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Magnetometer measures the derivative of the magnetic field, or dB/dt, with an output in microvolts (mV). The Sampling rate is 128 Hz, so if we collect data for 2 minutes, $2 \times 60 \times 128=15360$ points (discrete case). When I perform an FFT on these time series, what will the units of amplitude density be after the transform?

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    $\begingroup$ Your raw data is in mv vs 1/time thus after FFT the ordinate still has units of mv and the abscissa units of time. $\endgroup$ – porphyrin Dec 6 '16 at 8:45
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In short: if you directly use the voltage output (here in microvolts), your FFT amplitude should be in mV (microvolts) as well. If you have access to the sensor calibration curve to convert units, the FFT amplitude should be in Teslas per second (T/s), as you look at a derivative.

If you look at a power density spectrum (squared), then the units above should be squared as well (mV$^2$ or T$^2$/s$^2$).

Now in general, for a given unit U, albeit in volt (V), tesla (T), whatever. The FFT sums samples $x_k$ in the original units (U) multiplied by unitless complex values (due to discretization) $e^{-2\pi j\cdot}$. Thus the units after FFT remain the same as for the original signal, i.e. U.

If you take the absolute value, the same again. For instance, the $0$-frequency index gives you the DC, or average value of your signal (or at least it is proportional, with a factor of the number of samples, or its square root depending on the normalization).

The continuous Fourier transform, in contrast, "sums" the samples times complex cisoids times a time differential (the $\mathrm{d}t$): $$\int e^{-2\pi j\cdot}x(t)\mathrm{d}t $$

Its units would be U.s (s for seconds). Unless you normalize it by something with a unit, like in

$$\frac{1}{T}\int e^{-2\pi j\cdot}x(t)\mathrm{d}t $$

where to "average" over some period or time span $T$ (in time units), and then the results would have the same units as $x(t)$ again.

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  • $\begingroup$ Can you take a look at now,it is dicrete,what should be the amplitude unit? $\endgroup$ – MikiBelavista Dec 2 '16 at 17:41
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    $\begingroup$ Yes I also had problem to grasp this.It should be the voltage from the sensor. $\endgroup$ – MikiBelavista Dec 2 '16 at 18:25

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