Definitions

# 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, $t mapsto x\left(t + delta\right)$, results in a time-shifted output $t mapsto y\left(t + delta\right).$

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

$T\left(S_delta x\right) = S_delta \left(T x\right).$

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: $y\left(t\right) = tcdot x\left(t\right)$
• System B: $y\left(t\right) = 10cdot x\left(t\right)$

Since system A explicitly depends on t outside of $x\left(t\right)$ and $y\left(t\right)$ 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 $x_d\left(t\right) = ,!x\left(t + delta\right)$
$y_1\left(t\right) = t, x_d\left(t\right) = t, x\left(t + delta\right)$
Now delay the output by $delta$
$y\left(t\right) = t, x\left(t\right)$
$y_2\left(t\right) = ,!y\left(t + delta\right) = \left(t + delta\right) x\left(t + delta\right)$
Clearly $y_1\left(t\right) ,!ne y_2\left(t\right)$, therefore the system is not time-invariant.

System B:

Start with a delay of the input $x_d\left(t\right) = ,!x\left(t + delta\right)$
$y_1\left(t\right) = 10 ,x_d\left(t\right) = 10 ,x\left(t + delta\right)$
Now delay the output by $,!delta$
$y\left(t\right) = 10 , x\left(t\right)$
$y_2\left(t\right) = y\left(t + delta\right) = 10 ,x\left(t + delta\right)$
Clearly $y_1\left(t\right) = ,!y_2\left(t\right)$, 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 $mathbb\left\{T\right\}_r$ where $r$ is the amount by which a vector's index set should be shifted. For example, the "advance-by-1" system

$x\left(t+1\right) = ,!delta\left(t+1\right) * x\left(t\right)$

can be represented in this abstract notation by

$tilde\left\{x\right\}_1 = mathbb\left\{T\right\}_1 , tilde\left\{x\right\}$

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

$forall t in mathbb\left\{R\right\} tilde\left\{x\right\} = x\left(t\right)$

with the system yielding the shifted output

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

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

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

$forall r mathbb\left\{T\right\}_r , mathbb\left\{H\right\} = mathbb\left\{H\right\} , mathbb\left\{T\right\}_r$

If our system equation is given by

$tilde\left\{y\right\} = mathbb\left\{H\right\} , tilde\left\{x\right\}$

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

Applying the system operator first gives

$mathbb\left\{T\right\}_r , mathbb\left\{H\right\} , tilde\left\{x\right\} = mathbb\left\{T\right\}_r , tilde\left\{y\right\} = tilde\left\{y\right\}_r$

Applying the shift operator first gives

$mathbb\left\{H\right\} , mathbb\left\{T\right\}_r , tilde\left\{x\right\} = mathbb\left\{H\right\} , tilde\left\{x\right\}_r$

If the system is time-invariant, then

$mathbb\left\{H\right\} , tilde\left\{x\right\}_r = tilde\left\{y\right\}_r$