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<math>x(t) \longrightarrow x(t-t_0) \longrightarrow y(t)=(t-3)=x(t)=(t-3-t_0)\,</math> | <math>x(t) \longrightarrow x(t-t_0) \longrightarrow y(t)=(t-3)=x(t)=(t-3-t_0)\,</math> | ||
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If <math>x(t) \,</math> is first entered into the system, then time shifted: | If <math>x(t) \,</math> is first entered into the system, then time shifted: | ||
| − | <math>x(t) \longrightarrow y(t)=x(t-3) \longrightarrow \y(t-t_0)=x(t-t_0-3),</math> | + | <math>x(t) \longrightarrow y(t)=x(t-3) \longrightarrow y(t-t_0)=x(t-t_0-3)\,</math> |
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| + | Thus this system is T.I. | ||
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| + | == Time-Variant System == | ||
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| + | Consider the system: <math>y(t)=x(t^2-3) \,</math> | ||
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| + | If <math>x(t) \,</math> is first time shifted, then put into the system: | ||
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| + | <math>x(t) \longrightarrow x(t-t_0) \longrightarrow y(t)=x(t^2-3-t_0)\,</math> | ||
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| + | If <math>x(t) \,</math> is first entered into the system, then time shifted: | ||
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| + | <math>x(t) \longrightarrow y(t)=x(t^2-3) \longrightarrow y(t-t_0)=x((t-t_0)^2-3)\,</math> | ||
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| + | Thus this system isn't T.I. | ||
Latest revision as of 07:52, 11 September 2008
Time-Invariant System Definition
A time invariant system is a system that produces equivalent results for the following cases:
1. A time shifted input $ x(t+t_0) \, $ is entered into the system.
2. An input $ x(t) \, $ is entered into the system then time shifted by $ t_0 \, $.
Time-Invariant System
Consider the system: $ y(t)=x(t-3) \, $
If $ x(t) \, $ is first time shifted, then put into the system:
$ x(t) \longrightarrow x(t-t_0) \longrightarrow y(t)=(t-3)=x(t)=(t-3-t_0)\, $
If $ x(t) \, $ is first entered into the system, then time shifted:
$ x(t) \longrightarrow y(t)=x(t-3) \longrightarrow y(t-t_0)=x(t-t_0-3)\, $
Thus this system is T.I.
Time-Variant System
Consider the system: $ y(t)=x(t^2-3) \, $
If $ x(t) \, $ is first time shifted, then put into the system:
$ x(t) \longrightarrow x(t-t_0) \longrightarrow y(t)=x(t^2-3-t_0)\, $
If $ x(t) \, $ is first entered into the system, then time shifted:
$ x(t) \longrightarrow y(t)=x(t^2-3) \longrightarrow y(t-t_0)=x((t-t_0)^2-3)\, $
Thus this system isn't T.I.
