# Amplitude and phase spectra of fourier series

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?

• I did table for the amplitude spectra? Isn't that correct? I'm still not sure how you plot the phase spectra thought. Could give me a hint. – ELEC Oct 3 '14 at 21:25
• You have a slight mistake, $a_1 \approx 0.61$. And you can't say that $|a_3|=-0.14$, an absolute value can't be negative. It's simply $a_3$. However, I made a mistake in my previous comment, too. These harmonics have either a phase of 0° or 180° (0 and $\pi$ radians, respectively). If $a_n$ is positive or zero, the phase is 0. If it's negative, the phase is 180°. – hryghr Oct 3 '14 at 21:41
• I upvoted because of your pretty formatting. – HH- Apologize to Carole Baskin Oct 3 '14 at 22:37

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.
• Nice! Just a little extra question. In my case you can't use $\arctan$ to calculate the phase shift thought right? It will be zero on for all $n$ because $b_n$ is $0$. So to know what the phase spectra looks like you have to know that negative amplitude means $-\pi$ phase shift? Did I understand this correctly. – ELEC Oct 4 '14 at 10:21
• Ahhh! I think I got it. Found this from my old textbook. real valued and even signal will have phase shift $\arg(C_n)$ of only $0$ or if negative $\pm 180^o$. Odd signal would have $0$ or $\pm 90^o$. THANKS! – ELEC Oct 4 '14 at 10:29