# Time-invariant system

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:$^{:p.50}$
    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
    $\mathbb {H} {\tilde {x}}_{r}={\tilde {y}}_{r}$