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

A causal LTI system is described by the difference equation $$2y\left[ n \right] = ay\left[ {n - 2} \right] - 2x\left[ n \right] + \beta x\left[ {n - 1} \right].$$ The system is stable only if

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