1
GATE ECE 2008
+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)
2
GATE ECE 2008
+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)
3
GATE ECE 2007
+2
-0.6
The frequency response of a linear, time-invariant system is given by H(f) = $${5 \over {1 + j10\pi f}}$$. The step response of the system is
A
$$5(1 - {e^{ - 5t}})\,u(t)$$
B
$$5\left( {1 - {e^{ - {t \over 5}}}} \right)u(t)$$
C
$${1 \over 5}\left( {1 - {e^{ - 5t}}} \right)u(t)$$
D
$${1 \over 5}\left( {1 - {e^{ - {t \over 5}}}} \right)u(t)$$
4
GATE ECE 2006
+2
-0.6
Let g(t) = p(t) * p(t), where * denotes convolution and p(t) = u(t) - (t-1) with u(t) being the unit step function. The impulse response of filter matched to the singal s(t) = g(t) - $$[\delta (t - 2)*g(t)]$$ is given as
A
s(1 - t)
B
-s (1 - t)
C
-s(t)
D
s(t)
EXAM MAP
Medical
NEET