A

Going through the system first and then the time shift:

x[n] -> y[n] = (k + 1)^2x[n-(k+1)]

y[n] -> z[n] = y[n-1]

 = (k+1)^2x[n-1-(k+1)]


Going through the time shift first and then the system:

x[n] -> y[n] = x[n-1]

y[n] -> z[n] = (k+1)^2y[n-(k+1)]

 = (k+1)^2x[n-1-(k+1)]

Since the two results are the same, the system is time invariant.


B

The input would have to be u[n]

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang