Time-invariant System

: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: $y(t) = t\, x(t)$

System B: $\,\!b(t) = 10 x(t)$

Since system A explicitly depends on ''t'' outside of

x(t)

and

y(t)

then 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 why system A & B from above 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) = t\, x_d(t)

y_1(t) = t\, x_d(t) = t\, x(t + \delta)

:Now delay the output by

\delta

y(t) = t\, x_d(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(t) = 10 \, x_d(t)

y_1(t) = 10 \,x_d(t) = 10 \,x(t + \delta)

:Now delay the output by

\,\!\delta

y(t) = 10 \,x_d(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)

can be represented in this abstract notation by

:

ilde{x}_1 = \mathbb{T}_1 \, 	ilde{x}

where

ilde{x}

is a function given by

:

ilde{x} = x(t) \, orall \, t \in \mathbb{R}

with the system yielding the shifted output

:

ilde{x}_1 = x(t + 1) \, orall \, t \in \mathbb{R}

\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.,

:

\mathbb{T}_r \, \mathbb{H} = \mathbb{H} \, \mathbb{T}_r  \,\, orall \, 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}

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

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Time-invariant System