GATE ECE 2010 | GATE ECE Year Wise Previous Years Questions

GATE ECE 2010

MCQ (Single Correct Answer)

-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

$$\sqrt 3 $$

$${2 \over {\sqrt 3 }}$$

$${{\sqrt 3 } \over 2}$$

GATE ECE 2010

MCQ (Single Correct Answer)

-0.3

For the asymptotic Bode magnitude plot shown below, the system transfer function can be GATE ECE 2010 Control Systems - Frequency Response Analysis Question 57 English

GATE ECE 2010 Control Systems - Frequency Response Analysis Question 57 English

$${{10s + 1} \over {0.1s + 1}}$$

$${{100s + 1} \over {0.1s + 1}}$$

$${{100s} \over {10s + 1}}$$

$${{0.1s + 1} \over {10s + 1}}$$

GATE ECE 2010

MCQ (Single Correct Answer)

-0.6

A unity negative feedback closed loop system has a plant with the transfer function $$G(s) = {1 \over {{s^2} + 2s + 2}}$$ and a controller $${G_c}(s)$$ in the feed forward path. For a unit set input, the transfer function of the controller that gives minimum steady sate error is

$${G_C}\left( s \right) = {{s + 1} \over {s + 2}}$$

$${G_C}\left( s \right) = {{s + 2} \over {s + 1}}$$

$${G_C}\left( s \right) = {{\left( {s + 1} \right)\left( {s + 4} \right)} \over {\left( {s + 2} \right)\left( {s + 3} \right)}}$$

$${G_C}\left( s \right) = 1 + {2 \over s} + {3_s}$$

GATE ECE 2010

MCQ (Single Correct Answer)

-0.6

The signal flow graph of a system is shown below. GATE ECE 2010 Control Systems - State Space Analysis Question 25 English

The state variable representation of the system can be

$$\mathop x\limits^ \bullet = \left[ {\matrix{ 1 & 1 \cr { - 1} & 0 \cr } } \right]x + \left[ {\matrix{ 0 \cr 2 \cr } } \right]u$$
$$y = \left[ {\matrix{ 0 & {0.5} \cr } } \right]x$$

$$\eqalign{ & \mathop x\limits^ \bullet = \left[ {\matrix{ { - 1} & 1 \cr { - 1} & 0 \cr } } \right]x + \left[ {\matrix{ 0 \cr 2 \cr } } \right]u \cr & y = \left[ {\matrix{ 0 & {0.5} \cr } } \right]x \cr} $$

$$\eqalign{ & \mathop x\limits^ \bullet = \left[ {\matrix{ 1 & 1 \cr { - 1} & 0 \cr } } \right]x + \left[ {\matrix{ 0 \cr 2 \cr } } \right]u \cr & y = \left[ {\matrix{ {0.5} & {0.5} \cr } } \right]x \cr} $$

$$\eqalign{ & \mathop x\limits^ \bullet = \left[ {\matrix{ { - 1} & 1 \cr { - 1} & 0 \cr } } \right]x + \left[ {\matrix{ 0 \cr 2 \cr } } \right]u \cr & y = \left[ {\matrix{ {0.5} & {0.5} \cr } } \right]x \cr} $$

PREVIOUS NEXT

GATE ECE Papers

2024