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?

  • \$\begingroup\$ 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. \$\endgroup\$
    – ELEC
    Commented Oct 3, 2014 at 21:25
  • \$\begingroup\$ 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°. \$\endgroup\$
    – hryghr
    Commented Oct 3, 2014 at 21:41
  • 1
    \$\begingroup\$ I upvoted because of your pretty formatting. \$\endgroup\$ Commented Oct 3, 2014 at 22:37

1 Answer 1


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\$.

enter image description here

enter image description here

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.

  • \$\begingroup\$ 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. \$\endgroup\$
    – ELEC
    Commented Oct 4, 2014 at 10:21
  • \$\begingroup\$ 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! \$\endgroup\$
    – ELEC
    Commented Oct 4, 2014 at 10:29
  • \$\begingroup\$ Exactly. Glad I could help you. \$\endgroup\$
    – hryghr
    Commented Oct 4, 2014 at 12:09
  • \$\begingroup\$ @ELEC: Look for my recent edit, I made a mistake in the first graph. \$\endgroup\$
    – hryghr
    Commented Oct 4, 2014 at 16:03

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