## 4 Illustrative Example
### Recognition Process
#### 4.1.2 Second Recognition  
Only one of the recognition results (14.4% [\(\mathbf{a}\)H5\(\cup\)\(\mathbf{s}\)H5] and 4.4% [\(\mathbf{i}\)H5]) in conditions 1-2 is analyzed because of the above conclusion (the actual damage location is invariably near the recognition result or its symmetrical location), and the analysis process of other cases is similar. Figure 14 shows the comparison of the damage location and the element division. As described in Section 3, if the length of the selected unit is too large, the recognition result is in an extensive range, which affects the accuracy of recognition. Therefore, the bridge elements are re-divided, and the range of variables is limited. Take the first identification result 14.4% [\(\mathbf{a}\)H5\(\cup\)\(\mathbf{s}\)H5] of condition 1 as an example. The first identification result shows unit \(\mathbf{a}\) damaged, and the variable is limited to the range of [\(\mathbf{a}\)\(\mathbf{b}\)\(\mathbf{s}\)] in the second recognition,  
Figure 12: Distribution of first recognition results (\(SNR=14\)): (**a**) Condition 1; (**b**) Condition 2.  
Figure 13: Position distribution near the unit and its symmetry: (**a**) Unit \(\mathbf{a}\); (**b**) Unit \(\mathbf{s}\).  
corresponding to the second subdivision unit, which is [1 2 3 4 23 24 25 26]. Similarly, the first identification result shows unit \(s\) damaged, and the variable is limited to the range of [_a b c r s t_] in the second recognition, corresponding to the second subdivision unit, which is [1 2 3 4 5 6 21 22 23 24 25 26]. The total recognition rate of identifying damage locations twice is obtained according to Equation (27).  
\[IR_{\text{Z}}=IR_{\text{F}}\times IR_{\text{S}} \tag{27}\]  
where \(IR_{Z}\) is the total recognition rate, \(IR_{\text{F}}\) is the probability that the real damage is included in the range of the nearby and the symmetrical position of the first recognition result, and \(IR_{\text{S}}\) is the probability of the second recognition.  
From Figure 15, the overall locating effect of the two conditions is better, and the recognition rate is greater than 95%. However, in the quantization results, there is a slight error between the recognition results in Figure 15b and the actual damage, and most of them are identified as unit 13 (length 0.75 m) with 8% uniform damage. The actual damage is contained in unit 13, with a length of 0.6 m and damage of 10%. However, this error is relatively small and is acceptable for practical application.  
Figure 14: Comparison of damage location and element division.  
Figure 15: Distribution of second recognition results (\(SNR=14\)): (**a**) Condition 1; (**b**) Condition 2. **Note:** the location of the unit number in Figure 15 is different, as detailed in Figure 14.  
In terms of computational efficiency, because [2:2:10] is set as the recognition step and range of the second recognition, the search space for condition 1 is 2400. Plus, the search space in the first recognition is 3600, a total of 6000. If the traditional one-time recognition method is used, and the first recognition is divided according to a smaller length, the search space is as high as 40,000, which is 6.7 times that of this method. Thus, the search space of this method is much lower than that of the traditional one-time recognition method. Moreover, with the increase in the number of units damaged, the number of search spaces in the traditional one-time identification method exponentially increases; this method shows significant superiority, and the computing time is significantly reduced.