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I am recording data with a magnetometer of the background magnetic field in a building. I have applied the FFT algorithm to the data in order to look for the frequencies that appear in it. I would like to use this in order to identify (or at least make an educated guess) of the sources of the disturbances that I observe.

My question is: What is the meaning that I can attribute to the amplitude that I obtain from the FFT algorithm? Is there some unit that can be ascribed to it?

Looking at the formula for the continuous fourier transform (which I took from Wolfram Mathworld) : \begin{align} f(\nu)=\int\limits_{-\infty}^{+\infty}f(t)e^{-2\pi i \nu t} \mathrm{d}t \end{align} I do not really know how to accomodate the dimension of Tesla in there.

Thank you

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  • $\begingroup$ what are the units of your input data? $\endgroup$ – user28715 Jun 25 '17 at 18:57
  • $\begingroup$ The unit of the data is nanoTesla $\endgroup$ – Tarabostes Delectus Jun 25 '17 at 19:23
  • $\begingroup$ The t here refers to time? So you've measured time variation of the magnitude of the local magnetic field at a particular position coordinate? $\endgroup$ – Fat32 Jun 25 '17 at 20:04
  • $\begingroup$ @Fat32 That is indeed what I have measured $\endgroup$ – Tarabostes Delectus Jun 26 '17 at 6:07
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The continuous-time Fourier transform of a function $f(t)$ is in essence an integration of $f(t)$ multiplied with a complex exponential kernel: $$ F(\omega)=\int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt \tag{1}$$

Since the exponential function is unitless, the unit of the Fourier integral will be the multiplication of the units of the function $f(t)$ and the differential $dt$.

Assuming that the function $f(t)$ had a unit of micro Tesla, and its argument $t$ is time (in seconds), then the unit of $dt$ will be seconds. As a consequence, the unit of the Fourier transform, $F(\omega)$, will be micro Tesla second $$\mu T \cdot s \tag{2}$$

However, what you actually compute is the discrete-time Fourier transform, $F(e^{j\omega})$, of the samples $f[n]= f(nT_s)$ of the function $f(t)$, via the summation: $$F(e^{j\omega}) = \sum_{n=-\infty}^{\infty} f[n] e^{-j\omega n} \tag{3}$$ where $T_s$ is the sampling period in seconds.

Furthermore, instead of the continuous-argument function $F(e^{j\omega})$, you will compute its samples $F[k]$

$$ F[k] = \sum_{n=0}^{N-1} f[n] e^{-j \frac{2\pi}{N} n k} \tag{4}$$

through a DFT (discrete Fourier transform) of the samples $f[n]$ of length $N$, possibly implemented with an FFT algorithm.

The unit of the samples $f[n]$ is the same as the unit of $f(t)$, making the unit of $F(e^{j\omega})$ and $F[k]$ as micro Tesla. Therefore the unit of the FFT samples $F[k]$ of $F(e^{j\omega})$ will be micro Tesla.

Note that there is an (implicit) amplitude scaling by $1/T_s$ in the computed DFT samples $F[k]$, and when you want to display the continuous-time Fourier transform $F(\omega)$ from the samples $F[k]$, you multiply them with $T_s$, which corrects not only the amplitude scaling, but also the unit of it, by making it micro Tesla second as in (2).

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    $\begingroup$ Thank you for taking the time to write a detailed answer. $\endgroup$ – Tarabostes Delectus Jun 26 '17 at 6:09

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