## 4 Wavelet Energy Evolution Characteristics
### Wavelet Energy Analysis  
Wavelet decomposition was performed by using the tree algorithm. During the first-layer decomposition, to obtain a set of low-frequency coefficients and high-frequency coefficients, the original signal needs to be fed into a hybrid filter (including a low-pass filter and high-pass filter). The low-frequency coefficients are then used as a signal input filter in the second layer of decomposition, yielding the next set of low-frequency and high-frequency coefficients. In this way, computing halts when the specified grading level is reached. The mathematical expression can be shown as:  
\[\begin{split} a^{m}&=Ga^{m-1}\\ d^{m}&=Hd^{m-1}\end{split} \tag{7}\]  
where \(G\) represents the low-pass filter, and \(H\) represents the high-pass filter. The matrix is used to represent them. The original signal can be regarded as level 0 for the low-frequency coefficient \(a^{0}=(f_{0},f_{1},\ldots,f_{n})\).  
In this test, the effective AE signals in the frequency spectrum of 0-500 kHz were subjected to a five-layer wavelet decomposition operation and obtained six frequency band sub-signals (the frequency range is shown in Table 1). The wavelet basis function adopts the symmetric wavelet sym2. The wavelet decomposition structure tree is shown in Figure 9. Therein, CA is the low-frequency sub-signal of wavelet decomposition, and CD is the high-frequency sub-signal. Numbers 1 to 5 are the numbers of decomposition layers. For  
Figure 8: Removing zero processing to the original waveform. (**a**) Before processing; (**b**) After processing.  
example, CA\({}_{5}\) represents the low-frequency sub-signal of wavelet decomposition in layer 5, and CD\({}_{5}\) represents the high-frequency sub-signal of wavelet decomposition in layer 5.  
The wavelet decomposition coefficients were reconstructed to obtain the optimized signal \(S_{i}\) with different frequency ranges. The energy values of the reconstructed signal were quantified as per Formula (8) and the signal energy distribution coefficients were calculated for each frequency band. Meanwhile, the dominant frequency band was defined.  
\[K_{i}=\frac{E_{i}}{E}=\frac{\sum\limits_{k=1}^{m}S_{ik}^{2}}{\sum\limits_{i=1}^{6 }E_{i}}=\frac{\sum\limits_{k=1}^{m}S_{ik}^{2}}{\sum\limits_{k=1}^{6}\sum\limits _{k=1}^{m}S_{ik}^{2}}, \tag{8}\]  
where \(i=1\), \(2\), \(3\), \(4\), \(5\), \(6\). Here, \(S_{ik}\) is the amplitude of the \(k\) sampling point of the layer \(i\) signal; \(m\) is the number of the signal discrete sampling points; \(E_{i}\) is the corresponding energy of \(S_{i}\); \(K_{i}\) is the AE signal energy distribution coefficients of each frequency band and the dominant frequency band is established as the highest energy distribution coefficient's frequency band.  
Accordingly, the 3546 th effective acoustic emission signal was extracted during the whole test (see Figure 10a) and the signal was carried out as a five-layer wavelet decomposed and reconstructed (see Figure 10b). The reconstructed signal of each frequency band and its energy distribution coefficient was calculated as shown in Table 2.  
\begin{table}
\begin{tabular}{c c c} \hline
**Number \(i\)** & **Wavelet Sub-Signal** & **Frequency Range/kHz** \\ \hline
1 & CA\({}_{5}\) & 0–15.625 \\
2 & CD\({}_{5}\) & 15.625–31.250 \\
3 & CD\({}_{4}\) & 31.25–62.50 \\
4 & CD\({}_{3}\) & 62.5–125.0 \\
5 & CD\({}_{2}\) & 125–250 \\
6 & CD\({}_{1}\) & 250–500 \\ \hline \end{tabular}
\end{table}
Table 1: Frequency range of the wavelet sub-signal.  
Figure 9: Five-layer wavelet decomposition structure tree.  
As shown in Table 2, the dominant frequency band of the TL-3-3546 th acoustic emission signal is CD\({}_{2}\) (125\(\sim\)250 kHz) and its energy value is 0.2812. The energy distribution coefficient is 0.3990, which accounts for the largest proportion.