1
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 56 English Option 1
B
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 56 English Option 2
C
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 56 English Option 3
D
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 56 English Option 4
2
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
The input-output transfer function of a plant is h(s)=$${{100} \over {s{{\left( {s + 10} \right)}^2}}}$$. The plant is placed in a unity negative feedback configuration as shown in the figure below. GATE ECE 2011 Control Systems - Frequency Response Analysis Question 28 English The gain margin of the system under closed loop unity negative feedback is
A
0 dB
B
20 dB
C
26 dB
D
46 dB
3
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
The input-output transfer function of a plant is h(s)=$${{100} \over {s{{\left( {s + 10} \right)}^2}}}$$. The plant is placed in a unity negative feedback configuration as shown in the figure below. GATE ECE 2011 Control Systems - Frequency Response Analysis Question 29 English The signal flow graph that DOES NOT model the plant transfer function H(s) is
A
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 29 English Option 1
B
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 29 English Option 2
C
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 29 English Option 3
D
GATE ECE 2011 Control Systems - Frequency Response Analysis Question 29 English Option 4
4
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
The block diagram of a system with one input u and two outputs y1 and y2 is given below GATE ECE 2011 Control Systems - State Space Analysis Question 23 English

A state space model of the above system in terms of the state vector $$\underline x $$ and the output vector $$\underline y = {\left[ {\matrix{ {{y_1}} & {{y_2}} \cr } } \right]^\tau }$$ is

A
$$\mathop {\underline x }\limits^ \bullet = \left[ 2 \right]\underline x + \left[ 1 \right]u;\underline y = \left[ {\matrix{ 1 & 2 \cr } } \right]x$$
B
$$\mathop {\underline x }\limits^ \bullet = \left[ { - 2} \right]\underline x + \left[ 1 \right]u;\underline y = \left[ {\matrix{ 1 \cr 2 \cr } } \right]x$$
C
$$\mathop {\underline x }\limits^ \bullet = \left[ {\matrix{ { - 2} & 0 \cr 0 & { - 2} \cr } } \right]\underline x + \left[ {\matrix{ 1 \cr 1 \cr } } \right]u;\underline y = \left[ {\matrix{ 1 & 2 \cr } } \right]x$$
D
$$\mathop {\underline x }\limits^ \bullet = \left[ {\matrix{ 2 & 0 \cr 0 & 2 \cr } } \right]\underline x + \left[ {\matrix{ 1 \cr 1 \cr } } \right]u;\underline y = \left[ {\matrix{ 1 \cr 2 \cr } } \right]x$$
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