GATE ECE 2008 | GATE ECE Year Wise Previous Years Questions

GATE ECE 2008

MCQ (Single Correct Answer)

-0.6

The impulse response h(t) of a linear time invariant system is given by h(t) = $${e^{ - 2t}}u(t),$$ where u(t) denotes the unit step function.

The output of this system to the sinusoidal input x(t) = 2cos(t) for all time 't' is

$$0$$

$${2^{ - 0.25}}\cos \left( {2t - 0.125\pi } \right)$$

$${2^{ - 0.5}}\cos \left( {2t - 0.125\pi } \right)$$

$${2^{ - 0.5}}\cos \left( {2t - 0.25\pi } \right)$$

GATE ECE 2008

MCQ (Single Correct Answer)

-0.3

The pole-zero plot given below corresponds to a GATE ECE 2008 Control Systems - Compensators Question 19 English

Low pass filter

High pass filter

Band pass filter

notch filter

GATE ECE 2008

MCQ (Single Correct Answer)

-0.6

Group I gives two possible choices for the impedance Z in the diagram. The circuit elements in Z satisfy the condition $${R_2}{C_2} > {R_1}{C_{1.}}$$ The transfer $${\textstyle{{{V_0}} \over {{V_1}}}}$$ function represents a kind of controller. Match the impedances in Group I with the types of controllers in Group II. GATE ECE 2008 Control Systems - Compensators Question 9 English 1

GATE ECE 2008 Control Systems - Compensators Question 9 English 1

Group - I GATE ECE 2008 Control Systems - Compensators Question 9 English 2

Group - II
1. PID controller
2. Lead compensator
3. Lag compensator

Q - 1, R - 2

Q - 1, R - 3

Q - 2, R - 3

Q - 3, R - 2

GATE ECE 2008

MCQ (Single Correct Answer)

-0.6

A signal flow graph of a system is given below. GATE ECE 2008 Control Systems - State Space Analysis Question 25 English

The set of equations that correspond to this signal flow graph is

$${d \over {dt}}\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] = \left[ {\matrix{ \beta & { - \gamma } & 0 \cr \gamma & \alpha & 0 \cr { - \alpha } & { - \beta } & 0 \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] + \left[ {\matrix{ 1 & 0 \cr 0 & 0 \cr 0 & 1 \cr } } \right]\left( {\matrix{ {{u_1}} \cr {{u_2}} \cr } } \right)$$

$${d \over {dt}}\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] = \left[ {\matrix{ 0 & \alpha & \gamma \cr 0 & { - \alpha } & { - \gamma } \cr 0 & \beta & { - \beta } \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] + \left[ {\matrix{ 0 & 0 \cr 0 & 1 \cr 1 & 0 \cr } } \right]\left( {\matrix{ {{u_1}} \cr {{u_2}} \cr } } \right)$$

$${d \over {dt}}\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] = \left[ {\matrix{ { - \alpha } & \beta & 0 \cr { - \beta } & { - \gamma } & 0 \cr \alpha & \gamma & 0 \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] + \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr 0 & 0 \cr } } \right]\left( {\matrix{ {{u_1}} \cr {{u_2}} \cr } } \right)$$

$${d \over {dt}}\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] = \left[ {\matrix{ { - \gamma } & 0 & \beta \cr \gamma & 0 & \alpha \cr { - \beta } & 0 & { - \alpha } \cr } } \right]\left[ {\matrix{ {{x_1}} \cr {{x_2}} \cr {{x_3}} \cr } } \right] + \left[ {\matrix{ 0 & 1 \cr 0 & 0 \cr 1 & 0 \cr } } \right]\left( {\matrix{ {{u_1}} \cr {{u_2}} \cr } } \right)$$

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