Inverse Laplace

Chris0724 · #1 $Old$ October 15th, 2009, 08:48 AM

hi,

Inverse laplace

---(s + 9 )
------------
s^2 + 6s+ 9

-----(s + 3 )
--------------- + .........
(s + 3 )^ 2 + 4

{ e^-3t } * Inverse laplace { s / (s+4) } + .......

I have problem understand this step... how to take away the +3 in the equation ? i know there is a formula in the sheet to give a direct answer given but the example in the book really make my head crack

= e^-3t cos2t + ......

Thank you for any help

Mush · #2 $Old$ October 15th, 2009, 09:21 AM

Quote:

Originally Posted by Chris0724 $View Post$

hi,

Inverse laplace

---(s + 9 )
------------
s^2 + 6s+ 9

-----(s + 3 )
--------------- + .........
(s + 3 )^ 2 + 4

{ e^-3t } * Inverse laplace { s / (s+4) } + .......

I have problem understand this step... how to take away the +3 in the equation ? i know there is a formula in the sheet to give a direct answer given but the example in the book really make my head crack

= e^-3t cos2t + ......

Thank you for any help

Err, I think you've got your completing the square wrong.

$s^2 + 6s + 9 \neq (s+3)^2 + 4$

$(s+3)^2 + 4 = s^2 + 6s + 9 + 4 = s^2 + 6s + 13.$

The way I see it is:

$\frac{s+9}{s^2 + 6s + 9} = \frac{s+9}{(s+3)^2} = \frac{s+3}{(s+3)^2} + \frac{6}{(s+3)^2} = \frac{1}{s+3} + 6 \bigg(\frac{1}{(s+3)^2} \bigg)$

Then for the inverse, remember the following rules:

$\mathcal{L}^{-1} \bigg( \frac{1}{s+a}\bigg) = e^{-at}$

$\mathcal{L}^{-1} \bigg(\frac{n!}{(s+a)^{n+1}} \bigg)= t^n e^{-at}$

Chris0724 · #3 $Old$ October 15th, 2009, 10:41 PM

Quote:

Originally Posted by Mush $View Post$

Err, I think you've got your completing the square wrong.

$s^2 + 6s + 9 \neq (s+3)^2 + 4$

$(s+3)^2 + 4 = s^2 + 6s + 9 + 4 = s^2 + 6s + 13.$

The way I see it is:

$\frac{s+9}{s^2 + 6s + 9} = \frac{s+9}{(s+3)^2} = \frac{s+3}{(s+3)^2} + \frac{6}{(s+3)^2} = \frac{1}{s+3} + 6 \bigg(\frac{1}{(s+3)^2} \bigg)$

Then for the inverse, remember the following rules:

$\mathcal{L}^{-1} \bigg( \frac{1}{s+a}\bigg) = e^{-at}$

$\mathcal{L}^{-1} \bigg(\frac{n!}{(s+a)^{n+1}} \bigg)= t^n e^{-at}$

thanks mush,

i got the equation wrongly... it should be

$\frac{s+9}{s^2 + 6s + 13}$

Mush · #4 $Old$ October 16th, 2009, 03:22 AM

Quote:

Originally Posted by Chris0724 $View Post$

thanks mush,

i got the equation wrongly... it should be

$\frac{s+9}{s^2 + 6s + 13}$

Ah, well in that case, your completing the square was correct! Try this then:

$\frac{s+9}{s^2 + 6s + 13} = \frac{s+9}{(s+3)^2 + 2^2}$

$= \frac{s+3}{(s+3)^2 + 2^2} + \frac{6}{(s+3)^2 + 2^2}$

$= \frac{s+3}{(s+3)^2 + 2^2} + 3 \bigg(\frac{2}{(s+3)^2 + 2^2} \bigg)$

The inverse rules you must now remember is that:

$\mathcal{L}^{-1} \frac{s+a}{(s+a)^2 + \omega^2} = e^{-at} \cos(\omega t)$

$\mathcal{L}^{-1} \frac{\omega}{(s+a)^2 + \omega ^2} = e^{-at} \sin(\omega t)$