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How can I find the Fourier transform of

$$ f(x) = ( \cos(x) )^3$$

I know that for $ g(x) = \cos(x) $

$$ F \Big\{ g(x) \Big\} = F \Big\{ \cos(x) \Big\} = \pi \Big [ \delta(w-\pi / 2) + \delta(w+\pi / 2) \Big ]$$

But using this pair of Fourier transform how to obtain the $ F \Big\{ f(x) \Big\} $ ?? Is there a direct/simple way to do that?

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2 Answers 2

up vote 10 down vote accepted

One way would be to use the power-reduction trigonometric identity:

$$ \cos^3(x) = \frac{3 \cos(x) + \cos(3x)}{4} $$

Due to the linearity property of the Fourier transform, you can transform each term separately and take their weighted sum to get the transform of the entire expression. The relationship we will use (from line 304 here) is:

$$ \mathcal{F}\{\cos(ax)\} = \pi\left(\delta(\omega - a) + \delta(\omega + a)\right) $$

Which assumes that you're using the non-unitary, angular frequency definition of the Fourier transform:

$$ \mathcal{F}\{x(t)\} = X(\omega) = \int_{-\infty}^{\infty}x(t) e^{-j\omega t}dt $$

This would yield:

$$ \begin{align} \mathcal{F}\{\cos^3(x)\} &= 3 \mathcal{F}\{\cos(x)\} + \mathcal{F}\{\cos(3x)\} \\ &= \frac{\pi}{4}\left(3 \delta(\omega - 1) + 3\delta(\omega + 1) + \delta(\omega - 3) + \delta(\omega + 3) \right) \end{align} $$

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Shouldn't it be $\frac{1}{8}$ instead of $\frac{\pi}{4}$? –  Jim Clay Nov 19 '12 at 21:49
2  
It depends upon the version of the transform that you're using. I referenced the definition I used above, which uses angular frequency $\omega$. If you used ordinary frequency $f$, then you would get a factor of $\frac{1}{8}$, differing by a factor of $2\pi$. That transform definition is $X(f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt$. –  Jason R Nov 20 '12 at 2:58
    
+1. You cant say "Fourier" without @JasonR pouncing on it. :-) –  Mohammad Nov 20 '12 at 15:53

I prefer Jason's answer, but thought that I would present an alternate way of doing it anyway.

Multiplication in the time domain is equivalent to convolution in the frequency domain. You thus have the following-

$$ F\{cos^3(x)\} = F\{cos(x)cos(x)cos(x)\} = F\{cos(x)\} * F\{cos(x)\} * F\{cos(x)\} $$ If you use the fact that $F\{cos(x)\} = \frac{\delta(w - 1) + \delta(w + 1)}{2}$ you can get the same result as Jason's answer.

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Very nice, showing equivalence in convolution/multiplication between time&freq domains. –  Mohammad Nov 20 '12 at 15:54
    
+1 This way is also very nice because I cannot only compute for $ cos(x)^3 $ but for any power of it. Thanks. –  BRabbit27 Nov 20 '12 at 15:58

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