Find the inverse laplace transform of $F(s) = \frac{a(s^{2}-2a^{2})}{s^{4}+4a^{4}}$
I factored denominator by completing square but how do i proceed next .
Thanks
[Math] Find the inverse laplace transform of $F(s) = \frac{a(s^{2}-2a^{2})}{s^{4}+4a^{4}}$
laplace transform
Related Solutions
$$I(s)=\frac{6}{Ls^2 + Rs + \frac{1}{C}}$$
Having fixed $R$, $C$ and $L$ (for example $R=6$, $C=1/5$, $L=1$), we have:
$$I(s) = \frac{6}{s^2+6s+5}$$
First of all, find the zeros of denominator:
$$s^2 + 4s + 1= 0 \Rightarrow s = -5 \vee s = -1$$
Then $s^2+4s+1= (s+5)(s+1)$.
We can write $$I(s) = \frac{a}{s+5} + \frac{b}{s+1} = \frac{a(s+1)+b(s+5)}{(s+1)(s+5)} = \frac{s(a+b) + (a + 5b)}{s^2 + 6s + 5} = \frac{6}{s^2+6s+5}$$
Then:
$$\left\{\begin{array}{lcl}a+b & = & 0\\a + 5b & = & 6 \end{array}\right. \Rightarrow \left\{\begin{array}{lcl}a & = & -b\\-b + 5b & = & 6 \end{array}\right. \Rightarrow \left\{\begin{array}{lcl}a & = & -\frac{3}{2}\\b & = & \frac{3}{2} \end{array}\right.$$
Then:
$$I(s) = -\frac{\frac{3}{2}}{s+5} + \frac{\frac{3}{2}}{s+1}$$
The antitransformation yield to:
$$i(t) = -\frac{3}{2}e^{-5t} + \frac{3}{2}e^{-t}$$
Try this way:
$$\frac{1}{(s+1)(s^2+s+1)} = \frac{A}{s+1}+\frac{Bs+C}{s^2+s+1}$$
Notice that the second term has complex solutions, so it's a good idea if you try to transform it in something like $\frac{s+a}{(s+a)^2+b^2}$ and $\frac{b}{(s+a)^2+b^2}$
The solution is:
$$e^{-t}-\frac{e^{-t/2} (\sqrt3 \cos(\frac{\sqrt3 t}{2})-\sin(\frac{\sqrt3 t}{2}))}{\sqrt3}$$
$1)\frac{s+a}{(s+a)^2+b^2} \to e^{-a}\cos(bt)$
$2)\frac{b}{(s+a)^2+b^2} \to e^{-a}\sin(bt)$
Example: If $B=1$ and $C=1$
$$\frac{s+1}{s^2+s+1} = \frac{s+1}{(s+\frac12)^2+(1-\frac14)} = $$
$$\frac{s}{(s+\frac12)^2+(\frac34)} + \frac{\sqrt{\frac34}}{(s+\frac12)^2+(\frac34)}*(\sqrt{\frac43}) $$
In this case you have to add $\sqrt{\frac43}$, so the numerator can be equal to $1$, but you need $\sqrt{\frac34}$ to use the second transform (obviously you will add $\sqrt{\frac43}$ to the result of this transformation).
...
In your case is $A=1, B=-1, C=0$, so:
$$\frac{1}{s+1}-\frac{s}{(s+\frac12)^2+\frac34}$$
Now, to use the transformation $1)$ you need something to add to the fraction like that: $$\frac{{\frac12}}{(s+\frac12)^2+\frac34}$$
So:
$$\frac{1}{s+1}-(\frac{s}{(s+\frac12)^2+\frac34} + \frac{{\frac12}}{(s+\frac12)^2+\frac34} - \frac{{\frac12}}{(s+\frac12)^2+\frac34}) $$
So:
$$\frac{1}{s+1}-(\frac{s+{\frac12} }{(s+\frac12)^2+\frac34} - \frac{{\frac12}}{(s+\frac12)^2+\frac34}) $$
That becomes:
$$\frac{1}{s+1}-(\frac{s+{\frac12} }{(s+\frac{1}{2})^2+\frac34} - \frac{{\frac{\sqrt{3}}{2}}}{(s+\frac12)^2+\frac34}*\frac{2}{\sqrt{3}}*\frac12 )$$
Best Answer
Since: $$\mathcal{L}\left(\sinh(at)\right) = \frac{a}{s^2-a^2}\tag{1}$$ by computing the partial fraction decomposition of $F(s)$ it follows that: $$\mathcal{L}^{-1}(F(s)) = \cos(at)\sinh(at).\tag{2}$$