91Ó°ÊÓ

The given function is$f\left(t\right)=\{\begin{array}{l}3,0≤t≤2\\ 6-t,2<t\end{array}$

Using the Laplace transform definition, we get

$\begin{array}{c}L\left\{f\right(t\left)\right\}={∫}_0^{\mathrm{∞}}{e}^{-st}f\left(t\right)dt\\ ={∫}_0^{2}3e^{−st}dt+{∫}_2^{\mathrm{∞}}(6−t)e^{−st}dt\\ =3{[−\frac{e^{−st}}{s}]}_0^2+\underset{N→\mathrm{∞}}{\lim }{∫}_2^N(6−t)e^{−st}dt\\ =\frac{3}{s}(1−e^{−2s})+\underset{N→\mathrm{∞}}{\lim }{∫}_2^N(6−t)e^{−st}dt\end{array}$

Letrole="math" localid="1664044470656" $\left|\begin{array}{c}\begin{array}{ll}6−t=u& e^{−st}dt=dv\\ −dt=du& v=−\frac{e^{−st}}{s}\end{array}\\ \end{array}\right|$in second integral, then we can write as:

$\begin{array}{c}L\left\{f\right(t\left)\right\}=\frac{3}{s}(1−e^{−2s})+\underset{N→\mathrm{∞}}{\lim }({\left[−(6−t)\frac{e^{−st}}{s}\right]}_2^N−{∫}_2^N\frac{e^{−st}}{s}dt)\\ =\frac{3}{s}(1−e^{−2s})+\underset{N→\mathrm{∞}}{\lim }\left(\frac{4e^{−2s}}{s}−\frac{(6−N)e^{−sN}}{s}+{\left[\frac{e^{−st}}{s^2}\right]}_2^N\right)\\ =\frac{3}{s}(1−e^{−2s})+\underset{N→\mathrm{∞}}{\lim }\left(\frac{4e^{−2s}}{s}−\frac{(6−N)e^{−sN}}{s}+\frac{e^{−sN}}{s^2}−\frac{e^{−2s}}{s^2}\right)\\ =\frac{3}{s}(1−e^{−2s})+\underset{N→\mathrm{∞}}{\lim }\frac{4e^{−2s}}{s}−\underset{N→\mathrm{∞}}{\lim }\frac{(6−N)e^{−sN}}{s}+\underset{N→\mathrm{∞}}{\lim }\frac{e^{−sN}}{s^2}−\underset{N→\mathrm{∞}}{\lim }\frac{4e^{−2s}}{s^2}\end{array}$

Simplify further as:

$\begin{array}{c}L\left\{f\right(t\left)\right\}=\frac{3}{s}(1−e^{−2s})+\frac{4e^{−2s}}{s}−0+0−\frac{4e^{−2s}}{s^2}\\ =\frac{3}{s}(1−e^{−2s})+\frac{4e^{−2s}}{s}−\frac{4e^{−2s}}{s^2}\\ =\frac{3}{s}+\frac{e^{−2s}}{s}−\frac{e^{−2s}}{s^2}\end{array}$

Thus, the required Laplace transform is$L\left\{f\right(t\left)\right\}\left(s\right)=\frac{3}{s}+\frac{e^{-2s}}{s}-\frac{e^{-2s}}{s^2}$