# A Novel Method of Seismic Signal Detection Using Waveform Features
## 4 Implementation
### Extracting Features  
There are two main types of features, one is from the energy characteristics of seismic signals, and the other comes from the STA/LTA waveform features. We use different methods to extract various features. Firstly, we find features from the STA/LTA trace. In Figure 1, the STA/LTA amplitude of the signal is higher than the noise signal. The maximum amplitude of STA/LTA can be obtained. Then, the ratio between seismic signals and noise is also discriminate. Since the STA/LTA detector is sensitive to background noise, the spikes in the STA/LTA of noise change more drastically and richly, compared to seismic waves. The ratio of \(\rho\) is defined by  
Figure 4: Map of the historical earthquake events recorded by KSRS array from 2011 to 2014. \[\rho=\frac{\alpha}{\beta} \tag{20}\] \[\beta=\frac{\sum_{j=i}^{N}(\frac{\mathrm{STA}}{\mathrm{LTA}}(j))_{max}}{n}, \tag{21}\]  
where \(\alpha\) is the max amplitude, \(n\) is the number of spikes other than the maximum \(\alpha\), \(\beta\) is the average value of other peaks, and \(N\) is the quantity of STA/LTA short time windows. The spectrum energy characteristics obtained by Fast Fourier transform shows the property in the frequency domain. In Figure 2, the seismic energy mean is higher than noise in 1-2 Hz, which can distinguish seismic signals from noise. In time-frequency analysis, the length of the processing signal is identical. Due to the range of energy values being large, we need to perform log normalization for data. EMD is satisfactory for the study of non-linear and non-stationary signals, and it is based on the characteristics of the signal itself. The decomposed signal components are arranged from high to low frequency, determined by the principle of decomposition. Figure 3 is the EMD picture. Each element is arranged in descending order of frequency. Statistically speaking, most of the signal energy is spread in the first few components, whether it is seismic signal or noise. For the seismic signal, the most significant portion of the signal is located in IMF1, but the largest part of the noise is found in IMF2 or IMF3. The energy formula for calculating the IMF component is,  
\[E_{i}(t)=\int_{-\infty}^{+\infty}|\mathrm{IMF}_{i}(t)|^{2}dt,i=1,2,...,m. \tag{22}\]  
Figure 5: The relationship between distance and covariance. The blue line represents a location where no event has occurred (137\({}^{\circ}\) E, 35\({}^{\circ}\) N), the red line, green line, and yellow line represent three locations near to the historical events, in order (136\({}^{\circ}\) E, 34\({}^{\circ}\) N), (135.4\({}^{\circ}\) E, 37\({}^{\circ}\) N), (139.6\({}^{\circ}\) E, 36\({}^{\circ}\) N). It can be seen from the graph that covariance is related to whether an earthquake has occurred and its location. Furthermore, we can obtain the energy proportion of different IMF components,  
\[T_{i}(t)=\frac{E_{i}}{\sum_{i=1}^{5}E_{i}}. \tag{23}\]  
Most of the useful information about the original signal is concentrated in IMF1-IMF5. Therefore, the total energy of the signal is approximate to the former five components sum, using this to estimate the energy proportion of every IMF element, as shown in Equation (23). The statistical results of the above features are shown in Figures 6-9. In [38], researchers proposed a non-parametric test based on the Kolmogorov-Smirnov (KS) test, referred to as the KS Predictive Accuracy (KSPA) test, which can evaluate whether there exists a significant difference between two forecast models. In this paper, we use the common KS test to test whether the data set of features conforms to the theoretical distribution. We perform statistics on the data and observe the distribution characteristics of the data through the preliminary judgment of the histogram, and obtain the mean and variance of the data, and then use the KS test to determine whether each feature data set meets the obtained normal distribution. At this time, when the obtained \(p\)-value is higher than 0.05, it is considered to conform to the normal distribution.