex (s) plane:

6s

2s3s

4s

Re

Im

5s

1s

[2

8


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Question 2

(a) State any four of Root Locus Rules [4]

(b) Use the Root Locus method to determine the

stability of [7]

(c) Compensate the system of (b) above by

and discuss the change in stability [9]

)4(

)3)(2()( 3

ss

ssKsKGH

51

sC

9


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Solution 2

a) Any four (4) of the 12 Rules such as:

There are as many branches as there are open-

loop poles.

A branch begins at an open-loop pole and ends ata zero or infinity.

Each branch is asymptotic to a line whose angle

is given by;

Where x=number of open-loop poles, & y=number

of open-loop zeros

yxn

o

180)12(

[1

[1

[1

10


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Solution 2 Contd

All the asymptotes meet at one point on the real

axis and whose value is given by:

zerospolesloopopenofnoyxzeroloopopenofpartrealz

poleloopopenofpartrealpwhere

yx

zpm

b

a

x

a

y

bba

,,

;

1 1

[1

11


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Solution 2 Contd

(b) Given;

)4(

)3)(2()(

3

ss

ssKsKGH

4,0,0,0

3,2

4321

21

PPPP

ZZ

2&4; yxHence

[1

12

[1


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Solution 2 Contd

Recall;

Similarly;

2

1

24

324

1 1

m

yx

zp

m

x

a

y

b

ba

0

3

0

2

0

1 450,270,90

90)12(24

180)12(

180)12(

oo

o

nn

yx

n

[1

[113


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Solution 2 Contd

Sketch of root locus

System is unstable for all values of K.

x

-3 0.5-2-4 Re

IM

[2

[114


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Solution 2 Contd

(c) After compensation;

)5)(4(

)1)(3)(2()(

3

sss

sssKsKGH

5,4,0,0,0

1,3,2

54321

321

PPPPP

ZZZ

3&5; yxHence

[1

15

[1


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Solution 2 Contd

Recall;

Similarly;

2

3

35

13254

1 1

m

yx

zp

m

x

a

y

b

ba

0

3

0

2

0

1 450,270,90

90)12(35

180)12(

180)12(

oo

o

nn

yx

n

[1

[116


17/36


18/36

Question 3

(a) State the General Nyquist Stability Criterion. [5]

(b) Determine the stability of the system below

using the Nyquist Stability Criterion

[15])3)(2)(1(

)8()(

sss

sKsKGH

18


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Solution 3

a) NSC: A control system is stable provided thenumber of closed loop poles in RHS of s-plane is

equals to zero i.e P=0,

given that P=N+Z; whereZ= number of open-loop zeros in the RHS of s-

plane

P= number closed-loop poles in RHS of s-plane

N= number of encirclements of the point -1+j0

[1

[3

[1

19


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Solution 3 Contd

b) given:

alwimaginarypurelyw

thuswww

wjwwwKwjww

jwK

jwjwjwjwKjwKGH

sss

sKsKGH

Re:41&:25.1

;,)11()1(36

)}822(4837{

,

)11()1(6

)8(

)3)(2)(1()8()(

)2)(2)(1(

)8()(

22222

224

22

[1

[220


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Solution 3 Contd

It follows:

oo

o

oo

w

K

jwjwjw

jwKKGH

KKjjKGH

jKjKGH

KKK

jKGH

1800180))()((

)(

90734.0734.0)25.1(

)25.111(25.1)25.11(36

)}8225.1*2(25.1{)25.1(

033.103

4

)3)(2)(1(

8)0(

2

22

[1

21


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Solution 3 Contd

similarly

00

22

2

180033.01803

)41(

30)41(

)30(41)40(36

4841*3741)41(

KKjKGH

KjKGH

jKGH

[1

22


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Solution 3 Contd

Sketch

0P

0Q

I

R30K

[3

23


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Solution 3 Contd

From the transfer function;

Z=0

If the point -1+j0 is at Q; N=0

i.e

using P = N+Z = 0+0 = 0,

system stable for the above range

30,30

01 KK

j

[1

[1

[1

[1

24


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Solution 3 Contd

Recall Z=0 for the control system in question;

If the point -1+j0 is at P; N=2

i.e

using P = N+Z = 2+0 = 2,

system unstable for the above range

30,3001

K

K

j

[1

[1

[1

25


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Question 4

(a) Use the concept of a Unit Circle to explain

the stability, instability and oscillation of a

negative feedback control system. [5]

(b) Use the Unit Circle to determine the

stability of the system given by:

[7]

(c) Confirm (b) by Nyquist Criterion [8]

)5)(2(

)(

sss

KsKGH

26


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Solution 4

a) Consider the plot

below:

-1+j0P

g

Unitcircle

P

Q

Re

IM

[22nd quad 1st quad

27


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Solution 4 Contd

To use the concept of a unit circle; the KGH(jw)function is plotted for +w, and a circle whose

radius is one is superimposed on it.

The system is stable if the frequency responsedoes not cut the unit circle in the second

quadrant, and unstable otherwise.

From the figure above, the system is stable for the

continuous curve and unstable for the dotted

curve. It will oscillate if the curve passes through

the point -1+j0.

[1

[1

[128


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Solution 4 Contd

b) given:

alpurelyw

thuswww

wjwKwjwjw

KjwKGH

jwjwjwKjwKGH

sss

KsKGH

Re;10

;,49)10(

)}10(7{)710(

)(

)5)(2()(

)5)(2()(

222

2

2

[1

[129


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Solution 4 ContdNyquist plot of KGH(S)

Transfer function for +w:

Re

IM

7

K

Unitcircle2nd quad

[231


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Solution 4 Contd

It is clear that the system will be stable for

(the case shown in the diagram above)

since the curve does not cut the unit circle in 2

nd

Quadrant while for

the system will be unstable since it will cut the

unit circle in 2nd Quadrant.

70K

70K

[2

32


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Solution 4 Contd

c) Confirmation of the above results using

Nyquist:

In this approach, a complete Nyquist plot ofKGH(S) over the entire frequency is made

taking the frequency curve to be symmetrical

about the real axis. From the solution above

(using the unit circle); the complete frequencyresponse can be sketched as shown below.[1

33


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Solution 4 Contd

Complete Nyquist plot:

P QRe

IM

7

K

[3

34


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Solution 4 Contd

From the transfer function;

Z=0

If the point -1+j0 is at P; N=0

i.e

using P = N+Z = 0+0 = 0,

system stable for the above range

70,70

01 KK

j

[1

[135


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Solution 3 ContdRecall Z=0 for the control system in question;

If the point -1+j0 is at Q; N=2

i.e

using P = N+Z = 2+0 = 2,

system unstable for the above range

Therefore the system is conditionally stable, thus

confirming the results in (b) above.

END

70,30

01 K

K

j

[1

[136

ele 3202 cat 1-solutions [compatibility mode].pdf

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