Time-invariant system

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A time-invariant (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 time-shifted 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 time-invariant then the system block commutes with an arbitrary delay.

If a time-invariant system is also linear, it is the subject of LTI system theory (linear time-invariant) with direct applications in NMR spectroscopy, seismology, circuits, signal processing, control theory, and other technical areas. Nonlinear time-invariant systems lack a comprehensive, governing theory. Discrete time-invariant systems are known as shift-invariant systems. Systems which lack the time-invariant property are studied as time-variant systems.

Simple example[edit]

To demonstrate how to determine if a system is time-invariant then 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 time-invariant. System B, however, does not depend explicitly on t so it is time-invariant.

Formal example[edit]

A more formal proof of why system A & B from 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 time-invariant.

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 time-invariant. Although there are many other proofs, this is the easiest.

Abstract example[edit]

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

\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 time-invariant 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 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

See also[edit]