Time-invariant system: Definition from Answers.com

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A time-invariant system is one whose output does not depend explicitly on time. That is, treating time as the independent variable, it is an autonomous system.

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 $t \mapsto y(t + \delta).$

Formally, if $S$ is the shifting operator ( $S δ x (t) = x (t - δ)$ ), then the operator $T$ is called time-invariant, if

T (S δ x) = S δ (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.

1 Simple example
2 Formal example
3 Abstract example
4 See also

Simple example

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

System A: $y(t) = t\cdot x(t)$
System B: $y(t) = 10\cdot x(t)$
System C: $y(t) = x(10 \cdot t)$

Since system A explicitly depends on t outside of $x (t)$ and $y (t)$ it is not time-invariant (i.e. time-variant). System B does not depend explicitly on t, so it is time-invariant. System C is not time-invariant because a time shift will result in a scaled shift.

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

δ

$y(t) = t\, x(t)$

$y_2(t) = \,\!y(t + \delta) = (t + \delta) x(t + \delta)$

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

System B:

Start with a delay of the input

$y_1(t) = 10 \,x_d(t)= 10 \,x(t + \delta)$

Now delay the output by

δ

$y_2(t) = y(t + \delta) = 10 \,x(t+\delta)$

$y_1(t) = \,\!y_2(t)$ , therefore the system is time-invariant.

System C:

Start with a delay of the input

$y_1(t) = x_d(10 \cdot t) = x(10 \cdot t + \delta)$

Now delay the output by

δ

$y_2(t) = \,\!y(t + \delta) = x(10\cdot(t + \delta ))$

Clearly $y_1(t) \,\!\ne y_2(t)$ , therefore the system is not 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

$\tilde{x}_1 = \mathbb{T}_1 \, \tilde{x}$

where $\tilde{x}$ is a function given by

$\forall t \in \mathbb{R}\ \tilde{x} = x(t)$

with the system yielding the shifted output

$\forall t \in \mathbb{R}\ \tilde{x}_1 = x(t + 1)$

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

$\tilde{y} = \mathbb{H} \, \tilde{x}$

then it is time-invariant if we can apply the system operator $\mathbb{H}$ on $\tilde{x}$ 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

$\mathbb{T}_r \, \mathbb{H} \, \tilde{x} = \mathbb{T}_r \, \tilde{y} = \tilde{y}_r$

Applying the shift operator first gives

$\mathbb{H} \, \mathbb{T}_r \, \tilde{x} = \mathbb{H} \, \tilde{x}_r$

If the system is time-invariant, then

$\mathbb{H} \, \tilde{x}_r = \tilde{y}_r$

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Time-invariant system

Contents

Simple example

Formal example

Abstract example

See also