# Determine displaystyle{L}^{ -{{1}}}{left[{F}{left({s}right)}right]} for the given F. displaystyle{F}{left({s}right)}=frac{{{s}-{2}}}{{{s}^{2}+{2}{s}+{3}}}

Question
Laplace transform
Determine $$\displaystyle{L}^{ -{{1}}}{\left[{F}{\left({s}\right)}\right]}$$ for the given F.
$$\displaystyle{F}{\left({s}\right)}=\frac{{{s}-{2}}}{{{s}^{2}+{2}{s}+{3}}}$$

2021-01-11
Step 1
Given information:
\displaystyle{F}{\left({s}\right)}=\frac{{{s}-{2}}}{{{s}^{2}+{2}{s}+{3}}}\)
Linearity property of inverse Laplace Transform, for function f(s) and g(s) and constants a and b.
$$\displaystyle{L}^{ -{{1}}}{\left\lbrace{a}\cdot f{{\left({s}\right)}}+{b}\cdot g{{\left({s}\right)}}\right\rbrace}={a}\cdot{L}^{ -{{1}}}{\left\lbrace f{{\left({s}\right)}}\right\rbrace}+{b}\cdot{L}^{ -{{1}}}{\left\lbrace g{{\left({s}\right)}}\right\rbrace}$$
Step 2
First can be written as,
$$\displaystyle{L}^{ -{{1}}}{\left\lbrace\frac{{{s}-{2}}}{{{s}^{2}+{2}{s}+{3}}}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{{{s}+{1}}}{{{\left({s}+{1}\right)}^{2}+{2}}}-{3}\cdot\frac{1}{{{\left({s}+{1}\right)}^{2}+{2}}}\right\rbrace}$$
Apply the linearity property and inverse Laplace transform,
$$\displaystyle{L}^{ -{{1}}}{\left\lbrace\frac{{{s}+{1}}}{{{\left({s}+{1}\right)}^{2}+{2}}}-{3}\cdot\frac{1}{{{\left({s}+{1}\right)}^{2}+{2}}}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{{{s}+{1}}}{{{\left({s}+{1}\right)}^{2}+{2}}}\right\rbrace}-{3}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{\left({s}+{1}\right)}^{2}+{2}}}\right\rbrace}$$
$$\displaystyle={e}^{{-{t}}} \cos{{\left(\sqrt{{2}}{t}\right)}}-{3}{e}^{{-{t}}}\frac{1}{{\sqrt{{2}}}} \sin{{\left(\sqrt{{2}}{t}\right)}}$$
$$\displaystyle={e}^{{-{t}}} \cos{{\left(\sqrt{{2}}{t}\right)}}-\frac{3}{{\sqrt{{2}}}}{e}^{{-{t}}} \sin{{\left(\sqrt{{2}}{t}\right)}}$$

### Relevant Questions

Find the inverse Laplace transform $$f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace{F}{\left({s}\right)}\right\rbrace}$$ of each of the following functions.
$${\left({i}\right)}{F}{\left({s}\right)}=\frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}^{2}+{4}}}{{{\left({s}^{2}+{1}\right)}{\left({s}-{1}\right)}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
The inverse Laplace transform for
$$\displaystyle{F}{\left({s}\right)}=\frac{8}{{{s}+{9}}}-\frac{6}{{{s}^{2}-\sqrt{{3}}}}$$ is
a) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \sin{{h}}{{\left({3}{t}\right)}}$$
b) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
c) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \sin{{h}}{\left({3}{t}\right)}$$
d) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
Use the appropriate algebra and Table of Laplace's Transform to find the given inverse Laplace transform. $$L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}$$
Find inverse laplace
find y(t) for the following
$$\displaystyle{Y}{\left({s}\right)}=\frac{8}{{s}}+\frac{1}{{{s}-{3}}}\cdot{e}^{{-{2}{s}}}$$
Given that $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}$$
a) Find $${L}{\left[\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\right]}$$ by differentiating f(t) and then using the Laplace transform tables in lecture notes.
b) Find $${L}{\left[\frac{{{d} f{{\left({t}\right)}}}}{{{\left.{d}{t}\right.}}}\right]}$$ using the theorem for differentiation
c) Repeat a) and b) for the case that $$f{{\left({t}\right)}}={4}{e}^{{-{3}{\left({t}-{4}\right)}}}{u}{\left({t}-{4}\right)}$$
Find the Laplace transform of $$\displaystyle f{{\left({t}\right)}}={t}{e}^{{-{t}}} \sin{{\left({2}{t}\right)}}$$
Then you obtain $$\displaystyle{F}{\left({s}\right)}=\frac{{{4}{s}+{a}}}{{\left({\left({s}+{1}\right)}^{2}+{4}\right)}^{2}}$$
Please type in a = ?
Determine $$L^{-1}\left[\frac{(s-4)e^{-3s}}{s^2-4s+5}\right]$$
$$y"+y=f(t)$$
$$y(0)=0 , y'(0)=1$$ where
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$
Find inverse Laplace transform $$L^{-1}\left\{\frac{s-5}{s^2+5s-24}\right\}$$ Please provide supporting details for your answer
$${F}{\left({s}\right)}=\frac{{{2}{s}-{3}}}{{{s}^{2}-{4}}}$$