Find the inverse Laplace transform displaystylefrac{{{4}{s}^{3}-{s}^{2}+{5}{s}+{2}}}{{{s}^{4}}}

Question
Laplace transform
asked 2021-02-08
Find the inverse Laplace transform
\(\displaystyle\frac{{{4}{s}^{3}-{s}^{2}+{5}{s}+{2}}}{{{s}^{4}}}\)

Answers (1)

2021-02-09
Step 1
If \(\displaystyle{A}{\left({s}\right)}={L}{\left\lbrace{B}{\left({t}\right)}\right\rbrace}\), then the inverse transform of A(s) is defined as:
\(\displaystyle{L}^{{-{1}}}{\left[{A}{\left({s}\right)}\right]}={B}{\left({t}\right)}\)
Laplace transform of a function f(t) is given by:
\(\displaystyle{L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}\)
Step 2
\(\displaystyle{L}^{ -{{1}}}{\left(\frac{{{4}{s}^{3}-{s}^{2}+{5}{s}+{2}}}{{{s}^{4}}}\right)}\)
\(\displaystyle={L}^{ -{{1}}}{\left(\frac{4}{{s}}-\frac{1}{{s}^{2}}+\frac{5}{{s}^{3}}+\frac{2}{{s}^{4}}\right)}\)
\(\displaystyle={L}^{ -{{1}}}{\left(\frac{4}{{s}}\right)}-{L}^{ -{{1}}}{\left(\frac{1}{{s}^{2}}\right)}+{L}^{ -{{1}}}{\left(\frac{5}{{s}^{3}}\right)}+{L}^{ -{{1}}}{\left(\frac{2}{{s}^{4}}\right)}\)
\(\displaystyle={4}{H}{\left({t}\right)}-{t}+\frac{{{5}{t}^{2}}}{{2}}+\frac{{{t}^{3}}}{{3}}\)
0

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