# Fourier Series Coefficients of a Full-wave Rectified Cosine

I need to calculate the Fourier Coefficients $a_k$ of $\left | \cos(2 \pi f_c) \right |$ (a full-wave rectified cosine) so that $\left | \cos(2 \pi f_c) \right | = a_0 + \displaystyle\sum\limits_{k=1}^{\inf} a_k \cos(2 \pi k f_c t + \phi_k)$ and prove that for any odd value of $k$, $a_k=0$.

Applying (for $k\not=0$) the Fourier Cosine Series Coefficients Formula $a_k = \cfrac{1}{T} \displaystyle\int\limits_{T} f(x) \cos(k\,\pi\,x) \,\mathrm{d}x$, I get $a_k = 8\,f_c\displaystyle\int\limits_{0}^{\frac{1}{4f_c}} \cos(2 \pi f_c t)\,\cos(2\pi\,k\,f_c\,t) \, \mathrm{d}{x} = \frac{8\,f_c\,\cos(\frac{k\,\pi}{2})}{2\,f_c\,\pi-2\,f_c\,k^2\,\pi}$ (result of the integral confirmed with Mathematica).

Substituting $k=1$, we get an indeterminate form. If we calculate the limit at $k=1$, we get $1$. For any other odd $k$, $a_k=0$ as expected, with no indeterminate form.

When we use the Exponential Fourier Series Coefficients Formula, I end up with $m_k = \cfrac{2\,\sin(\frac{\pi}{2}\,(1-k))}{\pi\,(1-k)}$ and we also get an indeterminate form at $k=1$ with a limit of $1$.

I've also tried to convolve a cosine with a square wave (so that their multiplication would be the same as a full-wave rectified cosine) and I get the same formulas as above, with the same indeterminate form.

After a couple hours trying to solve this, I'm completely stumped. I have no idea how I can prove that $a_1=0$ when all attempts to calculate it result in an indeterminate form. What am I doing wrong? Thanks in advance for any help you can provide!

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Have you tried treating the case $k = 1$ seperately? By setting $k = 1$ in your first formula, I find that I need to calculate $\int_0^{\frac{1}{4 f_c}} ]cos^2(2\pi f_c t) \operatorname{d}t$. According to Mathematica, this is $\frac{1}{8 f_c} \cdot \left(1 - \frac{1}{\pi} \right)$. – Johannes Kloos Oct 20 '12 at 19:39
Yeah, I tried to do so, the result always equals the limit for $k=1$ (in that case, $1$). The integral does not seem to be wrong tough, but we tried with different integration limits (including just telling Mathematica to integrate the whole thing directly with the absolute cosine) and some have a limit of $1$, and some a limit of $0$, but they all match for all integer values of $k$ when $k\not=1$! – rpinheiro Oct 20 '12 at 19:41
I just noticed however you seem get a different result than me (altough it's not 0 either). What did you input in Mathematica? 8*fcIntegrate[Cos[2*Pifc*t]^2, {t, 0, 1/(4*fc)}] returns "1" – rpinheiro Oct 20 '12 at 19:47
I just calculated an anti-derivative of $\cos^2 (2\pi f_c t)$, but I propably made a typo somewhere. Anyway, if this integral is 1, it can't be 0 at the same time. – Johannes Kloos Oct 20 '12 at 20:01
I've figured it out, it was a mistake when simplifying the integral. Thanks for your help! – rpinheiro Oct 20 '12 at 21:17