# Determine the Laplace transforms of these functions: a)\ f(t)=(t-4)u(t-2) b)\ g(t)=2e^{-4t}u(t-1)

Determine the Laplace transforms of these functions:
$$\displaystyle{a}{)}\ {f{{\left({t}\right)}}}={\left({t}-{4}\right)}{u}{\left({t}-{2}\right)}$$
$$\displaystyle{b}{)}\ {g{{\left({t}\right)}}}={2}{e}^{{-{4}{t}}}{u}{\left({t}-{1}\right)}$$

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Squairron
a) We can express f(t) as
$$\displaystyle{f{{\left({t}\right)}}}={\left({t}-{2}-{2}\right)}{u}{\left({t}-{2}\right)}$$
$$\displaystyle={\left({t}-{2}\right)}{u}{\left({t}-{2}\right)}-{2}{u}{\left({t}-{2}\right)}$$
By the linearly and times shift properties,
$$\displaystyle{F}{\left({x}\right)}={L}{\left[{\left({t}-{2}\right)}{u}{\left({t}-{2}\right)}\right]}-{2}{L}{\left[{u}{\left({t}-{2}\right)}\right]}$$
$$\displaystyle={\frac{{{1}}}{{{s}^{{2}}}}}{e}^{{-{2}{s}}}-{\frac{{{2}}}{{{s}}}}{e}^{{-{2}{s}}}$$
$$\displaystyle={\left({\frac{{{1}}}{{{s}^{{2}}}}}-{\frac{{{2}}}{{{s}}}}\right)}{e}^{{-{2}{s}}}$$
b) We can express g(t) as
$$\displaystyle{g{{\left({t}\right)}}}={2}{e}^{{-{4}{\left({t}-{1}+{1}\right)}}}{u}{\left({t}-{1}\right)}$$
$$\displaystyle={2}{e}^{{-{4}}}{e}^{{-{4}{\left({t}-{1}\right)}}}{u}{\left({t}-{1}\right)}$$
By the time shift property,
$$\displaystyle{G}{\left({s}\right)}={2}{e}^{{-{4}}}{L}{\left[{e}^{{-{4}{\left({t}-{1}\right)}}}{u}{\left({t}-{1}\right)}\right]}$$
$$\displaystyle={2}{e}^{{-{4}}}{\frac{{{e}^{{-{s}}}}}{{{s}+{4}}}}$$
$$\displaystyle={\frac{{{2}{e}^{{-{\left({s}+{4}\right)}}}}}{{{s}+{4}}}}$$