Laplace Transform: (t sin(at) cos(bt))^2

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Laplace Transform: (t sin(at) cos(bt))^2

How can I laplace transform this (t sin(at) cos(bt))2

fitzmerl duron

To get the Laplace transform of $(t\sin(at)\cos(bt))^2$, start by modifying $(\sin(at)\cos(bt))^2$ to an equation where it is easier to get its Laplace transform.

$$(\sin(at))^2 (\cos(bt))^2 = (\sin(at)\cos(bt))^2$$ $$= \left(\frac{1}{2}\sin((a-b)t) + \frac{1}{2}\sin((a+b)t)\right)^2$$ $$= \left(\frac{1}{2}\sin((a-b)t)\right)^2 + \frac{1}{4}\sin((a+b)t)\sin((a-b)t) + \frac{1}{4}\sin((a+b)t)\sin((a-b)t) + \left(\frac{1}{2}\sin((a+b)t)\right)^2$$ $$= \left(\frac{1}{2}\sin((a-b)t)\right)^2 + \frac{1}{2}\sin((a+b)t)\sin((a-b)t) + \left(\frac{1}{2}\sin((a+b)t)\right)^2$$ $$= \frac{1}{4}\sin^2((a-b)t) + \frac{1}{2}\sin((a+b)t)\sin((a-b)t) + \frac{1}{4}\sin^2((a+b)t)$$ $$= \frac{1}{4}\left(\frac{1-\cos(2(a-b)t)}{2}\right) + \frac{1}{2}\sin((a+b)t)\sin((a-b)t) + \frac{1}{4}\left(\frac{1-\cos(2(a+b)t)}{2}\right)$$ $$= \frac{1}{8}(1-\cos(2(a-b)t))+\frac{1}{2}\left(\frac{1}{2}\cos(((a-b)-(a+b))t) - \frac{1}{2}\cos(((a-b)+(a+b))t) \right)+\frac{1}{8}(1-\cos(2(a+b)t)$$ $$= \frac{1}{8}(1-\cos(2(a-b)t)) + \frac{1}{4}(\cos((-2b)t) -\cos((2a)t) + \frac{1}{8}(1-\cos(2(a+b)t))$$ $$= \frac{1}{4}-\frac{1}{8}\cos(2(a-b)t) + \frac{1}{4}\cos((-2b)t) -\frac{1}{4}\cos((2a)t) - \frac{1}{8}(\cos(2(a+b)t))$$

The Laplace transform of $\cos \omega_ot \space u(t)$ is $\frac{s}{s^2 + \omega_o^2}$. The Laplace transform of $u(t)$ or $1$ is $\frac{1}{s}$. Then using the Linearity principle, stating that the Laplace transform of $af_1(t) \pm bf_2(t)$ is $aF_1(s) \pm bF_2(s)$, which is basically getting the individual Laplace transforms of individual terms on the equation.

Armed with these...the Laplace transform of $\frac{1}{4}-\frac{1}{8}\cos(2(a-b)t) + \frac{1}{4}\cos((-2b)t) -\frac{1}{4}\cos((2a)t) - \frac{1}{8}(\cos(2(a+b)t))$ would be:

$$\frac{1}{4s} - \frac{s}{8(s^2 + (2(a-b))^2)} + \frac{s}{4(s^2 + (2(b))^2)} - \frac{s}{4(s^2 + (2(a))^2)} + \frac{s}{8(s^2 + (2(a+b))^2)}$$

To finally get the Laplace transform of $(t\sin(at)\cos(bt))^2$, notice that the equation $t^2\sin^2(at)\cos^2(bt)$ has a factor $t^2$. Using the property of Laplace transform involving an equation multiplied by $t^n$, where $n = 1,2,3,...$, the Laplace transform of equations $f(t)$ being multiplied by $t^n$ would be $(-1)^n \frac{d^n F(s)}{ds^n}$.

In this case, the Laplace transform of $t^2\sin^2(at)\cos^2(bt)$ would be $(-1)^2 \frac{d^2 F(s)}{ds^2}$. We will get the expression for $(-1)^2 \frac{d^2 F(s)}{ds^2}$. Now getting the second derivatives of $$\frac{1}{4s} - \frac{s}{8(s^2 + (2(a-b))^2)} + \frac{s}{4(s^2 + (2(b))^2)} - \frac{s}{4(s^2 + (2(a))^2)} + \frac{s}{8(s^2 + (2(a+b))^2)}$$,
it was found out that the second derivative of the preceding equation would be:

$$\frac{1}{2s^3} + \frac{1}{8}\left(\frac{8s^3}{(s^2+(2(a-b))^2)^3} + \frac{-6s}{(s^2+(2(a-b))^2)^2}\right) + \frac{1}{4}\left(\frac{8s^3}{(s^2+(2(b))^2)^3} + \frac{-6s}{(s^2+(2(b))^2)^2}\right) - \frac{1}{4}\left(\frac{8s^3}{(s^2+(2(a))^2)^3} + \frac{-6s}{(s^2+(2(a))^2)^2}\right) + \left(\frac{8s^3}{(s^2+(2(a+b))^2)^3} + \frac{-6s}{(s^2+(2(a+b))^2)^2}\right)$$

I've skipped the step where I get the second derivative. It's too long.

Therefore, the Laplace transform of $t^2\sin^2(at)\cos^2(bt)$ is $(-1)^2 \frac{d^2 F(s)}{ds^2}$:

$$\displaystyle L\{t^2\sin^2(at)\cos^2(bt)\}=(-1)^2 \left(\frac{1}{2s^3} + \frac{1}{8}\left(\frac{8s^3}{(s^2+(2(a-b))^2)^3} + \frac{-6s}{(s^2+(2(a-b))^2)^2}\right) + \frac{1}{4}\left(\frac{8s^3}{(s^2+(2(b))^2)^3} + \frac{-6s}{(s^2+(2(b))^2)^2}\right) - \frac{1}{4}\left(\frac{8s^3}{(s^2+(2(a))^2)^3} + \frac{-6s}{(s^2+(2(a))^2)^2}\right) - \frac{1}{8}\left(\frac{8s^3}{(s^2+(2(a+b))^2)^3} + \frac{-6s}{(s^2+(2(a+b))^2)^2}\right)\right)$$ $$=\frac{1}{2s^3} + \frac{1}{8}\left(\frac{8s^3}{(s^2+(2(a-b))^2)^3} + \frac{-6s}{(s^2+(2(a-b))^2)^2}\right) + \frac{1}{4}\left(\frac{8s^3}{(s^2+(2(b))^2)^3} + \frac{-6s}{(s^2+(2(b))^2)^2}\right) - \frac{1}{4}\left(\frac{8s^3}{(s^2+(2(a))^2)^3} + \frac{-6s}{(s^2+(2(a))^2)^2}\right) -\frac{1}{8} \left(\frac{8s^3}{(s^2+(2(a+b))^2)^3} + \frac{-6s}{(s^2+(2(a+b))^2)^2}\right)$$.

Alternate solutions are encouraged......

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