A time-invariant system is one whose output does not depend explicitly on time.

If the input signal $x(t)$ produces an output $y(t)$ then any time shifted input, $x(t\; +\; \backslash delta)$, results in a time-shifted output $y(t\; +\; \backslash delta)$

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: $y(t)\; =\; t\backslash ,\; x(t)$
• System B: $\backslash ,\backslash !b(t)\; =\; 10\; x(t)$

Since system A explicitly depends on t outside of $x(t)$ and $y(t)$ then 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 why system A & B from above is now presented. To perform this proof, the second definition will be used.

System A:

Start with a delay of the input $x\_d(t)\; =\; \backslash ,\backslash !x(t\; +\; \backslash delta)$

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

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

Now delay the output by $\backslash delta$

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

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

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

System B:

Start with a delay of the input $x\_d(t)\; =\; \backslash ,\backslash !x(t\; +\; \backslash delta)$

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

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

Now delay the output by $\backslash ,\backslash !\backslash delta$

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

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

Clearly $y\_1(t)\; =\; \backslash ,\backslash !y\_2(t)$, 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 $\backslash mathbb\{T\}\_r$ where $r$ is the amount by which a vector's index set should be shifted. For example, the "advance-by-1" system

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

can be represented in this abstract notation by

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

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

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

with the system yielding the shifted output

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

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.,

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

If our system equation is given by

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

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

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

Applying the shift operator 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 time-invariant, then

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

See also

---

• Finite impulse response
• LTI system theory
• Sheffer sequence
• State space (controls)
• System analysis
• Time-variant system

Control theory Signal processing

Système invariant | Sistema linear invariante no tempo