I calculated the FFT of the Az acceleration sampled with MPU6050 connected to Arduino.

Wikipedia definition: "White noise" is a random signal having equal intensity at different frequencies, giving it a constant power spectral density

  1. Can I say from my FFT that the reason of the variance of the signal, in my test condition, is caused by white noise?

Additional question on FFT

  1. If my sampling frequency is 56Hz are my bin_frequency 28?
  2. My frequency bin is influenced by the length of the input vector or only by my sampling frequency?

Test condition:

  • I recorded of the z value in static condition on a table, without any perturbation
  • The signal was nonuniformly sampled so I resampled it


Sampling frequency 55.556

Az 718X1 Vector

Std_Input_Vector_Resampled 0,0385

enter image description here Az Input  Resampled

FFT Output

FFT 718X1 Vector

Without Zoom FFT Without Zoom ZoomX3 FFT Zoom x3


FFT Zoom X4

This is my first time I approach the FFT(I've been studying this argument from one month so I am very noob), and I have a civil engineer background, so is my first time with signal analysis, never studied Signal Theory before.

My interest for the FFT is to define the best filter to identify the pothole and road anomalies

Thanks for your time, thanks for your patience.


1 Answer 1


These are just some hints on how to proceed which are too long for a comment. We cannot give a definite answer from just looking at the plots.

First, you should plot the magnitude of the FFT in dB, i.e. plot(20*log10(fft(signal))). Then, you would not need to plot different zoomed versions.

To refer to question 1) Is this the whole signal or just a part of it? Try analyzing a longer measurement cycle to become more certain. Certainly, there is some addition of white noise to the signal. But, additionally, it seems to me there is an additional sine-wave of period length 4 contained (referring to your first plot).

To do some more analysis:

  1. Assuming you measure a constant acceleration, remove the mean from the signal. Then, the remaining signal just contains the measurement error.

  2. Have a look at the (circular) autocorrelation of this error. In case of white noise, it should be ideally a single peak at delay 0. If the autocorrelation exhibits extra (significant) peaks, there probably is something extra on top of the white noise. (Looking at your FFT-plots, I actually expect some extra peaks).

To answer questions 2), 3) (I dont really understand your question 2)):

If your sampling frequency is 56Hz, your FFT output will range from 0 to 56Hz, where the distance between the bins is 56Hz/FFT-length. Furthermore, as your input signal is real, the frequencies above 56/2Hz will be a mirror of the first half of the frequency domain output, they do not contain additional information. As a general rule for Fourier Analysis: The frequency resolution $\Delta_f$ after FFT is only determined by the time duration $T$ of the original signal, given by $\Delta_f=1/T$. You can have a look at this article I wrote on my website about the FFT and frequency domain resolution as an introduction to the topic.

  • $\begingroup$ Matthè thanks! In the following days (by now) i am going to study the article :) . How can i study Circular Autocorrelation on Matlab? Is a good way with this Matlab function? The mean is 9,7255, i just simply subtract this value for each element of my FFT input Vector? I plotted the whole signal cleaned from inital seconds and the ending, when i switched on/off the device, total rec time cleaned 13 seconds. Thanks a lot :) $\endgroup$ Commented Mar 16, 2017 at 8:28
  • $\begingroup$ For the questions 2) 3), you allay my doubts, so the distance between the frequency bins decrease with the FFT length and i will get information on my frequency domain only from 0 to 56/2 Hz $\endgroup$ Commented Mar 16, 2017 at 8:33
  • $\begingroup$ @uomodellamansarda circular autocorrelation in Matlab can be done my ifft(fft(x).*conj(fft(x))), using the convolution theorem, knowing that correlation is convolution with one function time-reversed and time-reversal amounts to complex conjugation in frequency domain. $\endgroup$ Commented Mar 16, 2017 at 8:56
  • $\begingroup$ Matthè i am going deep with the theory behind the FFT. My peak at 0.1 Hz is my DC offset or the DC offset is perfectly at 0.0 Hz? Thanks! $\endgroup$ Commented Mar 23, 2017 at 10:06
  • $\begingroup$ A DC offset is at 0Hz, at the first FFT bin. At 0.1 Hz it is not the DC-bin. $\endgroup$ Commented Mar 23, 2017 at 12:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.