Homework 2 Ben Horst: A :: B :: C :: D :: E
Time Invariance Explanation
A time invariant system is one that can be shifted freely. It is not bound by times in any way. It does not matter how much one shifts the input or output, it will remain of the same form.
Time Invariance Example
Consider the system where: x(t) -> system -> y(t) = 2x(t) + 5
Given this system and the definition of time variance
If the cascade of a system followed by a time delay t0 yields the same as a time delay followed by a system for any time t, the system is "Time Invariant."
Start with x(t)
Put it through the system
y(t) = 2x(t) + 5
Then delay it by time $ t_0 $ [hint: replace t with (t - $ t_0 $)]
z(t) = 2x(t - $ t_0 $) + 5
Repeat in reverse order...
Start with a time delay
y(t) = x(t - $ t_0 $) + 5
Then put that through the system
z(t) = 2x(n) + 5, where n equals (t - $ t_0 $)
Therefore we have
2x(t - $ t_0 $) + 5 = 2x(t - $ t_0 $) + 5
which is clearly true. Thus, the system is Time Invariant.
Given the counter-example of a system where: x(t) -> system -> y(t) = t*x(t)
Start with x(t)
Put it through the system
y(t) = t*x(t)
Then delay it by time $ t_0 $ [hint: replace t with (t-$ t_0 $)]
z(t) = (t- $ t_0 $)*x(t- $ t_0 $)
Repeat in reverse order...
Start with a time delay
y(t) = x(t-$ t_0 $)
Then put that through the system
z(t) = t*x(n), where n equals (t-$ t_0 $)
Therefore we have:
(t-$ t_0 $)*x(t-$ t_0 $) = t*x(t-$ t_0 $)
which is quite obviously untrue. Thus, the system is not Time Invariant. (conversely, it could be called Time Variant)
