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Simple example

To demonstrate how to determine if a system is time-invariant, consider the two systems:

• System A: y ( t ) = t x ( t ) {\displaystyle y(t)=tx(t)}
• System B: y ( t ) = 10 x ( t ) {\displaystyle y(t)=10x(t)}

Since the System Function y ( t ) {\displaystyle y(t)} for system A explicitly depends on t outside of x ( t ) {\displaystyle x(t)} , it is not time-invariant because the time-dependence is not explicitly a function of the input function.

In contrast, system B's time-dependence is only a function of the time-varying input x ( t ) {\displaystyle x(t)} . This makes system B time-invariant.

The Formal Example below shows in more detail that while System B is a Shift-Invariant System as a function of time, t, System A is not.

Formal example

A more formal proof of why systems A and B 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 + δ ) {\displaystyle x_{d}(t)=x(t+\delta )} y ( t ) = t x ( t ) {\displaystyle y(t)=tx(t)} y 1 ( t ) = t x d ( t ) = t x ( t + δ ) {\displaystyle y_{1}(t)=tx_{d}(t)=tx(t+\delta )} Now delay the output by δ {\displaystyle \delta } y ( t ) = t x ( t ) {\displaystyle y(t)=tx(t)} y 2 ( t ) = y ( t + δ ) = ( t + δ ) x ( t + δ ) {\displaystyle y_{2}(t)=y(t+\delta )=(t+\delta )x(t+\delta )} Clearly y 1 ( t ) ≠ y 2 ( t ) {\displaystyle y_{1}(t)\neq y_{2}(t)} , therefore the system is not time-invariant. System B: Start with a delay of the input x d ( t ) = x ( t + δ ) {\displaystyle x_{d}(t)=x(t+\delta )} y ( t ) = 10 x ( t ) {\displaystyle y(t)=10x(t)} y 1 ( t ) = 10 x d ( t ) = 10 x ( t + δ ) {\displaystyle y_{1}(t)=10x_{d}(t)=10x(t+\delta )} Now delay the output by δ {\displaystyle \delta } y ( t ) = 10 x ( t ) {\displaystyle y(t)=10x(t)} y 2 ( t ) = y ( t + δ ) = 10 x ( t + δ ) {\displaystyle y_{2}(t)=y(t+\delta )=10x(t+\delta )} Clearly y 1 ( t ) = y 2 ( t ) {\displaystyle y_{1}(t)=y_{2}(t)} , therefore the system is time-invariant.

More generally, the relationship between the input and output is

y ( t ) = f ( x ( t ) , t ) , {\displaystyle y(t)=f(x(t),t),}

and its variation with time is

d y d t = ∂ f ∂ t + ∂ f ∂ x d x d t . {\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} t}}={\frac {\partial f}{\partial t}}+{\frac {\partial f}{\partial x}}{\frac {\mathrm {d} x}{\mathrm {d} t}}.}

For time-invariant systems, the system properties remain constant with time,

∂ f ∂ t = 0. {\displaystyle {\frac {\partial f}{\partial t}}=0.}

Applied to Systems A and B above:

f A = t x ( t ) ⟹ ∂ f A ∂ t = x ( t ) ≠ 0 {\displaystyle f_{A}=tx(t)\qquad \implies \qquad {\frac {\partial f_{A}}{\partial t}}=x(t)\neq 0} in general, so it is not time-invariant, f B = 10 x ( t ) ⟹ ∂ f B ∂ t = 0 {\displaystyle f_{B}=10x(t)\qquad \implies \qquad {\frac {\partial f_{B}}{\partial t}}=0} so it is time-invariant.

Abstract example

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

x ( t + 1 ) = δ ( t + 1 ) ∗ x ( t ) {\displaystyle x(t+1)=\delta (t+1)*x(t)}

can be represented in this abstract notation by

x ~ 1 = T 1 x ~ {\displaystyle {\tilde {x}}_{1}=\mathbb {T} _{1}{\tilde {x}}}

where x ~ {\displaystyle {\tilde {x}}} is a function given by

x ~ = x ( t ) ∀ t ∈ R {\displaystyle {\tilde {x}}=x(t)\forall t\in \mathbb {R} }

with the system yielding the shifted output

x ~ 1 = x ( t + 1 ) ∀ t ∈ R {\displaystyle {\tilde {x}}_{1}=x(t+1)\forall t\in \mathbb {R} }

So T 1 {\displaystyle \mathbb {T} _{1}} is an operator that advances the input vector by 1.

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

T r H = H T r ∀ r {\displaystyle \mathbb {T} _{r}\mathbb {H} =\mathbb {H} \mathbb {T} _{r}\forall r}

If our system equation is given by

y ~ = H x ~ {\displaystyle {\tilde {y}}=\mathbb {H} {\tilde {x}}}

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

Applying the system operator first gives

T r H x ~ = T r y ~ = y ~ r {\displaystyle \mathbb {T} _{r}\mathbb {H} {\tilde {x}}=\mathbb {T} _{r}{\tilde {y}}={\tilde {y}}_{r}}

Applying the shift operator first gives

H T r x ~ = H x ~ r {\displaystyle \mathbb {H} \mathbb {T} _{r}{\tilde {x}}=\mathbb {H} {\tilde {x}}_{r}}

If the system is time-invariant, then

H x ~ r = y ~ r {\displaystyle \mathbb {H} {\tilde {x}}_{r}={\tilde {y}}_{r}}

See also

• Finite impulse response
• Sheffer sequence
• State space (controls)
• Signal-flow graph
• LTI system theory
• Autonomous system (mathematics)

References

1. Oppenheim, Alan; Willsky, Alan (1997). Signals and Systems (second ed.). Prentice Hall. ↩