(→6A) |
(→6B) |
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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> |
'''Proof:''' | '''Proof:''' | ||
| − | <math>x(t) \to System \to y(t)= | + | <math>x(t-k) \to System \to y(t)=(k+1)^{2}x(t-k-1) \to Time Shift(t0) \to z(t)=y(t-t0)</math> |
| − | <math>\, = | + | <math>\, =(k+1)^{2}x(t-t0-k-1)\,</math> |
| − | <math>x(t) \to Time Shift(t0) \to y(t)=x(t-t0) \to System \to z(t)= | + | <math>x(t-k) \to Time Shift(t0) \to y(t)=x(t-t0-k) \to System \to z(t)=(k1+1)^{2}y(t-k1-1)</math> |
| − | <math>\, = | + | <math>\, =(t0+k+1)^{2}x(t-t0-k-1)\,</math> |
| + | It is not time invariant. | ||
| − | + | ||
| + | == 6B == | ||
| + | If superposition works, integrating input will yield integration of output, | ||
| + | integration of <math> u[n-1] </math> is <math>\delta[n-1]</math>, thus u[n] as an input will yield the desired output. | ||
Latest revision as of 17:33, 12 September 2008
6A
$ \,x(t-k) \to System \to y(t)=(k+1)^{2}x(t-k-1)\, $
Proof:
$ x(t-k) \to System \to y(t)=(k+1)^{2}x(t-k-1) \to Time Shift(t0) \to z(t)=y(t-t0) $
$ \, =(k+1)^{2}x(t-t0-k-1)\, $
$ x(t-k) \to Time Shift(t0) \to y(t)=x(t-t0-k) \to System \to z(t)=(k1+1)^{2}y(t-k1-1) $
$ \, =(t0+k+1)^{2}x(t-t0-k-1)\, $
It is not time invariant.
6B
If superposition works, integrating input will yield integration of output, integration of $ u[n-1] $ is $ \delta[n-1] $, thus u[n] as an input will yield the desired output.
