# Time-invariant system

A time-invariant (TIV) 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::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:

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: $y(t)=10x(t)$ 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

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 $y_{1}(t)\,\!\neq y_{2}(t)$ , therefore the system is not time-invariant.

System B:

Start with a delay of the input $x_{d}(t)=\,\!x(t+\delta )$ $y(t)=10\,x(t)$ $y_{1}(t)=10\,x_{d}(t)=10\,x(t+\delta )$ 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

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

For time-invariant systems, the system properties remain constant with time, $\partial f/\partial t=0$ . 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 not time-invariant
$f_{B}=10x(t)\qquad \implies \qquad {\frac {\partial f_{B}}{\partial t}}=0$ so 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

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