(New page: 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 th...) |
(No difference)
|
Revision as of 14:41, 12 September 2008
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.
