# Why does the Fourier series not include phase information?

From my understanding, the Fourier series is a way to describe an arbitrary continuous signal in terms of sinusoids of varying frequencies and amplitudes, as shown here in Equation 1.

Why are the phases of the sinusoids not considered?

• The phases are considered. – pichenettes Jun 28 '14 at 7:49
• Where did you get the (incorrect) notion that a Fourier transform does not consider and include phase information? – hotpaw2 Jun 29 '14 at 23:25
• @hotpaw2: thefouriertransform.com/series/coefficients.php (Although I think I confused the transform with the series... the real Fourier Series looks like it has phase information discarded..at least according to the link above..see Equation 1). – Joebevo Jun 30 '14 at 3:55
• That is not a Fourier Transform but Fourier Series question then! FS include phase information in coefficients as well as magnitude info. amplitude is given as $c_n=\sqrt{a_n^2+b_n^2}$ and phase: $\theta_n= \mathtt{atan} (-b_n / a_n)$. So your question is basically wrong right from a start. – jojek Jun 30 '14 at 8:35
• @Joebevo: Please edit your question accordingly, placing link you posted and changing the topic. – jojek Jun 30 '14 at 10:33

(The following is the answer to the original version of the question asking about the Fourier Transform. But exactly the same argument also applies to Fourier series).

They are. Assume that $F(\omega)$ is the Fourier transform of the signal $f(t)$:

$$F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt$$

In general, $F(\omega)$ is a complex-valued function and can be written in polar form

$$F(\omega)=|F(\omega)|e^{j\phi(\omega)}$$

where $\phi(\omega)$ is the phase of $F(\omega)$. The signal $f(t)$ can be written in terms of $F(\omega)$ by virtue of the inverse Fourier transform:

$$f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{j\omega t}d\omega= \frac{1}{2\pi}\int_{-\infty}^{\infty}|F(\omega)|e^{j(\omega t+\phi(\omega))}d\omega\tag{1}$$

From (1) it can be seen that $f(t)$ is represented by complex sinusoids with varying amplitudes $|F(\omega)|$, phases $\phi(\omega)$, and frequencies $\omega$.

• @Downvoter: This answer was written when the question was still "Why does the Fourier Transformation ...". Anyway, the same applies to Fourier series. – Matt L. Jun 30 '14 at 15:05

The Fourier transform does include phase information. The result of the transform is complex data, a real and imaginary component for each bin. The real and imaginary components contain both magnitude and phase information.

However, in some applications the phase information may get discarded. If the end user is primarily interested in the magnitude information then the system may automatically convert the real and imaginary components into magnitude data and discard the phase information. This way the system produces an array of magnitude data that can be more easily plotted as a spectrum (i.e., less computations required for the user). And the magnitude-only data can be represented with half the data of the real and imaginary components so if the data is going to be transferred or stored then the magnitude-only data requires half the bandwidth and storage space. These are a couple reasons why the phase information may get discarded in some applications.

Note that after you discard the phase information, you can not reproduce the original waveform from the magnitude-only data. You need the real and imaginary components in order to reproduce the waveform.

Now when your question is updated and corrected, here is the explanation. The Fourier Series do contain phase information. As you are getting two types of coefficients $a_n$ and $b_n$ corresponding to correlation with sinusoids and cosines, then you can very easily extract phase information. Both magnitude $c_n$ and phase $\theta_n$ are given by:

$$c_n = \sqrt{a_n^2+b_n^2}$$

$$\theta_n=\tan^{-1}\left(-\frac{b_n}{a_n} \right)$$

And apart from standard representation of Fourier Series which you probably know:

$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n \cos( n\omega x)+ b_n \sin(n\omega x)\right)$$

You can also express it with usage of phase and magnitude:

$$f(x) = \frac{a_0}{2}+\sum_{n=1}^{\infty}c_n\cos(n \omega x+\theta_n)$$

Or even if you wish, complex exponential notation can be used. In case of link you provided, think of $a_n$ and $b_n$ as of real and imaginary part of the Fourier Transform coefficients (very naive, but intuitive), from which you can extract the magnitude and phase information.

The link you provided doesn't use sines with phases, but sums of sines and cosines. Since sin(f*t + phi) = sin(f*t) * cos(phi) + cos(f*t) * sin(phi) the two are fundamentally equivalent. The phase phi just ends up in the coefficients of the sine and cosine terms.