# Inverse Laplace Transform

• Jul 13th 2008, 04:17 PM
hercules
Inverse Laplace Transform
I need help finding the inverse laplace transform of

$\displaystyle \frac {8s^2-4s+12}{s(s^2+4)}$

and

$\displaystyle \frac {1-2s}{s^2+4s+5}$

Thank You
• Jul 13th 2008, 04:42 PM
Jhevon
Quote:

Originally Posted by hercules
I need help finding the inverse laplace transform of

$\displaystyle \frac {8s^2-4s+12}{s(s^2+4)}$

By partial fractions, we can write this as:

$\displaystyle \frac 3s + \frac {5s - 4}{s^2 + 4}$

or, in other words:

$\displaystyle 3 \cdot {\color{red} \frac 1s} + 5 \cdot {\color{red} \frac s{s^2 + 2^2}} - 2 \cdot { \color{red}\frac 2{s^2 + 2^2}}$

now just look up the rules in the table of your text. what are the inverse Laplace transforms for the guys in red?

Quote:

and

$\displaystyle \frac {1-2s}{s^2+4s+5}$

Thank You

with some algebraic manipulation (completing the square of the denominator, factoring out a minus 1 from the numerator and writing -1 as 4 - 5) we find that this expression is:

$\displaystyle -2 \cdot {\color{red}\frac {s + 2}{(s + 2)^2 + 1}} + 5 \cdot {\color{red} \frac 1{(s + 2)^2 + 1}}$

now this is the same story as the last one. look up the transforms in the table in your text
• Jul 13th 2008, 04:49 PM
ThePerfectHacker
Quote:

Originally Posted by hercules
I need help finding the inverse laplace transform of

$\displaystyle \frac {8s^2-4s+12}{s(s^2+4)}$

You can write, $\displaystyle \tfrac{3}{s} + \tfrac{5s}{s^2+4} - 2\cdot \tfrac{2}{s^2+4}$.
Now just recognize the forms, $\displaystyle 3+5\cos (2x) - 2 \sin (2x)$.
• Jul 13th 2008, 04:50 PM
TheEmptySet
Quote:

Originally Posted by hercules
I need help finding the inverse laplace transform of

$\displaystyle \frac {8s^2-4s+12}{s(s^2+4)}$

and

$\displaystyle \frac {1-2s}{s^2+4s+5}$

Thank You

by partial fractions (work not shown) we get

$\displaystyle \frac{8s^2-4s+12}{s(s^2+4)}=\frac{3}{s}+\frac{5s-4}{(s^2+4)}=3\frac{1}{s}+5\frac{s}{s^2+4}+2\frac{2 }{s^2+4}$

now taking the inverse transform we get

$\displaystyle \mathcal{L}^{-1}\{ 3\frac{1}{s}+5\frac{s}{s^2+4}+2\frac{2}{s^2+4}\}=$

by the linearity property we get

$\displaystyle 3\mathcal{L}^{-1}\{ \frac{1}{s}\} +5\mathcal{L}^{-1}\{ \frac{s}{s^2+4}\}+2\mathcal{L}^{-1}\{ \frac{2}{s^2+4}\} =3+5\cos(2t)+2\sin(2t)$

Edit: late again
• Jul 13th 2008, 04:57 PM
cyph1e
Quote:

Originally Posted by hercules
I need help finding the inverse laplace transform of

$\displaystyle \frac {8s^2-4s+12}{s(s^2+4)}$

and

$\displaystyle \frac {1-2s}{s^2+4s+5}$

Thank You

Hi! I'll be omitting $\displaystyle \theta(t)$, the Heaviside step function, in this post.

Assuming you know how to do partial fraction expansion we have:

$\displaystyle \frac{8s^2-4s+12}{s(s^2+4)} = 3\frac1s + \frac{5s-4}{s^2-2^2} = \frac3s + 5\frac{s}{s^2-2^2} - 2\frac2{s^2-2^2}$

and knowing these transformation pairs:

$\displaystyle \mathcal{L}^{-1}\left\{\frac1s\right\}=1$
$\displaystyle \mathcal{L}^{-1}\left\{\frac{\omega}{s^2+\omega^2}\right\}=\sin( \omega t)$
$\displaystyle \mathcal{L}^{-1}\left\{\frac{s}{s^2+\omega^2}\right\}=\cos(\omeg a t)$

we have

$\displaystyle \mathcal{L}^{-1}\left\{3\frac1s + 5\frac{s}{s^2-2^2} - 2\frac2{s^2-2^2}\right\} = 3\mathcal{L}^{-1}\left\{\frac1s\right\} +5\mathcal{L}^{-1}\left\{\frac{s}{s^2-2^2}\right\} -2\mathcal{L}^{-1}\left\{\frac2{s^2-2^2}\right\}$$\displaystyle =3+5\cos(2t)-2\sin(2t)$

Edit: Wow, some guys are too fast for me!