Question: 1 -
The open loop transfer function of a unity feedback system is given by G(s) = 3e-2s/s(s+2) The gain and phase crossover frequencies in rad/sec are, respectively
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0.632 and 1.26
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1.26 and 0.632
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0.485 and 0.632
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0.632 and 0.485
Answer:
1.26 and 0.632
Solution:
|G(s)| = 1
ωg = 1.26
∠ GH = 180°
ωφ = 0.632
|G(s)| = 1
ωg = 1.26
∠ GH = 180°
ωφ = 0.632
Question: 2 -
The Nyquist plot for the open-loop transfer function G(s) of a unity negative feedback system is shown in the figure, if G(s) has no pole in the right-half of s-plane, the number of roots of the system characteristic equation in the right-half of s-plane is
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1
-
0
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2
-
3
Answer:
0
Solution:
N = 0 (1encircle mention CW direction and other in CCW)
P=0 (no pole in right half)
So, N = P − Z
Z = P − N = 0
∴ No roots on RH of s-plane.
N = 0 (1encircle mention CW direction and other in CCW)
P=0 (no pole in right half)
So, N = P − Z
Z = P − N = 0
∴ No roots on RH of s-plane.
Question: 3 -
The system with the open loop transfer function G(s)H(s) = 1/s(s2+s+1) has a gain margin of
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3.5dB
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6dB
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0dB
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-6dB
Answer:
0dB
Solution:
use ωφ when ∠of G(s)H(s) = −180°
after solving: ωφ = 1 rad/sec
and
G.M. = −20log1 = 0
use ωφ when ∠of G(s)H(s) = −180°
after solving: ωφ = 1 rad/sec
and
G.M. = −20log1 = 0
Question: 4 -
The gain margin and the phase margin of a feedback system with G(s)H(s) = s/(s+100)3 are
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∞,0°
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88.5 dB,∞
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∞,∞
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0 dB,0°
Answer:
∞,∞
Solution:
G.M. and P.M. of the system cannot be determined.
G.M. and P.M. of the system cannot be determined.
Question: 5 -
A system has poles at 0.01 Hz, 1 Hz and 80Hz; zeros at 5 Hz, 100 Hz and 200 Hz. The approximate phase of the system response at 20 Hz is
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0°
-
−90°
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−180°
-
90°
Answer:
−90°
Solution:
Pole at 0.01 and 1 Hz gives −180° phase. Zero at 5 Hz gives +90° phase
∴ at 20Hz −90° phase shift is provided.
Pole at 0.01 and 1 Hz gives −180° phase. Zero at 5 Hz gives +90° phase
∴ at 20Hz −90° phase shift is provided.