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Find the inverse Laplace

  • Thread starter Umayer
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  • #1
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Homework Statement



I have some problem finding the inverse laplace transformation of the function: [tex]\frac{s}{s^2+2s+5}[/tex]

Homework Equations



http://math.fullerton.edu/mathews/c2003/laplacetransform/LaplaceTransformMod/Images/Table.12.2.jpg [Broken]

The Attempt at a Solution



I tried to factorise the denominator: [tex]\frac{s}{(s+1)^2+2^2}[/tex]
 
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Answers and Replies

  • #2
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This is a good start. Now write the s in the numerator as (s+1)-1, and then separate the expression into two fractions.

Chet
 
  • #3
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You mean like this?
[tex]\frac{s+1}{(s+1)^2+2^2} - \frac{1}{(s+1)^2+2^2}[/tex]

The inverse laplace of the first term would be I think then: [tex]e^{-t}*cos(2t)[/tex]

But I'm not so sure what to do then since I don't recognize the term in the table, would is be something like this?
[tex] - \frac{0s+1}{(s+1)^2+2^2}[/tex]

So the second term doesn't have an inverse? Oh and thanks for responding!
 
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  • #4
gneill
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You mean like this?
[tex]\frac{s+1}{(s+1)^2+2^2} - \frac{1}{(s+1)^2+2^2}[/tex]

The inverse laplace of the first term would be I think then: [tex]e^{-t}*cos(2t)[/tex]
Looks good.
But I'm not so sure what to do then since I don't recognize the term in the table, would is be something like this?
[tex] - \frac{0s+1}{(s+1)^2+2^2}[/tex]

So the second term doesn't have an inverse? Oh and thanks for responding!
Would you recognize it if it were written:
$$-\frac{1}{2} \frac{2}{(s + 1)^2 + 2^2}$$
?
 
  • #5
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Yes now I see it, thanks!!!
 
  • #6
rude man
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Or, you could have factored the denominator s^2 + 2s + 5 = (s + a)(s + b), then done a partial fraction expansion into
F = s/(s^2 + 2s + 5) = A/(s+a) + B/(s+b).

We all know 1/(s+a) transforms to exp(-at).
 
  • #7
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The discriminant is a negative number so it cannot be factorised. At least to my knowledge.
 
  • #8
rude man
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The discriminant is a negative number so it cannot be factorised. At least to my knowledge.
F(s) = s/(s2 + 2s + 5)

The factors for the denominator are a,b = 1 +/- j2.
So we get A/(s + 1 + j2) + B/(s + 1 - j2) → Aexp(-1 + j2)t + Bexp(-1 - j2)t

with A = (-1 + j2)/j4 and B = -(-1 - j2)/j4

and by using Euler's formula plus some algebra you can reduce this to the real answer
f(t) = (1/2)exp(-t){2cos(2t) - sin(t)}.

It's a bit messy but how did you work with 2/[(s+1)2 + 22] without looking it up?
 
  • #9
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I did thought of doing that way but I felt that it would take more time doing that method plus my book and my teacher never showed doing that way. But thanks for showing that it's possible to work with imaginary numbers!
 

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