# Solving the second-order differential equation for an RLC circuit using Laplace Transform

I'm trying to solve this second order differential equation for a RLC series circuit using Laplace Transform. The Laplace transform of the equation is as follows:

$$I(s) = \frac{E}{s^2+ \frac{R}{L}s + \frac{1}{LC}}$$

I'm having trouble trying to bring it back to the time domain. Should I be using partial fractions with quadratic factors or there a easier method to go abut this? And is the damping factor to be considered? If it is, how do I go about dealing with it?

## 5 Answers

I will give it a try:

let $D = \frac{R}{2L}$ and $\omega^2 = \frac{1}{LC}$

for $D^2 \not= \omega^2$:

$$I(s) = \frac{E}{s^2+ 2Ds + \omega^2}=$$

$$=\frac{E}{\left(s+\left(-D+\sqrt{D^2 - \omega^2}\right)\right)\left(s+\left(-D-\sqrt{D^2 - \omega^2}\right)\right)} =$$ (partial fraction) $$=\frac{E}{-2\sqrt{D^2 - \omega^2}} \frac{1}{\left(s+\left(-D+\sqrt{D^2 - \omega^2}\right)\right)} + \frac{E}{-2\sqrt{D^2 - \omega^2}} \frac{1}{\left(s+\left(-D-\sqrt{D^2 - \omega^2}\right)\right)}$$

Looking up the Laplace transform $$\mathcal{L}^{-1}\left[\frac{1}{s+a}\right] = e^{-at}$$

$$\mathcal{L}^{-1}[I(s)] = \frac{E}{-2\sqrt{D^2 - \omega^2}} \left( e^{t\left(-D+\sqrt{D^2 - \omega^2}\right)} - e^{t\left(-D-\sqrt{D^2 - \omega^2}\right)} \right)$$

for $D^2 = \omega^2$:

$$I(s) = \frac{E}{s^2+ 2Ds + D^2}=\frac{E}{(s+D)^2}$$

Looking up the Laplace transform $$\mathcal{L}^{-1}\left[\frac{1}{(s+a)^{n+1}}\right] = \frac{t^n}{n!} e^{-at}$$

$$\mathcal{L}^{-1}[I(s)] = E \cdot t \cdot e^{-D t}$$

• Hmm. It looks right, but why do I get the sense that something isn't right for the D^2 > w^2 case? Maybe I'm just expecting a particular solution to oscillate in all cases, and my intuition is wrong. If your math is right, then there's also a special case for D^2 = w^2 (critically damped) of the form t*e^(-at) Feb 7 '12 at 15:07
• @JasonS Indeed; if the oscillating period is big enough, the exponential function should be predominant and cause a discharging exponential function...if I'm not wrong Feb 7 '12 at 15:44
• I expanded on the case D^2 = w^2. Feb 7 '12 at 16:17
• @PetPaulsen- Thanks a lot PetPaulsen, this was a really good way of showing me how to do it. The problem I have is that when I worked it out, I got the A term in D^2 > w^2 should be positive, not negative. Feb 8 '12 at 23:12

Massage it into a form that looks like something in this table. You should come up with something like sine or cosine whose frequency depends on L and C, and multiplied by a damping function that looks like a decaying exponential depending on R, L, and C intuitively.

• @vicatcu- Thanks for the link to the Laplace Tables and the advice and the damping function :) Feb 8 '12 at 23:15

You have a second order function at the denominator, that can be solved to give two factors in the form $(s-a)(s-b)$. This is the equivalent of

$${1 \over (s-a)} \cdot {1 \over (s-b)}$$

This is a multiplication of two functions, that, for Laplace transform as well as Fourier transform, gives a convolution of functions when anti-transformed. So you can obtain the convolution of two functions in the time domain, and you can manipulate that to obtain a nicer function.

Should I be using partial fractions with quadratic factors or there a easier method to go abut this?

That's the way to go, if you have to use the Laplace transform. (If you don't have to use the Laplace transform, look the answer up. :-)

$$ax^2+bx=a \left[ (x+\frac{b}{2a})^2 -(\frac{b}{2a}) \right]$$ so we can write : $$s^2+\frac{R}{L}s=(s+\frac{R}{2L})^2 - (\frac{R}{2L})^2$$ $$I(s)=\frac{E}{s^2+\frac{R}{L}s+\frac{1}{LC} }=\frac{E}{(s+\frac{R}{2L})^2 +\frac{1}{LC}- (\frac{R}{2L})^2}$$ we know: $$\mathcal{L} \left[ e^{-at}sin(wt) \right] = \frac{w}{(s+a)^2+w^2}$$ so we have : $$i(t)=\frac{E}{\sqrt{\frac{1}{LC}-(\frac{R}{2L})^2}} e^{\frac{-R}{2L}t} . sin(\sqrt{\frac{1}{LC}-(\frac{R}{2L})^2}\space.t)$$