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A time-invariant (TIV) system is one whose output does not depend explicitly on time.

If the input signal $x(t)$ produces an output $y(t)$ then any time shifted input, $x(t+\delta )$, results in a time-shifted output $y(t+\delta )$

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. 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: $y(t)=tx(t)$
• 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 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 $x_d(t)=x(t+\delta )$

$y(t)=tx(t)$

$y_1(t)=t{x}_d(t)=tx(t+\delta )$

Now delay the output by $\delta$

$y(t)=tx(t)$

$y_2(t)=y(t+\delta )=(t+\delta )x(t+\delta )$

Clearly $y_1(t)\ne y_2(t)$, 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. Although there are many other proofs, this is the easiest.

Abstract example

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

$x(t+1)=\delta (t+1)\ast x(t)$

can be represented in this abstract notation by

${\tilde{x}}_1={𝕋}_1\tilde{x}$

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

$\tilde{x}=x(t)\mathrm{\forall }t\in ℝ$

with the system yielding the shifted output

${\tilde{x}}_1=x(t+1)\mathrm{\forall }t\in ℝ$

So ${𝕋}_1$ is an operator that advances the input vector by 1.

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

${𝕋}_rℍ=ℍ{𝕋}_r\mathrm{\forall }r$

If our system equation is given by

$\tilde{y}=ℍ\tilde{x}$

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

Applying the system operator first gives

${𝕋}_rℍ\tilde{x}={𝕋}_r\tilde{y}={\tilde{y}}_r$

Applying the shift operator first gives

$ℍ{𝕋}_r\tilde{x}=ℍ{\tilde{x}}_r$

If the system is time-invariant, then

$ℍ{\tilde{x}}_r={\tilde{y}}_r$

See also

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