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
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) $${e}^{{-{2}{t}}}{u}{\left({t}\right)}$$

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.