# Laplace Transformation

• Oct 30th 2010, 09:08 AM
Musab
Laplace Transformation
Hi,

I want to find the Laplace inverse of the following function:

$1/(s^2 +2zws + w^2)$ , where z and w are constants.

I know that the result should have an exponintial part and a sin or cos, but I don't know how to factor the function to use the table.

Thanks,
• Oct 30th 2010, 09:14 AM
Musab
This is what I get after completing the square:
$
\frac {1}{(s+ \frac{zw}{2})^2 - (\frac{zw}{2})^2 + w^2}$

I still don't know what to do next
• Oct 30th 2010, 09:24 AM
Jhevon
Quote:

Originally Posted by Musab
This is what I get after completing the square:
$
\frac {1}{(s+ \frac{zw}{2})^2 - (\frac{zw}{2})^2 + w^2}$

I still don't know what to do next

Um, that's not completing the square...

if you complete the square, you get

$\displaystyle \frac 1{(s + zw)^2 + w^2(1 - z^2)}$

now what?
• Oct 30th 2010, 10:14 AM
Musab
Thx

I think you got the signs wrong, it should be $\displaystyle \frac 1{(s + zw)^2 + w^2(-z^2 + 1)}$

Now I should do the following:

$\frac{1}{(w^2(-z^2+1))^{0.5}} . \frac{(w^2(-z^2+1))^{0.5}}{(s + zw)^2 + w^2(-z^2 + 1)}$

and use the laplace table
is that right ?
• Oct 30th 2010, 08:42 PM
Jhevon
Quote:

Originally Posted by Musab
Thx

I think you got the signs wrong, it should be $\displaystyle \frac 1{(s + zw)^2 + w^2(-z^2 + 1)}$

yes

Quote:

Now I should do the following:

$\frac{1}{(w^2(-z^2+1))^{0.5}} . \frac{(w^2(-z^2+1))^{0.5}}{(s + zw)^2 + w^2(-z^2 + 1)}$

and use the laplace table
is that right ?
and yes