If I have even and periodic signal \$x(t)\$ that has cosine fourier series
$$ x(t)\sim\underbrace{\frac 12}_{a_0}+\sum_{n=1}^{\infty} \underbrace{\left(\frac{6\cos \left(\frac{n\pi }{3}\right)-6\cos \left(\frac{2n\pi }{3}\right)}{n^2 \pi^2} \right)}_{a_n} \cos \left(\frac{n\pi t}{3}\right)$$
because for even function my \$ b_n=0 \$ coefficient vanishes so \$ C_n = a_n \$
If I want to construct amplitude spectra I plot \$a_n\$ with \$n=0, \pm 1\, \pm 2\,…,\$ right? Like this?
$$\begin{array}{c|c}
n & a_n \\ \hline
0 & 0.5 \\ \hline
\pm1 & 0.61 \\ \hline
\pm2& 0 \\ \hline
\pm3& 0.14
\end{array}$$
But how about the phase spectra. Phase or \$ \theta=\arg(C_n)=\frac{\operatorname{Im(C_n)}}{\operatorname{Re}(C_n)} =\arctan \frac {-b_n}{a_n} \$. But in my case there is no \$b_n\$. Doesn't my fourier serie have phase spectra?
Best Answer
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.