Time-invariant system

A system that is time-invariant is a system that has an output that does not depend explicitly on time.

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

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 communative with an arbitrary delay.

Simple example

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

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

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

$y(t)=t\,x_{d}(t)$

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

Now delay the output by $\delta$

$y(t)=t\,x_{d}(t)$

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

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

System B:

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

$y(t)=10x_{d}(t)$

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

Now delay the output by $\delta$

$y(t)=10x_{d}(t)$

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

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

See also

• Finite impulse response
• State space (controls)
• System analysis
• Time-variant