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

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

Since the System Function ${\displaystyle y(t)}$  for system A explicitly depends on t outside of ${\displaystyle x(t)}$ , 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 ${\displaystyle x(t)}$ . 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.

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 ${\displaystyle x_{d}(t)=x(t+\delta )}$ 

${\displaystyle y(t)=tx(t)}$ 

${\displaystyle y_{1}(t)=tx_{d}(t)=tx(t+\delta )}$ 

Now delay the output by ${\displaystyle \delta }$ 

${\displaystyle y(t)=tx(t)}$ 

${\displaystyle y_{2}(t)=y(t+\delta )=(t+\delta )x(t+\delta )}$ 

Clearly ${\displaystyle y_{1}(t)\neq y_{2}(t)}$ , therefore the system is not time-invariant.

System B: Start with a delay of the input ${\displaystyle x_{d}(t)=x(t+\delta )}$ 

${\displaystyle y(t)=10x(t)}$ 

${\displaystyle y_{1}(t)=10x_{d}(t)=10x(t+\delta )}$ 

Now delay the output by ${\displaystyle \delta }$ 

${\displaystyle y(t)=10x(t)}$ 

${\displaystyle y_{2}(t)=y(t+\delta )=10x(t+\delta )}$ 

Clearly ${\displaystyle y_{1}(t)=y_{2}(t)}$ , therefore the system is time-invariant.

More generally, the relationship between the input and output is

${\displaystyle y(t)=f(x(t),t),}$ 

and its variation with time is

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

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

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

Applied to Systems A and B above:

${\displaystyle f_{A}=tx(t)\qquad \implies \qquad {\frac {\partial f_{A}}{\partial t}}=x(t)\neq 0}$  in general, so it is not time-invariant,

${\displaystyle f_{B}=10x(t)\qquad \implies \qquad {\frac {\partial f_{B}}{\partial t}}=0}$  so it is time-invariant.

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

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

can be represented in this abstract notation by

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

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

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

with the system yielding the shifted output

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

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

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

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

If our system equation is given by

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

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

Applying the system operator first gives

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

Applying the shift operator first gives

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

If the system is time-invariant, then

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

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