Time-invariant system

Time-invariant system

A time-invariant system is one whose output does not depend explicitly on time.:If the input signal x produces an output y then any time shifted input, t mapsto x(t + delta), results in a time-shifted output t mapsto y(t + delta).

Formal: If S is the shifting operator (S_delta x(t) = x(t-delta)),then the operator T is called time-invariant, if

:T(S_delta x) = S_delta (T x).

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.This property can also be stated in another way in terms of a schematic

:If a system is time-invariant then the system block is commutative with an arbitrary delay.

Simple example

To demonstrate how to determine if a system is time-invariant then consider the two systems:
* System A: y(t) = tcdot x(t)
* System B: y(t) = 10cdot x(t)

Since system A explicitly depends on "t" outside of x(t) and y(t) it is time-variant. System B, however, does not depend explicitly on "t", so it is time-invariant.

Formal example

A more formal proof of the previous example is now presented.For this proof, the second definition will be used.

System A::Start with a delay of the input x_d(t) = ,!x(t + delta)::y_1(t) = t, x_d(t) = t, x(t + delta):Now delay the output by delta::y(t) = t, x(t)::y_2(t) = ,!y(t + delta) = (t + delta) x(t + delta):Clearly y_1(t) ,! e y_2(t), therefore the system is not time-invariant.

System B::Start with a delay of the input x_d(t) = ,!x(t + delta)::y_1(t) = 10 ,x_d(t) = 10 ,x(t + delta):Now delay the output by ,!delta::y(t) = 10 , x(t)::y_2(t) = y(t + delta) = 10 ,x(t + delta):Clearly y_1(t) = ,!y_2(t), therefore the system is time-invariant. Although there are many other proofs, this is the easiest.

Abstract example

We can denote the shift operator by mathbb{T}_r where r is the amount by which a vector's index set should be shifted. For example, the "advance-by-1" system

:x(t+1) = ,!delta(t+1) * x(t)

can be represented in this abstract notation by

: ilde{x}_1 = mathbb{T}_1 , ilde{x}

where ilde{x} is a function given by

:forall t in mathbb{R} ilde{x} = x(t)

with the system yielding the shifted output

:forall t in mathbb{R} ilde{x}_1 = x(t + 1)

So mathbb{T}_1 is an operator that advances the input vector by 1.

Suppose we represent a system by an operator mathbb{H}. This system is time-invariant if it commutes with the shift operator, i.e.,

:forall r mathbb{T}_r , mathbb{H} = mathbb{H} , mathbb{T}_r

If our system equation is given by

: ilde{y} = mathbb{H} , ilde{x}

then it is time-invariant if we can apply the system operator mathbb{H} on ilde{x} followed by the shift operator mathbb{T}_r, or we can apply the shift operator mathbb{T}_r followed by the system operator mathbb{H}, with the two computations yielding equivalent results.

Applying the system operator first gives

:mathbb{T}_r , mathbb{H} , ilde{x} = mathbb{T}_r , ilde{y} = ilde{y}_r

Applying the shift operator first gives

:mathbb{H} , mathbb{T}_r , ilde{x} = mathbb{H} , ilde{x}_r

If the system is time-invariant, then

:mathbb{H} , ilde{x}_r = ilde{y}_r

See also

*Finite impulse response
*LTI system theory
*Sheffer sequence
*State space (controls)
*System analysis
*Time-variant system

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