Find the Laplace transforms of the given functions. displaystyle g{{left({t}right)}}={4} cos{{left({4}{t}right)}}-{9} sin{{left({4}{t}right)}}+{2} cos{{left({10}{t}right)}}

Question
Laplace transform
asked 2021-02-21
Find the Laplace transforms of the given functions.
\(\displaystyle g{{\left({t}\right)}}={4} \cos{{\left({4}{t}\right)}}-{9} \sin{{\left({4}{t}\right)}}+{2} \cos{{\left({10}{t}\right)}}\)

Answers (1)

2021-02-22
Step 1 : Introduction
Given function is,
\(\displaystyle g{{\left({t}\right)}}={4} \cos{{\left({4}{t}\right)}}-{9} \sin{{\left({4}{t}\right)}}+{2} \cos{{\left({10}{t}\right)}}\)
We have to evaluate the Laplace transform of given function g(t)
step 2
\(\displaystyle g{{\left({t}\right)}}={4} \cos{{\left({4}{t}\right)}}-{9} \sin{{\left({4}{t}\right)}}+{2} \cos{{\left({10}{t}\right)}}\)
using Laplace transform properties
\(\displaystyle{L}{\left\lbrace \sin{{a}}{t}\right\rbrace}=\frac{a}{{{s}^{2}+{a}^{2}}}\)
\(\displaystyle{L}{\left\lbrace \cos{{a}}{t}\right\rbrace}=\frac{s}{{{s}^{2}+{a}^{2}}}\)
Applying Laplace transform on g(t) function
\(\displaystyle\therefore{L}{\left\lbrace g{{\left({t}\right)}}\right\rbrace}={L}{\left\lbrace{4} \cos{{\left({4}{t}\right)}}-{9} \sin{{\left({4}{t}\right)}}+{2} \cos{{\left({10}{t}\right)}}\right\rbrace}\)
\(\displaystyle{L}{\left\lbrace g{{\left({t}\right)}}\right\rbrace}={4}{L}{\left\lbrace \cos{{\left({4}{t}\right)}}\right\rbrace}-{9}{L}{\left\lbrace \sin{{\left({4}{t}\right)}}\right\rbrace}+{2}{L}{\left\lbrace \cos{{\left({10}{t}\right)}}\right\rbrace}\)
\(\displaystyle={4}{\left(\frac{s}{{{s}^{2}+{4}^{2}}}\right)}-{9}{\left(\frac{4}{{{s}^{2}+{4}^{2}}}\right)}+{2}{\left(\frac{s}{{{s}^{2}+{10}^{2}}}\right)}\)
\(\displaystyle=\frac{{{4}{s}}}{{{s}^{2}+{16}}}-\frac{36}{{{s}^{2}+{16}}}+\frac{{{2}{s}}}{{{s}^{2}+{100}}}\)
Thus , Laplace transform of given function g(t) is
\(\displaystyle{L}{\left\lbrace g{{\left({t}\right)}}\right\rbrace}=\frac{{{4}{s}}}{{{s}^{2}+{16}}}-\frac{36}{{{s}^{2}+{16}}}+\frac{{{2}{s}}}{{{s}^{2}+{100}}}\)
0

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