1
GATE ECE 2013
MCQ (Single Correct Answer)
+1
-0.3
The Bode plot of a transfer function G (s) is shown in the figure below. GATE ECE 2013 Control Systems - Frequency Response Analysis Question 51 English The gain (20 log $$\left| {G(s)} \right|$$ ) is 32 dB and -8dB at 1rad/s and 10rad/s respectively. The phase is negative for all $$\omega .$$ Then G(s) is
A
$${\textstyle{{39.8} \over s}}$$
B
$${\textstyle{{39.8} \over {{s^2}}}}$$
C
$${\textstyle{{32} \over {{s}}}}$$
D
$${\textstyle{{32} \over {{s^2}}}}$$
2
GATE ECE 2012
MCQ (Single Correct Answer)
+1
-0.3
A system with transfer function g(s) = $${{\left( {{s^2} + 9} \right)\left( {s + 2} \right)} \over {\left( {s + 1} \right)\left( {s + 3} \right)\left( {s + 4} \right)}},$$ is excited by $$\sin \left( {\omega t} \right).$$ The steady-state output of the system is zero at
A
$$\omega = 1rad/\sec $$
B
$$\omega = 2rad/\sec $$
C
$$\omega = 3rad/\sec $$
D
$$\omega = 4rad/\sec $$
3
GATE ECE 2011
MCQ (Single Correct Answer)
+1
-0.3
For the transfer function G$$\left( {j\omega } \right) = 5 + j\omega ,$$ the corresponding Nyquist plot for positive frequency has the form
A
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 53 English Option 1
B
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 53 English Option 2
C
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 53 English Option 3
D
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 53 English Option 4
4
GATE ECE 2010
MCQ (Single Correct Answer)
+1
-0.3
A system with the transfer function $${{Y(s)} \over {X(s)}} = {s \over {s + p}},$$ has an output y(t)=$$\cos \left( {2t - {\pi \over 3}} \right),$$ for input signal x(t)=$$p\cos \left( {2t - {\pi \over 2}} \right).$$ Then the system parameter 'p' is
A
$$\sqrt 3 $$
B
$${2 \over {\sqrt 3 }}$$
C
1
D
$${{\sqrt 3 } \over 2}$$
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