Find inverse laplace find y(t) for the following displaystyle{Y}{left({s}right)}=frac{8}{{s}}+frac{1}{{{s}-{3}}}cdot{e}^{{-{2}{s}}}

Find inverse laplace find y(t) for the following displaystyle{Y}{left({s}right)}=frac{8}{{s}}+frac{1}{{{s}-{3}}}cdot{e}^{{-{2}{s}}}

Question
Laplace transform
asked 2020-11-30
Find inverse laplace
find y(t) for the following
\(\displaystyle{Y}{\left({s}\right)}=\frac{8}{{s}}+\frac{1}{{{s}-{3}}}\cdot{e}^{{-{2}{s}}}\)

Answers (1)

2020-12-01
Step 1
Consider the given function,
\(\displaystyle{Y}{\left({s}\right)}=\frac{8}{{s}}+{e}^{{-{2}{s}}}\cdot\frac{1}{{{s}-{3}}}\)
use the inverse transform rule which states that,
\(\displaystyle\text{if }\ {L}^{ -{{1}}}{\left\lbrace{Y}{\left({s}\right)}\right\rbrace}= f{{\left({t}\right)}}\ \text{ then }\ {L}^{ -{{1}}}{\left\lbrace{e}^{{-{a}{s}}}{Y}{\left({s}\right)}\right\rbrace}={U}{\left({t}-{a}\right)} f{{\left({t}-{a}\right)}}\)
\(\displaystyle{L}^{ -{{1}}}{\left\lbrace\frac{1}{{s}}\right\rbrace}={1}\)
\displaystyle{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{a}}}\right\rbrace}={e}^{{-{a}{t}}}\)
Step 2
by using the above rules the inverse Laplace transform of the function will be,
\(\displaystyle{L}^{ -{{1}}}{\left\lbrace{Y}{\left({s}\right)}\right\rbrace}={L}^{ -{{1}}}{\left\lbrace\frac{8}{{s}}+{e}^{{-{2}{s}}}\frac{1}{{{s}-{3}}}\right\rbrace}\)
\(\displaystyle f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace\frac{8}{{s}}\right\rbrace}+{L}^{ -{{1}}}{\left\lbrace{e}^{{-{2}{s}}}\frac{1}{{{s}-{3}}}\right\rbrace}\)
\(\displaystyle={8}+{U}{\left({t}-{2}\right)}{L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{3}}}\right\rbrace}{\left({t}-{2}\right)}\)
\(\displaystyle={8}+{U}{\left({t}-{2}\right)}{e}^{{{2}{\left({t}-{2}\right)}}}\)
hence the inverse Laplace transform is \(\displaystyle f{{\left({t}\right)}}={8}+{U}{\left({t}-{2}\right)}{e}^{{{2}{\left({t}-{2}\right)}}}\)
0

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