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:
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?
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.