1
GATE ECE 2011
MCQ (Single Correct Answer)
+1
-0.3
The differential equation $$100{{{d^2}y} \over {dt}} - 20{{dy} \over {dt}} + y = x\left( t \right)$$ describes a system with an input x(t) and output y(t). The system, which is initially relaxed, is excited by a unit step input. The output y(t) can be represented by the waveform
A
GATE ECE 2011 Signals and Systems - Continuous Time Linear Invariant System Question 40 English Option 1
B
GATE ECE 2011 Signals and Systems - Continuous Time Linear Invariant System Question 40 English Option 2
C
GATE ECE 2011 Signals and Systems - Continuous Time Linear Invariant System Question 40 English Option 3
D
GATE ECE 2011 Signals and Systems - Continuous Time Linear Invariant System Question 40 English Option 4
2
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
An input x(t) = exp( -2t) u(t) + $$\delta $$(t-6) is applied to an LTI system with impulse response h(t) = u(t). The output is
A
[ 1- exp( -2t)] u(t) + u(t+6)
B
[ 1- exp( -2t)] u(t) + u(t-6)
C
0.5 [1 - exp( -2t)] u(t) + u(t+6)
D
0.5 [1- exp( -2t)] u(t) + u(t-6)
3
GATE ECE 2011
MCQ (Single Correct Answer)
+1
-0.3
A system is defined by its impulse response $$h\left( n \right) = {2^n}\,u\left( {n - 2} \right).$$ The system is
A
stable and causal.
B
causal but not stable.
C
stable but not causal.
D
unstable and non causal.
4
GATE ECE 2011
MCQ (Single Correct Answer)
+2
-0.6
Two system $${H_1}\left( z \right)$$ and $${H_2}\left( z \right)$$ are connected in cascade as shown below. The overall output $$y\left( n \right)$$ is the same as the input $$x\left( n \right)$$ with a one unit delay. The transfer function of the second system $${H_2}\left( z \right)$$ is GATE ECE 2011 Signals and Systems - Discrete Time Linear Time Invariant Systems Question 10 English
A
$${{\left( {1 - 0.6\,{z^{ - 1}}} \right)} \over {{z^{ - 1}}\left( {1 - 0.4\,{z^{ - 1}}} \right)}}\,$$
B
$${{{z^{ - 1}}\left( {1 - 0.6\,{z^{ - 1}}} \right)} \over {\left( {1 - 0.4\,{z^{ - 1}}} \right)}}$$
C
$${{{z^{ - 1}}\left( {1 - 0.4\,{z^{ - 1}}} \right)} \over {\left( {1 - 0.6\,{z^{ - 1}}} \right)}}$$
D
$${{\left( {1 - 0.4\,{z^{ - 1}}} \right)} \over {{z^{ - 1}}\left( {1 - 0.6\,{z^{ - 1}}} \right)}}$$
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