System invariant

A system invariant by temporal shift is a system whose exit explicitly does not depend on time.

Definition

If the entry signal $x\; (T)\; \backslash$ produces an exit $y\; (T)\; \backslash ,$, then some is the entry shifted temporally $x\; (T\; +\; \backslash \; delta)\; \backslash$, the exit is it also shifted $y\; (T\; +\; \backslash \; delta)\; \backslash$.

This property can be satisfied (but not necessarily) if the Transfer function transfer of the system is not a function of time, if it is not in the expression of the entry and the exit.

Equivalent definition: If the system is invariant, then the block of the system is commutative with an arbitrary block time.

Examples

Basic example

To know how to determine if a system is invariant, consider the two systems:

• System a: $y\; (T)\; =\; T\; \backslash ,\; X\; (T)$
• System b: $\backslash ,\; \backslash !\; B\; (T)\; =\; 10\; X\; (T)$

As the system has explicitly depends on time T apart from $x\; (T)\; \backslash ,$ and $y\; (T)\; \backslash ,$, then the system is not invariant. The system B, it, do not depend explicitly on time T and are thus invariant.

Formal example

A more formal proof of the invariance (or not) of the systems has and B Ci above is presented here. To carry out this proof, the second definition will be used.

Système has :

From the entry with a shift $x\_d\; (T)\; =\; \backslash ,\; \backslash !\; X\; (T\; +\; \backslash \; delta)$

$y\; (T)\; =\; T\; \backslash ,\; x\_d\; (T)$

$y\_1\; (T)\; =\; T\; \backslash ,\; x\_d\; (T)\; =\; T\; \backslash ,\; X\; (T\; +\; \backslash \; delta)$

Maintenant let us delay the exit by $\backslash \; delta$

$y\; (T)\; =\; T\; \backslash ,\; x\_d\; (T)$

$y\_2\; (T)\; =\; \backslash ,\; \backslash !\; there\; (T\; +\; \backslash \; delta)\; =\; (T\; +\; \backslash \; delta)\; X\; (T\; +\; \backslash \; delta)$

Clearly $y\_1\; (T)\; \backslash ,\; \backslash !\; \backslash \; y\_2\; (T)$, this is why the system is not invariant.

System B :

From the entry with a shift $x\_d\; (T)\; =\; \backslash ,\; \backslash !\; X\; (T\; +\; \backslash \; delta)$

$y\; (T)\; =\; 10\; \backslash ,\; x\_d\; (T)$

$y\_1\; (T)\; =\; 10\; \backslash ,\; x\_d\; (T)\; =\; 10\; \backslash ,\; X\; (T\; +\; \backslash \; delta)$

Maintenant let us delay the exit by $\backslash ,\; \backslash !\; \backslash \; delta$

$y\; (T)\; =\; 10\; \backslash ,\; x\_d\; (T)$

$y\_2\; (T)\; =\; there\; (T\; +\; \backslash \; delta)\; =\; 10\; \backslash ,\; X\; (T\; +\; \backslash \; delta)$

Clearly $y\_1\; (T)\; =\; \backslash ,\; \backslash !\; y\_2\; (T)$, this is why the system is invariant

Abstract example

Let us note the operator delay by $\backslash \; mathbb\; \{T\}\; \_r$ where $r$ is the quantity per which the vectorial parameter must be delayed. For example, the system " 1" advances; :

$x\; (t+1)\; =\; \backslash ,\; \backslash !\; \backslash \; delta\; (t+1)\; *\; X\; (T)$

can be represented by the abstract notation:

$\backslash \; tilde\; \{X\}\; \_1\; =\; \backslash \; mathbb\; \{T\}\; \_1\; \backslash ,\; \backslash \; tilde\; \{X\}$

where $\backslash \; tilde\; \{X\}$ is the function given by

$\backslash \; tilde\; \{X\}\; =\; X\; (T)\; \backslash ,\; \backslash \; forall\; \backslash ,\; T\; \backslash \; in\; \backslash \; mathbb\; \{R\}$

the system producing the shifted exit

$\backslash \; tilde\; \{X\}\; \_1\; =\; X\; (T\; +\; 1)\; \backslash ,\; \backslash \; forall\; \backslash ,\; T\; \backslash \; in\; \backslash \; mathbb\; \{R\}$

Thus $\backslash \; mathbb\; \{T\}\; \_1$ is an operator who advances the vectorial entry of 1.

Let us suppose that we represent the system by an operator $\backslash \; mathbb\; \{H\}$. This system is invariant if it commutates with the operator delay, i.e.:

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

If the equation of the system is given by:

$\backslash \; tilde\; \{there\}\; =\; \backslash \; mathbb\; \{H\}\; \backslash ,\; \backslash \; tilde\; \{X\}$

Then it is a system invariant if one can apply the operator $\backslash \; mathbb\; \{H\}$ to $\backslash \; tilde\; \{X\}$ followed by the operator delay $\backslash \; mathbb\; \{T\}\; \_r$, or apply the operator delay $\backslash \; mathbb\; \{T\}\; \_r$ followed by the system operator $\backslash \; mathbb\; \{H\}$, 2 calculations producing an equivalent result.

Let us apply the system operator in first:

$\backslash \; mathbb\; \{T\}\; \_r\; \backslash ,\; \backslash \; mathbb\; \{H\}\; \backslash ,\; \backslash \; tilde\; \{X\}\; =\; \backslash \; mathbb\; \{T\}\; \_r\; \backslash ,\; \backslash \; tilde\; \{there\}\; =\; \backslash \; tilde\; \{there\}\; \_r$

To apply the operator delay in first gives:

$\backslash \; mathbb\; \{H\}\; \backslash ,\; \backslash \; mathbb\; \{T\}\; \_r\; \backslash ,\; \backslash \; tilde\; \{X\}\; =\; \backslash \; mathbb\; \{H\}\; \backslash ,\; \backslash \; tilde\; \{X\}\; \_r$

If the system is invariant, then

$\backslash \; mathbb\; \{H\}\; \backslash ,\; \backslash \; tilde\; \{X\}\; \_r\; =\; \backslash \; tilde\; \{there\}\; \_r$

See too

• Impulse response
• Automatic
• linear System

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