1
GATE EE 2012
+2
-0.6
L et y[n] denote the convolution of h[n] and g[n], where $$h\left[n\right]=\left(1/2\right)^nu\left[n\right]$$ and g[n] is a causal sequence. If y[0] = 1 and y[1] = 1/2, then g[1] equals
A
0
B
1/2
C
1
D
3/2
2
GATE EE 2012
+2
-0.6
The input x(t) and output y(t) of a system are related as $$\int_{-\infty}^tx\left(\tau\right)\cos\left(3\tau\right)d\tau$$.The system is
A
time-invariant and stable
B
stable and not time-invariant
C
time-invariant and not stable
D
not time-invariant and not stable
3
GATE EE 2011
+2
-0.6
The response h(t) of a linear time invariant system to an impulse $$\delta\left(t\right)$$, under initially relaxed condition is $$h\left(t\right)=e^{-t}\;+\;e^{-2t}$$. The response of this system for a unit step input u(t) is
A
$$u\left(t\right)\;+\;e^{-t}\;+\;e^{-2t}$$
B
$$\left(e^{-t}\;+\;e^{-2t}\right)u\left(t\right)$$
C
$$\left(1.5\;-\;e^{-t}\;-\;0.5e^{-2t}\right)u\left(t\right)$$
D
$$\;e^{-t}\delta\left(t\right)\;+\;e^{-2t}u\left(t\right)$$
4
GATE EE 2010
+2
-0.6
Given the finite length input x[n] and the corresponding finite length output y[n] of an LTI system as shown below, the impulse response h[n] of the system is
A
$$\begin{array}{l}h\left[n\right]=\left\{1,\;0,\;0,\;1\right\}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\uparrow\end{array}$$
B
$$\begin{array}{l}h\left[n\right]=\left\{1,\;0,\;1\right\}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\uparrow\end{array}$$
C
$$\begin{array}{l}h\left[n\right]=\left\{1,\;1,\;1,\;1\right\}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\uparrow\end{array}$$
D
$$\begin{array}{l}h\left[n\right]=\left\{1,\;1,\;1\right\}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\uparrow\end{array}$$
EXAM MAP
Medical
NEET