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

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 , $$f{{\left({t}\right)}}={\int_{{0}}^{{t}}}{e}^{{-{\left({t}-\tau\right)}}} \cos{{2}}\tau{d}\tau$$

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