Question 2

Repeat Question 1 for the same x(n) but

$$h(k) = \left \{ \begin{array}{lll} 1 & \mbox{for $k=0$ }\\ ... ...ox{for $k=1$ }\\ 0 & \mbox{otherwise} \end {array} \right.$$

Solution.

The response y(n) can be expressed as the same way in Question 1.

$$\begin{eqnarray*}y \left (n \right) & = & \sum_{ k= 0 }^{ 1 } h(k)x(n-k)\\ & = & h(0)x(n)+h(1)x(n-1)\\ & = & x(n)-x(n-1) \end{eqnarray*}$$

n	x(n)	x(n-1)	y(n)
0	1	0	1
1	1	1	0
2	0	1	-1
3	0	0	0