| Line 11: | Line 11: | ||
== part b == | == part b == | ||
| − | since it is linear | + | since it is linear |
| + | input would be X[n] = u[n] | ||
Latest revision as of 19:36, 11 September 2008
part a
it can't be a time invariant because output is not same.
$ X_k[n] = \delta[n-k] $
$ \delta[n-k]\rightarrow time delay\rightarrow\delta[n-k-n_0]\rightarrow system \rightarrow (k+n_0+1)^2\delta[n-(k+n_0+1)] $
$ \delta[n-k]\rightarrow system \rightarrow Y_k[n] \rightarrow time delay \rightarrow (k+1)^2\delta[n-(k+n_0+1)] $
part b
since it is linear input would be X[n] = u[n]
