# No idea on how to start finding inverse Laplace Transform

• abc617
So now you have X(s)= 1/(s+ 2/3)@ Now do you know any laplace transform that looks like this?Well I have this function:\mathcal{L}^{-1}(\frac{1}{s + a}) = e^{-at}So I assume the answer should be:x(t) = \mathcal{L}^{-1}(\frac{1}{s+\frac{2}{3}}) = e^{\frac{-2}{3}t}?

#### 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:
$$X(s) = \frac{3(3s+2)}{9s^{2}+12s+4}$$

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

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

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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

Last edited:
Use partial fractions on X(s)

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

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

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

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 $$s$$. Would I factor out a $$3$$ and get this:

$$\frac{3}{3s+2} => \frac{3}{3} (\frac{1}{s+ \frac{2}{3}})$$

So now you have

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

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

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

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

abc617 said:
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:
$$X(s) = \frac{3(3s+2)}{9s^{2}+12s+4}$$

Attempt
All I've been able to figure out so far is:
-> $$X(s) = \frac{3}{(3s+2)^{2}} * \frac{1}{(3s+2)}}$$
No. This doesn't follow from the equation above it.
$$\frac{3(3s+2)}{9s^{2}+12s+4} = \frac{3(3s + 2)}{(3s + 2)^2}$$
Now simplify.

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

abc617 said:
But after that I'm unsure. The table of Laplace Transforms I have doesn't have anything resembling this form:
$$\frac{1}{(as+b)^{2}}$$

-=-=-=-=-==-=-=-=-=-=-
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

Last edited:
abc617 said:
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 =>
abc617 said:
-Resimplified the equation
$$X(s) \frac{3(3s+2)}{9s^{2}+12s+4} => \frac{3(3s+2)}{(3s+2)(3s+2)} => \frac{3}{3s+2}$$
This is what you should have.
$$X(s) = \frac{3(3s+2)}{9s^{2}+12s+4} = \frac{3(3s+2)}{(3s+2)(3s+2)} = \frac{3}{3s+2}$$

abc617 said:
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 $$s$$. Would I factor out a $$3$$ and get this:

$$\frac{3}{3s+2} => \frac{3}{3} (\frac{1}{s+ \frac{2}{3}})$$