Question

Find the Laplace transform of the function.

β(t)={0,0t<2π.3sin3t,t2π\beta(t)= \begin{cases}0, & 0 \leq t<2 \pi . \\ 3 \sin 3 t, & t \geq 2 \pi\end{cases}

Solution

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Answered 1 year ago
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We want to find the Laplace transform of the function

β(t)=3u2π(t)sin3t.\beta(t) = 3u_{2\pi}(t)\sin{3t}.

To determine the Laplace transform we use the following results

(1)L(sinat)=as2+a2(2)L(ua(t)f(ta))=easL(f(t)).\begin{align*} & (1) \: \: \mathcal{L}(\sin {at}) = \frac{a}{s^2+a^2} \\ & (2) \: \: \mathcal{L}\left(u_a(t)f(t-a)\right) = e^{-as}\mathcal{L}\left(f(t)\right). \end{align*}

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