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Given the function: $$x(t)=A_1\cos(\omega_1 t+\phi)$$ its Fourier transform is: $$X(\omega)=\sqrt{\frac{\pi}{2}}A_1 e^{-i\phi}\delta(\omega_1-\omega)+\sqrt{\frac{\pi}{2}}A_1 e^{i\phi}\delta(\omega_1+\omega)$$ More in general, given the function: $$x(t)=\prod_{k=1}^NA_k \cos(\omega_kt+\phi_k)$$ what is the formula of Fourier transform $X(\omega)$ of $x(t)$?

Thank you in advance.

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Any product of cosines can be written as a sum of cosines using $2 \cos a \cos b = \cos(a+b) + \cos(a-b)$. – Willie Wong Jul 13 '12 at 11:04
Maybe it helps to write it as $\prod A_k (e^{-i\omega_k t+\varphi_k}+e^{i\omega_k t+\varphi_k})/2$, then factor out the product and sum in the exponent. FT is then done on a sum of $\exp$'s, which should be easy for you. – draks ... Jul 13 '12 at 11:06
up vote 2 down vote accepted

If you extend Willie Wong's comment on the Product-to-sum identity (works better than mine) $$ \cos \theta_1 \cos \theta_2= {\cos(\theta_1 + \theta_2) + \cos(\theta_1 - \theta_2) \over 2} $$ (with $\omega_kt+\varphi_k\color{black}{=\theta_k)}$ to $3$ cosines
$$ \begin{eqnarray} \cos \theta_1 \cos \theta_2 \cos \theta_3&=& {\cos(\theta_1 + \theta_2) + \cos(\theta_1 - \theta_2) \over 2}\cos \theta_3\\ &=&\scriptstyle{\cos(\theta_1 + \theta_2+\theta_3) +\cos(\theta_1 + \theta_2-\theta_3) + \cos(\theta_1 - \theta_2+\theta_3)+\cos(\theta_1 - \theta_2-\theta_3) \over 4}, \end{eqnarray} $$ you recognize that all $2^{3-1}$ $"\pm"$ combinations of $\theta_k$ appear. So we get $$ x(t)=\prod_{k=1}^NA_k \cos(\omega_kt+\phi_k)= \frac A{2^{N-1}}\sum_{b=0}^{2^{N-1}-1}\cos\left(\sum_{k=1}^N (-1)^{b_k}\theta_k\right), $$ with $\left(\prod_{k=1}^NA_k\right)$ and $b=\sum_{k=1}^{N} 2^{k-1}b_k$. Then use what you already know about the Fourier Transform and you're done.

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