1314sem1finalexam_soln-cntrl

7/26/2019 1314sem1finalexam_soln-cntrl

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SEE 3113

Question 1

a)

i) The exponential term increases indefinitely

ii) No. That will only be the case if there is no sign change from ROZ

downwards.

iii) The system is unstable

b)

i) The system is unstable because of the negative sign occurred in the

characteristic equation.

ii)

S5 1 4 3

S4 -1 -4 -2

S3 1 0

S2

41

-2 0

S1

41

412 2

+

0 0

S

0

-2 0 0

3 right hand-plane, 2 left hand-plane. The systemis unstable.

c) Since the system oscillates, it is marginally stable. The characteristic equation

of the system becomes:

1 +( + 2)

3 +

2 + 3

+ 2

= 0

Or 3 + 2 + ( + 3) + 2( + 1) = 0

Rouths

s3

1 K+3

s2

p 2(K+1)

s1

(

+ 3)

2(

+ 1)

0

s0

2(K+1)

1

1

1

3

3

1

2

2

1

1

2


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SEE 3113

At marginal value of K (+3)2(+1) = 0Or

=2(+1)+3

Again at this value of p,

() =2 + 2( + 1) = 0Or

2 =2( + 1) = 2( + 1)

Given =2.5rad/sec. Therefore:

2(+1) = 2.5[1mark]

= 2( + 1)6.25

Substitute equation (1) into (2)2(+1)+3 =2(+1)6.25 K=3.25Therefore

p=2(3.25+1)/6.25= 1.36

(2)

1

1

2

2

1


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SEE 3113

Question 2

a)

By looking at the locations of the open loop poles and zeros. If there are some

on the LHP and some on the RHP, so the root locus will cross the imaginary

axis.

b)

()() = (+1)(+5)(+1)+(62)+5+2Then,

s2 K + 1 5K + 2

s1 6K 2 0

s0 5K + 2 0

Therefore K > 0.333

From auxiliary equation and substituting K = 0.333,

43 2 +113 = 0 = 1.66

c)

i. K = 0: closed loop poles = 1+j ; 1-j

ii. K -> infinity: closed loop poles = -1 ; -5

iii. 2 branches

iv. symmetrical about real axis

v. no asymptote angle

vi. no asymptotes crossings

vii locus on real axis : between -5 and -1

viii Departure angles = 180 90 + 26.6 + 9.5 = 126.1

ix jw-crossings : from part (b), s =1.66x

. (+1)(+5)(2+2) = 0solving this will obtain 1.325 and -2.075so the break-in point is at -2.075

therefore the root locus:

2

1

1.5

1.5

2

1

1

2

1

2

1

2


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SEE 3113

d)

From the root locus, if the damping ratio is 0.6

the dominant poles are at

1.13

1.52

and the value of K = 1.14

-5 -4 -3 -2 -1 0 1-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

Root Locus

Real Axis

ImaginaryAxis

2

2

3


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SEE 3113

Question 3

a)

b) A vehicle speed control system is shown in Figure Q3b.

2

11

2

2

0.5

1

0.5


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SEE 3113

c) Roughly sketch the root locus of the uncompensated and compensated systems

in part (b) above.

5

2

2

1


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SEE 3113

2.5

2.5


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SEE 3113

Question 4

a)

i) A measurement of output response of a system in frequency domain. The

output response consists of the magnitude and phase values as the

frequency is varied.

ii) By obtaining the gain margin and phase margin. The system is said to

stable if the gain and phase margins are positive.

b)

i) By using the straight line approximation methods, plot the Bode plot of the

open-loop system for K= 1.

ii) Gain Margin = 53.2 dB

Phase Margin = 91.6

iii)Stable. Positive values of GM and PM

iv)The value of gain Kneeded is 40.3 dB

-200

-150

-100

-50

0

50

Magnitude(dB)

10-2

10-1

100

101

102

103

104

-270

-225

-180

-135

-90

-45

Phase(deg)

Bode Diagram

Frequency (rad/sec)

1.5

1.5

7

7

1

1

2

2


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SEE 3113

v)

Bode Diagram

Frequency (rad/sec)

10-2

10-1

100

101

102

103

104

-270

-225

-180

-135

-90

-45

System: G1

Phase Margin (deg): 44.6

Delay Margin (sec ): 0.0652

At frequency (r ad/sec): 11.9

Closed Loop Stable? Yes

Phase(de

g)

-150

-100

-50

0

50

100

System: G1

Gain Margin (dB): 12.9

At frequency (r ad/sec): 27.2

Closed Loop Stable? Yes

Magnitude(dB) 2


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SEE 3113

Question 5

a)


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SEE 3113

b)

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1314sem1finalexam_soln-cntrl