time-invariant

In control theory, a time-invariant (TI) system has a time-dependent system function that is not a direct function of time. Such systems are regarded as a class of systems in the field of system analysis. The time-dependent system function is a function of the time-dependent input function. If this function depends ''only'' indirectly on the time-domain (via the input function, for example), then that is a system that would be considered time-invariant. Conversely, any direct dependence on the time-domain of the system function could be considered as a "time-varying system".

Mathematically speaking, "time-invariance" of a system is the following property:

:''Given a system with a time-dependent output function , and a time-dependent input function , the system will be considered time-invariant if a time-delay on the input directly equates to a time-delay of the output function. For example, if time is "elapsed time", then "time-invariance" implies that the relationship between the input function and the output function is constant with respect to time ''
::$y(t) = f( x(t), t ) = f( x(t)).$

In the language of signal processing, this property can be satisfied if the transfer function of the system is not a direct function of time except as expressed by the input and output.

In the context of a system schematic, this property can also be stated as follows, as shown in the figure to the right:

:''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 linear time-invariant 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

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

* System A: $y(t) = t x(t)$
* System B: $y(t) = 10 x(t)$

Since the System Function $y(t)$ for system A explicitly depends on ''t'' outside of $x(t)$, it is not time-invariant because the time-dependence is not explicitly a function of the input function.

In contrast, system B's time-dependence is only a function of the time-varying input $x(t)$. This makes system B time-invariant.

The Formal Example below shows in more detail that while System B is a Shift-Invariant System as a function of time, ''t'', System A is not.

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 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.

More generally, the relationship between the input and output is

:$y(t) = f(x(t), t),$

and its variation with time is

:$\frac = \frac + \frac \frac.$

For time-invariant systems, the system properties remain constant with time,

:$\frac =0.$

Applied to Systems A and B above:

:$f_A = t x(t) \qquad \implies \qquad \frac = x(t) \neq 0$ in general, so it is not time-invariant,
:$f_B = 10 x(t) \qquad \implies \qquad \frac = 0$ so it is time-invariant.

Abstract example

We can denote the shift operator by $\mathbb_r$ 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

:$\tilde_1 = \mathbb_1 \tilde$

where $\tilde$ is a function given by

:$\tilde = x(t) \forall t \in \R$

with the system yielding the shifted output

:$\tilde_1 = x(t + 1) \forall t \in \R$

So $\mathbb_1$ is an operator that advances the input vector by 1.

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

:$\mathbb_r \mathbb = \mathbb \mathbb_r \forall r$

If our system equation is given by

:$\tilde = \mathbb \tilde$

then it is time-invariant if we can apply the system operator $\mathbb$ on $\tilde$ followed by the shift operator $\mathbb_r$, or we can apply the shift operator $\mathbb_r$ followed by the system operator $\mathbb$, with the two computations yielding equivalent results.

Applying the system operator first gives

:$\mathbb_r \mathbb \tilde = \mathbb_r \tilde = \tilde_r$

Applying the shift operator first gives

:$\mathbb \mathbb_r \tilde = \mathbb \tilde_r$

If the system is time-invariant, then

:$\mathbb \tilde_r = \tilde_r$

See also

* Finite impulse response
* Sheffer sequence
*State space (controls)
* Signal-flow graph
*LTI system theory
* Autonomous system (mathematics)

References

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Control theory
Signal processing