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I have search my answer in questions like this one and this one but I have not figured out how to solve this.

I have an RLC circuit in series, from which I have computed the voltage across its only capacitor using Laplace transform with zero initial conditions, giving me the following (huge) expression in time: $${v_C}_1=6.26\cdot10^3\cos(314t)+80\sin(314t)\;+$$ $$e^{-7.81\cdot t}\cdot(6.26\cdot10^3\cos(314t)-234\sin(314t))$$

Using Matlab to graph it I get the following: Matlab simulation for voltage across capacitor

My question is if there is a mathematical way to know how much time the transient last until the voltage reach its steady-state. My calculus is a bit rusty and I can't figure it out since the expression seems to tend to 0 because of the negative exponentials when t tends to infinity, right?

Martín
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  • How is this related to mathematics? – copper.hat Mar 20 '18 at 21:02
  • @copper.hat Because what I am asking is a mathematical problem, to know at which time the function converges. Other people have asked similar questions, so maybe if it is solved it will help others – Martín Mar 20 '18 at 21:04
  • The expression you have there is not what appears in the graph.The expression converges to zero. – copper.hat Mar 20 '18 at 21:06
  • @copper.hat Thanks for your time, I've considered the first exponential as 1 since it had a very small exponent, so maybe now is computable – Martín Mar 20 '18 at 21:35
  • The transient will never disappear entirely. Engineers often take a particular number of time constants so that the output is 'close enough' for some practical purpose (see https://electronics.stackexchange.com/questions/65104/4-or-5-time-constants-to-reach-practical-potential, for example). – copper.hat Mar 20 '18 at 22:01
  • @copper.hat Thank you for the information :) – Martín Mar 21 '18 at 00:04

1 Answers1

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Consider the exponential $$e^{-at},$$

where $a > 0$. Mathematics says that this exponential will never reach $0$. Anyway, as $t$ grows, $e^{-at}$ approaches to $0$. From a numerical, practical and engineering point of view, given $k>0$, but very close to $0$ (i.e. $k= 10^{-6}$) you can say that, when:

$$e^{-at} = k,$$

then the exponential term "disappeared", and the steady state (in this case you are in sinusoidal regime) is reached.

Then, fix the $k$ that you prefer, and find $t_k$ such that

$$e^{-at} = k.$$

Specifically, you get:

$$t_k = -\frac{1}{a}\log(k),$$

where $\log(\cdot)$ is the natural logarithm.

Therefore, you can say that the effect of the exponential "disappears" when $t > t_k$, with respect to the threshold $k$ that you have chosen. Indeed, for $t > t_k$, you can say that:

$$e^{-at} < k.$$

As a side remark, since the number $k$ is very close to $0$, then it is also smaller than $1$. Recalling that the logarithm of a number smaller than $1$ is negative, then the formula of $t_k$ will provide you a positive number, as expected.

the_candyman
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