Find the Laplace transform for the functions: f(t)=tu_2(t) , f(t)=u_(pi)(t)sin(t)

Aneeka Hunt 2021-09-26 Answered
Find the Laplace transform for the functions:
\(\displaystyle{f{{\left({t}\right)}}}={t}{u}_{{2}}{\left({t}\right)}\)
\(\displaystyle{f{{\left({t}\right)}}}={u}_{{\pi}}{\left({t}\right)}{\sin{{\left({t}\right)}}}\)

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Expert Answer

hesgidiauE
Answered 2021-09-27 Author has 16356 answers

Step 1
Use the laplace transformation table to compute
Step 2
We know that
\(L\left\{u_c(t)f(t)\right\}=e^{\frac{-c}{s}}L\left\{f(t+c)\right\}\)
\(\displaystyle\therefore{f{{\left({t}\right)}}}={t}{u}_{{2}}{\left({t}\right)}\)
\(L\left\{f(t)\right\}=L\left\{u_2(t )\cdot t\right\}\)
\(=e^{-\frac{2}{s}}L\left\{(t+2)\right\}\)
\(=e^{-\frac{2}{s}}(L\left\{t\right\}+2L\left\{1\right\})\)
\(=e^{-\frac{2}{s}}\left\{\frac{11}{s^2}+\frac{2}{s}\right\}\)
\(\displaystyle={e}^{{-{\frac{{{2}}}{{{s}}}}}}{\left({\frac{{{2}{s}+{1}}}{{{s}^{{2}}}}}\right)}\)
\(\displaystyle\therefore{f{{\left({t}\right)}}}={u}_{{\pi}}{\left({t}\right)}{\sin{{\left({t}\right)}}}\)
\(L\left\{f(t)\right\}=L\left\{u_{\pi}(t)\sin(t)\right\}\)
\(=e^{\frac{-\pi}{s}}L\left\{\sin(t+\pi)\right\}\)
\(=e^{\frac{-\pi}{s}}L\left\{-\sin t\right\}\)
\(\displaystyle=-{\frac{{{e}^{{{\frac{{-\pi}}{{{s}}}}}}}}{{{s}^{{2}}+{1}}}}\)

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