Exercise: Automatic Control Systems

Questions for: Automatic Control Systems

For the system in the given figure, the characteristic equation is
A:
s2 + s(3 + 6a) + 6 = 0
B:
s2 + s(3 + 12a) + 12 = 0
C:
s2 + 3s = 0
D:
s2 + 6s = 0
Answer: A

Transfer function to inner loop = .

Then .

The characteristic equation is s2 + 3s + 6as + 6 = 0.

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For the system in the given figure, the transfer function C(s)/R(s) is
A:
G1 + G2 + G3
B:
G1 G2 + G3
C:
G1 G2 G3
D:
(G1 + G2) G3
Answer: B

G1 and G2 are cascaded.

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The polar plot of a transfer function passes through (-1, 0) point. The gain margin is
A:
zero
B:
- 1 dB
C:
1 dB
D:
infinity
Answer: A

When polar plot passes through (-1, 0) the system is on limit of stability and gain margin is zero.

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The transient response of a second order system is given by for 5% criterion the settling time is
A:
0.25 sec
B:
0.75 sec
C:
1.25 sec
D:
4 sec
Answer: B

For 5% criterion settling time = 3T = .

Since ΞΎΟ‰n = 4, and 3T = 0.75 secs.

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For G(jω) =
A:
magnitude is 1 at all frequencies and phase angle varies from 0 to - 180Β° as Ο‰ varies from 0 to ∞
B:
magnitude is T at all frequencies and phase angle 0
C:
magnitude is (1 + Ο‰2T2)0.5 and phase angle varies from0 to 0 - 180Β° as Ο‰ varies from 0 to ∞
D:
none of the above
Answer: A

When Ο‰ = 0, phase angle is 0.

For high values of phase angle = - 90Β° - 90Β° = 180Β°.

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