A time-invariant (TIV) system is one whose output does not depend explicitly on time.
- If the input signal
produces an output
then any time shifted input,
, results in a time-shifted output ![{\displaystyle y(t+\delta )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/326a77521df914c5e0e346d84e748c8214733c7c)
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:
Since system A explicitly depends on t outside of
and
, it is not time-invariant. 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 differ is now presented.
To perform this proof, the second definition will be used.
System A:
- Start with a delay of the input
![{\displaystyle y(t)=t\,x(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6481c5648020491664114380e416e05b373f6d59)
![{\displaystyle y_{1}(t)=t\,x_{d}(t)=t\,x(t+\delta )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cc4c3e80e7adc14ec57106a358984fe3e757ad1)
- Now delay the output by
![{\displaystyle y(t)=t\,x(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6481c5648020491664114380e416e05b373f6d59)
![{\displaystyle y_{2}(t)=\,\!y(t+\delta )=(t+\delta )x(t+\delta )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cec9ea0a4336188fc20a3b277f5aef5ad1e787db)
- Clearly
, therefore the system is not time-invariant.
System B:
- Start with a delay of the input
![{\displaystyle y(t)=10\,x(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21a5bb2366d35d2405fa4c3384bf5e5febe1bccb)
![{\displaystyle y_{1}(t)=10\,x_{d}(t)=10\,x(t+\delta )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/906af5167bfc85f14f351f8bd768b48f04c86bb7)
- Now delay the output by
![{\displaystyle y(t)=10\,x(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21a5bb2366d35d2405fa4c3384bf5e5febe1bccb)
![{\displaystyle y_{2}(t)=y(t+\delta )=10\,x(t+\delta )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90812b5493d1eef0dd694c53814594ad297fa9eb)
- Clearly
, 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
where
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a32a9100ebbcca784e5fdcb75b556393c2f73b90)
can be represented in this abstract notation by
![{\displaystyle {\tilde {x}}_{1}=\mathbb {T} _{1}\,{\tilde {x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2090aa70d03a374bf21433d0ea20165b17dc615b)
where
is a function given by
![{\displaystyle {\tilde {x}}=x(t)\,\forall \,t\in \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb7ddf558ca8cd87496167e697b726a30d5e94ed)
with the system yielding the shifted output
![{\displaystyle {\tilde {x}}_{1}=x(t+1)\,\forall \,t\in \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/64874c53c05a899a6742903805efe8a298ab3f00)
So
is an operator that advances the input vector by 1.
Suppose we represent a system by an operator
. 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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a2cee7f141f3ee7723eea7f46a5b1c880f865f2d)
If our system equation is given by
![{\displaystyle {\tilde {y}}=\mathbb {H} \,{\tilde {x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73baddc615b3f352155e27ca39e6201f86307176)
then it is time-invariant if we can apply the system operator
on
followed by the shift operator
, or we can apply the shift operator
followed by the system operator
, 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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65931e30892883ef3e98398b7f17b5fdc9318256)
Applying the shift operator first gives
![{\displaystyle \mathbb {H} \,\mathbb {T} _{r}\,{\tilde {x}}=\mathbb {H} \,{\tilde {x}}_{r}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf129c39b955eda06b84d5c375774ede7e73f5d1)
If the system is time-invariant, then
![{\displaystyle \mathbb {H} \,{\tilde {x}}_{r}={\tilde {y}}_{r}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff85035a0d6fd0e91537806d460ba4b7efcfad02)
See also