1
GATE ECE 1991
+2
-0.6
The voltage across an impedance in a network is V(s) = Z(s) I(s), where V(s), Z(s) and $${\rm I}$$(s) are the Laplace Transforms of the corresponding time functions V(t), z(t) and i(t).

The voltage v(t) is

A
$$v\left( t \right) = z\left( t \right)\,.\,i\left( t \right)$$
B
$$v\left( t \right) = \int\limits_0^t {i\left( \tau \right)\,z\left( {t - \tau } \right)d\tau }$$
C
$$v\left( t \right) = \int\limits_0^t {i\left( \tau \right)z\left( {t + \tau } \right)d\tau }$$
D
$$v\left( t \right) = z\left( t \right) + i\left( t \right)$$
2
GATE ECE 1991
+2
-0.6
An excitation is applied to a system at $$t = T$$ and its response is zero for $$- \infty < t < T$$. Such a system is a
A
non-causal system
B
stable system
C
causal system
D
unstable system
3
GATE ECE 1990
+2
-0.6
The response of an initially relaxed linear constant parameter network to a unit impulse applied at $$t = 0$$ is $$4{e^{ - 2t}}u\left( t \right).$$ The response of this network to a unit step function will be:
A
$$2\left[ {1 - {e^{ - 2t}}} \right]u\left( t \right)$$
B
$$4\left[ {{e^{ - t}} - {e^{ - 2t}}} \right]u\left( t \right)$$
C
$$\sin 2t$$
D
$$\left( {1 - 4{e^{ - 4t}}} \right)u\left( t \right)$$
4
GATE ECE 1990
+2
-0.6
The impulse response and the excitation function of a linear time invariant casual system are shown in Fig. a and b respectively. The output of the system at t = 2 sec. is equal to
A
0
B
1/2
C
3/2
D
1
EXAM MAP
Medical
NEET