Find the Laplace transform for displaystyle f{{left({t}right)}}={2}{u}{left({t}-{3}right)}{{cos}^{2}{2}}{t}-{6}{e}^{{{2}{t}+{7}}}delta{left({t}+{3}right)}

Question
Laplace transform
asked 2021-01-05
Find the Laplace transform for
\(\displaystyle f{{\left({t}\right)}}={2}{u}{\left({t}-{3}\right)}{{\cos}^{2}{2}}{t}-{6}{e}^{{{2}{t}+{7}}}\delta{\left({t}+{3}\right)}\)

Answers (1)

2021-01-06
Step 1
Calculate the Laplace transform:
\(\displaystyle{\int_{ -{\infty}}^{\infty}} f{{\left({t}\right)}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}={\int_{ -{\infty}}^{\infty}}{\left({2}{u}{\left({t}-{3}\right)}{{\cos}^{2}{2}}{t}-{6}{e}^{{{2}{t}+{7}}}\delta{\left({t}+{3}\right)}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}\)
\(\displaystyle={2}{\int_{ -{\infty}}^{\infty}}{u}{\left({t}-{3}\right)}{{\cos}^{2}{2}}{t}\cdot{e}^{{-{s}{t}}}{\left.{d}{t}\right.}-{6}{\int_{ -{\infty}}^{\infty}}{e}^{{{2}{t}+{7}}}\delta{\left({t}+{3}\right)}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}\)
\(\displaystyle={2}{\int_{{3}}^{\infty}}{e}^{{-{s}{t}}}{{\cos}^{2}{2}}{t}{\left.{d}{t}\right.}-{6}{e}^{{{3}{s}+{1}}}\)
\(\displaystyle={\int_{{3}}^{\infty}}{e}^{{-{s}{t}}}{\left( \cos{{4}}{t}+{1}\right)}{\left.{d}{t}\right.}-{6}{e}^{{{3}{s}+{1}}}\)
\(\displaystyle={\int_{{3}}^{\infty}}{e}^{{-{s}{t}}} \cos{{4}}{t}{\left.{d}{t}\right.}+{\int_{{3}}^{\infty}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}-{6}{e}^{{{3}{s}+{1}}}\)
\(\displaystyle={\int_{{3}}^{\infty}}{e}^{{-{s}{t}}} \cos{{4}}{t}{\left.{d}{t}\right.}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}\)
\(\displaystyle={\int_{{3}}^{\infty}}{e}^{{-{s}{t}}}\frac{{{e}^{{{4}{j}{t}}}+{e}^{{-{4}{j}{t}}}}}{{2}}{\left.{d}{t}\right.}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}\)
\(\displaystyle={\int_{{3}}^{\infty}}\frac{{{e}^{{-{s}{t}+{4}{j}{t}}}+{e}^{{-{s}{t}-{4}{j}{t}}}}}{{2}}{\left.{d}{t}\right.}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}\)
\(\displaystyle=\frac{{{1}-{e}^{{-{3}{s}+{12}{j}}}}}{{{2}{s}-{8}{j}}}+\frac{{{1}-{e}^{{-{3}{s}-{12}{j}}}}}{{{2}{s}+{8}{j}}}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}\)
\(\displaystyle=\frac{{{2}{s}+{8}{j}-{\left({2}{s}+{8}{j}\right)}{e}^{{-{3}{s}+{12}{j}}}+{\left({2}{s}-{8}{j}\right)}{e}^{{-{3}{s}-{12}{j}}}}}{{{4}{s}^{2}+{64}}}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}\)
\(\displaystyle=\frac{{{4}{s}-{2}{e}^{{-{3}{s}}}{\left({2}{s} \cos{{12}}-{8} \sin{{12}}\right)}}}{{{4}{s}^{2}+{64}}}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}\)
Step 2
Thus, Laplace transform is \(\displaystyle\frac{{{4}{s}-{2}{e}^{{-{3}{s}}}{\left({2}{s} \cos{{12}}-{8} \sin{{12}}\right)}}}{{{4}{s}^{2}+{64}}}+\frac{{{1}-{e}^{{-{3}{s}}}}}{{s}}-{6}{e}^{{{3}{s}+{1}}}\)
0

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