Question

Prove {F}{left({s}right)}=frac{1}{{{s}-{2}}} text{ then } f{{left({t}right)}} is a) {e}^{{{2}{t}}}{u}{left({t}right)} b) u(t+2) c) u(t-2) d) {e}^{{-{2}{t}}}{u}{left({t}right)}

Laplace transform
ANSWERED
asked 2021-01-27
Prove \({F}{\left({s}\right)}=\frac{1}{{{s}-{2}}}\ \text{ then }\ f{{\left({t}\right)}}\) is
a) \({e}^{{{2}{t}}}{u}{\left({t}\right)}\)
b) \(u(t+2)\)
c) \(u(t-2)\)
d) \({e}^{{-{2}{t}}}{u}{\left({t}\right)}\)

Answers (1)

2021-01-28
Step 1
Given \({F}{\left({s}\right)}=\frac{1}{{{s}-{2}}}\)
We have to find the f(t)
Use definition of inverse Laplace transform, which is given below
\(f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace{F}{\left({s}\right)}\right\rbrace}\ldots{\left({1}\right)}\)
Step 2
Take inverse Laplace transform of \({F}{\left({s}\right)}=\frac{1}{{{s}-{2}}}\)
Hence, \({L}^{ -{{1}}}{\left\lbrace{F}\right\rbrace}{\left({s}\right)}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{2}}}\right\rbrace}…{\left({2}\right)}\)
From equation (1) and equation (2)
\(f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace\frac{1}{{{s}-{2}}}\right\rbrace}\)
\(={e}^{{{2}{t}}}{u}{\left({t}\right)}\)
Therefore, \(f{{\left({t}\right)}}={e}^{{{2}{t}}}{u}{\left({t}\right)}\)
Hence, option (a) is correct.
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