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:[1]:p. 50

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 :

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 spectroscopyseismologycircuitssignal processingcontrol 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 exampleEdit

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

  • System A: {\displaystyle 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 exampleEdit

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 {\displaystyle x_{d}(t)=x(t+\delta )}
{\displaystyle y(t)=tx(t)}
{\displaystyle y_{1}(t)=tx_{d}(t)=tx(t+\delta )}
Now delay the output by \delta
{\displaystyle y(t)=tx(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 )}
y(t)=10x(t)
{\displaystyle y_{1}(t)=10x_{d}(t)=10x(t+\delta )}
Now delay the output by \delta
y(t)=10x(t)
{\displaystyle y_{2}(t)=y(t+\delta )=10x(t+\delta )}
Clearly {\displaystyle y_{1}(t)=y_{2}(t)}, therefore the system is time-invariant.

More generally, the relationship between the input and output is

{\displaystyle y(t)=f(x(t),t),}

and its variation with time is

{\displaystyle {\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,

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

Applied to Systems A and B above:

{\displaystyle 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,
{\displaystyle f_{B}=10x(t)\qquad \implies \qquad {\frac {\partial f_{B}}{\partial t}}=0} so it is time-invariant.

Abstract exampleEdit

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

{\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 {\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 {\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.,

{\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 \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

{\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}}