Question

Find the inverse laplace trans. displaystyle{F}{left({s}right)}=frac{10}{{{s}{left({s}^{2}+{9}right)}}}

Laplace transform
ANSWERED
asked 2021-02-19
Find the inverse laplace trans.
\(\displaystyle{F}{\left({s}\right)}=\frac{10}{{{s}{\left({s}^{2}+{9}\right)}}}\)

Answers (1)

2021-02-20
Step 1
Given,
\(\displaystyle{F}{\left({s}\right)}=\frac{10}{{{s}{\left({s}^{2}+{9}\right)}}}\)
Find the inverse Laplace transform of this function.
Step 2
\(\displaystyle{F}{\left({s}\right)}=\frac{10}{{{s}{\left({s}^{2}+{9}\right)}}}\)
\(\displaystyle=\frac{10}{{{9}{s}}}-\frac{{{10}{s}}}{{{9}{\left({s}^{2}+{9}\right)}}}\)
Taking inverse Laplace transform of both sides,
\(\displaystyle{L}^{ -{{1}}}{\left[{F}{\left({s}\right)}\right]}={L}^{ -{{1}}}{\left[\frac{10}{{{9}{s}}}-\frac{{{10}{s}}}{{{9}{\left({s}^{2}+{9}\right)}}}\right]}\)
Then, \(\displaystyle f{{\left({t}\right)}}={L}^{ -{{1}}}{\left[\frac{10}{{{9}{s}}}\right]}-{L}^{ -{{1}}}{\left[\frac{{{10}{s}}}{{{9}{\left({s}^{2}+{9}\right)}}}\right]}\)
\(\displaystyle=\frac{10}{{9}}{L}^{ -{{1}}}{\left[\frac{1}{{s}}\right]}-\frac{10}{{9}}{L}^{ -{{1}}}{\left[\frac{s}{{{s}^{2}+{9}}}\right]}\)
\(\displaystyle=\frac{10}{{9}}{L}^{ -{{1}}}{\left[\frac{1}{{s}}\right]}-\frac{10}{{9}}\cdot{L}^{ -{{1}}}{\left[\frac{s}{{{s}^{2}+{3}^{2}}}\right]}\)
Step 3
Use the formula such that
\(\displaystyle{L}^{ -{{1}}}{\left[\frac{1}{{s}}\right]}={t}\)
\(\displaystyle{L}^{ -{{1}}}{\left[\frac{s}{{{s}^{2}+{a}^{2}}}\right]}= \cos{{\left({a}{t}\right)}}\)
\(\displaystyle f{{\left({t}\right)}}=\frac{10}{{9}}\cdot{t}-\frac{10}{{9}}\cdot \cos{{\left({3}{t}\right)}}\)
\(\displaystyle=\frac{10}{{9}}{\left[{t}- \cos{{\left({3}{t}\right)}}\right]}\)
Step 4
Hence,
\(\displaystyle f{{\left({t}\right)}}=\frac{10}{{9}}{\left[{t}- \cos{{\left({3}{t}\right)}}\right]}\)
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