1
GATE EE 2013
MCQ (Single Correct Answer)
+1
-0.3
The impulse response of a system is h(t) = tu(t). For an input u(t − 1), the output is
A
$$\frac{t^2}2u\left(t\right)$$
B
$$\frac{t\left(t-1\right)}2u\left(t-1\right)$$
C
$$\frac{\left(t-1\right)^2}2u\left(t-1\right)$$
D
$$\frac{t^2-1}2u\left(t-1\right)$$
2
GATE EE 2013
MCQ (Single Correct Answer)
+1
-0.3
Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is GATE EE 2013 Signals and Systems - Linear Time Invariant Systems Question 40 English
A
u(t)
B
tu(t)
C
$$\frac{t^2}2u\left(t\right)$$
D
$$e^{-t}u\left(t\right)$$
3
GATE EE 2013
MCQ (Single Correct Answer)
+1
-0.3
Two systems 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 h2(t) and h1(t)
4
GATE EE 2011
MCQ (Single Correct Answer)
+1
-0.3
Given two continuous time signals $$x\left(t\right)=e^{-t}$$ and $$y\left(t\right)=e^{-2t}$$ which exist for t > 0, the convolution z(t) = x(t)*y(t) is
A
$$e^{-t}-e^{-2t}$$
B
$$e^{-3t}$$
C
$$e^{+t}$$
D
$$e^{-t}\;+\;e^{-2t}$$
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