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

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

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mhalmantus

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: $$L\left\{f(t) \cdot u(t-a)\right\}=e^{-as}L\left\{f(t+a)\right\}$$
$$L\left\{f(t)\right\}=e^{-\pi s}L\left\{\sin (t+\pi)\right\}-e^{-2\pi s} L\left\{\sin (t+2\pi)\right\}$$
$$=-e^{-\pi s} L\left\{\sin (t)\right\}-e^{-2 \pi s} L\left\{\sin (t)\right\}$$
$$\displaystyle=-{e}^{{-\pi{s}}}{\left({\frac{{{1}}}{{{s}^{{2}}+{1}}}}\right)}-{e}^{{-{2}\pi{s}}}{\left({\frac{{{1}}}{{{s}^{{2}}+{1}}}}\right)}$$
$$\displaystyle=-{\frac{{{1}}}{{{\left({s}^{{2}}+{1}\right)}}}}{\left({e}^{{-\pi{s}}}+{e}^{{-{2}\pi{s}}}\right)}$$