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My question is : find the function $f(t)$ that has the following Laplace transform $$ F(s) = \frac{s+1}{(s^2 + 1)(s^2 +4s+13)} $$ thanks

Shuhao Cao
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user82922
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2 Answers2

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The partial fraction decomposition is

$$\frac{1}{20} \cdot \left[ \frac{s+2}{s^2 + 1} \ - \ \frac{s+6}{[s+2]^2 + 3^2 }\right] \ , $$

omitting the somewhat tedious linear algebra $ ^* $ and completing the square in the denominator. Arranging this for interpretation produces

$$\frac{1}{20} \cdot \left[ \frac{s}{s^2 + 1} \ + \ \frac{2 \cdot 1}{s^2 + 1} \ - \ \frac{s+2}{[s+2]^2 + 3^2 } \ - \ \frac{\frac{4}{3} \cdot 3}{[s+2]^2 + 3^2 }\right] \ , $$

from which we can read off the inverse Laplace transform as

$$\frac{1}{20} \cdot \left[ \ \cos t \ + \ 2 \sin t \ - \ e^{-2t} \ \cdot \left( \cos 3t \ + \ \frac{4}{3} \cdot \sin 3t \right) \right] \ . $$

$ ^* $ all right, I did check this afterwards with WolframAlpha...

$$ \\ $$

ADDENDUM (6/19) --

As for setting up the decomposition, we note that the two quadratic factors in the denominator are "irreducible over the real numbers" (both have only complex conjugate zeroes). The sum of "fractions" (actually rational functions) that we set up involve linear functions divided by the irreducible quadratic ones thusly,

$$ \frac{As + B}{s^2 + 1} \ + \ \frac{Cs + D}{s^2 + 4s + 13} \ = \ \frac{s + 1}{(s^2 + 1 ) \cdot (s^2 + 4s + 13)} \ . $$

This leaves us to solve for the four coefficients satisfying

$$ (As + B) \cdot (s^2 + 4s + 13) \ + \ (Cs + D) \cdot (s^2 + 1) \ = \ s + 1 , $$

(so, for instance, the cubic and quadratic terms must "zero out"), which prove to be

$$ A = \frac{1}{20} \ , \ B = \frac{1}{10} \ , \ C = - \frac{1}{20} \ , \ D = - \frac{3}{10} \ . $$

colormegone
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I am going to evaluate this using residues. If you have no idea of what these are, then I will just give you an easy-to-understand intermediate result: if $f(s) = p(s)/q(s)$ and $q$ has a zero at $s_0$, then the residue of $f$ at the pole $s=s_0$ is

$$\frac{p(s_0)}{q'(s_0)}$$

The inverse LT of the given

$$\hat{f}(s) = \frac{s+1}{(s^2+1)(s^2+4 s+13)}$$

is simply the sum of the residues of $\hat{f}(s) e^{s t}$ at the poles. The poles here are at $s_1=i$, $s_2=-i$, $s_3=-2+i 3$, and $s_4=-2-i 3$. I will allow you to apply the above formula to compute the residues of $\hat{f}(s) e^{s t}$ and add them up. I get for the final result,

$$f(t) = - e^{-2 t}\left[ \frac{1}{15} \sin{3 t}+\frac{1}{20} \cos{3 t}\right] + \frac{1}{20} (\cos{t}+2 \sin{t})$$

Ron Gordon
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  • Beautiful solution, +1. I'm especially pleased by this since I've never seen this method of evaluating an inverse Laplace transform. Thanks for sharing! – Alex Wertheim Jun 19 '13 at 01:20
  • I certainly won't dispute your method, but I believe the zeroes of $ \ s^2 + 4s + 13 \ $ are a little simpler than what you've written (but are complex)... [I see the issue now: the discriminant is -36 , not +68 .] – colormegone Jun 19 '13 at 04:12
  • @RecklessReckoner: you caught a dumb mistake - thanks. Fixed. – Ron Gordon Jun 19 '13 at 04:19
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    I tell students that "sign errors" are the #1 error made in mathematics from beginners to professionals. Wish I had a dime for each of mine... – colormegone Jun 19 '13 at 04:20