How to separate Transient and Steady-State Expression from Periodic Summation Response?

Question

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?

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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) $

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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}]

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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.

Analytic expression is what I hope for.

Answer 1

Let $x_0(t)$ be the part of the input signal $x(t)$ in the interval $[0,T]$:

$$x_0(t)=\begin{cases}x(t),&t\in [0,T]\\0,&\textrm{otherwise}\end{cases}$$

The pseudo-periodic input signal is then given by

$$x(t)=\sum_{n=0}^{\infty}x_0(t-nT)\tag{1}$$

If $h(t)$ is the impulse response of a causal and stable LTI system, and if $y_0(t)=(x_0\star h)(t)$ is its response to $x_0(t)$, the response to $x(t)$ is given by

$$y(t)=\sum_{n=0}^{\infty}y_0(t-nT)\tag{2}$$

Since $x(t)$ starts at $t=0$, the response $y(t)$ is composed of a steady-state component and of a transient component. The latter decays to zero because we have assumed that the LTI system is stable. The steady-state response is the response that would be observed if the input were periodic, i.e., if it had been switched on at $t=-\infty$. Consequently, the steady-state response is

$$y_s(t)=u(t)\sum_{n=-\infty}^{\infty}y_0(t-nT)\tag{3}$$

where $u(t)$ denotes the unit step function. From $(2)$ and $(3)$, the transient response must be

$$y_t(t)=-u(t)\sum_{n=-\infty}^{-1}y_0(t-nT)\tag{4}$$

such that

$$y(t)=y_s(t)+y_t(t)\tag{5}$$

holds.

Note that in $(3)$ and $(4)$ we don't actually need to evaluate infinite sums. For all practical purposes, the number of relevant past periods (i.e., the number of negative indices $n$) can be chosen to correspond to a few time constants of the system.

For the input signal and the LTI system given in the question (and with time constants $\tau=RC=20$ and $\tau=RC=5$, respectively), we obtain the following decompositions of the output signal:

Answer 2

From the impulse response (the output if there was only an impulse at the input) we can see directly that it never completely vanishes, but it will get insignificantly small. What you need to do is first determine what “significant” is and then from that determine when the impulse decays to this value (such as 1%, or 0.1%, or 0.00001% for example, more commonly expressed in dB (1% is -40 dB), and then that can be the duration from when a steady input can be considered “steady” at the output.

Answer 3

Just add trivial details after Matt L. answers.

This is the closed-form expression of transient response as per my questions.

Since n is negative, it will make all of the unit step functions returns 1, so it's a lot easier to evaluate.
The sum of the constants may look like diverge.
Just try it and you'll get through.

$y_{tr}(t)=-u(t)\frac{2\left(\left(\tau-5\right)e^{\left(5/\tau\right)}-\tau\right)}{\left(1-e^{\left(T/\tau\right)}\right)}e^{\left(-t/\tau\right)}$

But I couldn't find the closed-form of $y(t)$ and $y_{ss}(t)$, just the analytic expressions of the infinite series.