Given the following system: Input $ \to $ Output

$ X0[n] = \delta[n] \to Y0[n] = \delta[n-1] $

$ X1[n] = \delta[n-1] \to Y1[n] = 4\delta[n-2] $

$ X2[n] = \delta[n-2] \to Y2[n] = 9\delta[n-3] $

$ X3[n] = \delta[n-3] \to Y3[n] = 16\delta[n-4] $

...

$ Xk[n] = \delta[n-k] \to Yk[n] = (k+1)^{2}\delta[n-(k+1)] $

Part A

Is this system Time Invariant?

Using $ X1[n] = \delta[n-1] $


Delay by 1: $ Y1[n] = \delta[n-2] $

Feed into system: $ Z1[n] = 9\delta[n-3] $


This time, feed into the system first: $ Y1[n] = 4\delta[n-2] $ Delay by 1: $ Z1[n] = 4\delta[n-3] $

Since $ 9\delta[n-3] \ne 4\delta[n-3] $ the system is NOT Time Invariant.

Part B

Assuming that this system is linear, what input $ X[n] $ would yield the output $ Y[n]=u[n-1] $?

Since the system is linear, the input would be $ X[n] = u[n] $ based on the system we were given.

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood