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'''Time invariant Signal'''
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'''Time Invariant Signal'''
--
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<math>S_1(t) = 5x(t)</math>
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<math>S_2(t) = t - t_0</math>
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<math>x(t) -> S_1(t) -> S_2(t) -> 5x(t - t_0)</math>
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<math>x(t) -> S_2(t) -> S_1(t) -> 5x(t - t_0)</math>
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This means that <math>S_1(t) = 5x(t)</math> is a time invariant signal.
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'''Time Variant Signal'''
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<math>S_1(t) = x(1-t)</math>
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<math>S_2(t) = t - t_0</math>
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<math>x(t) -> S_1(t) -> S_2(t) -> 5x(1 - t + t_0)</math>
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<math>x(t) -> S_2(t) -> S_1(t) -> 5x(1 - t - t_0)</math>
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This means that <math>S_1(t) = x(1-t)</math> is a time variant signal.

Latest revision as of 17:22, 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.

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin