1
GATE EE 2013
+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
+1
-0.3
Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is 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
+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
+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
$$e^{-t}\;+\;e^{-2t}$$
GATE EE Subjects
Electromagnetic Fields
Signals and Systems
Engineering Mathematics
General Aptitude
Power Electronics
Power System Analysis
Analog Electronics
Control Systems
Digital Electronics
Electrical Machines
Electric Circuits
Electrical and Electronics Measurement
EXAM MAP
Joint Entrance Examination