1
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
$${1 \over 5}\left( {1 - {e^{ - {t \over 5}}}} \right)u(t)$$
2
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)
3
GATE ECE 2005
+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
$$(1 - 0.5\cos \omega T){e^{ - j\omega {t_d}}}$$
4
GATE ECE 2004
+2
-0.6
A system described by the differential equation: $${{{d^2}y} \over {d{t^2}}} + 3{{dy} \over {dt}} + 2y = x(t)$$ is initially at rest. For input x(t) = 2u(t), the output y(t) is
A
$$(1 - 2{e^{ - t}} + {e^{ - 2t}})\,u(t)$$
B
$$(1 + 2{e^{ - t}} - 2\,{e^{ - 2t}})\,u(t)$$
C
$$(0.5 + {e^{ - t}} + 1.5\,{e^{ - 2t}})\,u(t)$$
D
$$(0.5 + 2{e^{ - t}} + 2\,\,{e^{ - 2t}})\,u(t)$$
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