No idea on how to start finding inverse Laplace Transform

[abc617]

So my professor is really bad and I've had to try to learn to learn this stuff myself. But I have no idea where to start and where to go.

Problem:
[tex] X(s) = \frac{3(3s+2)}{9s^{2}+12s+4}[/tex]Attempt
All I've been able to figure out so far is:
-> [tex] X(s) = \frac{3}{(3s+2)^{2}} * \frac{1}{(3s+2)}}[/tex]

But after that I'm unsure. The table of Laplace Transforms I have doesn't have anything resembling this form:
[tex]\frac{1}{(as+b)^{2}}[/tex]-=-=-=-=-==-=-=-=-=-=-
On a side note, I still have this other thread [https://www.physicsforums.com/showthread.php?t=465600] that I still am unsure of. Help on this thread would be greatly appreciated

 

[hunt_mat]

Use partial fractions on X(s)

 

[rock.freak667]

If you know that 9s²+12s+4 factors as (3s+2)², then how did you reduce X(s) to that?

 

[abc617]

Alright so I tried redoing it and this is what I redid:

-Resimplified the equation
[tex] X(s) \frac{3(3s+2)}{9s^{2}+12s+4}
=> \frac{3(3s+2)}{(3s+2)(3s+2)}
=> \frac{3}{3s+2} [/tex]

The problem here is that I don't know if there is an inverse transformation for the denominator because there is a coefficient in front of the [tex]s[/tex]. Would I factor out a [tex]3[/tex] and get this:

[tex]\frac{3}{3s+2} => \frac{3}{3} (\frac{1}{s+ \frac{2}{3}}) [/tex]

 

[rock.freak667]

So now you have

X(s)= 1/(s+ 2/3)

Now do you know any laplace transform that looks like this?

 

[abc617]

Well I have
[tex]\mathcal{L}^{-1}(\frac{1}{s + a}) = e^{-at}[/tex]

So I assume the answer should be:
[tex] x(t) = \mathcal{L}^{-1}(\frac{1}{s+\frac{2}{3}}) = e^{\frac{-2}{3}t}[/tex]?

 

[Mark44]

So my professor is really bad and I've had to try to learn to learn this stuff myself. But I have no idea where to start and where to go.

Problem:
[tex] X(s) = \frac{3(3s+2)}{9s^{2}+12s+4}[/tex]


Attempt
All I've been able to figure out so far is:
-> [tex] X(s) = \frac{3}{(3s+2)^{2}} * \frac{1}{(3s+2)}}[/tex]

No. This doesn't follow from the equation above it.
[tex]\frac{3(3s+2)}{9s^{2}+12s+4} = \frac{3(3s + 2)}{(3s + 2)^2}[/tex]
Now simplify.

EDIT: I see that you corrected this in your later post.

But after that I'm unsure. The table of Laplace Transforms I have doesn't have anything resembling this form:
[tex]\frac{1}{(as+b)^{2}}[/tex]


-=-=-=-=-==-=-=-=-=-=-
On a side note, I still have this other thread [https://www.physicsforums.com/showthread.php?t=465600] that I still am unsure of. Help on this thread would be greatly appreciated

 

[Mark44]

Alright so I tried redoing it and this is what I redid:

This is better, but your notation is confusing. You should not be using =>

-Resimplified the equation
[tex] X(s) \frac{3(3s+2)}{9s^{2}+12s+4}
=> \frac{3(3s+2)}{(3s+2)(3s+2)}
=> \frac{3}{3s+2} [/tex]

This is what you should have.
[tex] X(s) = \frac{3(3s+2)}{9s^{2}+12s+4}
= \frac{3(3s+2)}{(3s+2)(3s+2)}
= \frac{3}{3s+2} [/tex]

The problem here is that I don't know if there is an inverse transformation for the denominator because there is a coefficient in front of the [tex]s[/tex]. Would I factor out a [tex]3[/tex] and get this:

[tex]\frac{3}{3s+2} => \frac{3}{3} (\frac{1}{s+ \frac{2}{3}}) [/tex]