I need help from the gurus

Frogger888

New member
Joined
Feb 2, 2006
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39
I am working on transforms of integrals
This is the problem
Find the inverse transform of {1/[(s) * (s-a)^2]

this is what I am getting

I get (1/s) * [1/(s-a)] * [1/(s-a)] which yields the next step
integral from 0 to t [1 + e^ta + e^ta]dt then I would just solve the simple integral.

Am I doing this right or messing up
 
I am guessing from your s's and a's that this is a Laplace Transform?

There is a simple chart you can follow:

\(\displaystyle \L
f(t) \rightarrow f(s)\\
1\rightarrow\frac{1}{s}\\
t\rightarrow\frac{1}{s^2}\\
ce^{at}\rightarrow\frac{c}{s-a}\\\)

I don't recall the rest of them, but they're out there on the web somewhere...

Anyway, although I don't remember the setup of the integral, I can tell you that

\(\displaystyle \frac{1}{s}\frac{1}{(s-a)^2}\) does not imply the result you've given. It looks like you confused multiplication with addition.

I believe you will have to use partail fractions:

\(\displaystyle \L\frac{1}{s}\frac{1}{(s-a)^2} =\frac{A}{s} + \frac{B}{(s-a)} + \frac{C}{(s-a)^2}\\ \Rightarrow 1 = A(s-a)^2 + Bs(s-a) + Cs\)

Choosing simple values for s will give you solutions for A, B, C. Then you may integrate.
 
I understand the partial fractions no problem but is there another way you could attempt this problem????
 
The PFD of your Laplace is

\(\displaystyle \L\\\frac{1}{s(s-a)^{2}}=\frac{1}{a(s-a)^{2}}-\frac{1}{a^{2}(s-a)}+\frac{1}{a^{2}s}\)

Now, this leaves us with:

\(\displaystyle \L\\\frac{1}{a}L^{-1}[\frac{1}{(s-a)^{2}}]-\frac{1}{a^{2}}L^{-1}[\frac{1}{(s-a)}]+\frac{1}{a^{2}}L^{-1}[\frac{1}{s}]\)

Now, use the table of Laplace transforms to find the inverses of the enclosed brackets.
 
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