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== 6A == | == 6A == | ||
| − | <math>\,y(t)=( | + | <math>\,x(t-k) \to System \to y(t)=(k+1)^{2}x(t-k-1)\,</math> |
Revision as of 17:23, 12 September 2008
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.
