1
GATE ECE 2015 Set 1
+1
-0.3
The result of the convolution $$x\left( { - t} \right) * \delta \left( { - t - {t_0}} \right)$$ is
A
$$x\left( {t + {t_0}} \right)\,$$
B
$$x\left( {t - {t_0}} \right)\,$$
C
$$x\left( { - t + {t_0}} \right)$$
D
$$\,x\left( { - t - {t_0}} \right)$$
2
GATE ECE 2013
+1
-0.3
Two system with impulse responses h1(t) and h2(t) are connected in cascade. Then the overall impulse response of the cascaded system is given by
A
Product of h1(t) and h2(t).
B
Sum of h1(t) and h2(t).
C
Convolution of h1(t) and h2(t).
D
Subtraction of h1(t) and h2(t)
3
GATE ECE 2013
+1
-0.3
The impulse response of a system is h(t) = t u(t). For an input u(t - 1), the output is
A
$${{{t^2}} \over 2}u\left( t \right)$$
B
$${{t\left( {t - 1} \right)} \over 2}u\left( {t - 1} \right)$$
C
$${{{{\left( {t - 1} \right)}^2}} \over 2}u\left( {t - 1} \right)\,$$
D
$${{{t^2} - 1} \over 2}u\left( {t - 1} \right)$$
4
GATE ECE 2011
+1
-0.3
The differential equation $$100{{{d^2}y} \over {dt}} - 20{{dy} \over {dt}} + y = x\left( t \right)$$ describes a system with an input x(t) and output y(t). The system, which is initially relaxed, is excited by a unit step input. The output y(t) can be represented by the waveform
A
B
C
D
EXAM MAP
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