ele 3202 cat 1-solutions [compatibility mode].pdf
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EE 3202: CAT 1
Questions and Answers
13 March 2010
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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
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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
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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
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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
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Solution 1 Contd
complex (s) plane:
6s
2s3s
4s
Re
Im
5s
1s
[2
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Solution 3 Contd
Sketch
0P
0Q
I
R30K
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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
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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
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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
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Solution 4
a) Consider the plot
below:
-1+j0P
g
Unitcircle
P
Q
Re
IM
[22nd quad 1st quad
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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
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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
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Solution 4 ContdNyquist plot of KGH(S)
Transfer function for +w:
Re
IM
7
K
Unitcircle2nd quad
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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
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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
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Solution 4 Contd
Complete Nyquist plot:
P QRe
IM
7
K
[3
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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
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