A timeinvariant (TIV) system is a system whose output does not depend explicitly on time. Such systems are regarded as a class of systems in the field of system analysis. Lack of time dependence is captured in the following mathematical property of such a system:

If the input signal x(t) produces an output y(t) then any time shifted input, x(t + \delta), results in a timeshifted output y(t + \delta)
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.
In the context of a system schematic, this property can also be stated as follows:

If a system is timeinvariant then the system block commutes with an arbitrary delay.
If a timeinvariant system is also linear, it is the subject of LTI system theory (linear timeinvariant) with direct applications in NMR spectroscopy, seismology, circuits, signal processing, control theory, and other technical areas. Nonlinear timeinvariant systems lack a comprehensive, governing theory. Discrete timeinvariant systems are known as shiftinvariant systems. Systems which lack the timeinvariant property are studied as timevariant systems.
Contents

Simple example 1

Formal example 2

Abstract example 3

See also 4
Simple example
To demonstrate how to determine if a system is timeinvariant, consider the two systems:

System A: y(t) = t\, x(t)

System B: y(t) = 10 x(t)
Since system A explicitly depends on t outside of x(t) and y(t), it is not timeinvariant. System B, however, does not depend explicitly on t so it is timeinvariant.
Formal example
A more formal proof of why systems A and B above differ is now presented. To perform this proof, the second definition will be used.
System A:

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

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

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) \,\!\ne y_2(t), therefore the system is not timeinvariant.
System B:

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

y(t) = 10 \, x(t)

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 timeinvariant.
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 "advanceby1" 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

\tilde{x} = x(t) \, \forall \, t \in \mathbb{R}
with the system yielding the shifted output

\tilde{x}_1 = x(t + 1) \, \forall \, t \in \mathbb{R}
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 timeinvariant if it commutes with the shift operator, i.e.,

\mathbb{T}_r \, \mathbb{H} = \mathbb{H} \, \mathbb{T}_r \,\, \forall \, r
If our system equation is given by

\tilde{y} = \mathbb{H} \, \tilde{x}
then it is timeinvariant 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 timeinvariant, then

\mathbb{H} \, \tilde{x}_r = \tilde{y}_r
See also
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