Time invariant Green's function (inpulse response)

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In a time-invariant system, the response to a delta input at time t=0 generates an impulse response function h(t). If the delta input is delayed by tau, the system's response is g(t) = h(t-tau), confirming the time invariance as the response is merely shifted. For time-invariant systems, the impulse response can be expressed as h(t-t0), while time-variant systems would require a function h(t, t-t0), indicating dependence on both time variables. The discussion clarifies that once time invariance is established, the variable t can be simplified to h(t) without the need for t0. Overall, the key point is that time invariance allows for a simplified expression of the impulse response.
fisico30
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Hello Forum,

given a input=delta located at time t=0, the system will respond generating a function h(t).

If the delta is instead located at t=t0 (delayed by tau), the system will respond with a function g(t)=h(t-tau), just a shifted version of the response for the delta a t=0...

If this is the case, the system is time invariant and the impulse response is said to be only a function of t-t0...: h(t-t0)

If the system was time variant instead, it will be a function h(t,t-t0), that is, a function of both time t and t-t0...it is as if it was a function of two variables...

I am not clear on this: isn't the function h, the impulse response, a function of time t also in the case of time invariant system?
To be time invariant, does the variable t need to always occur with t0 in a subtraction?

thanks,
Fisico30
 
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My understanding is that once you have proved that the system is time invariant, i.e. h(t) = h(t-t0), you can safely drop t0 in the expression and simply state the impulse response is h(t).
 
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