Rookie mistakes to avoid

As I mentioned above, you can express PSD as V/$\sqrt{{Hz}}$. One of my first measurements involving PSD called for just that (measuring eigenmode shapes of MEMS devices by thermal noise excitation. The mode shape was proportional to V not V²). So, rather than square a bunch of FFTs, average them together, and then take the square root, what I did was to just average together a bunch of FFTs. I figured this saved me a lot of computation time and would give the same answer. I was wrong. Squares and square roots are not linear operators. Now if your data is pretty clean it will be close and you might not notice. For example $\sqrt{{\frac{1}{3}(1^{2}+1^{2}+1^{2})}}$ = 1 and ${\frac{{1}}{{3}}}$(1 + 1 + 1) = 1 so it seems like this works, but with noisy data $\sqrt{{\frac{1}{3}(1.1^{2}+1.0^{2}+0.9^{2})}}$ = 1.0033 while ${\frac{{1}}{{3}}}$(1.1 + 1.0 + 0.9) = 1.

Now to drive home an earlier point, what happens if we calculate the PSD of a periodic signal? This is shown in 25. As you can see, the FFT magnitude is correct for each different length of signal, but the PSD is not. To reiterate: PSD (and only PSD) applies to random vibration. FFT magnitude (and only FFT magnitude) applies to periodic vibration. Do not confuse the two!

Also, if you've done a lot of vibration analysis, you know that for a periodic signal, you can convert between displacement, velocity, and acceleration in frequency (Fourier) domain by simplying multiplying or dividing by $\omega$. You can do a similar operation for PSD as well, but the relations are now:

P_disp = ${\frac{{P_{vel}}}{{\omega^{2}}}}$ to convert (m/s)²/Hz to m²/Hz or P_accel = $\omega^{{2}}_{}$P_vel, to convert (m/s)²/Hz to (m/s²)²/Hz.

The extra $\omega$ comes from the fact that the units are squared.

http://www.desy.de/~sahoo/WebPage/Ground%20Motion.htm

Figure 25:

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