# What does the Phase part of a FFT of a received signal represent?

Imagine we've receive a signal, $R$, at a distance of $L$ from a transmitter. Then, we do FFT on the received signal. This is what FFT gives us:

\begin{align} R(f) &= A e^{j\phi} \tag1\\ \ \end{align}

where, $A$ is the amplitude, we can take it by abs() function over the FFT. And, $\phi$ is the phase and we can take it by angle() function over FFT.

Now this is my confusion: If the initial phase, I mean the phase of the transmitted signal, would be $\phi_0$, then, Is it correct if I say:

$\phi = \phi_0 + \theta$ ?

$\theta$ is a number that is added to the transmitted phase over the link.

or I can also write this:

$\phi_{received} = \phi + \phi_0$ ?

$\phi_{received}$ is the phase of the received signal.

Which one of those is correct?

Please take this into account that I have no information about the transmitted signal. Just assume a transmitted signal with initial phase $\phi_0$.

Also just consider a very simple case, like a line-of-sight signal. And, I'm using MATLAB for FFT, and angle().

EDIT: I have two received signals at the distances of $L_1$ and $L_2$, with the phases of $\phi_1$ and $\phi_2$. I calculated the difference between these two phases. And, plotted them. And, the phase shift between these two received signal should be

$\phi_2 - \phi_1 = 2\pi f \tau$

So, Can I just go and consider the two signals in the time domain and calculate the time difference, $\tau$, between them? Any thoughts? What am I missing here?

• you're not saying what your signal is. I think you really should think about that – your eq. $(2)$ is wrong in general. Please, start by defining your transmit signal (in time domain!) first. Afterwards, we can look at the receive signal (both in time and frequency domain). We can make wild assumptions on your transmit signal, but I'd really prefer if you actually gave us a formula for what you're sending – that would mean we actually answer the question that you have, and also, it would encourage you to bring clarity into your notation and mind about what signals we're talking about! Mar 1 '18 at 23:07
• Also, are you related to the Chili of this question? it's not clever to have multiple accounts, because then we can't factor in your other questions to help you. If you are the same person: please try to explain what you're doing, by giving the whole background. Mar 1 '18 at 23:11
• I don't have the transmit signal or a formula for it. My transmitter transmit a signal in a very high frequency band,(0.1 - 2 THz), and I receive it by a receiver device. So I have the received signal in time domain. Then I can do FFT on it. My goal is to model the phase shifting for this system. So, I have to make sure that I understand the phase from FFT correctly. The equation 2 works only for a single LOS signal. You're right, in general it's not correct. Mar 1 '18 at 23:38
• What is "a single line of sight signal" to you? To me, this does not work for arbitrary "single" signals, whatever "single" means here. Mar 1 '18 at 23:40
• Really, what I'm asking you is to come up with a signal model. You seem to have a specific class of signals in mind, but that's not "all signals". As you can see from hotpaw2's general answer, you won't be getting specific information if you're not even trying to restrict what you're talking about! Mar 1 '18 at 23:43

The phases of an FFT result are a phase measurement relative to the time that the FFT input sample window was taken. Each phase measurement is fairly useless unless you know the absolute time that the sample window was taken, or know the offset of the window from another measurement window (another FFT window with a known overlap/offset/delay/etc., or another window sampled at the same time but offset in space/etc., or some combination of the two), or are comparing or using phases across FFT bins (for reconstruction, convolution, etc.)

Whether the phase is positive or negative depends on how the cosine (or evenness or real) and sine (oddness or "imaginary") components of each DFT result bin relate to each other within the FFT aperture or sample window. Usually a negative phase indicates the opposite oddness component (hi/low versus low/hi asymmetry) than a positive phase.

If you do an FFT-shift before the FFT, then the phase measurement (the evenness/oddness ratio represented by the atan2 function) is relative to the middle of the vector of samples of the window fed to the FFT, independant of whether the signal is exactly integer periodic in aperture or not.

To answer line of sight problem, you have to define an antenna with a time dependent input or feed then solve the wave equation. The solution gives you the time dependency of received field at every position.

Now if you compare transmitted and received signals, each frequency of received signal is weakened and also an additional phase added. If the magnitude of all frequencies weakened by same amount and the additional phase were linear(with respect to frequency) then the received signal is a weaker and delayed version of transmitted signal. But generally attenuation is a function of frequency and phase is not linear.

Also I have to say "phase" is meaningful only for a single sinusoid. if you solve wave equation in time domain you have to solve it for each input signal, but if the system were linear (which is the case) you solve wave equation in frequency domain which is called Helmholtz equation, the solution gives you effect of transmission on each frequency so by using Fourier analysis on input signal you could find effect of transmission of whole signal.

Also remember when you are using Fourier transform and obtain phase from that, you have to unwrap the phase.

• Because of my background, It's a little bit difficult for me understand everything that you're talking about. But, I appreciate it. That'd be great if you give me a link of some tutorials about this stuff. And, yeah my phases are linear with respect to the frequency. I'll edit my questions and will add some of my plots. Mar 2 '18 at 1:50
• You're welcome and I will do so if I find a good tutorial. But if you're not concerned with 'line of sight' and the phase is linear, the only thing that remains is delay time which you can find it by just looking at signals (e.g. if your signal has a peak then look how much this peak shifted) Mar 2 '18 at 9:42
• Ok, I added a plot and edited my questions. Anyway, if I find the delay time, based on what you recommend, then you think I'm gonna get the same plot as the above one? Mar 2 '18 at 19:55
• the slope of phase difference gives you the delay. Mar 2 '18 at 20:21
• if you ploted the phase difference correctly, the delay is about 3 picoseconds, and multiply it by speed of light gives you about 100 micrometers which is the path difference. Mar 2 '18 at 20:33