Fourier coefficients double half-wave rectifier

Question

I’d like to find the Fourier coefficients of

$$ x(t)=|A \cos(2 \pi f_0 t )| $$ The period is $ T_0 / 2 $ so i applied the definition $$\frac{1}{T}\int_{0}^{L} |A \cos(2\pi f_0 t )| \cos \left(\frac{k \pi t }{T}\right)dt $$

Now i applied the Euler formulas and I transformed cos into exp

$$ \frac{|A|}{4T}\int_{-T/4}^{T/4} |(e^{i2\pi f_0 t} + e^{-2i\pi f_0 t})|(e^{\frac{ik\pi t}{T}}+e^{-\frac{ik\pi t}{T}}) $$

Can you now please help me with the calculation. The result should be $ \frac{-2A}{\pi}\frac{(-1)^k}{4k^2 -1 } $ But the solution I obtained is always wrong

Thank you !

Answer 1

The solution can be obtained directly from the definition, but I'd suggest revising the one you provided to be a bit more precise: $$a_k=\frac{1}{T_0/2}\int_{-T_0/4}^{T_0/4}A\cos(2\pi f_0t)\cos(2\pi k2f_0t)dt.$$

Note that the fundamental ($k=1$) basis function here is twice the frequency of $x(t)$. From here, there are a couple ideas than can be tricky to spot. In the order you need them, they are:

1. $\cos(\alpha)\times\cos(\beta) = 1/2\cos(\alpha-\beta) + 1/2\cos(\alpha+\beta)$
2. $\sin((2k-1)\pi/2) = (-1)^{k+1}$
3. $\sin((2k+1)\pi/2) = (-1)^{k}$
4. $(-1)^{k+1}=(-1)^{k-1}$
5. $\frac{1}{2k-1}+\frac{1}{2k+1}=\frac{2}{4k^2-1}$

Give it another go, and if you're still having trouble provide your MWE so we can do a better job of helping.

Also, see this excellent YouTube video from Darryl Morrell showing a much more elegant way to get to your required answer.