Revision as of 18:23, 12 September 2008 by Lee251 (Talk)

6A

$ \,x(t-k) \to System \to y(t)=(k+1)^{2}x(t-k-1)\, $


Proof:

$ x(t) \to System \to y(t)=e^{x(t)} \to Time Shift(t0) \to z(t)=y(t-t0) $

$ \, =e^{x(t-t0)}\, $


$ x(t) \to Time Shift(t0) \to y(t)=x(t-t0) \to System \to z(t)=e^{y(t)} $

$ \, =e^{x(t-t0)}\, $


Both cascades yielded the same outputs, thus $ \,y(t)=e^{x(t)}\, $ is time invariant.

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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood