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| A '''time-invariant''' (TIV) system is one whose output does not depend explicitly on time.
| | The title of the writer is Luther. Meter studying is where my main income comes from but quickly I'll be on my own. I currently live in Arizona but now I'm contemplating other options. The favorite hobby for him and his children is to drive and now he is trying to make cash with it.<br><br>Feel free to visit my weblog [http://Ipstube.Ips.k12.in.us/users/GPGK http://Ipstube.Ips.k12.in.us] |
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| :If the input signal <math>x(t)</math> produces an output <math>y(t)</math> then any time shifted input, <math>x(t + \delta)</math>, results in a time-shifted output <math>y(t + \delta)</math>
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| This property can be satisfied if the [[transfer function]] of the system is not a function of time except expressed by the input and output.
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| This property can also be stated in another way in terms of a schematic
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| :If a system is time-invariant then the system block is [[commutative]] with an arbitrary delay.
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| <!-- Insert picture showing this 2nd definition pictorially as a block diagram -->
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| == Simple example ==
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| To demonstrate how to determine if a system is time-invariant then consider the two systems:
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| * System A: <math>y(t) = t\, x(t)</math>
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| * System B: <math>y(t) = 10 x(t)</math>
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| Since system A explicitly depends on ''t'' outside of <math>x(t)</math> and <math>y(t)</math>, it is not time-invariant. System B, however, does not depend explicitly on ''t'' so it is time-invariant.
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| == Formal example ==
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| A more formal proof of why system A & B from above differ is now presented.
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| To perform this proof, the second definition will be used.
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| System A:
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| :Start with a delay of the input <math>x_d(t) = \,\!x(t + \delta)</math>
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| ::<math>y(t) = t\, x(t)</math>
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| ::<math>y_1(t) = t\, x_d(t) = t\, x(t + \delta)</math>
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| :Now delay the output by <math>\delta</math>
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| ::<math>y(t) = t\, x(t)</math>
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| ::<math>y_2(t) = \,\!y(t + \delta) = (t + \delta) x(t + \delta)</math>
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| :Clearly <math>y_1(t) \,\!\ne y_2(t)</math>, therefore the system is not time-invariant.
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| System B:
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| :Start with a delay of the input <math>x_d(t) = \,\!x(t + \delta)</math>
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| ::<math>y(t) = 10 \, x(t)</math> | |
| ::<math>y_1(t) = 10 \,x_d(t) = 10 \,x(t + \delta)</math>
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| :Now delay the output by <math>\,\!\delta</math>
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| ::<math>y(t) = 10 \,x(t)</math>
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| ::<math>y_2(t) = y(t + \delta) = 10 \,x(t + \delta)</math>
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| :Clearly <math>y_1(t) = \,\!y_2(t)</math>, therefore the system is time-invariant. Although there are many other proofs, this is the easiest.
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| == Abstract example ==
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| We can denote the '''[[shift operator]]''' by <math>\mathbb{T}_r</math> where <math>r</math> is the amount by which a vector's [[parameter|index set]] should be shifted. For example, the "advance-by-1" system
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| :<math>x(t+1) = \,\!\delta(t+1) * x(t)</math>
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| can be represented in this abstract notation by
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| :<math>\tilde{x}_1 = \mathbb{T}_1 \, \tilde{x}</math>
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| where <math>\tilde{x}</math> is a function given by
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| :<math>\tilde{x} = x(t) \, \forall \, t \in \mathbb{R}</math> | |
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| with the system yielding the shifted output
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| :<math>\tilde{x}_1 = x(t + 1) \, \forall \, t \in \mathbb{R}</math>
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| So <math>\mathbb{T}_1</math> is an operator that advances the input vector by 1.
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| Suppose we represent a system by an [[Operator (mathematics)|operator]] <math>\mathbb{H}</math>. This system is '''time-invariant''' if it [[Commutative operation|commutes]] with the shift operator, i.e.,
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| :<math>\mathbb{T}_r \, \mathbb{H} = \mathbb{H} \, \mathbb{T}_r \,\, \forall \, r</math>
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| If our system equation is given by
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| :<math>\tilde{y} = \mathbb{H} \, \tilde{x}</math>
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| then it is time-invariant if we can apply the system operator <math>\mathbb{H}</math> on <math>\tilde{x}</math> followed by the shift operator <math>\mathbb{T}_r</math>, or we can apply the shift operator <math>\mathbb{T}_r</math> followed by the system operator <math>\mathbb{H}</math>, with the two computations yielding equivalent results.
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| Applying the system operator first gives
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| :<math>\mathbb{T}_r \, \mathbb{H} \, \tilde{x} = \mathbb{T}_r \, \tilde{y} = \tilde{y}_r</math>
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| Applying the shift operator first gives
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| :<math>\mathbb{H} \, \mathbb{T}_r \, \tilde{x} = \mathbb{H} \, \tilde{x}_r</math>
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| If the system is time-invariant, then
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| :<math>\mathbb{H} \, \tilde{x}_r = \tilde{y}_r</math>
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| == See also ==
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| *[[Finite impulse response]]
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| *[[LTI system theory]]
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| *[[Sheffer sequence]]
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| *[[State space (controls)]]
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| *[[System analysis]]
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| *[[Time-variant system]]
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| *[[Shift invariant system]]
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| [[Category:Control theory]]
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| [[Category:Signal processing]]
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The title of the writer is Luther. Meter studying is where my main income comes from but quickly I'll be on my own. I currently live in Arizona but now I'm contemplating other options. The favorite hobby for him and his children is to drive and now he is trying to make cash with it.
Feel free to visit my weblog http://Ipstube.Ips.k12.in.us