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I have trouble understanding how to find the Fourier transform of $|\cos(2\pi f_0 t)|$. If I have $\cos(2\pi f_0 t)$, I can see it as $\frac{1}{2}\cdot[e^{i2\pi f_0 t}+e^{-i2\pi f_0 t}] $ and its transform is the sum of two Dirac deltas, but what if I have the function written above?

$|\cos(2\pi f_0 t)|$ should be periodic in $ t $ with period $ \frac{1}{2f_0}=\frac{T_0}{2} $ . From the definition, $ x(t)= \sum_{k=-\infty}^{+\infty} X_k \cdot e^{-i2\pi f_0 kt} $ and $ X_k= \frac{1}{T_0} \cdot \int_{-\frac{T_0}{2}}^{+\frac{T_0}{2}}x(t) \cdot e^{-i2\pi f_0 kt} dt $ where $ T_0 $ is the generic period. So $X_k= \frac{2}{T_0} \cdot \int_{-\frac{T_0}{4}}^{+\frac{T_0}{4}} cos(2\pi f_0 t) \cdot e^{-i2\pi f_0 kt} dt $ . If I try to solve this integral, the result I get is $ 0 $ . What am I doing wrong?

Maghreb_1911
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    Up to a rescaling in time, you are looking for the Fourier transform of the absolute value of a cosine a question which is already answered in the given link. There is also this answer which might be helpful. – cs89 Apr 06 '23 at 10:02
  • @cs89 Hello and thank you for your suggestions, but I still don't understand how to proceed. Regarding the answers linked in your previous comment, why do we have to consider the whole $ 2\pi $ interval when the absolute value is periodic with period $ \pi $ in $ x $ ? – Maghreb_1911 Apr 07 '23 at 12:05
  • You don't "have to", you "can". If a function is periodic of period $P$, then you can represent it as the Fourier series $f(x) = \sum c_n e^{\frac{2i\pi n x}{P}}$. It can happen that your function is also, at the same time, by chance, periodic of period $P/2$, or $P/7$, or whatever and thus you can also represent it as another Fourier series $f(x) = \sum c_n' e^{\frac{2i\pi n x}{P/2}}$. – cs89 Apr 07 '23 at 12:36
  • I see, thank you. Going back to my initial question, we have a period of $ \frac{T_0}{2} $ and I am stuck to that point. – Maghreb_1911 Apr 07 '23 at 12:59
  • Don't you get $0$ because you forgot the absolute value around the cosine inside the integral? – cs89 Apr 07 '23 at 15:42
  • No, I don't think so. Maybe I finally found my mistake: I thought that since the period of $ |\cos(2\pi f_0 t)| $ was $ \frac{1}{2f_0}= \frac{T_0}{2} $ , then the $ f_0 $ inside the cosine was equal to $ \frac{2}{T_0} $. If the value of $ f_0 $ remains $ \frac{1}{T_0} $ then the integral is not 0 anymore and $ X_{2p} = \frac{(-1)^{p+1}}{\pi} \cdot \frac{2}{4p^2 -1 } $, because the odd terms are 0. What do you think? – Maghreb_1911 Apr 08 '23 at 07:56
  • It looks consistent with the other links indeed. – cs89 Apr 08 '23 at 11:38
  • That's good then. Thank you for your help. – Maghreb_1911 Apr 08 '23 at 13:20

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