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## Summary

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". Block diagram illustrating the time invariance for a deterministic continuous-time single-input single-output system. The system is time-invariant if and only if y2(t) = y1(t – t0) for all time t, for all real constant t0 and for all input x1(t). Click image to expand it.

Mathematically speaking, "time-invariance" of a system is the following property:: p. 50

Given a system with a time-dependent output function $y(t)$ , and a time-dependent input function $x(t)$ , the system will be considered time-invariant if a time-delay on the input $x(t+\delta )$ directly equates to a time-delay of the output $y(t+\delta )$ function. For example, if time $t$ is "elapsed time", then "time-invariance" implies that the relationship between the input function $x(t)$ and the output function $y(t)$ is constant with respect to time $t:$ $y(t)=f(x(t),t)=f(x(t)).$ 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: $y(t)=tx(t)$
• System B: $y(t)=10x(t)$

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 $x_{d}(t)=x(t+\delta )$
$y(t)=tx(t)$
$y_{1}(t)=tx_{d}(t)=tx(t+\delta )$
Now delay the output by $\delta$
$y(t)=tx(t)$
$y_{2}(t)=y(t+\delta )=(t+\delta )x(t+\delta )$
Clearly $y_{1}(t)\neq y_{2}(t)$ , therefore the system is not time-invariant.
System B: Start with a delay of the input $x_{d}(t)=x(t+\delta )$
$y(t)=10x(t)$
$y_{1}(t)=10x_{d}(t)=10x(t+\delta )$
Now delay the output by $\delta$
$y(t)=10x(t)$
$y_{2}(t)=y(t+\delta )=10x(t+\delta )$
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

${\frac {\mathrm {d} y}{\mathrm {d} t}}={\frac {\partial f}{\partial t}}+{\frac {\partial f}{\partial x}}{\frac {\mathrm {d} x}{\mathrm {d} t}}.$

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

${\frac {\partial f}{\partial t}}=0.$

Applied to Systems A and B above:

$f_{A}=tx(t)\qquad \implies \qquad {\frac {\partial f_{A}}{\partial t}}=x(t)\neq 0$  in general, so it is not time-invariant,
$f_{B}=10x(t)\qquad \implies \qquad {\frac {\partial f_{B}}{\partial t}}=0$  so it is time-invariant.

## 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

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

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

$\mathbb {H} {\tilde {x}}_{r}={\tilde {y}}_{r}$