# Time-invariant system

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 ${\displaystyle x(t)}$ produces an output ${\displaystyle y(t)}$ then any time shifted input, ${\displaystyle x(t+\delta )}$, results in a time-shifted output ${\displaystyle y(t+\delta )}$

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

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

Since system A explicitly depends on t outside of ${\displaystyle x(t)}$ and ${\displaystyle y(t)}$, 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 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 ${\displaystyle x_{d}(t)=\,\!x(t+\delta )}$
${\displaystyle y(t)=t\,x(t)}$
${\displaystyle y_{1}(t)=t\,x_{d}(t)=t\,x(t+\delta )}$
Now delay the output by ${\displaystyle \delta }$
${\displaystyle y(t)=t\,x(t)}$
${\displaystyle y_{2}(t)=\,\!y(t+\delta )=(t+\delta )x(t+\delta )}$
Clearly ${\displaystyle y_{1}(t)\,\!\neq y_{2}(t)}$, therefore the system is not time-invariant.

System B:

Start with a delay of the input ${\displaystyle x_{d}(t)=\,\!x(t+\delta )}$
${\displaystyle y(t)=10\,x(t)}$
${\displaystyle y_{1}(t)=10\,x_{d}(t)=10\,x(t+\delta )}$
Now delay the output by ${\displaystyle \,\!\delta }$
${\displaystyle y(t)=10\,x(t)}$
${\displaystyle y_{2}(t)=y(t+\delta )=10\,x(t+\delta )}$
Clearly ${\displaystyle y_{1}(t)=\,\!y_{2}(t)}$, therefore the system is time-invariant. Although there are many other proofs, this is the easiest.

## Abstract example

We can denote the shift operator by ${\displaystyle \mathbb {T} _{r}}$ where ${\displaystyle r}$ is the amount by which a vector's index set should be shifted. For example, the "advance-by-1" system

${\displaystyle x(t+1)=\,\!\delta (t+1)*x(t)}$

can be represented in this abstract notation by

${\displaystyle {\tilde {x}}_{1}=\mathbb {T} _{1}\,{\tilde {x}}}$

where ${\displaystyle {\tilde {x}}}$ is a function given by

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

with the system yielding the shifted output

${\displaystyle {\tilde {x}}_{1}=x(t+1)\,\forall \,t\in \mathbb {R} }$

So ${\displaystyle \mathbb {T} _{1}}$ is an operator that advances the input vector by 1.

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

${\displaystyle \mathbb {T} _{r}\,\mathbb {H} =\mathbb {H} \,\mathbb {T} _{r}\,\,\forall \,r}$

If our system equation is given by

${\displaystyle {\tilde {y}}=\mathbb {H} \,{\tilde {x}}}$

then it is time-invariant if we can apply the system operator ${\displaystyle \mathbb {H} }$ on ${\displaystyle {\tilde {x}}}$ followed by the shift operator ${\displaystyle \mathbb {T} _{r}}$, or we can apply the shift operator ${\displaystyle \mathbb {T} _{r}}$ followed by the system operator ${\displaystyle \mathbb {H} }$, with the two computations yielding equivalent results.

Applying the system operator first gives

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

Applying the shift operator first gives

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

If the system is time-invariant, then

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