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...for any and all choices of <math>t_0</math>. | ...for any and all choices of <math>t_0</math>. | ||
| − | In other words, let's say you get output for some input. If the input | + | In other words, let's say you get output for some input. If the graph of the input is shifted by some amount of time, the corresponding output is just the previous output, shifted over that same amount. |
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==Example time invariant system== | ==Example time invariant system== | ||
Latest revision as of 19:22, 10 September 2008
Definition of Time Invariance
I don't like the diagrammed version of time invariance much; I would much rather use the mathematical definition. As I mentioned here, I think of time invariance like this:
Define $ S_{t_0} $ as the shifting operator $ S_{t_0}(x(t))=x(t-t_0). $ (In other words, $ S_{t_0} $ introduces a time delay of $ t_0 $ onto the function/signal x(t).) A function ("system") f is considered time invariant iff $ f(S_{t_0}(x))=S_{t_0}(f(x))\ \forall x(t), t_0. $
In layman's terms, that means that a system (call it f) is time invariant if a function (call it x) can be sent through the system in either one of these two ways and come out with the same result:
- x is time-shifted by some number of seconds $ t_0 $, then sent through the system.
- x is sent through the system, then is time-shifted by $ t_0 $ seconds.
...for any and all choices of $ t_0 $.
In other words, let's say you get output for some input. If the graph of the input is shifted by some amount of time, the corresponding output is just the previous output, shifted over that same amount.
Example time invariant system
Let system f be defined for any function x as follows: f(x) = 2x. (In class, we would say "x --> system --> y = 2x".) We want to show f is time invariant.
Say we take any function $ x(t) $ and any time delay $ t_0 $. (Pick one!)
$ f(S_{t_0}(x)) = f(x(t-t_0)) = 2x(t-t_0) $
$ S_{t_0}(f(x)) = S_{t_0}(2x(t)) = 2x(t-t_0) $
$ f(S_{t_0}(x))=S_{t_0}(f(x)) $
Since I was never explicit in my choice of $ x \text{ or } t_0 $ (I even let you choose them!), we can conclude that the above applies to every combination of $ x\text{ and } t_0 $ (within the constraints of the definition):
$ f(S_{t_0}(x))=S_{t_0}(f(x))\ \forall x(t), t_0. $
Thus, f(x) = 2x is time invariant.
Example time variant system
Let system f be defined for any function x as follows: $ f(x(t)) = x(t) + t $. We want to show f is time variant. (In class, we would say "x(t) --> system --> y(t) = x(t) + t".)
Say we take any function $ x(t) $ and any time delay $ t_0 $. (This time, I'll choose them; we'll save that for later.)
$ f(S_{t_0}(x)) = f(x(t-t_0)) = x(t-t_0) + t $
$ S_{t_0}(f(x)) = S_{t_0}(x(t)-t) = x(t-t_0) + (t-t_0) $
Let me leave $ x $ alone; I don't care what it is. However, I would like for $ t_0 = 1 $.
$ f(S_{1}(x)) = x(t-1) + t $
$ S_{1}(f(x)) = x(t-1) + (t-1) $
$ f(S_{t_0}(x)) \neq S_{t_0}(f(x)) $
Recall the definition for time invariance:
A function ("system") f is considered time invariant iff $ f(S_{t_0}(x))=S_{t_0}(f(x))\ \forall x(t), t_0. $
Since, in this case, the definition is not true for all functions and constants (The one above didn't work, for instance.), I can conclude that the system is not time invariant, i.e., it is time variant.
