Find the inverse Laplace transform of the given function by using the convolution theorem. {F}{left({s}right)}=frac{s}{{{left({s}+{1}right)}{left({s}^{2}+{4}right)}}}

Find the inverse Laplace transform of the given function by using the convolution theorem. {F}{left({s}right)}=frac{s}{{{left({s}+{1}right)}{left({s}^{2}+{4}right)}}}

Question
Laplace transform
asked 2021-01-13
Find the inverse Laplace transform of the given function by using the convolution theorem. \({F}{\left({s}\right)}=\frac{s}{{{\left({s}+{1}\right)}{\left({s}^{2}+{4}\right)}}}\)

Answers (1)

2021-01-14

Step 1
the given function is \({F}{\left({s}\right)}=\frac{s}{{{\left({s}+{1}\right)}{\left({s}^{2}+{4}\right)}}}\)
Let \({G}{\left({s}\right)}=\frac{1}{{{s}+{1}}}\ \text{ and }\ {H}{\left({s}\right)}=\frac{s}{{{s}^{2}+{4}}}\)
Since \({L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{a}}}\right\rbrace}={e}^{{{a}{t}}}\ \text{ and }\ {L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}^{2}+{a}^{2}}}\right\rbrace}= \cos{{a}}{t}\)
step 2
That implies,
\({L}^{ -{{1}}}{\left\lbrace{G}{\left({s}\right)}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}+{1}}}\right\rbrace}={e}^{{-{t}}}\) and
\({L}{\left\lbrace{H}{\left({s}\right)}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}^{2}+{4}}}\right\rbrace}= \cos{{2}}{t}\)
\(g{{\left({t}\right)}}={e}^{{-{t}}}\ \text{ and }\ {h}{\left({t}\right)}= \cos{{2}}{t}\)
Now, \({F}{\left({s}\right)}=\frac{s}{{{\left({s}+{1}\right)}{\left({s}^{2}+{4}\right)}}}\)
Step 3
By using the convolution theorem,
If \({F}{\left({s}\right)}={L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}\ \text{ and }\ {G}{\left({s}\right)}={L}{\left\lbrace g{{\left({t}\right)}}\right\rbrace}\) both exists for \({s}>{a}\ge{0}\) ,then
\(H(s)=F(s)G(s)\)
\(={L}{\left\lbrace{h}{\left({t}\right)}\right\rbrace},{s}>{a}\)
Where \({h}{\left({t}\right)}={\int_{{0}}^{{t}}} f{{\left({t}-\tau\right)}} g{{\left(\tau\right)}}{d}\tau={\int_{{0}}^{{t}}} f{{\left(\tau\right)}} g{{\left({t}-\tau\right)}}{d}\tau\)
Step 4
That implies,
\(F(s)=G(s)H(s)\)
\(=\frac{s}{{{\left({s}+{1}\right)}{\left({s}^{2}+{4}\right)}}}\)
\(={L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}\)
\(={L}{\left\lbrace{\int_{{0}}^{{t}}} g{{\left({t}-\tau\right)}}{h}{\left(\tau\right)}{d}\tau\right\rbrace}\)
Thus , \({F}{\left({s}\right)}={L}{\left\lbrace{\int_{{0}}^{{t}}} g{{\left({t}-\tau\right)}}{h}{\left(\tau\right)}{d}\tau\right\rbrace}\)
Step 5
Taking the inverse Laplace transform,
\({L}^{ -{{1}}}{\left\lbrace{F}{\left({s}\right)}\right\rbrace}={\int_{{0}}^{{t}}} g{{\left({t}-\tau\right)}}{h}{\left(\tau\right)}{d}\tau\)
\(={\int_{{0}}^{{t}}}{e}^{{-{\left({t}-\tau\right)}}} \cos{{2}}\tau{d}\tau\)
Therefore , \(f{{\left({t}\right)}}={\int_{{0}}^{{t}}}{e}^{{-{\left({t}-\tau\right)}}} \cos{{2}}\tau{d}\tau\)

0

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