Find the inverse Laplace transform f(t)=L^(-1){F(s)} of the function F(s)=frac(8e^(-4s))(s^2+64)

Cabiolab 2021-09-25 Answered

Find the inverse Laplace transform \(f(t)=L^{-1}\left\{F(s)\right\}\) of the function \(\displaystyle{F}{\left({s}\right)}={\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}\)
Use h(t-c) for the Heaviside function h_c(t) if necessary.
\(f(t)=L^{-1}\left\{\frac{8e^{-4s}}{s^2+64}\right\}=\)

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Expert Answer

StrycharzT
Answered 2021-09-26 Author has 12279 answers

Step 1
Consider the provided function,
\(\displaystyle{F}{\left({s}\right)}={\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}\)
The inverse laplace transform of F(s) is find as,
\(\displaystyle{f{{\left({t}\right)}}}={L}^{{-{1}}}{\left({F}{\left({s}\right)}\right)}={L}^{{-{1}}}{\left({\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}\right)}\)
Apply inverse tranform rule,
if \(\displaystyle{L}^{{-{1}}}{\left({F}{\left({s}\right)}\right)}={f{{\left({t}\right)}}}\ {t}{h}{e}{n}\ {L}^{{-{1}}}{\left({e}^{{-{c}{s}}}{F}{\left({s}\right)}\right)}={h}{\left({t}-{c}\right)}{f{{\left({t}-{c}\right)}}}\)
where \(\displaystyle{h}_{{c}}{\left({t}\right)}\) is Heaviside funtion.
For \(\displaystyle{\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}.{F}{\left({s}\right)}={\frac{{{8}}}{{{s}^{{2}}+{64}}}},{c}={4}\)
Step 2
Use the inverse Laplace transform table,
\(\displaystyle{L}^{{-{1}}}{\left({\frac{{{a}}}{{{s}^{{2}}+{a}^{{2}}}}}\right)}={\sin{{\left({a}{t}\right)}}}\)
So, \(\displaystyle{L}^{{-{1}}}{\left({\frac{{{8}}}{{{s}^{{2}}+{64}}}}\right)}={L}^{{-{1}}}{\left({\frac{{{8}}}{{{s}^{{2}}+{8}^{{2}}}}}\right)}={\sin{{\left({8}{t}\right)}}}\)
Thus, \(\displaystyle{f{{\left({t}\right)}}}={L}^{{-{1}}}{\left({\frac{{{8}{e}^{{-{4}{s}}}}}{{{s}^{{2}}+{64}}}}\right)}={h}{\left({t}-{4}\right)}{\sin{{\left({8}{\left({t}-{4}\right)}\right)}}}\)

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