Question

What is the Laplace transform of displaystyle{x}{left({t}right)}={e}^{{-{t}}} cos{{left(piright)}}{t}?

Laplace transform
ANSWERED
asked 2020-10-19
What is the Laplace transform of \(\displaystyle{x}{\left({t}\right)}={e}^{{-{t}}} \cos{{\left(\pi\right)}}{t}\)?

Expert Answers (1)

2020-10-20
Data analysis
To find the Laplace transform of ,
\(\displaystyle{x}{\left({t}\right)}={e}^{{-{t}}} \cos{\pi}{t}\)
Here the proof for standard form has been shown,
Let standard form be,
\(\displaystyle{x}{\left({t}\right)}={e}^{{{a}{t}}} \cos{{b}}{t}\)
Derivation and Solution
\(\displaystyle{L}{\left\lbrace{e}^{{-{a}{t}}} \cos{{b}}{t}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{{a}{t}}} \cos{{b}}{t}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}=\)
\(\displaystyle={\int_{{0}}^{\infty}}{e}^{{-{s}{\left({s}+{a}\right)}{t}}} \cos{{b}}{t}{\left.{d}{t}\right.}\)
Let \(\displaystyle{u}= \cos{{b}}{t}\ \ \ {d}{v}={e}^{{-{s}{\left({s}+{a}\right)}{t}}}\)
\(\displaystyle{d}{u}=-{b} \sin{{b}}{t}{\left.{d}{t}\right.}\ \ \ \ {v}=-\frac{1}{{{s}+{a}}}{e}^{{-{\left({s}+{a}\right)}{t}}}\)
(using integration by parts)
\(\displaystyle\Rightarrow{y}={{\left|-\frac{{ \cos{{b}}{t}}}{{{s}+{a}}}{e}^{{-{\left({s}+{a}\right)}{t}}}\right|}_{{0}}^{\infty}}-\frac{b}{{{s}+{a}}}{\int_{{0}}^{\infty}}{e}^{{-{\left({s}+{a}\right)}{t}}} \sin{{b}}{t}{\left.{d}{t}\right.}\)
Formula: \(\displaystyle\int{u}{d}{v}={u}{v}-\int{v}'{\left({x}\right)}{d}{u}\)
\(\displaystyle\Rightarrow{y}={\left({0}-{\left(-\frac{1}{{{s}+{a}}}\right)}\right)}-\frac{b}{{{s}+{a}}}{\int_{{0}}^{\infty}}{e}^{{-{\left({s}+{a}\right)}{t}}} \sin{{b}}{t}{\left.{d}{t}\right.}\)
let \(\displaystyle{u}= \sin{{b}}{t}\ \ \ \ {d}{v}={e}^{{-{\left({s}+{a}\right)}{t}}}{\left.{d}{t}\right.}\)
\(\displaystyle{d}{u}={b} \cos{{b}}{t}{\left.{d}{t}\right.}\ \ \ \ {v}=-\frac{1}{{{s}+{a}}}{e}^{{-{\left({s}+{a}\right)}{t}}}\)
Again integration by parts
\(\displaystyle\Rightarrow{y}=\frac{1}{{{s}+{a}}}-\frac{b}{{{s}+{a}}}{\left[\frac{{-{e}^{{-{\left({s}+{a}\right)}{t}}} \sin{{b}}{t}}}{{{s}+{a}}}{{|}_{{0}}^{\infty}}+\frac{b}{{{s}+{a}}}{\int_{{0}}^{\infty}}{e}^{{-{\left({s}+{a}\right)}{t}}} \cos{{b}}{t}\right]}\)
\(\displaystyle\Rightarrow{y}=\frac{1}{{{s}+{a}}}-\frac{b}{{{s}+{a}}}{\left[{0}+\frac{b}{{{s}+{a}}}{y}\right]}\)
\(\displaystyle=\frac{1}{{{s}+{a}}}-\frac{{{b}^{2}}}{{\left({s}+{a}\right)}^{2}}{y}\)
\(\displaystyle\Rightarrow\frac{{{\left({s}+{a}\right)}^{2}+{b}^{2}}}{{\left({s}+{a}\right)}^{2}}{y}=\frac{1}{{{s}+{a}}}\)
\(\displaystyle\Rightarrow{y}=\frac{{{s}+{a}}}{{{\left({s}+{a}\right)}^{2}+{b}^{2}}}\)
\(\displaystyle\Rightarrow{L}{\left\lbrace{e}^{{-{a}{t}}} \cos{{b}}{t}\right\rbrace}=\frac{{{s}+{a}}}{{{\left({s}+{a}\right)}^{2}+{b}^{2}}}\)
Substituting \(\displaystyle{a}={1},{b}=\pi\Rightarrow{L}{\left\lbrace{e}^{{-{t}}} \cos{\pi}{t}\right\rbrace}=\frac{{{s}+{1}}}{{{\left({s}+{1}\right)}^{2}+\pi^{2}}}\)
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