Consider a discrete-time signal
$$x\left[ n \right] = \left\{ {\matrix{ {n\,\,for\,\,0 \le n \le 10} \cr {0\,\,otherwise} \cr } } \right.$$

If $$y\left[ n \right]$$ is the convolution of $$x\left[ n \right]$$ with itself, the value of $$y\left[ 4 \right]$$ is

Let $$y\left[ n \right]$$ denote the convolution of $$h\left[ n \right]$$ and $$g\left[ n \right]$$, where $$h\left[ n \right]$$ $$ = \,{\left( {1/2} \right)^2}\,\,u\left[ n \right]$$ and $$g\left[ n \right]\,$$ is a causal sequence. If $$y\left[ 0 \right]\,$$ $$ = \,1$$ and $$y\left[ 1 \right]\,$$ $$ = \,1/2,$$ then $$g\left[ 1 \right]$$ equals

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

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?