1
GATE EE 2010
MCQ (Single Correct Answer)
+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 GATE EE 2010 Signals and Systems - Linear Time Invariant Systems Question 30 English
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
$$\begin{array}{l}h\left[n\right]=\left\{1,\;1,\;1\right\}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\uparrow\end{array}$$
2
GATE EE 2009
MCQ (Single Correct Answer)
+2
-0.6
A cascade of 3 Linear Time Invariant systems is casual and unstable. From this, we conclude that
A
each system in the cascade is individually casual and unstable
B
at least one system is unstable and atleast one system is casual
C
at least one system is casual and all systems are unstable
D
the majority are unstable and the majority are casual
3
GATE EE 2009
MCQ (Single Correct Answer)
+2
-0.6
The $$z$$$$-$$ transform of a signal $$x\left[ n \right]$$ is given by $$4{z^{ - 3}} + 3{z^{ - 1}} + 2 - 6{z^2} + 2{z^3}.$$ It is applied to a system, with a transfer function $$H\left( z \right) = 3{z^{ - 1}} - 2.$$ Let the output be $$y(n)$$. Which of the following is true?
A
$$y\left( n \right)$$ is non causal with finite support
B
$$y\left( n \right)$$ is causal with infinite support
C
$$y\left( n \right)$$ $$ = 0;\,|n| > 3$$
D
$$\eqalign{ & {\mathop{\rm Re}\nolimits} {\left[ {Y\left( z \right)} \right]_{z = {e^{j0}}}} = - {\mathop{\rm Re}\nolimits} {\left[ {Y\left( z \right)} \right]_{z = {e^{j0}}}}; \cr & {\rm I}m{\left[ {Y\left( z \right)} \right]_{z = {e^{j0}}}}\, = {\rm I}m{\left[ {Y\left( z \right)} \right]_z} = {e^{j0}};\,\, - \pi \le \theta < \pi \cr} $$
4
GATE EE 2008
MCQ (Single Correct Answer)
+2
-0.6
A system with input $$x(t)$$ and output $$y(t)$$ is defined by the input $$-$$ output relation:
$$y\left( t \right) = \int\limits_{ - \infty }^{ - 2t} {x\left( \tau \right)} d\tau .$$ The system will be
A
causal, time $$-$$ invariant and unstable
B
causal, time $$-$$ invariant and stable
C
non $$-$$ causal, time $$-$$ invariant and unstable
D
non $$-$$ causal, time $$-$$ variant and unstable
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