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?'s question at Yahoo! Answers regarding Laplace Transforms

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Chris L T521

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Jan 26, 2012
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Chris L T521

Well-known member
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Jan 26, 2012
995
Hello ?,

We first use the trig identity $\sin(2\theta)=2\sin\theta\cos\theta$ to rewrite $\sin(3t)\cos(3t)$; in particular, we have that $\sin(3t)\cos(3t)=\frac{1}{2}\sin(2\cdot(3t)) =\frac{1}{2}\sin(6t)$.

Now, I'm not sure exactly what formulas you can use, but if you can use certain formulas for Laplace Transforms, then you're done since we know $\mathcal{L}\left\{\frac{1}{2}\sin(6t)\right\} =\frac{1}{2}\mathcal{L}\{\sin(6t)\} =\dfrac{1}{2}\cdot\dfrac{6}{s^2+6^2} =\dfrac{3}{s^2+36}$.

Otherwise, we need to compute the Laplace Transform by first principles; to make this easier for us, let us first compute $\mathcal{L}\{e^{at}\}$ where $a\in\mathbb{R}$.

\[\begin{aligned}\mathcal{L}\{e^{at}\} &= \int_0^{\infty}e^{-st}e^{at}\,dt\\ &= \int_0^{\infty}e^{-(s-a)t}\,dt\\ &= \lim_{b\to\infty} \left.\left[-\frac{e^{-(s-a)t}}{s-a}\right]\right|_{0}^{b}\\ &= \lim_{b\to\infty} -\frac{e^{-(s-a)b}}{s-a} + \frac{1}{s-a}\end{aligned}\]
Note that $\displaystyle\lim_{b\to\infty}-\frac{e^{-(s-a)b}}{s-a}=0$ if $s>a$. Thus, we have that $\mathcal{L}\{e^{at}\}=\dfrac{1}{s-a}$ for $s>a$.

Now, how does this help us with computing $\mathcal{L}\{\sin(6t)\}$? Well, we know from complex variables that $\sin(kx)=\dfrac{e^{ikx}-e^{-ikx}}{2i}$, so it follows that
\[\begin{aligned}\mathcal{L}\{\sin(6t)\} &= \frac{1}{2i}\left[\mathcal{L}\left\{e^{6it}\right\} - \mathcal{L}\left\{e^{-6it}\right\}\right] \\ &= \frac{1}{2i}\left[\frac{1}{s-6i} - \frac{1}{s+6i}\right]\\ &= \frac{1}{2i}\left[\frac{12i}{s^2-(6i)^2}\right] \\ &= \frac{6}{s^2+36}\end{aligned}\]
And thus $\mathcal{L}\{\sin(3t)\cos(3t)\} =\frac{1}{2}\mathcal{L}\{\sin(6t)\}=\dfrac{3}{s^2+36}$.

I hope this makes sense!