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'''Time Invariant Signal''' | '''Time Invariant Signal''' | ||
| − | <math>S_1(t) = | + | <math>S_1(t) = 5x(t)</math> |
<math>S_2(t) = t - t_0</math> | <math>S_2(t) = t - t_0</math> | ||
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<math>x(t) -> S_2(t) -> S_1(t) -> 5x(t - t_0)</math> | <math>x(t) -> S_2(t) -> S_1(t) -> 5x(t - t_0)</math> | ||
| − | This means that <math>S_1(t) = | + | This means that <math>S_1(t) = 5x(t)</math> is a time invariant signal. |
| + | |||
| + | '''Time Variant Signal''' | ||
| + | |||
| + | <math>S_1(t) = x(1-t)</math> | ||
| + | |||
| + | <math>S_2(t) = t - t_0</math> | ||
| + | |||
| + | <math>x(t) -> S_1(t) -> S_2(t) -> 5x(1 - t + t_0)</math> | ||
| + | |||
| + | <math>x(t) -> S_2(t) -> S_1(t) -> 5x(1 - t - t_0)</math> | ||
| + | |||
| + | This means that <math>S_1(t) = x(1-t)</math> is a time variant signal. | ||
Latest revision as of 16:22, 12 September 2008
A system is time invariant if for any time shifted input signal the system produces a shifted output such that if an input $ x(t) $ produced an output $ y(t) $ then the input $ x(t + t_0) $ would produced the output $ y(t + t_0) $
Time Invariant Signal
$ S_1(t) = 5x(t) $
$ S_2(t) = t - t_0 $
$ x(t) -> S_1(t) -> S_2(t) -> 5x(t - t_0) $
$ x(t) -> S_2(t) -> S_1(t) -> 5x(t - t_0) $
This means that $ S_1(t) = 5x(t) $ is a time invariant signal.
Time Variant Signal
$ S_1(t) = x(1-t) $
$ S_2(t) = t - t_0 $
$ x(t) -> S_1(t) -> S_2(t) -> 5x(1 - t + t_0) $
$ x(t) -> S_2(t) -> S_1(t) -> 5x(1 - t - t_0) $
This means that $ S_1(t) = x(1-t) $ is a time variant signal.
