(New page: == 6A == <math>\,y(t)=e^{x(t)}\,</math> '''Proof:''' <math>x(t) \to System \to y(t)=e^{x(t)} \to Time Shift(t0) \to z(t)=y(t-t0)</math> <math>\, ...) |
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| − | <math>\,y(t)= | + | <math>\,y(t)=(a+1)^2x(t-a)}\,</math> |
Revision as of 17:21, 12 September 2008
6A
$ \,y(t)=(a+1)^2x(t-a)}\, $
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.
