time-invariant system

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

• System A: $y(t)\; =\; t\; x(t)$
• System B: $y(t)\; =\; 10\; x(t)$

Since the System Function $y(t)$ for system A explicitly depends on t outside of $x(t)$, 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 $x(t)$. 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 $x\_d(t)\; =\; x(t\; +\; \backslash delta)$

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

::$y\_1(t)\; =\; t\; x\_d(t)\; =\; t\; x(t\; +\; \backslash delta)$

:Now delay the output by $\backslash delta$

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

::$y\_2(t)\; =\; y(t\; +\; \backslash delta)\; =\; (t\; +\; \backslash delta)\; x(t\; +\; \backslash delta)$

:Clearly $y\_1(t)\; \backslash ne\; y\_2(t)$, therefore the system is not time-invariant.

:System B: Start with a delay of the input $x\_d(t)\; =\; x(t\; +\; \backslash delta)$

::$y(t)\; =\; 10\; x(t)$

::$y\_1(t)\; =\; 10\; x\_d(t)\; =\; 10\; x(t\; +\; \backslash delta)$

:Now delay the output by $\backslash delta$

::$y(t)\; =\; 10\; x(t)$

::$y\_2(t)\; =\; y(t\; +\; \backslash delta)\; =\; 10\; x(t\; +\; \backslash delta)$

:Clearly $y\_1(t)\; =\; y\_2(t)$, therefore the system is time-invariant.

More generally, the relationship between the input and output is

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

and its variation with time is

:$\backslash frac\{\backslash mathrm\{d\}\; y\}\{\backslash mathrm\{d\}\; t\}\; =\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; t\}\; +\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; x\}\; \backslash frac\{\backslash mathrm\{d\}\; x\}\{\backslash mathrm\{d\}\; t\}.$

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

:$\backslash frac\{\backslash partial\; f\}\{\backslash partial\; t\}\; =0.$

Applied to Systems A and B above:

:$f\_A\; =\; t\; x(t)\; \backslash qquad\; \backslash implies\; \backslash qquad\; \backslash frac\{\backslash partial\; f\_A\}\{\backslash partial\; t\}\; =\; x(t)\; \backslash neq\; 0$ in general, so it is not time-invariant,

:$f\_B\; =\; 10\; x(t)\; \backslash qquad\; \backslash implies\; \backslash qquad\; \backslash frac\{\backslash partial\; f\_B\}\{\backslash partial\; t\}\; =\; 0$ so it is time-invariant.

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

:$x(t+1)\; =\; \backslash delta(t+1)\; *\; x(t)$

can be represented in this abstract notation by

:$\backslash tilde\{x\}\_1\; =\; \backslash mathbb\{T\}\_1\; \backslash tilde\{x\}$

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

:$\backslash tilde\{x\}\; =\; x(t)\; \backslash forall\; t\; \backslash in\; \backslash R$

with the system yielding the shifted output

:$\backslash tilde\{x\}\_1\; =\; x(t\; +\; 1)\; \backslash forall\; t\; \backslash in\; \backslash R$

So $\backslash mathbb\{T\}\_1$ is an operator that advances the input vector by 1.

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

:$\backslash mathbb\{T\}\_r\; \backslash mathbb\{H\}\; =\; \backslash mathbb\{H\}\; \backslash mathbb\{T\}\_r\; \backslash forall\; r$

If our system equation is given by

:$\backslash tilde\{y\}\; =\; \backslash mathbb\{H\}\; \backslash tilde\{x\}$

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

Applying the system operator first gives

:$\backslash mathbb\{T\}\_r\; \backslash mathbb\{H\}\; \backslash tilde\{x\}\; =\; \backslash mathbb\{T\}\_r\; \backslash tilde\{y\}\; =\; \backslash tilde\{y\}\_r$

Applying the shift operator first gives

:$\backslash mathbb\{H\}\; \backslash mathbb\{T\}\_r\; \backslash tilde\{x\}\; =\; \backslash mathbb\{H\}\; \backslash tilde\{x\}\_r$

If the system is time-invariant, then

:$\backslash mathbb\{H\}\; \backslash tilde\{x\}\_r\; =\; \backslash tilde\{y\}\_r$

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

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Category:Control theory

Category:Signal processing