# Solved Problems-2

Problems-2

Consider a system

*S*with input*x*[*n*] and output*y*[*n*] related by,*y*[

*n*] =

*x*[

*n*]{

*g*[

*n*] +

*g*[

*n*â€“ 1]}

- If
*g*[*n*] = 1, for all*n*, show that*S*is time-invariant. - If
*g*[*n*] =*n*, show that*S*is not time-invariant. - If
*g*[*n*] = 1 + (â€“1), show that^{n}*S*is time-invariant.

Solution

- If
*g*[*n*] = 1, for all*n*, then

For input

*x*[*n*] =*x*_{1}[*n*], output â€¦(i)For input

*x*[*n*] =*x*_{1}[*n*â€“*n*_{0}], output, â€¦(ii)From the condition of time-invariance, the output should be,

â€¦(iii)

From equations (ii) and (iii),

*y*_{2}[*n*] =*y*_{1}[*n*â€“*n*_{0}]Hence, the system is time-invariant.

- If
*g*[*n*] =*n,*then

For input

*x*[*n*] =*x*_{1}[*n*], output â€¦(i)For input

*x*[*n*] =*x*_{1}[*n*â€“*n*_{0}], output, â€¦(ii)From the condition of time-invariance, the output should be,

â€¦(iii)

From equations (ii) and (iii),

*y*_{2}[*n*] â‰*y*_{1}[*n*â€“*n*_{0}]Hence, the system is not time-invariant.

- If
*g*[*n*] = 1+ (â€“1), then^{n}

This relation is same as that of part (a).
Hence the system is time-invariant.