1
GATE ECE 2011
+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
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
$${{\left( {1 - 0.4\,{z^{ - 1}}} \right)} \over {{z^{ - 1}}\left( {1 - 0.6\,{z^{ - 1}}} \right)}}$$
2
GATE ECE 2010
+2
-0.6
The transfer function of a discrete time LTI system is given by
$$H\left( z \right) = {{2 - {3 \over 4}{z^{ - 1}}} \over {1 - {3 \over 4}{z^{ - 1}} + {1 \over 8}{z^{ - 2}}}}$$

Consider the following statements:
S1: The system is stable and causal for $$ROC:\,\,\,\left| z \right| > \,1/2$$
S2: The system is stable but not causal for $$ROC:\,\,\,\left| z \right| < \,1/4$$
S3: The system is neither stable nor causal for $$ROC:\,\,1/4\, < \,\left| z \right| < \,{\raise0.5ex\hbox{\scriptstyle 1} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{\scriptstyle 2}}$$

Which one of the following statements is valid?

A
Both S1 and S2 are true
B
Both S2 and S3 are true
C
Both S1 and S3 are true
D
S1, S2 and S3 are all true
3
GATE ECE 2008
+2
-0.6
A discrete time linear shift - invariant system has an impulse response $$h\left[ n \right]$$ with $$h\left[ 0 \right]$$ $$= 1,\,\,h\left[ 1 \right]\,\, = - 1,\,\,h\left[ 2 \right]\,\, = \,2$$, and zero otherwise. The system is given an input sequence $$x\left[ n \right]$$ with $$x\left[ 0 \right]$$ $$= \,x\left[ 2 \right]\, = \,1,$$ and zero otherwise. The number of nonzero samples in the output sequence $$y\left[ n \right]$$, and the value of $$y\left[ 2 \right]$$ are, respectively
A
5, 2
B
6, 2
C
6, 1
D
5, 3
4
GATE ECE 2006
+2
-0.6
A system with input $$x\left( n \right)$$ and output $$y\left( n \right)$$ is given as $$y\left( n \right)$$ $$= \left( {\sin {5 \over 6}\,\pi \,n} \right)x\left( n \right).$$ The system is
A
linear, stable and invertible.
B
non-linear, stable and non-invertible.
C
linear, stable and non-invertible.
D
linear, unstable and invertible.
EXAM MAP
Medical
NEET