F(t)=0 . =sin (t). 0<= t< pi . pi<t<2pi. Find L(f(t))=?

Rivka Thorpe 2021-09-26 Answered

\(F(t)=0\)
\(\displaystyle={\sin{{\left({t}\right)}}}\)
\(\displaystyle{0}\leq{t}{<}\pi\)
\(\displaystyle\pi{<}{t}{<}{2}\pi\)
Find \(L(f(t))=\)?

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Expert Answer

mhalmantus
Answered 2021-09-27 Author has 8658 answers

Step 1
We have to find the Laplace transform given step function.
Step 2
\(f(x)=\begin{cases}0& \text{ it }0 \leq t \leq \pi\\ \sin t & \text{ it } x\leq t \leq 2\pi\end{cases}\)
Now,
\(f(t)=0\left[u(t)-u(t-\pi)\right]+\sin (t) \left[u(t-\pi)-u(t-2\pi)\right]\)
\(\displaystyle{f{{\left({t}\right)}}}={\sin{{\left({t}\right)}}}{u}{\left({t}-\pi\right)}-{\sin{{\left({t}\right)}}}{u}{\left({t}-\pi\right)}\)
\(\therefore L\left\{f(t)\right\}=L\left\{\sin t u(t-\pi)\right\}-L\left\{\sin (t) u(t-2\pi)\right\}\)
Using formula: \(L\left\{f(t) \cdot u(t-a)\right\}=e^{-as}L\left\{f(t+a)\right\}\)
\(L\left\{f(t)\right\}=e^{-\pi s}L\left\{\sin (t+\pi)\right\}-e^{-2\pi s} L\left\{\sin (t+2\pi)\right\}\)
\(=-e^{-\pi s} L\left\{\sin (t)\right\}-e^{-2 \pi s} L\left\{\sin (t)\right\}\)
\(\displaystyle=-{e}^{{-\pi{s}}}{\left({\frac{{{1}}}{{{s}^{{2}}+{1}}}}\right)}-{e}^{{-{2}\pi{s}}}{\left({\frac{{{1}}}{{{s}^{{2}}+{1}}}}\right)}\)
\(\displaystyle=-{\frac{{{1}}}{{{\left({s}^{{2}}+{1}\right)}}}}{\left({e}^{{-\pi{s}}}+{e}^{{-{2}\pi{s}}}\right)}\)

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