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 $.
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.
