# Thread: Quick Laplace Transform Question

1. ## Quick Laplace Transform Question

This is my first post

I've got (s^2 + 5) / (s - 2)^4. I need to find the inverse laplace transform of this.

I can't use partial fractions i believe, since the only thing in the denominator is s-2, and I have a higher power in the numerator, so my A would need an S.

I really can't find anything on any of my tables or anything on line to get an s^2 in my numerator and a simple s in my denominator (except f^n(t), but I couldn't get the values to line up)... please help! And MANY thanks ahead of time, I have been doing this homework all night :\

-Andrew

2. Originally Posted by drewkx152
This is my first post

I've got (s^2 + 5) / (s - 2)^4. I need to find the inverse laplace transform of this.

I can't use partial fractions i believe, since the only thing in the denominator is s-2, and I have a higher power in the numerator, so my A would need an S.

I really can't find anything on any of my tables or anything on line to get an s^2 in my numerator and a simple s in my denominator (except f^n(t), but I couldn't get the values to line up)... please help! And MANY thanks ahead of time, I have been doing this homework all night :\

-Andrew
Actually, the degree of the denominator is 4, which is larger than 2. What you can do here is manipulate the fraction:

$\frac{s^2+5}{(s-2)^4}=\frac{s^2-4+9}{(s-2)^4}=\frac{s+2}{(s-2)^3}+\frac{9}{(s-2)^4}$ $=\frac{s-2+4}{(s-2)^3}+\frac{9}{(s-2)^4}=\frac{1}{(s-2)^2}+4\frac{1}{(s-2)^3}+9\frac{1}{(s-2)^4}$

Now, by the translation theorem, $\mathcal{L}^{-1}\!\left\lbrace F\left(s-a\right)\right\rbrace=e^{at}f\!\left(t\right)$.

Thus, $\mathcal{L}^{-1}\!\left\lbrace\frac{1}{(s-2)^2}+4\frac{1}{(s-2)^3}+9\frac{1}{(s-2)^4}\right\rbrace$ $=\mathcal{L}^{-1}\!\left\lbrace\frac{1}{(s-2)^2}\right\rbrace+2\mathcal{L}^{-1}\!\left\lbrace \frac{2!}{(s-2)^3}\right\rbrace+\tfrac{3}{2}\mathcal{L}^{-1}\!\left\lbrace \frac{3!}{(s-2)^4}\right\rbrace=e^{2t}t+2e^{2t}t^2+\tfrac{3}{2} e^{2t}t^3$ $=\boxed{e^{2t}\left(\tfrac{3}{2}t^3+2t^2+t\right)}$

Does this make sense?

3. Wow, i tried some fraction manipulation but i never thought of eliminating s from the numerator. That should help with the future problems I've been having trouble with! Thank you VERY much Chris!

Oh and yes it makes sense, I recognize all those rules i just couldn't break the problem apart into the necessary parts.

4. ## So much trouble with these.. any help is appreciated

Allright, I feel like a complete idiot.

If you have better things to do, that's fine, I'm just going to ask because I've spent nearly an hour ripping this apart.

I have (s^2) / (s^3 +8).. I need to find the Inverse Transform again...

Could you provide me with this so I have another example to reference (book and teacher provided none ). I have about 8 of these to do, I won't ask for any more help, I certainly do not want to hog the forums.

so... is there a specific way you look at these problems to find out how to solve them? I'm very sorry to ask again, its just I've had no problems with any transforms until these... Thanks ahead of time.

5. Originally Posted by drewkx152
Allright, I feel like a complete idiot.

If you have better things to do, that's fine, I'm just going to ask because I've spent nearly an hour ripping this apart.

I have (s^2) / (s^3 +8).. I need to find the Inverse Transform again...

Could you provide me with this so I have another example to reference (book and teacher provided none ). I have about 8 of these to do, I won't ask for any more help, I certainly do not want to hog the forums.

so... is there a specific way you look at these problems to find out how to solve them? I'm very sorry to ask again, its just I've had no problems with any transforms until these... Thanks ahead of time.
Partial fraction decomposition:

$\frac{s^2}{3^3 + 8} = \frac{A}{s + 2} + \frac{B s}{s^2 - 2s + 4} + \frac{C}{s^2 - 2s + 4}$

(the values of A, B and C are left for you to get)

Complete the square:

$= \frac{A}{s + 2} + \frac{B s}{(s-1)^2 + 3} + \frac{C}{(s-1)^2 + 3}$.

Now apply a couple of theorems and some standard results.

6. Thanks so much guys.

To try to help the community (just to show I'm not someone leeching off you guys) I posted a solution on something that I was more capable of: http://www.mathhelpforum.com/math-he...annuities.html.

Thanks again for all your help, I really appreciate it.

-Drew

7. Originally Posted by mr fantastic
Partial fraction decomposition:

$\frac{s^2}{3^3 + 8} = \frac{A}{s + 2} + \frac{B s}{s^2 - 2s + 4} + \frac{C}{s^2 - 2s + 4}$

(the values of A, B and C are left for you to get)

Complete the square:

$= \frac{A}{s + 2} + \frac{B s}{(s-1)^2 + 3} + \frac{C}{(s-1)^2 + 3}$.

Now apply a couple of theorems and some standard results.

I ran with that and got:

$
\frac{s^2}{s^3+8} = \frac{s}{3[(s-1)^2+3]} - \frac{2}{3[(s-1)^2+3]} + \frac{1}{3(s+2)}
$

Unable to do anything with this... did i mess up my partial fractions? (I don't believe I did, got values A=1/3, C=-2/3, B= 1/3) or can i just still not figure out how to get things into a form I can inverse transform?

Any tips on how to do these in general, and can someone got me thru the rest of this?

NOTE:

could it be that the partial fraction for B should be:
$
\frac{B(s-1)}{(s-1)^2+3}
$

if we were to change the bottom of the fraction from
$\frac{s^2}{(s+2)(s^2-2s+4)}$ to $\frac{s^2}{(s-1)^2+3}$

before we started the partial fractions process, that would change my coefficient of B to s-1 wouldn't it?

8. Originally Posted by drewkx152
I ran with that and got:

$
\frac{s^2}{s^3+8} = \frac{{\color{red}2}s}{3[(s-1)^2+3]} - \frac{2}{3[(s-1)^2+3]} + \frac{1}{3(s+2)}
$
Mr F says: Note the correction in red, which makes a big difference.

Unable to do anything with this... did i mess up my partial fractions? (I don't believe I did, got values A=1/3, C=-2/3, B= 1/3) or can i just still not figure out how to get things into a form I can inverse transform?

Any tips on how to do these in general, and can someone got me thru the rest of this?

NOTE:

could it be that the partial fraction for B should be:
$
\frac{B(s-1)}{(s-1)^2+3}
$

if we were to change the bottom of the fraction from
$\frac{s^2}{(s+2)(s^2-2s+4)}$ to $\frac{s^2}{(s-1)^2+3}$

before we started the partial fractions process, that would change my coefficient of B to s-1 wouldn't it?
From standard tables and shift theorem:

$LT[e^{bt} \cos (at)] = \frac{s - b}{(s - b)^2 + a^2}$.
This takes care of the first two terms ( $b = 1$ and $a = \sqrt{3}$) .....

$LT[e^{at}] = \frac{1}{s - a}$.
This takes care of the third term ....

You have to re-arrange the partial fractions given to fit what you need ....