Question

Find the laplace transform of F(s)=\frac{11s+28}{(s+2)^2(s+5)} \rightarrow f(t)=L^{-1}\left[F(s)\right]=?

Laplace transform
ANSWERED
asked 2021-09-10
Find the laplace transform of
\(\displaystyle{F}{\left({s}\right)}={\frac{{{11}{s}+{28}}}{{{\left({s}+{2}\right)}^{{2}}{\left({s}+{5}\right)}}}}\rightarrow{f{{\left({t}\right)}}}={L}^{{-{1}}}{\left[{F}{\left({s}\right)}\right]}=?\)

Expert Answers (1)

2021-09-11
Solution:
Given: \(\displaystyle{F}{\left({s}\right)}={\frac{{{11}{s}+{28}}}{{{\left({s}+{2}\right)}^{{2}}{\left({s}+{5}\right)}}}}\)
Find \(\displaystyle{f{{\left({t}\right)}}}={L}^{{-{1}}}{\left[{F}{\left({s}\right)}\right]}\)
i.e. Find the inverse Laplace Transform of \(\displaystyle{\frac{{{11}{s}+{28}}}{{{\left({s}+{2}\right)}^{{2}}{\left({s}+{5}\right)}}}}\)
use the partial fraction
\(\displaystyle{\frac{{{11}{s}+{28}}}{{{\left({s}+{2}\right)}^{{2}}{\left({s}+{5}\right)}}}}={\frac{{{A}}}{{{\left({s}+{2}\right)}}}}+{\frac{{{B}}}{{{\left({s}+{2}\right)}^{{2}}}}}+{\frac{{{C}}}{{{\left({s}+{5}\right)}}}}\)
\(\displaystyle\Rightarrow{\frac{{{11}{s}+{28}}}{{{\left({s}+{2}\right)}^{{2}}{\left({s}+{5}\right)}}}}={\frac{{{A}{\left({s}+{2}\right)}{\left({s}+{5}\right)}+{B}{\left({s}+{5}\right)}+{C}{\left({s}+{2}\right)}^{{2}}}}{{{\left({s}+{2}\right)}^{{2}}{\left({s}+{5}\right)}}}}\)
\(\displaystyle\Rightarrow{11}{s}+{28}={A}{\left({s}+{2}\right)}{\left({s}+{5}\right)}+{B}{\left({s}+{5}\right)}+{C}{\left({s}+{2}\right)}^{{2}}\)
At \(\displaystyle{s}=-{2},{11}\times-{2}+{28}={A}{\left({0}\right)}{\left({3}\right)}+{B}{\left({3}\right)}+{C}\cdot{0}^{{2}}\)
\(\displaystyle\Rightarrow{28}-{22}={0}+{3}{B}+{0}\Rightarrow{3}{B}={6}\)
\(\displaystyle\Rightarrow{B}={2}\)
\(\displaystyle\therefore{B}={2}\)
At \(\displaystyle{s}=-{5},{11}\times-{5}+{28}={A}{\left(-{3}\right)}{0}+{B}\cdot{0}+{C}{\left(-{3}\right)}^{{2}}\)
\(\displaystyle\Rightarrow{28}-{55}={0}+{0}+{9}{C}\Rightarrow{9}{C}=-{27}\)
\(\displaystyle\Rightarrow{C}=-{3}\)
\(\displaystyle\therefore{C}=-{3}\)
At s=0,
\(\displaystyle{11}\times{0}+{28}={A}{\left({2}\right)}{\left({5}\right)}+{B}\cdot{5}+{C}\cdot{2}^{{2}}\)
\(\displaystyle\Rightarrow{28}={10}{A}+{5}{B}+{4}{C}\)
\(\displaystyle\Rightarrow{28}={10}{A}+{5}\times{2}+{4}\times-{3}{\left[\because{B}={2},{C}=-{3}\right]}\)
\(\displaystyle\Rightarrow{28}={10}{A}+{10}-{12}={10}{A}-{2}\)
\(\displaystyle\Rightarrow{10}{A}={28}+{2}\Rightarrow{10}{A}={30}\Rightarrow{A}={3}\)
\(\displaystyle\therefore{A}={3}\)
\(\displaystyle\therefore{\frac{{{11}{s}+{28}}}{{{\left({s}+{2}\right)}^{{2}}{\left({s}+{5}\right)}}}}={\frac{{{3}}}{{{\left({s}+{2}\right)}}}}+{\frac{{{2}}}{{{\left({s}+{2}\right)}^{{2}}}}}+{\frac{{-{3}}}{{{\left({s}+{5}\right)}}}}\)
So, \(\displaystyle{f{{\left({t}\right)}}}={L}^{{-{1}}}{\left[{\frac{{{3}}}{{{\left({s}+{2}\right)}}}}+{\frac{{{2}}}{{{\left({s}+{2}\right)}^{{2}}}}}-{\frac{{{3}}}{{{\left({s}+{5}\right)}}}}\right]}\)
\(\displaystyle={3}{L}^{{-{1}}}{\left[{\frac{{{1}}}{{{\left({s}+{2}\right)}}}}\right]}+{2}{L}^{{-{1}}}{\left[{\frac{{{1}}}{{{\left({s}+{2}\right)}^{{2}}}}}\right]}-{3}{L}^{{-{1}}}{\left[{\frac{{{1}}}{{{\left({s}+{5}\right)}}}}\right]}\)
\(\displaystyle={3}{e}^{{-{2}{t}}}+{2}{t}{e}^{{-{2}{t}}}-{3}{e}^{{-{5}{t}}}\)
Thus
\(\displaystyle{f{{\left({t}\right)}}}={3}{e}^{{-{2}{t}}}+{2}{t}{e}^{{-{2}{t}}}-{3}{e}^{{-{5}{t}}}\)
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