1
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)
2
GATE ECE 2007
MCQ (Single Correct Answer)
+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
$${1 \over 5}\left( {1 - {e^{ - {t \over 5}}}} \right)u(t)$$
3
GATE ECE 2006
MCQ (Single Correct Answer)
+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)
4
GATE ECE 2005
MCQ (Single Correct Answer)
+2
-0.6
The output y(t) of a linear time invariant system is related to its input x(t) by the following equation: y(t) = 0.5 x $$(t - {t_d} + T) + \,x\,(t - {t_d}) + 0.5\,x(t - {t_d} - T)$$. The filter transfer function $$H(\omega )$$ of such a system is given by
A
$$(1 + \cos \omega T){e^{ - j\omega {t_d}}}$$
B
$$(1 + 0.5\cos \omega T){e^{ - j\omega {t_d}}}$$
C
$$(1 + \cos \omega T){e^{j\omega {t_d}}}$$
D
$$(1 - 0.5\cos \omega T){e^{ - j\omega {t_d}}}$$
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