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 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 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

The impulse response $$h\left[ n \right]$$ of a linear time invariant system is given as
$$h\left[ n \right] = \left\{ {\matrix{ { - 2\sqrt 2 ,} & {n = 1, - 1} \cr {4\sqrt 2 ,} & {n = 2, - 2} \cr {0,} & {otherwise} \cr } } \right.$$

If the input to the above system is the sequence $${e^{j\pi n/4}},$$ then the output is