1
GATE EE 2006
+2
-0.6
$$x\left[ n \right] = 0;\,n < - 1,\,n > 0,\,x\left[ { - 1} \right] = - 1,\,x\left[ 0 \right]$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2$$ is the input and
$$y\left[ n \right] = 0;\,n < - 1,\,n > 2,\,y\left[ { - 1} \right] = - 1,\, = y\left[ 1 \right],\,y\left[ 0 \right] = 3,\,y\left[ 2 \right]$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =- 2$$ is the output of a discrete-time $$LTI$$ system. The system impulse response $$h\left[ n \right]$$ will be
A
\eqalign{ & h\left[ n \right] = 0;\,\,n < 0,\,\,n > 2, \cr & h\left[ 0 \right] = 1,\,h\left[ 1 \right] = h\left[ 2 \right] = - 1 \cr}
B
\eqalign{ & h\left[ n \right] = 0;\,\,n < - 1,\,\,n > 1, \cr & h\left[ { - 1} \right] = 1,\,h\left[ 0 \right] = h\left[ 1 \right] = 2 \cr}
C
\eqalign{ & h\left[ n \right] = 0;\,\,n < 0,\,\,n \ge 3,\,h\left[ 0 \right] = - 1, \cr & h\left[ 1 \right] = 2,\,h\left[ 2 \right] = 1 \cr}
D
\eqalign{ & h\left[ n \right] = 0;\,\,n < - 2,\,\,n > 1,\, \cr & h\left[ { - 2} \right] = h\left[ 1 \right] = - 2,\,h\left[ { - 1} \right] = - h\left[ 0 \right] = 3 \cr}
2
GATE EE 2004
+2
-0.6
In the system shown in Fig. the input $$x\left( t \right) = \sin t.$$ In the steady-state, the response $$y(t)$$ will be
A
$${1 \over {\sqrt 2 }}\,\sin \left( {t - {{45}^ \circ }} \right)$$
B
$${1 \over {\sqrt 2 }}\,\sin \left( {t + {{45}^ \circ }} \right)$$
C
$$\sin \left( {t - {{45}^ \circ }} \right)$$
D
$$\sin \left( {t + {{45}^ \circ }} \right)$$
3
GATE EE 2001
+2
-0.6
Given the relationship between the input $$u(t)$$ and the output $$y(t)$$ to be
$$y\left( t \right) = \int\limits_0^t {\left( {2 + t - \tau } \right){e^{ - 3\left( {t - \tau } \right)}}} u\left( \tau \right)d\tau$$
the transfer function $$Y\left( s \right)/U\left( s \right)$$ is
A
$${{2{e^{ - 2s}}} \over {s + 3}}$$
B
$${{s + 2} \over {{{\left( {s + 3} \right)}^2}}}$$
C
$${{2s + 5} \over {s + 3}}$$
D
$${{2s + 7} \over {{{\left( {s + 3} \right)}^2}}}$$
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