1
GATE EE 2006
MCQ (Single Correct Answer)
+2
-0.6
$$y\left[ n \right]$$ denotes the output and $$x\left[ n \right]$$ denotes the input of a discrete-time system given by the difference equation $$y\left[ n \right] - 0.8y\left[ {n - 1} \right] = x\left[ n \right] + 1.25\,x\left[ {n + 1} \right].$$ Its right-sided impulse response is
A
causal
B
unbounded
C
periodic
D
non-negative
2
GATE EE 2006
MCQ (Single Correct Answer)
+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
$$\eqalign{ & h\left[ n \right] = 0;\,\,n < - 2,\,\,n > 1,\, \cr & h\left[ { - 2} \right] = h\left[ 1 \right] = - 2,\,h\left[ { - 1} \right] = - h\left[ 0 \right] = 3 \cr} $$
3
GATE EE 2006
MCQ (Single Correct Answer)
+2
-0.6
A continuous-time system is described by $$y\left( t \right) = {e^{ - |x\left( t \right)|}},$$ where $$y(t)$$ is the output and $$x(t)$$ is the input. $$y(t)$$ is bounded
A
only when $$x(t)$$ is bounded
B
only when $$x(t)$$ is non-negative
C
only for $$t \ge 0$$ if $$x(t)$$ is bounded for $$t \ge 0$$
D
even when $$x(t)$$ is not bounded
4
GATE EE 2006
MCQ (Single Correct Answer)
+1
-0.3
$$x(t)$$ is a real valued function of a real variable with period $$T.$$ Its trigonometric. Fourier Series expansion contains no terms of frequency
$$\omega = 2\pi \left( {2k} \right)/T;\,\,k = 1,2,........$$ Also, no sine terms are present. Then $$x(t)$$ satisfies the equation
A
$$x\left( t \right) = - x\left( {t - T} \right)$$
B
$$x\left( t \right) = x\left( {T - t} \right) = - x\left( { - t} \right)$$
C
$$x\left( t \right) = x\left( {T - t} \right) = - x\left( {t - T/2} \right)$$
D
$$x\left( t \right) = x\left( {t - T} \right) = - x\left( {t - T/2} \right)$$
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