I have been running into this issue more and more frequently, and I'm sure there is a way to do this or use a different method (i.e. not partial fractions), but I don't recall it. Basically, I need to use inverse Laplace Transforms on functions that look similar to this: [math]F(s)=\frac{1}{(s^2+25)(s^2+4)}[/math] or this: [math]F(s)=\frac{1}{(s^2+4)^2}[/math] Now, I've tried using partial fractions on the first equation, but I always end up with something like A=B=0 (I assume that's because the factors contain variables of the second degree?). I sort of want to try setting [imath]r=s^2[/imath] but logic tells me that wouldn't work during the inverse Laplace transform because that function is only valid if r is not negative.

Any help, please?

## Inverse Laplace transforms and partial fractions

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### Re: Inverse Laplace transforms and partial fractions

Poohblah wrote: [math]F(s)=\frac{1}{(s^2+25)(s^2+4)}[/math]

Any help, please?

Doesn't this give you your first fraction?

[math]\frac{1}{21}\frac{1}{s^2+4}-\frac{1}{21}\frac{1}{s^2+25}[/math]

I think you may just be making an algebra mistake. The second one actually does look like it won't have a (real?) partial fraction decomposition though.

Maybe it would help to take a look at this example: http://en.wikipedia.org/wiki/Partial_fr ... enominator

### Re: Inverse Laplace transforms and partial fractions

For the second one, you can use the fact that the inverse Laplace transform takes products to convolution. It does have a complex partial fraction decomposition if you want to do it that way; you'll end up with terms of the form [imath]\frac{A}{(s + 2i)^2} + \frac{B}{(s - 2i)^2}[/imath] which have pretty nice inverses.

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### Re: Inverse Laplace transforms and partial fractions

The inverse Laplace Transform of [imath]\frac{1}{s^2+a^2}[/imath] has an elementary solution that you can easily look up in a table. There's no need to decompose it further.

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### Re: Inverse Laplace transforms and partial fractions

Thanks, that was the ticket.csrjjsmp wrote:Maybe it would help to take a look at this example: http://en.wikipedia.org/wiki/Partial_fr ... enominator

Of course, it's [imath]\frac{1}{a}sin(as)[/imath]. But neither of my equations are of that form.Mathmagic wrote:The inverse Laplace Transform of [imath]\frac{1}{s^2+a^2}[/imath] has an elementary solution that you can easily look up in a table. There's no need to decompose it further.

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### Re: Inverse Laplace transforms and partial fractions

Poohblah wrote:Thanks, that was the ticket.csrjjsmp wrote:Maybe it would help to take a look at this example: http://en.wikipedia.org/wiki/Partial_fr ... enominatorOf course, it's [imath]\frac{1}{a}sin(as)[/imath]. But neither of my equations are of that form.Mathmagic wrote:The inverse Laplace Transform of [imath]\frac{1}{s^2+a^2}[/imath] has an elementary solution that you can easily look up in a table. There's no need to decompose it further.

As csrjjsmp pointed out, your expression decomposes into [imath]\frac{1}{21(s^2+4)}-\frac{1}{21(s^2+25)}[/imath], and you can do the simple inverse Laplace Transform from there.

Axman: That, and have you played DX 10 games? It's like having your corneas swabbed with clits made out of morphine.

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