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No idea on how to start finding inverse Laplace Transform

  • Thread starter abc617
  • Start date
  • #1
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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
 
Last edited:

Answers and Replies

  • #2
hunt_mat
Homework Helper
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Use partial fractions on X(s)
 
  • #3
rock.freak667
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If you know that 9s2+12s+4 factors as (3s+2)2, then how did you reduce X(s) to that?
 
  • #4
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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]
 
  • #5
rock.freak667
Homework Helper
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So now you have

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

Now do you know any laplace transform that looks like this?
 
  • #6
11
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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]?
 
  • #7
33,505
5,191
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
 
Last edited:
  • #8
33,505
5,191
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]
 

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