Question

Find f(t) , L^{-1}\left\{\frac{(3s-1)}{s^2(s+1)^3}\right\}

Laplace transform
ANSWERED
asked 2021-09-18

Find f(t)
\(L^{-1}\left\{\frac{(3s-1)}{s^2(s+1)^3}\right\}\)

Expert Answers (1)

2021-09-19

Step 1
From the given information it is needed to calculate:
\(L^{-1}\left\{\frac{(3s-1)}{s^2(s+1)^3}\right\}\)
Step 2
Now, first take the partial fractions of the given expression and solve:
\(\displaystyle{\frac{{{\left({3}{s}-{1}\right)}}}{{{s}^{{2}}{\left({s}+{1}\right)}^{{3}}}}}={\frac{{{a}_{{0}}}}{{{s}}}}+{\frac{{{a}_{{1}}}}{{{s}^{{2}}}}}+{\frac{{{a}_{{2}}}}{{{\left({s}+{1}\right)}}}}+{\frac{{{a}_{{3}}}}{{{\left({s}+{1}\right)}^{{2}}}}}+{\frac{{{a}_{{4}}}}{{{\left({s}+{1}\right)}^{{3}}}}}\)
\(\displaystyle={\frac{{{a}_{{0}}{\left({s}\right)}{\left({s}+{1}\right)}^{{3}}+{a}_{{1}}{\left({s}+{1}\right)}^{{3}}+{a}_{{2}}{\left({s}^{{2}}\right)}{\left({s}+{1}\right)}^{{2}}+{a}_{{3}}{\left({s}^{{2}}\right)}{\left({s}+{1}\right)}+{a}_{{4}}{\left({s}^{{2}}\right)}}}{{{s}^{{2}}{\left({s}+{1}\right)}^{{3}}}}}\)
\(\displaystyle{3}{s}-{1}={a}_{{0}}{\left({s}\right)}{\left({s}+{1}\right)}^{{3}}+{a}_{{1}}{\left({s}+{1}\right)}^{{3}}+{a}_{{2}}{\left({s}+{1}\right)}^{{2}}+{a}_{{3}}{\left({s}^{{2}}\right)}{\left({s}+{1}\right)}+{a}_{{4}}{\left({s}^{{2}}\right)}\)
when \(\displaystyle{s}={0}\Rightarrow{3}{\left({0}\right)}-{1}={a}_{{1}}{\left({1}\right)}^{{3}}\Rightarrow{a}_{{1}}=-{1}\)
when \(\displaystyle{s}=-{1}\Rightarrow{3}{\left(-{1}\right)}-{1}={a}_{{4}}{\left(-{1}\right)}^{{2}}\Rightarrow-{4}={a}_{{4}}\)
Step 3
Now calculate rest of the unknowns by putting the value of the \(\displaystyle{a}_{{1}}\ \text{ and }\ {a}_{{4}}\):
\(\displaystyle{3}{s}-{1}={a}_{{0}}{\left({s}\right)}{\left({s}+{1}\right)}^{{3}}+{\left(-{1}\right)}{\left({s}+{1}\right)}^{{3}}+{a}_{{2}}{\left({s}^{{2}}\right)}{\left({s}+{1}\right)}^{{2}}+{a}_{{3}}{\left({s}^{{2}}\right)}{\left({s}+{1}\right)}+{\left(-{4}\right)}{\left({s}^{{2}}\right)}\)
\(\displaystyle{3}{s}-{1}={a}_{{0}}{s}^{{4}}+{a}_{{2}}{s}^{{4}}+{3}{a}_{{0}}{s}^{{3}}+{2}{a}_{{2}}{s}^{{3}}+{a}_{{3}}{s}^{{3}}-{s}^{{3}}+{3}{a}_{{0}}{s}^{{2}}+{a}_{{2}}{s}^{{2}}+{a}_{{3}}{s}^{{2}}-{7}{s}^{{2}}+{a}_{{0}}{s}-{3}{s}-{1}\)
\(\displaystyle{3}{s}-{1}={s}^{{4}}{\left({a}_{{0}}+{a}_{{2}}\right)}+{s}^{{3}}{\left({3}{a}_{{0}}+{2}{a}_{{2}}+{a}_{{3}}-{1}\right)}+{s}^{{2}}{\left({3}{a}_{{0}}+{a}_{{2}}+{a}_{{3}}-{7}\right)}+{s}{\left({a}_{{0}}-{3}\right)}-{1}\)
now by comparing:
\(\displaystyle{a}_{{0}}+{a}_{{2}}={0}\)
\(\displaystyle{3}{a}_{{0}}+{2}{a}_{{2}}+{a}_{{3}}-{1}={0}\)
\(\displaystyle{3}{a}_{{0}}+{a}_{{2}}+{a}_{{3}}-{7}={0}\)
\(\displaystyle{a}_{{0}}-{3}={3}\Rightarrow{a}_{{0}}={6}\)
so , \(\displaystyle{a}_{{0}}+{a}_{{2}}={0}\Rightarrow{6}+{a}_{{2}}={0}\Rightarrow{a}_{{2}}=-{6}\)
and , \(\displaystyle{3}{a}_{{0}}+{2}{a}_{{2}}+{a}_{{3}}-{1}={0}\Rightarrow{3}{\left({6}\right)}+{2}{\left(-{6}\right)}+{a}_{{3}}-{1}={0}\)
\(\displaystyle{18}-{12}-{1}+{a}_{{3}}={0}\Rightarrow{a}_{{3}}=-{5}\)
Step 4
Now, put the value of unknowns:
\(\displaystyle{\frac{{{\left({3}{s}-{1}\right)}}}{{{s}^{{2}}{\left({s}+{1}\right)}^{{3}}}}}={\frac{{{6}}}{{{s}}}}+{\frac{{-{1}}}{{{s}^{{2}}}}}+{\frac{{{\left(-{6}\right)}}}{{{\left({s}+{1}\right)}}}}+{\frac{{-{5}}}{{{\left({s}+{1}\right)}^{{2}}}}}+{\frac{{-{4}}}{{{\left({s}+{1}\right)}^{{3}}}}}\)
\(L^{-1}\left\{\frac{(3s-1)}{s^2(s+1)^3}\right\}=L^{-1}\left\{\frac{6}{s}+\frac{-1}{s^2}+\frac{(-6)}{(s+1)}+\frac{-5}{(s+1)^2}+\frac{-4}{(s+1)^3}\right\}\)
\(=L^{-1}\left\{\frac{6}{s}\right\}+L^{-1}\left\{\frac{-1}{s^2}\right\}+L^{-1}\left\{\frac{(-6)}{(s+1)}\right\}+L^{-1}\left\{\frac{-5}{(s+1)^2}\right\}+L^{-1}\left\{\frac{-4}{(s+1)^3}\right\}\)
\(\displaystyle={6}{\left({1}\right)}-{t}-{6}{\left({e}^{{-{t}}}\right)}-{5}{\left({e}^{{-{t}}}{t}\right)}-{4}{\left({2}{e}^{{-{t}}}{t}^{{2}}\right)}\)
\(\displaystyle={6}-{t}-{6}{e}^{{-{t}}}-{5}{e}^{{-{t}}}{t}-{8}{e}^{{-{t}}}{t}^{{2}}\)
Step 5
The formula used in the above inverse laplace transform are:
\(\displaystyle{L}^{{-{1}}}{\left({\frac{{{1}}}{{{s}}}}\right)}={1}\)
\(\displaystyle{L}^{{-{1}}}{\left({\frac{{{n}!}}{{{s}^{{{n}+{1}}}}}}\right)}={t}^{{n}}\)
\(\displaystyle{L}^{{-{1}}}{\left({\frac{{{1}}}{{{s}-{a}}}}\right)}={e}^{{{a}{t}}}\)

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