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 $$K$$

$$X\left( z \right) = 1 - 3\,\,{z^{ - 1}},\,\,Y\left( z \right) = 1 + 2\,\,{z^{ - 2}}$$ 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 $$\delta \left[ {n - 1} \right]$$

A signal is processed by a causal filter with transfer function $$G(s).$$ For a distortion free output signal waveform, $$G(s)$$ must.

$$G\left( z \right) = a{z^{ - 1}} + \beta \,\,{z^{ - 3}}$$ is a low-pass digital filter with a phase characteristic same as that of the above question if

A signal is processed by a causal filter with transfer function $$G(s).$$ For a distortion free output signal waveform, $$G(s)$$ must