Electronic – Fourier Series Expansion of Periodic Signal

If you calculate some more of them, you will come to the conclusion that almost every odd coefficient \$a_n\$ is a real positive number, except for \$a_3, a_9, a_{15}...\$, \$a_{3+6k}\$, where \$k\$ is a non-negative integer. These are all negative. (And every even coefficient is zero.) For those, which are positive numbers, the phase is 0 degrees/radians; for those, which are negative, the phase is -180°/\$-\pi\$ radians, because negating a periodic signal is equal to shifting its phase by -180°. Be careful using arctan, as it has a value range of \$\pm 90°\$ but both 0° and -180° has the same tangent value, zero.

Here I plotted the values and the phase spectrum from \$n=0\$ to \$n=21\$.

Edit: The first image is obviously not the amplitude (as it can't be negative), just the values of \$a_n\$ from 0 to 21. My bad.

Let’s say you have an interharmonic of 1.5. If you scale your assumed fundamental frequency by 0.5 then you will have two integer harmonic sinusoids contributing to the signal - a 2nd harmonic and a 3rd harmonic. Equivalent, of course, to the original signal - just an integer way to look at it.

First case: Fundamental = f1

$$ x(t) = \cos{(1*2 \pi f_1 t + \theta_n)} + \cos{(1.5*2 \pi f_1t + \theta_m )} $$

Second case: Fundamental = f2 = 0.5f1

$$ x(t) = \cos{(2*2 \pi f_2 t + \theta_n)} + \cos{(3*2 \pi f_2t + \theta_m)} $$

These are equivalent signals (and periodic), I just chose to assume a different fundamental frequency for my analysis so i would have integer harmonics.