$$X\left( z \right) = 1 - 3\,\,{z^{ - 1}},\,\,Y\left( z \right) = 1 + 2\,\,{z^{ - 2}}$$ are $$Z$$-transforms of two signals $$x\left[ n \right],\,\,y\left[ n \right]$$ respectively. A linear time invariant system has the impulse response $$h\left[ n \right]$$ defined by these two signals as $$h\left[ n \right] = x\left[ {n - 1} \right] * y\left[ n \right]$$ where $$ * $$ denotes discrete time convolution. Then the output of the system for the input $$\delta \left[ {n - 1} \right]$$

Consider the discrete-time system shown in the figure where the impulse response of $$G\left( z \right)$$ is
$$g\left( 0 \right) = 0,\,\,g\left( 1 \right) = g\left( 2 \right) = 1,\,g\left( 3 \right) = g\left( 4 \right) = .... = 0$$

This system is stable for range of values of $$K$$

$$y\left[ n \right]$$ denotes the output and $$x\left[ n \right]$$ denotes the input of a discrete-time system given by the difference equation $$y\left[ n \right] - 0.8y\left[ {n - 1} \right] = x\left[ n \right] + 1.25\,x\left[ {n + 1} \right].$$ Its right-sided impulse response is

A continuous-time system is described by $$y\left( t \right) = {e^{ - |x\left( t \right)|}},$$ where $$y(t)$$ is the output and $$x(t)$$ is the input. $$y(t)$$ is bounded