Find the inverse Laplace transform of the function: F(s)=frac(2s^3+9s^2+11s+3)((s+1)^3s)

permaneceerc 2021-09-27 Answered
Find the inverse Laplace transform of the function:
\(\displaystyle{F}{\left({s}\right)}={\frac{{{2}{s}^{{3}}+{9}{s}^{{2}}+{11}{s}+{3}}}{{{\left({s}+{1}\right)}^{{3}}{s}}}}\)

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Expert Answer

Gennenzip
Answered 2021-09-28 Author has 11665 answers

Step 1: Given:
\(\displaystyle{F}{\left({s}\right)}={\frac{{{2}{s}^{{3}}+{9}{s}^{{2}}+{11}{s}+{3}}}{{{\left({s}+{1}\right)}^{{3}}{s}}}}\)
Step 2 To determine:
Inverse Laplace transform of the given function:=?
Step 3 Method used:
We use the partial fraction method to solve the given function.
First of all,
We find the partial fraction of the given function.
Then use some formulas of inverse Laplace transformations:
\(L^{-1}\left\{\frac{1}{s+1}\right\}=e^{-t}\)
\(L^{-1}\left\{\frac{2}{(s+1)^2}\right\}=2te^{-t}\)
\(L^{-1}\left\{\frac{1}{(s+1)^3}\right\}=\frac{e^{-t}t^2}{2}\)
\(L^{-1}\left\{\frac{3}{s}\right\}=3H(t)\)
Step 4 Solution to the question:
Take the partial fraction of:
\(\displaystyle{\frac{{{2}{s}^{{3}}+{9}{s}^{{2}}+{11}{s}+{3}}}{{{\left({s}+{1}\right)}^{{3}}{s}}}}=-{\frac{{{1}}}{{{s}+{1}}}}+{\frac{{{2}}}{{{\left({s}+{1}\right)}^{{2}}}}}+{\frac{{{1}}}{{{\left({s}+{1}\right)}^{{3}}}}}+{\frac{{{3}}}{{{s}}}}\)
Now:
Taking Laplace inverse of the resultant function:
\(L^{-1}\left\{-\frac{1}{s+1}+\frac{2}{(s+1)^2}+\frac{1}{(s+1)^3}+\frac{3}{s}\right\}\)
Step 5
Now:
Solving this expression:
\(L^{-1}\left\{-\frac{1}{s+1}+\frac{2}{(s+1)^2}+\frac{1}{(s+1)^3}+\frac{3}{s}\right\}\)
\(=-L^{-1}\left\{\frac{1}{s+1}\right\}+2L^{-1}\left\{\frac{1}{(s+1)^2}\right\}+L^{-1-1}\left\{\frac{1}{(s+1)^3}\right\}+3L^{-1}\left\{\frac{1}{s}\right\}\)
Substituting all the values as:
\(L^{-1}\left\{\frac{1}{s+1}\right\}=e^{-t}\)
\(L^{-1}\left\{\frac{2}{(s+1)^2}\right\}=2te^{-t}\)
\(L^{-1}\left\{\frac{1}{(s+1)^3}\right\}=\frac{e^{-t}t^2}{2}\)
\(L^{-1}\left\{\frac{3}{s}\right\}=3H(t)\)
Hence,
We get:
\(\displaystyle{L}^{{-{1}}}{\left({F}{\left({s}\right)}\right)}={\frac{{{2}{s}^{{3}}+{9}{s}^{{2}}+{11}{s}+{3}}}{{{\left({s}+{1}\right)}^{{3}}{s}}}}\)
\(\displaystyle=-{e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}+{\frac{{{e}^{{-{t}}}{t}^{{2}}}}{{{2}}}}+{3}{H}{\left({t}\right)}\)
So,
This is the Laplace inverse of the given function.

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