# Background

My question comes from here, it's a response of 1st order LPF RC circuit from an arbitrary periodic input.

How to determine the transient response of a circuit to causal periodic inputs?

# Problem

Suppose if I have input signal with period of $$T = 10s$$

$$\displaystyle u(t) = 2t (\theta(t) - \theta(t - 5)) + 0 (\theta(t - 5) - \theta(t - 10) )$$

For t from 0 to infinity, there'll be $$1-e^{-sT}$$ in the denominator. and then the system transfer function

$$H(s) = \dfrac{1/sC}{R + 1/sC}$$

In order to find it's output we need F(s) which is a transfer function for one single period response from the input signal. Inverse it. Multiplies it with unit step since it's a causal signal and system. And then time shifted it by $$nT$$.

$$F(s) = H(s) U(s)$$

$$f(t) \theta(t) =$$ \displaystyle \begin{align} f(t-nT) \theta(t-nT) &= 2\ \theta(t - nT) \left( RC (e^{-(t-nT)/(RC)} -1) + (t - nT)\right) \\ &+ 2\ \theta(t-5 - nT) \left( (5 - RC) e^{-(t-5-nT)/(RC)} + RC - (t - nT) \right) \end{align}

Then from periodic summation properties of laplace transform we get $$\displaystyle y(t) = \mathcal{L}^{-1}\left[\frac{1}{1-e^{-sT}} F(s) \right] = \sum_{n=0}^{\infty} f(t-nT) \theta(t-nT)$$

Assume if $$R = 50k$$ and $$C = 100uF$$, thus time constant of $$5s$$. This is the plot

Sum[2UnitStep[t-10n] ( 5 ( e^(-(t-10n)/(5) ) - 1) + t - 10n ) + 2UnitStep[t - 5 -10n] ( (5-5) e^(-(t - 5 - 10n)/(5) ) + 5 - (t - 10n) ), {n, 0, 5}] Now if I change the time constant into $$20s$$. This is the plot if we sum it from n = 0 to 20.

Sum[2UnitStep[t-10n] ( 20 ( e^(-(t-10n)/(20) ) - 1) + t - 10n ) + 2UnitStep[t - 5 -10n] ( (5-20) e^(-(t - 5 - 10n)/(20) ) + 20 - (t - 10n) ), {n, 0, 20}] # Question

How to separate its transient and steady state response? Such as,

$$\displaystyle y(t) = y_{tr}(t) + y_{ss}(t)$$

At what time t such that the transient vanishes?

How many period T of input signal does it take for it to be vanished?

I couldn't find any approach since it's difficult to find when the overlapping magnitude became steady. And that depends on the time constant, which results in asymptotic to 0 as t goes to infinity on each summation term.