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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) $

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

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

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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:

$$ \begin{align} \mathcal{F}\{\cos^3(x)\} &= \frac 34 \mathcal{F}\{\cos(x)\} + \frac 14 \mathcal{F}\{\cos(3x)\} \\ &= \frac{\pi}{4}\left(3 \delta(\omega - 1) + 3\delta(\omega + 1) + \delta(\omega - 3) + \delta(\omega + 3) \right) \end{align} $$

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  • $\begingroup$ Shouldn't it be $\frac{1}{8}$ instead of $\frac{\pi}{4}$? $\endgroup$
    – Jim Clay
    Commented Nov 19, 2012 at 21:49
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    $\begingroup$ 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 $X(f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt$. $\endgroup$
    – Jason R
    Commented Nov 20, 2012 at 2:58
  • $\begingroup$ +1. You cant say "Fourier" without @JasonR pouncing on it. :-) $\endgroup$
    – Spacey
    Commented Nov 20, 2012 at 15:53
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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-

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

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  • $\begingroup$ Very nice, showing equivalence in convolution/multiplication between time&freq domains. $\endgroup$
    – Spacey
    Commented Nov 20, 2012 at 15:54
  • $\begingroup$ +1 This way is also very nice because I cannot only compute for $ cos(x)^3 $ but for any power of it. Thanks. $\endgroup$
    – BRabbit27
    Commented Nov 20, 2012 at 15:58

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