1
GATE EE 2006
MCQ (Single Correct Answer)
+2
-0.6
The Bode magnitude plot of $$H\left( {j\omega } \right) = {{{{10}^4}\left( {1 + j\,\omega } \right)} \over {\left( {10 + j\,\omega } \right){{\left( {100 + j\omega } \right)}^2}}}$$ is
A
GATE EE 2006 Control Systems - Polar Nyquist and Bode Plot Question 12 English Option 1
B
GATE EE 2006 Control Systems - Polar Nyquist and Bode Plot Question 12 English Option 2
C
GATE EE 2006 Control Systems - Polar Nyquist and Bode Plot Question 12 English Option 3
D
GATE EE 2006 Control Systems - Polar Nyquist and Bode Plot Question 12 English Option 4
2
GATE EE 2005
MCQ (Single Correct Answer)
+2
-0.6
If the compensated system shown in the figure has a phase margin of $${60^ \circ }$$ at the crossover frequency of $$1 rad/sec,$$ the value of the gain $$K$$ is GATE EE 2005 Control Systems - Polar Nyquist and Bode Plot Question 28 English
A
$$0.366$$
B
$$0.732$$
C
$$1.366$$
D
$$2.738$$
3
GATE EE 2005
MCQ (Single Correct Answer)
+2
-0.6
In the $$GH(s)$$ plane, the Nyquist plot of the loop transfer function $$G\left( s \right)\,H\left( s \right) = {{\pi {e^{ - 0.25s}}} \over s}$$ passes through the negative real axis at the point
A
$$\left( { - 0.25,j0} \right)$$
B
$$\left( { - 0.5,j0} \right)$$
C
$$\left( { - 1,j0} \right)$$
D
$$\left( { - 2,j0} \right)$$
4
GATE EE 2004
MCQ (Single Correct Answer)
+2
-0.6
In the system shown in figure, the input $$x(t)$$ $$=$$ $$sin$$ $$t.$$ In the steady-state, the response $$y(t)$$ will be GATE EE 2004 Control Systems - Polar Nyquist and Bode Plot Question 31 English
A
$${K \over {\sqrt 2 }}\sin \left( {t - {{45}^0}} \right)$$
B
$${K \over {\sqrt 2 }}\sin \left( {t + {{45}^0}} \right)$$
C
$$K\,\sin \left( {t - {{45}^0}} \right)$$
D
$$K\,\sin \left( {t + {{45}^0}} \right)$$
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