(New page: == Time Invariance == If the cascade x(t)--->[time delay by t0]----->[system]-----z(t) ---(1) yields the same output as the reverse of (a);x(t)--->[system]--->[time delay by t0]---y(t)...) |
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x(t)--->[time delay by t0]--->y(t)=x(t-t0)----->[system]---->z(t)=2x(t-t0) ---(1) | x(t)--->[time delay by t0]--->y(t)=x(t-t0)----->[system]---->z(t)=2x(t-t0) ---(1) | ||
| − | x(t)--->[system]--->y(t)=2x(t)---->[time delay by y0]---->2x(t-t0) ---(2) | + | x(t)--->[system]--->y(t)=2x(t)---->[time delay by y0]---->z(t)=2x(t-t0) ---(2) |
The results of (1) and (2) are the same. Thus, it is time invariant. | The results of (1) and (2) are the same. Thus, it is time invariant. | ||
Latest revision as of 05:20, 12 September 2008
Time Invariance
If the cascade
x(t)--->[time delay by t0]----->[system]-----z(t) ---(1)
yields the same output as the reverse of (a);x(t)--->[system]--->[time delay by t0]---y(t), it is called Time invariant.
Prove
x(t)--->[system]-->y(t)=2x(t)
x(t)--->[time delay by t0]--->y(t)=x(t-t0)----->[system]---->z(t)=2x(t-t0) ---(1)
x(t)--->[system]--->y(t)=2x(t)---->[time delay by y0]---->z(t)=2x(t-t0) ---(2)
The results of (1) and (2) are the same. Thus, it is time invariant.
