(→6A) |
(→6B) |
||
| (2 intermediate revisions by the same user not shown) | |||
| Line 20: | Line 20: | ||
== 6B == | == 6B == | ||
If superposition works, integrating input will yield integration of output, | If superposition works, integrating input will yield integration of output, | ||
| − | integration of <math> u[n-1] </math> is \delta [n-1], thus u[n] as an input will yield the desired 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.
