a) $ Xk[n]=d[n-k] $ -> syst. -> $ Yk[n]=(k+1)^2 d[n-(k+1)] $

To see if the system is time invariant we must run it through time delay.

$ Xk[n]=d[n-k] $ -> syst. -> $ Yk[n]=(k+1)^2 d[n-(k+1)] $ -> delay -> $ Zk[n] = Yk[n+n0] = (k+1)^2 d[n-(k+1)+n0] $

Alternately: $ Xk[n]=d[n-k] $ -> delay -> $ Yk[n]=d[n-k+no] $ -> syst. -> $ Zk[n]=(k+n0+1)^2 d[n-(k+no+1)+no] $

The outputs are not the same, therefore the system is not time invariant.

b) k would have to be 0 because only then is d[2n-(k+1)] not affected.

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood