1
GATE ECE 2010
MCQ (Single Correct Answer)
+2
-0.6
A continuous time LTI system is described by $${{{d^2}y(t)} \over {d{t^2}}} + 4{{dy(t)} \over {dt}} + 3y(t)\, = 2{{dx(t)} \over {dt}} + 4x(t)$$.

Assuming zero initial conditions, the response y(t) of the above system for the input x(t) = $${e^{ - 2t}}$$ u(t) is given by

A
$$({e^t} - {e^{3t}})\,u(t)$$
B
$$({e^{ - t}} - {e^{ - 3t}})\,u(t)$$
C
$$({e^{ - t}} + {e^{ - 3t}})\,u(t)$$
D
$$({e^t} + {e^{3t}})\,u(t)$$
2
GATE ECE 2009
MCQ (Single Correct Answer)
+2
-0.6
Consider a system whose input x and output y are related by the equation $$$y(t) = \int\limits_{ - \infty }^\infty {x(t - \tau )\,h(2\tau )\,d\tau } $$$

Where h(t) is shown in the graph.

GATE ECE 2009 Signals and Systems - Continuous Time Linear Invariant System Question 15 English

Which of the following four properties are possessed by the system?


BIBO: Bounded input gives a bounded output.
Causal: The system is casual.
LP: The system is low pass.
LTI: The system is linear and time- invariant.
A
Causal, LP
B
BIBO, LTI
C
BIBO, Causal, LTI
D
LP, LTI
3
GATE ECE 2008
MCQ (Single Correct Answer)
+2
-0.6
Let x(t) be the input and y(t) be the output of a continuous time system. Match the system properties P1, P2 and P3 with system relations R1, R2, R3, R4.

Properties


P1 : Linear but NOT time-invariant
P2: Time-invariant but NOT linear
P3: Linear and time-invariant

Relations


R1: y(t) = $${t^2}$$ x(t)
R2: y(t) = t$$\left| {x(t)} \right|$$
R3: y(t) = $$\left| {x(t)} \right|$$
R4: y(t) = x(t-5)
A
(P1, R1), (P2, R3), (P3, R4)
B
(P1, R2), (P2, R3), (P3, R4)
C
(P1, R3), (P2, R1), (P3, R2)
D
(P1, R1), (P2, R2), (P3, R3)
4
GATE ECE 2008
MCQ (Single Correct Answer)
+2
-0.6
A linear, time-invariant, causal continuous time system has a rational transfer function with simple poles at s = - 2 and s = - 4, and one simple zero at s = - 1. A unit step u(t) is applied at the input of the system. At steady state, the output has constant value of 1. The impulse response of this system is
A
[exp( -2t) + exp( -4t)]u(t)
B
[ -4exp( -2t) + 12 exp( -4t) - exp( -t)]u(t)
C
[ -4 exp( -2t) +12 exp(-4t)]u(t)
D
[-0.5 exp(-2t) +1.5 exp(-4t)]u(t)
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