# 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.

Analytic expression is what I hope for.

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: • This is the part that confuse me the most. 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 $t= −∞$. The system is causal, yet we have the knowledge of assumption for the signals to be exist in the future. How could $n=-∞$ to be exist is what I haven't been able to figure out for years. I see that you multiply it with unit step in the end for the output signal. Without that, you will get y(t) exist before x(t). So, is it true that we should firstly ignore the causality of the system? Jun 19 at 10:48
• @Unknown123: This has nothing to do with the causality of the system. We just assume that the input signal has existed forever. Hence, the output signal at any finite time is in its steady state. Jun 19 at 11:20
• Wait, how did you know that you should assume that the input signal has to be existed forever in order to find the steady state response? Where did you learn that? If the input signal has existed forever, then how does it exist in the first place? Jun 19 at 12:43
• @Unknown123: That's the definition of a steady-state response: the response after all transients have subsided. For the given example that means that we use the periodic continuation of the input signal because in that case there are no effects of switching the signal on. Jun 19 at 13:23
• If a signal starts at $t=-\infty$ then the signal is already on steady state at $t > -\infty$. That is clear to me. So, what is the definition of signal on transient state? When does it switched on? When does it became transient? The mathematics of $(4)$ such that $(5)$ holds does makes sense. But the abstract intuition is what I haven't understand. Jun 20 at 10:30

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.

• I was hoping for analytic expression. But thanks for giving me an approximation. Jun 20 at 5:55

$$y_{tr}(t)=-u(t)\frac{2\left(\left(\tau-5\right)e^{\left(5/\tau\right)}-\tau\right)}{\left(1-e^{\left(T/\tau\right)}\right)}e^{\left(-t/\tau\right)}$$
But I couldn't find the closed-form of $$y(t)$$ and $$y_{ss}(t)$$, just the analytic expressions of the infinite series.