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I have the function:

$$F(s) = \frac{s^4+3s^3+2s^2+4s+4}{(s+3)(s^2+1)}$$

and I have to make inverse Laplace. I tried to collect $s^3$ from the first and second element of the numerator in order to obtain:

$$\frac{2s^2+4s+4}{(s+3)(s^2+1)} + \frac{s^3}{s^2+1}$$

for the first, I calculated residues and for the second, I made a polynomial division. I obtained

$$\frac{1}{s+3} + \frac{1}{s^2+1} + s - \frac{s}{s^2+1}$$

When I calculated the inverse I obtained: $$e^{-3t}H(t) + \sin(t)H(t) + \delta_0' - \cos(t)H(t)$$

But, in the solution there isn't $\cos(t)H(t)$

Adrian Keister
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Lorenzoi
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1 Answers1

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Because you have a mistake when computing the Residues, you should get $$F(s) = \frac{s^4+3s^3+2s^2+4s+4}{(s+3)(s^2+1)} = \frac{1}{s+3} + \frac{1}{s^2+1} + s \tag{1} $$ which answers your question on why you have an extra $\cos(t)H(t)$ appearing. To see why you have done a calculation mistake, take a look here: $$\frac{1}{s+3} + \frac{1}{s^2+1} + s = \frac{s^2+1+s+3+s(s+3)(s^2+1)}{(s+3)(s^2+1)}$$ which upon arranging terms you get equation $(1)$.

Ahmad Bazzi
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  • Thank you! I make an error because I tried to calculate the Residues of $s^2+1$ as if it is $s+1$. But so, how I can calculate the Residues of $\frac{2s^2+4s+4}{(s+2)(s^2+1)}$ in $s^2=-1$? – Lorenzoi Sep 19 '18 at 20:24
  • You'll have to decompose it to a simpler term, i.e. $s^2 + 1 =(s + i)(s - i)$. Then get a residue for $i$, then $-i$. – Ahmad Bazzi Sep 19 '18 at 20:24
  • Ok, thank you ! – Lorenzoi Sep 19 '18 at 20:26