# Help me get the inverse laplace transform

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#### Kral

I need the Inverse LaPlace transform of G(s) = s/((s-a)^2 + b^2), where a, b are constants. I can't find it in the tables. However I can find the Inverse transform of G(s) = 1/((s-a)^2 + b^2). It's (1/b)(exp(at)(sin(bt)). I know that multiplication by s in the s domain is equivalent to differentiating in the time domain. So I found the inverse transform by differentiating (1/b)(exp(at)(sin(bt)). Is this valid?
Thanks

#### _Eduardo_

##### Full Member level 5
Re: Laplace Transform

Kral said:
... However I can find the Inverse transform of G(s) = 1/((s-a)^2 + b^2). It's (1/b)(exp(at)(sin(bt)). I know that multiplication by s in the s domain is equivalent to differentiating in the time domain. So I found the inverse transform by differentiating (1/b)(exp(at)(sin(bt)). Is this valid?
Yes.

If you have problems with the inverse transforms, try WolframAlpha

https://www.wolframalpha.com/input/?i=InverseLaplaceTransform[s%2F%28%28s+-+a%29^2+%2B+b^2%29%2C+s%2C+t]

### Kral

Points: 2

#### Kral

Re: Laplace Transform

_Eduardo,
Thanks for the tip. I'm sure that this site will be useful to me in the future.
Regards,
Kral

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