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) ,!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_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

tilde{x}_1 = mathbb{T}_1 , tilde{x}

where tilde{x} is a function given by

forall t in mathbb{R} tilde{x} = x(t)

with the system yielding the shifted output

forall t in mathbb{R} tilde{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

tilde{y} = mathbb{H} , tilde{x}

then it is time-invariant if we can apply the system operator mathbb{H} on tilde{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} , tilde{x} = mathbb{T}_r , tilde{y} = tilde{y}_r

Applying the shift operator first gives

mathbb{H} , mathbb{T}_r , tilde{x} = mathbb{H} , tilde{x}_r

If the system is time-invariant, then

mathbb{H} , tilde{x}_r = tilde{y}_r

See also

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