Find the inverse Laplace transform f{{left({t}right)}}={L}^{ -{{1}}}{leftlbrace{F}{left({s}right)}rightrbrace} of each of the following functions. {left({i}right)}{F}{left({s}right)}=frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}} Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms. {left({i}{i}right)}{F}{left({s}right)}=frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}} Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms. {left({i}{i}{i}right)}{F}{left({s}right)}=frac{{{3}{s}^{2}+{4}}}{{{left({s}^{2}+{1}right)}{left({s}-{1}right)}}} Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.

Find the inverse Laplace transform f{{left({t}right)}}={L}^{ -{{1}}}{leftlbrace{F}{left({s}right)}rightrbrace} of each of the following functions. {left({i}right)}{F}{left({s}right)}=frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}} Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms. {left({i}{i}right)}{F}{left({s}right)}=frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}} Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms. {left({i}{i}{i}right)}{F}{left({s}right)}=frac{{{3}{s}^{2}+{4}}}{{{left({s}^{2}+{1}right)}{left({s}-{1}right)}}} Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.

Question
Laplace transform
asked 2020-11-01
Find the inverse Laplace transform \(f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace{F}{\left({s}\right)}\right\rbrace}\) of each of the following functions.
\({\left({i}\right)}{F}{\left({s}\right)}=\frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}}\)
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
\({\left({i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}}\)
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
\({\left({i}{i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}^{2}+{4}}}{{{\left({s}^{2}+{1}\right)}{\left({s}-{1}\right)}}}\)
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.

Answers (1)

2020-11-02
\({\left({i}\right)}{F}{\left({s}\right)}=\frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}}\)
\(\frac{{{2}{s}+{1}}}{{\left({s}-{1}\right)}^{2}}=\frac{A}{{{s}-{1}}}+\frac{B}{{\left({s}-{1}\right)}^{2}}\)
\({2}{s}+{1}=\frac{A}{{{s}-{1}}}+\frac{B}{{\left({s}-{1}\right)}^{2}}\)
\(2s+1=A(s-1)+B\)
\(A=2, -A+B=1\)
\(B=3\)
\({F}{\left({s}\right)}=\frac{2}{{{s}-{1}}}+\frac{3}{{\left({s}-{1}\right)}^{2}}\)
\(f{{\left({t}\right)}}={2}{e}^{t}+{3}{t}{e}^{t}
\({\left({i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}}\)
\(\frac{{{3}{s}+{2}}}{{{\left({s}-{2}\right)}{\left({s}-{1}\right)}}}=\frac{A}{{{s}-{2}}}+\frac{B}{{{s}-{1}}}\)
\(3s+2=A(s-1)+B(s-2)\)
\(A+B=3\)
\(A=8 , B=-5\)
\(-A-2B=2\)
\({F}{\left({s}\right)}=\frac{8}{{{s}-{2}}}-\frac{5}{{{s}-{1}}}\)
\(f{{\left({t}\right)}}={8}{e}^{{{2}{t}}}-{5}{e}^{{{t}}}\)
\({\left({i}{i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}^{2}+{4}}}{{{\left({s}^{2}+{1}\right)}{\left({s}-{1}\right)}}}\)
\(\frac{{{3}{s}^{2}+{4}}}{{{\left({s}-{1}\right)}{\left({s}^{2}+{1}\right)}}}=\frac{{{A}{s}+{B}}}{{{s}^{2}+{1}}}+\frac{C}{{{s}-{1}}}\)
\(\Rightarrow{3}{s}^{2}+{4}={A}{\left({s}^{2}-{s}\right)}+{B}{\left({s}-{1}\right)}+{C}{\left({s}^{2}+{1}\right)}\)
\(A+B=3\)
\(-A+B=0\)
\(-B+C=4\)
\({A}=-\frac{1}{{2}},{B}=-\frac{1}{{2}},{C}=\frac{7}{{2}}\)
\({F}{\left({s}\right)}=\frac{{-{1}{s}}}{{{2}{\left({s}^{2}+{1}\right)}}}-\frac{1}{{{2}{\left({s}^{2}+{1}\right)}}}+\frac{7}{{{2}{\left({s}-{1}\right)}}}\)
\(f{{\left({t}\right)}}=-\frac{1}{{2}} \cos{{t}}-\frac{1}{{2}} \sin{{t}}+\frac{7}{{2}}{e}^{t}\)
0

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