What is the Fourier transform of this function:
$\frac{1}{t^2} \sin( at)^{2}\cos (at) \sin (2 a t)$
I tried to solve it with trigonometric identities but i got a strage result..
1 Answer
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Up to scale change and multiplication by a constant, the function is
$\frac 1 {x^2} \sin^2 x \cos x \sin 2x = \frac {\sin x} x \frac {\sin x} x \cos x \sin 2x$
We shall use the convolution theorem, i.e. that the Fourier transform of $f(x)g(x)$ is $\frac 1 {2\pi} \hat f(\nu) * \hat g(\nu)$.
The Fourier transform of $\frac {\sin x} x$ is $\pi \cdot \text{rect}(\nu/2)$. The transform of $\frac {\sin^2 x} {x^2}$ is thus $1/2\pi$ times the convolution of two such functions, which I will denote by $T(\nu)$, because it looks like a triangle. ;-)
The Fourier transform of $\cos(x)$ is $\pi \delta(\nu-1)+\pi \delta(\nu+1)$. For $\sin 2x$, it is $-i\pi \delta(\nu-2)+i\pi \delta(\nu+2)$. Their convolution multiplied by $1/2\pi$ is $\frac {i\pi} 2(-\delta(\nu-1)+\delta(\nu+1)-\delta(\nu-3)+\delta(\nu+3))$. The same result could be obtained by expanding into complex exponentials and then Fourier-transforming.
Finally, the transform of the original function is the convolution of these two functions divided by $2\pi$, which is $\frac i 4 (-T(x-3)-T(x-1)+T(x+1)+T(x+3))$. The function is purely imaginary, as it should be for an odd function of $x$.
This is the triangle function, $T(x)$.
And this is the imaginary part of the Fourier transform (the real part is obviously zero).
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$\begingroup$ Might it be easier to rewrite your integrand as $$\frac{\sin(x)\cdot\sin^2(2x)}{2x^2}.$$ Or alternatively maybe $$\frac{2\sin^3(x)}{x^2}-\frac{2\sin^5(x)}{x^2}.$$ $\endgroup$ Commented Dec 5, 2020 at 21:59
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