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This is accomplished using almost the same method as linearity. The system is: | This is accomplished using almost the same method as linearity. The system is: | ||
| − | <math>y(t)=2x(t)cos(t)</math> | + | <math>\,\!y(t)=2x(t)cos(t)</math> |
First we will take a signal and go through the system, then time delay it: | First we will take a signal and go through the system, then time delay it: | ||
| − | <math>y(t)=2x(t)cos(t)</math> , now time delay it, | + | <math>\,\!y(t)=2x(t)cos(t)</math> , now time delay it, |
| − | <math>y(t-k)=2x(t-k)cos(t-k | + | <math>\,\!z(t)=y(t-k)=2x(t-k)cos(t-k)</math> |
Next, we do the time shift first, then the system: | Next, we do the time shift first, then the system: | ||
| − | <math>y(t)= | + | <math>\,\!y(t)=x(t-k)</math> , now the system |
| − | <math> | + | <math>\,\!z(t)=2y(t)cos(t)=2x(t-k)cos(t-k)</math> |
| + | These results are equal, so it is time invariant | ||
=== Example of Time Variant function === | === Example of Time Variant function === | ||
| + | |||
| + | The system is: | ||
| + | |||
| + | <math>\,\!y(t)=t^2x(t)</math> | ||
| + | |||
| + | Do the same method: | ||
| + | |||
| + | <math>\,\!y(t)=t^2x(t)</math> | ||
| + | |||
| + | <math>\,\!z(t)=y(t-k)=(t-k)^2x(t-k)</math> | ||
| + | |||
| + | 2nd part: | ||
| + | |||
| + | <math>\,\!y(t)=x(t-k)</math> | ||
| + | |||
| + | <math>\,\!z(t)=t^2y(t)=t^2x(t-k)</math> | ||
| + | |||
| + | These results are not equal, so it is time variant, or non-time-invariant | ||
| + | |||
| + | Citing help with this concept from Sourabh Ranka's homework example | ||
Latest revision as of 15:05, 11 September 2008
Contents
Time Invariance
Background
A time invariant system refers to a system where the time has no affect on the systematic outputs of the function. In other words, if the function x is put through a system to create y, and that system then time delayed, the result should be the same as if that function was time delayed, then put through the system.
Example of Time Invariant function
This is accomplished using almost the same method as linearity. The system is:
$ \,\!y(t)=2x(t)cos(t) $
First we will take a signal and go through the system, then time delay it:
$ \,\!y(t)=2x(t)cos(t) $ , now time delay it,
$ \,\!z(t)=y(t-k)=2x(t-k)cos(t-k) $
Next, we do the time shift first, then the system:
$ \,\!y(t)=x(t-k) $ , now the system
$ \,\!z(t)=2y(t)cos(t)=2x(t-k)cos(t-k) $
These results are equal, so it is time invariant
Example of Time Variant function
The system is:
$ \,\!y(t)=t^2x(t) $
Do the same method:
$ \,\!y(t)=t^2x(t) $
$ \,\!z(t)=y(t-k)=(t-k)^2x(t-k) $
2nd part:
$ \,\!y(t)=x(t-k) $
$ \,\!z(t)=t^2y(t)=t^2x(t-k) $
These results are not equal, so it is time variant, or non-time-invariant
Citing help with this concept from Sourabh Ranka's homework example
