4
$\begingroup$

I'm trying to solve this second order differential equation 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 is there a easier method to go abut this? Any help would be much appreciated.

$\endgroup$
1
  • $\begingroup$ Yes, you should be using partial fraction decomposition and then reverse the Laplace transform. You might as well use $A=R/L$ and $B=1/(LC)$ if it helps you in the intermediate steps. $\endgroup$
    – anon
    Feb 7, 2012 at 10:54

2 Answers 2

2
$\begingroup$

Looking at the table here you will recognize three different possible behaviors. Let us see why. Consider the denominator. This is can be rewritten as

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

where

$$\alpha=\frac{R}{2L} \qquad \beta=\sqrt{\frac{1}{LC}-\frac{R}{2L}}.$$

So, when $\frac{1}{LC}>\frac{R}{2L}$ you will recognize an exponentially decaying sine wave. When $\frac{1}{LC}=\frac{R}{2L}$ you will get just an exponential decay. When $\frac{1}{LC}<\frac{R}{2L}$ you will get an exponential decay multiplied by a hyperbolic cosine. All this can be deduced from the table I linked at the beginning of this answer.

$\endgroup$
0
$\begingroup$

I think the problem you are having is that none of the elements are suitably defined for your problem. Perhaps you need to treat the denominator as a quadratic function with three possible solutions: two distinct real roots, two repeating roots, and complex conjugates. Give that a try.

$\endgroup$
1

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .