Time-invariant system

A time-invariant system is one whose output does not depend explicitly on time.:If the input signal $x$ produces an output $y$ then any time shifted input, $t\; mapsto\; x(t\; +\; delta)$, results in a time-shifted output $t\; mapsto\; y(t\; +\; delta).$

Formal: If $S$ is the shifting operator ($S\_delta\; x(t)\; =\; x(t-delta)$),then the operator $T$ is called time-invariant, if

:$T(S\_delta\; x)\; =\; S\_delta\; (T\; x).$

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.This property can also be stated in another way in terms of a schematic

:If a system is time-invariant then the system block is commutative with an arbitrary delay.

Simple example

To demonstrate how to determine if a system is time-invariant then consider the two systems:
* System A: $y(t)\; =\; tcdot\; x(t)$
* System B: $y(t)\; =\; 10cdot\; x(t)$

Since system A explicitly depends on "t" outside of $x(t)$ and $y(t)$ it is time-variant. System B, however, does not depend explicitly on "t", so it is time-invariant.

Formal example

A more formal proof of the previous example is now presented.For this proof, the second definition will be used.

System A::Start with a delay of the input $x\_d(t)\; =\; ,!x(t\; +\; delta)$::$y\_1(t)\; =\; t,\; x\_d(t)\; =\; t,\; x(t\; +\; delta)$:Now delay the output by $delta$::$y(t)\; =\; t,\; x(t)$::$y\_2(t)\; =\; ,!y(t\; +\; delta)\; =\; (t\; +\; delta)\; x(t\; +\; delta)$:Clearly $y\_1(t)\; ,!\; e\; y\_2(t)$, therefore the system is not time-invariant.

System B::Start with a delay of the input $x\_d(t)\; =\; ,!x(t\; +\; delta)$::$y\_1(t)\; =\; 10\; ,x\_d(t)\; =\; 10\; ,x(t\; +\; delta)$: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. Although there are many other proofs, this is the easiest.

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

:$ilde\{x\}\_1\; =\; mathbb\{T\}\_1\; ,\; ilde\{x\}$

where $ilde\{x\}$ is a function given by

:$forall\; t\; in\; mathbb\{R\}\; ilde\{x\}\; =\; x(t)$

with the system yielding the shifted output

:$forall\; t\; in\; mathbb\{R\}\; ilde\{x\}\_1\; =\; x(t\; +\; 1)$

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.,

:$forall\; r\; mathbb\{T\}\_r\; ,\; mathbb\{H\}\; =\; mathbb\{H\}\; ,\; mathbb\{T\}\_r$

If our system equation is given by

:$ilde\{y\}\; =\; mathbb\{H\}\; ,\; ilde\{x\}$

then it is time-invariant if we can apply the system operator $mathbb\{H\}$ on $ilde\{x\}$ 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

:$mathbb\{T\}\_r\; ,\; mathbb\{H\}\; ,\; ilde\{x\}\; =\; mathbb\{T\}\_r\; ,\; ilde\{y\}\; =\; ilde\{y\}\_r$

Applying the shift operator first gives

:$mathbb\{H\}\; ,\; mathbb\{T\}\_r\; ,\; ilde\{x\}\; =\; mathbb\{H\}\; ,\; ilde\{x\}\_r$

If the system is time-invariant, then

:$mathbb\{H\}\; ,\; ilde\{x\}\_r\; =\; ilde\{y\}\_r$

See also

*Finite impulse response
*LTI system theory
*Sheffer sequence
*State space (controls)
*System analysis
*Time-variant system