Jamarcus Lindsey

2022-10-10

How do I compute the following transform?
$\frac{s-1}{2{s}^{2}+s+6}$
I've gotten this far:
$\frac{1}{2}\cdot \frac{s-1}{{\left(s+\frac{1}{4}\right)}^{2}+\frac{47}{16}}$

Jeremy Mayo

You can now use:
${\mathcal{L}}_{s}\left({\mathrm{e}}^{-\lambda t}{\mathrm{e}}^{it\omega }\right)={\int }_{0}^{\mathrm{\infty }}{\mathrm{e}}^{-st}{\mathrm{e}}^{-\lambda t}{\mathrm{e}}^{it\omega }\mathrm{d}t=\frac{1}{s+\lambda -i\omega }$
valid as long as $s+\lambda >0$, and $\omega \in \mathbb{R}$. From here, taking real and imaginary parts you conclude:
${\mathcal{L}}_{s}\left({\mathrm{e}}^{-\lambda t}\mathrm{cos}\left(\omega t\right)\right)=\frac{s+\lambda }{\left(s+\lambda {\right)}^{2}+{\omega }^{2}},\phantom{\rule{2em}{0ex}}{\mathcal{L}}_{s}\left({\mathrm{e}}^{-\lambda t}\mathrm{sin}\left(\omega t\right)\right)=\frac{\omega }{\left(s+\lambda {\right)}^{2}+{\omega }^{2}}$
Now you the decomposition you obtained and read off the coefficients, keeping in mind that the inverse Laplace transform has the form ${\mathrm{e}}^{-\lambda t}\left(\alpha \mathrm{cos}\left(\omega t\right)+\beta \mathrm{sin}\left(\omega t\right)\right)$ for some $\lambda ,\omega ,\alpha$ and $\beta$

Parker Pitts

$F\left(s\right)=\frac{s-1}{2{s}^{2}+s+6}=\frac{s-1}{2\left({s}^{2}+\frac{s}{2}+3\right)}$
$a{x}^{2}+bx=a\left[\left(x+\frac{b}{2a}{\right)}^{2}-\left(\frac{b}{2a}{\right)}^{2}\right]$
So:
$2\left({s}^{2}+\frac{s}{2}+3\right)=2\left[\left(s+\frac{1}{4}{\right)}^{2}-\left(\frac{1}{4}{\right)}^{2}+3\right]=2\left[\left(s+\frac{1}{4}{\right)}^{2}+\left(\frac{47}{15}\right)\right]$
therefore:
$F\left(s\right)=\left(\frac{1}{2}\right)\frac{s-1+\frac{5}{4}-\frac{5}{4}}{\left(s+\frac{1}{4}{\right)}^{2}+\left(\frac{47}{15}\right)}=\left(\frac{1}{2}\right)\left[\frac{s+\frac{1}{4}}{\left(s+\frac{1}{4}{\right)}^{2}+\left(\frac{47}{15}\right)}-\frac{\frac{5}{4}}{\left(s+\frac{1}{4}{\right)}^{2}+\left(\frac{47}{15}\right)}\right]$
We know:
$\mathcal{L}\left[{e}^{-at}cos\left(bt\right)\right]=\frac{s+a}{\left(s+a{\right)}^{2}+{b}^{2}}$
$\mathcal{L}\left[{e}^{-at}sin\left(bt\right)\right]=\frac{b}{\left(s+a{\right)}^{2}+{b}^{2}}$
therefore:
$f\left(t\right)=\left(\frac{1}{2}\right){e}^{\frac{-t}{4}}\left[cos\left(\frac{\sqrt{47}}{4}t\right)-\frac{5}{\sqrt{47}}sin\left(\frac{\sqrt{47}}{4}t\right)\right]$

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