The time-invariant system is defined as following:
If y(t) = (H(x))(t) and y(t − τ ) = (H(x))(t− τ )
Then H is a time-invariant system.
Has this definition taken system initial stats account? At t=0 and t=τ, the same system could have different states, due to a discharging capacitor and etc. Then y(t) and y(t-τ) could be different for a time-invariant system.
So when taking initial states into account, a time-invariant system could be no longer time-invariant. Is my understanding correct?
Best Answer
If it doesn't hold for any τ, it isn't time invariant! That's the definition, and it's enough to answer your question:
They are different, so the system MUST be time variant!
Really, that's it.
The story here is you are mixing a property of linear systems with time invariant systems:
A linear system with no initial state is time invariant. Because it is linear. A linear system with a initial state is time variant. Because the initial state doesn't shift with the input.
Yet, the system is linear. That means you can decompose the system in 2 parts:
The Homogeneous State Response (u=0,x(0)!=0)
The Forced State Response (u!=0,x(0)=0)
The output of the system is the sum of those two part (remember, it's linear!). The system is only fully time invariant if the state response is 0, otherwise it doesn't fit the definition!
Still, the second system is time-invariant, and you can always use time invariant techniques over it. You just have to remember to sum the response coming from 1