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$$F(S)=\frac{se^{-3s}}{s^2+22s+125}$$

My first reflex was to attempt a decomposition into partial fractions, but I am simply left with a fraction that is just as "complicated", instead of the usual decomposition into multiple simpler fractions separated by addition/subtraction/etc.

P.S. To save you some time, it is sufficient to show me how to get a decomposed fraction from this fraction, I can do the actual inverse laplace transforms from there.

Thank you!

Jessie
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user885431
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  • Do you want an expression in terms of complex exponentials or sines and cosines? The procedure is different depending on which one you want. – Ian Mar 17 '21 at 17:47
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    Assuming complex exponentials, you factor the denominator as $(s-s_1)(s-s_2)$ where $s_1,s_2=-11 \pm 2i$, and then you do partial fractions on the part without the $e^{-3s}$. Then you just multiply through by $e^{-3s}$ in that partial fraction result. Assuming sines and cosines, you just complete the square and then combine the results from the table for the cases $\frac{b}{(s+a)^2+b^2}$ and $\frac{s+a}{(s+a)^2+b^2}$, and then again take the $e^{-3s}$ into account. – Ian Mar 17 '21 at 18:30

1 Answers1

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Hint:

Solve $s^2+22s+125=0$. This yields $$s_1= -11 - 2i,\quad s_2=-11+2i$$

Now $$f(t)= \operatorname{Res}_{s=s_1}(F(s)\cdot e^{st})+\operatorname{Res}_{s=s_2}(F(s)\cdot e^{st}).$$

Calculate the residues by factoring the denominator with $s_1$ and $s_2$ and then calculate the limit, without any indeterminate form.

vitamin d
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  • Either you or the OP permuted the 125 and the 22 at some point. Also surely there is something else there besides just the residues themselves, or else $f$ is just constant... – Ian Mar 17 '21 at 18:28