Envelope analyis basic question

I have a vibration signal sampled at 25.6 kHz, and I want to perform Envelope Analysis on it.

"Vibration-Based Condition Monitoring of Wind Turbines" by T. Barszcz has the following figure:

1. I understand why the frequency is changing in the FFT plot during the HP and LP filters, but why is the frequency range changing from (b) -> (c) [0 -> 8 kHz] when performing the abs (rectifier) step? Cheers!

Edit: Here is an FFT for "Gearbox High Speed shaft" from my data.

• Frequency axes are difficult to follow: (b) is [4-10], (c) is [0-8]. – Laurent Duval Mar 6 '20 at 15:31
• @user10971344. Hi. If by "Envelope Analysis" you mean "envelope detection', then perhaps the material at the following web page will be of some interest to you: dsprelated.com/showarticle/938.php – Richard Lyons Mar 7 '20 at 11:32

sampled at 25.6 Hz, 25.6 kHz is much more likely.

1. Your signal contains a strong fundamental (probably the turbine itself) plus some "fuzzy" stuff on top which is the vibration signal the author is apparently after. You see this clearly in the spectrum: a strong peak at low frequency and a bit of fuzz between 4k and 8k
2. Apply a high pass filter to get rid of the fundamental. The time signal has lost the big sine wave. The 4k-8k region is the same, just "zoomed in". The low frequency peak is gone, but you can't see this because of the way the graph is scaled
3. Absolute Value: An envelope can only have positive numbers so you need a way to make this all positive. There are typical two choices: absolute value or squaring. In most case squaring is the better option since it has a good physical interpretation: it's simply the energy vs time. Taking the absolute value is highly non-linear operation so you are going to see more higher frequencies. It looks like the author made a fairly big mistake here: Most of the energy sits around 6kHz and the non-linearity will create energy around 12 kH, 18 kHz (less), 24 kHz, etc, so there is probably quite a bit of aliasing happening. To make matters worse, you can't see the aliasing easily since, given the sample rate, the second order harmonics falls on top of the down-mix products at low frequencies. I'm pretty sure the small peaks around 3k are pure aliasing. Squaring also makes it easier to manage aliasing, you only need to upsample by a factor of 2.
4. Lowpass filter: most people interpret this input signal as "energy versus time" with some high speed variations on top which cancel each over very short time periods so I'm not interested in those. So you use a lowpass filter to get rid of them . There are other options: lossy peak detectors, rms detectors, etc. The frequency graphs is the same as in step c) except all the high frequency fuzz is gone, but that's not shown as the graph is scaled differently

but why is the frequency range changing from (b) -> (c) [0 -> 8 kHz] when performing the abs (rectifier) step?

Mostly sloppiness and laziness of the author. It's also a mistake: The frequency range does indeed change, but not in the way intended or shown. They appeare to be taking the absolute value without up-sampling first, which incurs significant aliasing. The "true" frequency range of the signal after the absolute value operation is substantial larger than show and calculated.

• Thank you Hilmar, that was very helpful. Looking at my FFT (in the EDIT), could an envelope still show promising results? I could look at 4000-7000 Hz and maybe find something interesting. – meerkat Mar 9 '20 at 8:06
• Do I have to upsample by a factor of 2 if I square the signal? Why is this? – meerkat Mar 27 '20 at 12:38

A high-pass filtered signal has lost its low frequencies, so you can refocus the spectrum on a higher frequency range. When you rectify it, the first phenomenon is that the signal becomes positive. Hence, it is not zero-mean anymore, and thus recovers low-frequencies that were attenuated in the high-pass filtered signal.

A second is the presence of higher frequency terms. Typically, when a sinusoidal component has period $$T$$, its absolute value doubles its period and generates higher-order harmonics.