(2-5) Damping constant and logarithmic decay rate
In this issue, we will consider the relationship between the damping constant and the logarithmic damping rate.
Figure 3: Attenuation waveform
Let the adjacent peak values of the attenuated waveform in Figure 3 be x1, x2, x3, ...
Let the ratio be e^δ.
e^δ=x1/x2=x2/x3=・・・=xn-1/Xn ・・・（16）
Here, x1 = e^(-ζωoT1), x2 = e^(-ζωoT2), and T2 = T1 + Td.
Furthermore, Td is one period of the attenuation waveform.
Td=2π÷(ωo√(1-ζ^2)) ・・・(17)
(1-ζ^2 represents the area inside the square root, and the same applies below.)
e^δ=x1/x2=e^(-ζωoT1)/e^(-ζωo(T1+Td))
=e^(ζωoTd)
Take the natural logarithm.
δ=ζωoTd
=2πζ÷√(1-ζ^2) ・・・（18）
ζ≪1
δ≒2πζ ・・・（18‘）
Also
x1/xn=(x1/x2)(x2/x3)・・・(xn-1/xn) ・・・（19）
=e^(nδ)
Therefore
δ=(1/n)*Ln(x1/xn) ・・・（20）
Let n be the number of cycles until the amplitude is halved.
δ≒2πζ=(1/n)*Ln2=0.693/n
δ is called the logarithmic decay rate, and it has a relationship with the decay constant ζ given by equations (18), (18'), and (19). From the decay waveform measured in experiments, the logarithmic decay rate δ and the decay constant ζ can be determined using this relationship.
(2-6) Determine the damping constant ζ from the Hilbert transform.
From what we've learned so far, if we know the natural frequency ωo and the damping constant ζ, we can express it using mathematical formulas. So, how do we find ωo and ζ?
One method is to provide initial conditions as shown in section (2-2) and obtain the result from the naturally decaying waveform.
An FFT analyzer has a function that uses the Hilbert transform to obtain the envelope, and then calculates the logarithmic decay rate and decay constant from the graph shown in Figure 4, which is obtained by taking the common logarithm of the envelope. Let's review how this is done.
Figure 4: Calculating the logarithmic decay rate from the Hilbert transform in Figure 3.
example:
𝑥=1.0012e−𝑡cos(20𝑡−1/20) ・・・ （1）
From the envelope of this equation, we can determine the logarithmic decay rate and the decay constant.
The envelope y in equation (1) above is given by the following equation:
y=1.0012e-t
The graph obtained by taking 10log y² is shown in Figure 4.
The points x1 and x2 can be chosen arbitrarily; for example, the points at t=0.2s and 1.2s are shown in the table below.
Therefore,
Δy = -1.72676 -(- 10.4127) = 8.68589
Δx = 1.2 - 0.2=1
Also, the frequency fn is calculated from cos[20t-1/20]
2πfn=20 fn=20/2π
Therefore, the logarithmic decay rate δ is
δ=0.115Δ_y÷(Δ_x×f_n )=0.115×8.68589÷(1×20/2π)≅0.314
From the relationship between the logarithmic decay rate and the decay constant (3), the decay coefficient ζ can be found as follows: ζ≪1
δ=2πζ ・・・ （2）
Ζ=δ÷(2π)=0.314+(2π) ≅ 0.05
This value was the same as the ζ value explained in section (2-3) of the main text.
Comparing equation (1) with the general equation (2) as shown below makes it easier to understand the damping constant.
x=1.0012-tCOS(20t-1/20) ・・・ （1）
x=Ae-ω0t cos(√(1-ς2 ) α0 t-ψ) ・・・ （2）
In Figures 3 and 4,
n=Δx÷Td=Δx*fn (ωn=2πfn, fn: resonant frequency)
Furthermore, equation (20) is a formula for changing the base of the logarithm.
LogbM = LogaM ÷ Logab
(LogaM: logarithm of M with base a, M: argument of LogaM)
(If M = a^r, then r = LogaM)
Therefore, LnM is
LnM = LogM ÷ Loge
(Ln is the natural logarithm with base e, Log is the common logarithm with base 10)
When you use this to transform it
δ=(1/n)*Ln(x1/xn)
=(1/n)*Log(x1/xn)÷Loge
=(1/n)*(1/20)*{20Logx1−20Logxn}÷Loge
=(1/n)*{10Logx1^2−10Logxn^2}÷(20Loge)
=1/(Δx*fn)*ΔY÷(20Loge)
=(ΔY/Δx)*(1/fn)÷(20Loge)
=0.115*(ΔY/Δx)*(1/fn) ・・・（21）
From equation (21), we can see that δ can be obtained from the ratio of Δx to ΔY.
Note that ΔY is logarithmic.
Please refer to the explanation of the example in Figure 4.
Using equation (13') as an example, we tried to determine the logarithmic attenuation rate and the attenuation constant from the amplitude ratio of the attenuated waveform.
x=1.0012e^{-tcos(20t-Φ)} ・・・（13‘）
(2-7) Hilbert transform of FFT analyzer
I checked what happens with an actual FFT analyzer.
Figure 5 shows the screen displaying the results of calculating the logarithmic decay rate (Log.d) and the decay constant (Damp) using the Hilbert variable function of the DS2000 series FFT analyzer.
Figure 5: The damping coefficient of x = 1.0012e- t cos(20t-Φ) is found using the Hilbert transform.
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Top: Time waveform
Center: Power spectrum (fn=3.125)
Below: Hilbert transform, Y-axis Log, logarithmic damping rate Log.d, damping constant Damp, and vibration frequency Freq are calculated and displayed.
The analysis results
Log.d: 0.320 Damp: 5.095% (0.05) Free vibration frequency: 3.125Hz
And so the result was as shown in the formula.
Depending on the object being measured, the attenuation waveform may not be a single frequency, and the Hilbert transform graph may not be a straight line. This is likely due to nonlinear factors such as the influence of harmonics and amplitude dependence, where the attenuation rate changes depending on the amplitude. Evaluation methods should be considered, such as limiting the amplitude to a linear range, using a large Δx to obtain an averaged attenuation rate, applying frequency bandwidth limitations to the Hilbert transform, averaging multiple measured attenuation constants, or comparing with those obtained from the frequency response function.
The Hilbert transform process with frequency bandwidth limitations involves obtaining a waveform obtained by applying an inverse Fourier transform (IFFT) with frequency bandwidth limitations, and then performing a Hilbert transform on that waveform to determine the logarithmic decay rate and decay constant.
For reference, Figure 6 shows the waveform obtained using IFFT with frequency bandwidth limiting applied to the data in Figure 5. Because the original data is clean, the effect of frequency bandwidth limiting is not apparent. Please view this for reference only.
Figure 6: Hilbert transform obtained from the time-domain waveform after IFFT calculation with frequency bandwidth limiting.
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Top: Waveform obtained by band-limiting the data in Figure 5 across the resonant frequency and performing an IFF calculation.
Below: The IFFT waveform was subjected to a Hilbert transform in the same manner as in Figure 5 to determine the attenuation constant.
(Excerpt from the email newsletter published onApril20,2006)
