Time-invariant - Definition and Overview

A system that is time-invariant is a system that has an output that does not depend explicitly on time.

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>

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.

Contents

Simple example

To demostrate how to determine if a system is time-invariant then consider the two systems:

System A: <math>y(t) = t\, x(t)<math>
System B: <math>\,\!b(t) = 10 x(t)<math>

Since system A explicitly depends on t outside of <math>x(t)<math> and <math>y(t)<math> then 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 why system A & B from above is now presented. To perform this proof, the second definition will be used.

System A:

Start with a delay of the input <math>x_d(t) = \,\!x(t + \delta)<math>

<math>y_1(t) = t\, x_d(t) = t\, x(t + \delta)<math>

Now delay the output by <math>\delta<math>

<math>y_2(t) = \,\!y(t + \delta) = (t + \delta) x(t + \delta)<math>

Clearly <math>y_1(t) \,\!\ne y_2(t)<math>, therefore the system is not time-invariant.

System B:

Start with a delay of the input <math>x_d(t) = \,\!x(t + \delta)<math>

<math>y_1(t) = 10 \,x_d(t) = 10 \,x(t + \delta)<math>

Now delay the output by <math>\,\!\delta<math>

<math>y_2(t) = y(t + \delta) = 10 \,x(t + \delta)<math>

Clearly <math>y_1(t) = \,\!y_2(t)<math>, therefore the system is time-invariant.

Abstract example

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

<math>x(t+1) = \,\!\delta(t+1) * x(t)<math>

can be represented in this abstract notation by

<math>\tilde{x}_1 = \mathbb{T}_1 \, \tilde{x}<math>

where <math>\tilde{x}<math> is a function given by

<math>\tilde{x} = x(t) \, \forall \, t \in \mathbb{R}<math>

with the system yielding the shifted output

<math>\tilde{x}_1 = x(t + 1) \, \forall \, t \in \mathbb{R}<math>

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

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

<math>\mathbb{T}_r \, \mathbb{H} = \mathbb{H} \, \mathbb{T}_r \,\, \forall \, r<math>

If our system equation is given by

<math>\tilde{y} = \mathbb{H} \, \tilde{x}<math>

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.

Applying the system operator first gives

<math>\mathbb{T}_r \, \mathbb{H} \, \tilde{x} = \mathbb{T}_r \, \tilde{y} = y_r<math>

Applying the shift operator first gives

<math>\mathbb{H} \, \mathbb{T}_r \, \tilde{x} = \mathbb{H} \, \tilde{x}_r<math>

If the system is time-invariant, then

<math>\mathbb{H} \, \tilde{x}_r = \tilde{y}_r<math>