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
asked 2021-01-10
Determine \(\displaystyle{L}^{ -{{1}}}{\left[{F}{\left({s}\right)}\right]}\) for the given F.
\(\displaystyle{F}{\left({s}\right)}=\frac{{{s}-{2}}}{{{s}^{2}+{2}{s}+{3}}}\)

Answers (1)

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,
\(\displaystyle{L}^{ -{{1}}}{\left\lbrace\frac{{{s}+{1}}}{{{\left({s}+{1}\right)}^{2}+{2}}}-{3}\cdot\frac{1}{{{\left({s}+{1}\right)}^{2}+{2}}}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{{{s}+{1}}}{{{\left({s}+{1}\right)}^{2}+{2}}}\right\rbrace}-{3}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{\left({s}+{1}\right)}^{2}+{2}}}\right\rbrace}\)
\(\displaystyle={e}^{{-{t}}} \cos{{\left(\sqrt{{2}}{t}\right)}}-{3}{e}^{{-{t}}}\frac{1}{{\sqrt{{2}}}} \sin{{\left(\sqrt{{2}}{t}\right)}}\)
\(\displaystyle={e}^{{-{t}}} \cos{{\left(\sqrt{{2}}{t}\right)}}-\frac{3}{{\sqrt{{2}}}}{e}^{{-{t}}} \sin{{\left(\sqrt{{2}}{t}\right)}}\)
0

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