We know that many machines consist of a rotor and a stator. The rotor rotates while the stator remains stationary. However, due to factors such as dust accumulation, misalignment, and imbalance, the entire machine can vibrate.
Some vibrations we can hear, some we can touch or see, but no matter what, increased vibration is never a good thing. It may indicate some kind of malfunction, and if it continues to work for a long time, the machine will break down.
Of course, sometimes we also deliberately create vibrations, such as the vibration of a mobile phone or the vibration of a slimming machine!
For unintentional vibrations, we need to monitor them, as they contain a wealth of information that can help us analyze the machine's operating status.
Just as "temperature" is a numerical way to scientifically describe "cold" and "hot," vibration also has its own descriptive language, the two most basic values of which are amplitude and frequency.
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The vibrations of real machines are very complex, but they can often be decomposed into small, periodic vibrations with fixed amplitudes, and each small vibration can be mathematically represented by a "sine function".
Do you remember high school physics? Vibrations are also subject to Newton's laws of motion. Let me help you recall:
Okay, if the displacement of an object's vibration changes with time according to a sinusoidal function curve, then what are the changes in its velocity and acceleration? Please see the following diagram:
Did you guess correctly? Please note the following points:
The answer is, it depends on the frequency of the vibration.
The explanation is as follows:
To obtain the best signal-to-noise ratio, data with relatively flat frequency spectrum should be used as the basis for analysis. For example, if the vibration velocity is 7.6 mm/s, would you consider 1 kHz displacement data or 10 Hz acceleration data as noise if you did not use "velocity" as the basis for analysis?
The usual selection criteria are: for low frequencies, "displacement" is used as the basis for analysis; for mid frequencies, "velocity" is used as the basis for analysis; and for high frequencies, "acceleration" is used as the basis for analysis.
Displacement sensors are non-contact, making them suitable for measuring the vibration of rotating shafts. However, displacement sensors have a relatively low frequency response.
The internal mechanical elastic structure makes it difficult for the sensor to achieve a high frequency response. However, there are currently speed sensors based on piezoelectric technology, and even speed sensors based on laser Doppler technology. Of course, the latter are very expensive.
Accelerometers can achieve a relatively wide frequency response. The peak of the red curve in the figure below represents its resonant frequency. The effective bandwidth is generally one-third of the resonant frequency, while the sensitivity of the sensor will be greatly reduced in the part close to the resonant frequency.
The two waveforms describe the same signal, except that the horizontal axis in the time domain is time, and the horizontal axis in the frequency domain is frequency. For the vertical axis, the amplitude is the same, both being 1V (the vertical axis is represented by the root mean square, i.e., 70.7% of the peak amplitude).
Let's look at another case, this signal is more complex, it is composed of four sinusoidal signals superimposed in the time domain:
Look, the final waveform after superposition is so strange, it's crooked and twisted, and there's no regularity at all. If you weren't told beforehand that it's made up of four superimposed sine waves, could you guess it?
Many signals that appear complex in the time domain become very simple in the frequency domain because the Fourier transform decomposes the signal from the frequency domain.
In fact, Fourier told us that any periodic function can be regarded as the superposition of sine waves with different amplitudes and phases.
In practical applications, the objects we collect are finite non-periodic signals. So how do we perform Fourier transform?
Good question! Actually, we would repeatedly extend it forward and backward in the time domain to imagine that it has periodicity.
However, this leads to an interesting phenomenon. For example, if we sample a sine wave and obtain two signals, the sine waves in these two signals will be exactly the same. The only difference is that one signal contains a complete set of 6 sine wave cycles, while the other contains 6.5 sine wave cycles. So, what will their Fourier transform results be?
Was this result within your expectations?
After expansion, signal (a) becomes a normal sine wave, which, after Fourier transform, becomes a pulse in the frequency domain. However, after time-domain expansion, signal (b), although periodic, is no longer a sine wave; the transitions between periods are very abrupt. This transition effect requires the superposition of more sine waves to form, resulting in numerous pulses in the frequency domain.
However, we must understand that the result of signal (a) is what we want. The real signal is a sine wave. If we obtain the result of signal (b), then it is wrong.
To avoid this phenomenon, the time-domain signal is processed through a "weighted window" before performing a Fourier transform.
Let’s look at an interesting phenomenon. If you have two sine waves, one high frequency (a) and one low frequency (b), and assign them the same sampling rate, you will get two identical sampling results if you are lucky (the green line is the sampling period and the black dot is the sampling result).
If we use this sampling result to reproduce the real signal, would you say we get signal (a), or (a), or (a)?
We call this phenomenon the "stroboscopic effect," which is very common in daily life. For example, if you see a wheel turning very slowly or even in the opposite direction, it's because your eye's sampling rate can't keep up with the wheel's rotation speed. If it were really turning in the opposite direction, that would be strange, because the car is clearly moving forward!
The "flicker effect" can introduce significant errors into the Fourier transform results. For example, in the image below, there are four signals. The left side shows the time-domain waveform, and the right side shows its frequency-domain waveform. The sampling frequency is 6Hz. The signal waveforms are represented by the blue line, the sampling period by the green line, and the flicker effect by the red line. Looking at the image from bottom to top:
Can you find a pattern here? It seems that when the sampling frequency fs is greater than twice the highest frequency Fmax of the signal (Fs>2Fmax), the sampled signal can be used to recover the original real signal?
The person who discovered this pattern in history was Nyquist, and he named it the "Nyquist sampling theorem".
To avoid flickering, we first pass the signal through a low-pass filter. The cutoff frequency of this low-pass filter is half the sampling frequency (Fs/2). Signals with frequencies below this value are sampled, while signals with frequencies above this value are filtered out.
In fact, it is difficult to make the edge of a low-pass filter very steep. Therefore, the actual sampling frequency is generally 2.56 to 4 times the highest frequency of the signal (Fs > 2.56Fmax).
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