Editing Time-invariant system (section)

== Abstract example ==

We can denote the '''[[shift operator]]''' by <math>\mathbb{T}_r</math> where <math>r</math> is the amount by which a vector's [[parameter|index set]] should be shifted. For example, the "advance-by-1" system

:<math>x(t+1) = \delta(t+1) * x(t)</math>

can be represented in this abstract notation by

:<math>\tilde{x}_1 = \mathbb{T}_1 \tilde{x}</math>

where <math>\tilde{x}</math> is a function given by

:<math>\tilde{x} = x(t) \forall t \in \R</math>

with the system yielding the shifted output

:<math>\tilde{x}_1 = x(t + 1) \forall t \in \R</math>

So <math>\mathbb{T}_1</math> is an operator that advances the input vector by 1.

Suppose we represent a system by an [[Operator (mathematics)|operator]] <math>\mathbb{H}</math>. This system is '''time-invariant''' if it [[Commutative operation|commutes]] with the shift operator, i.e.,

:<math>\mathbb{T}_r \mathbb{H} = \mathbb{H} \mathbb{T}_r \forall r</math>

If our system equation is given by

:<math>\tilde{y} = \mathbb{H} \tilde{x}</math>

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

Applying the system operator first gives

:<math>\mathbb{T}_r \mathbb{H} \tilde{x} = \mathbb{T}_r \tilde{y} = \tilde{y}_r</math>

Applying the shift operator first gives

:<math>\mathbb{H} \mathbb{T}_r \tilde{x} = \mathbb{H} \tilde{x}_r</math>

If the system is time-invariant, then

:<math>\mathbb{H} \tilde{x}_r = \tilde{y}_r</math>