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Inverse Laplace Transform(s/((s^2)+1)^2

  • #1

Homework Statement


Find inverse Laplace transform of:
s/((s^2)+1)^2

Homework Equations





The Attempt at a Solution


The answer is:
(t/2)sin t
First shifting theorem or partial fractions don't work, I think. From a Laplace transforms table, sin at = a/((s^2)+(a^2)), which is almost the form that the problem is in.
Thanks
 

Answers and Replies

  • #2
hunt_mat
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Here is something that may help, take the sin transform and differentiate it w.r.t s and what do you get?

Mat
 
  • #3
Thanks for replying.
d/ds of sin transform = -2sa/(s^2+a^2)^2
This would give an inverse of the problem of:
(d/ds)-sin(t)/2
I'm not sure I did that right.
 
  • #4
hunt_mat
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Set a=-1 and what do you get?
 
  • #5
hunt_mat
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Not quite, we have:
[tex]
\frac{1}{s^{2}+a^{2}}=\int_{0}^{\infty}e^{-st}\sin atdt
[/tex]
Differentiate this w.r.t and tell me what you get?
 
  • #6
I'm getting:
1/(a-s)
with integration-by-parts twice.
 
  • #7
33,514
5,195
No, hunt_mat said to differentiate, not integrate.
 
  • #8
Oh, right, sorry.
(d/ds)1/(s^2+a^2)=-2s/(s^2+a^2)^2
 
  • #9
hunt_mat
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and on the RHS?
 
  • #10
RHS:
integral from 0 to infinity(-s(e^-st)sin atdt
 
  • #11
hunt_mat
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set a=-1 and what have you got?
 
  • #12
If a=-1:
-2s/(s^2+a^2)^2 = -s/(-s-1)
 
  • #13
Sorry:
-2s/(s^2+1)^2=-s/(-s-1)
 
  • #14
hunt_mat
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No, if a=-1:
[tex]
\frac{s}{(s^{2}+1)^{2}}=\int_{0}^{\infty}\frac{t}{2}e^{-st}\sin tdt
[/tex]
What does that tell you?
 
  • #15
I know then that:
L{tsin(at)/2}=s/(s^2+a^2)?
I'm not sure.
 
  • #16
hunt_mat
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Let me re-arrange the RHS:
[tex]
\int_{0}^{\infty}\frac{t\sin t}{2}e^{-st}dt
[/tex]
Does it make sense now?
 
  • #17
No, I'm sorry, I'm just not seeing it. Thanks for all your help, though.
 
  • #18
hunt_mat
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Think of it like this, you know that:
[tex]
\frac{1}{1-x}=1+x+x^{2}+x^{3}+x^{4}+\cdots
[/tex]
Now differentiate both sides to show that:
[tex]
\frac{1}{(1-x)^{2}}=1+2x+3x^{2}+4x^{3}+\cdots
[/tex]
and now you know the power series for [tex](1-x)^{2}[/tex] is just by differentiating the series.
 
  • #19
So I know that differentiating both sides of the sin(at) LT will give me the square that I need in the denominator, but then I'm stuck with the d/ds on the sin side. I'm not sure what to do after this.
Thanks
 
  • #20
hunt_mat
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when you differentiate the integral you take in differentiation under the integral sign:
[tex]
\frac{d}{ds}\int_{0}^{\infty}e^{-st}\sin tdt=\int_{0}^{\infty}\frac{d}{ds}\left( e^{-st}\sin t\right) dt
[/tex]
Differentiating the integrand:
[tex]
\frac{d}{ds}e^{-st}\sin t=-te^{-st}\sin t
[/tex]
As now we regard s as variable and t is fixed. So the above can be written as:
[tex]
(-t\sin t)e^{-st}
[/tex]
and so you're taking the Laplace transform of something different now. That something is the answer to your question. Is this making sense now?
 
  • #21
I think I got it now.
After differentiating both sides and dividing by -2:
s/(s^2+a^2)^2=L{tsin at/2}
Hence,
L-1{s/(s^2+1)^2}=tsint/2
However, how do we know that "a" isn't -1, since (-1)^2=1.
 
  • #22
hunt_mat
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well done, got there in the end.
 

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