## 3 Numerical Validations
### Continuous Beam Example  
A two-span continuous beam is established, as shown in Figure 1. The span length is 5 m. The cross-section is an I-section with an area of \(1.2\times 10^{3}\) mm\({}^{2}\) and an inertia moment of \(1.94\times 10^{6}\) mm\({}^{4}\). The Elastic modulus is 210 GPa, and the density is 7800 kg/m\({}^{3}\). A fixed periodic load is applied at 4.6 m from the left end. The displacement responses are obtained from D1 and D2, respectively. The Rayleigh damping coefficients are set as 0.0822 and 0.0046. A harmonic force and a square wave force is applied, respectively, and the responses of D1 and D2 are polluted by 5% white noises [8]:  
\[Y_{noise}=Y+level\times\frac{1}{\chi}\sum_{i=1}^{\chi}\lvert Y\rvert\times random \tag{20}\]  
where \(Y_{noise}\) is the responses with noise, \(level\) is the noise level, \(\chi\) is the element number of \(Y\). \(random\) is a standard normal distribution vector. The time history of the force and responses are shown in Figure 2. The total measuring time is 5 s, and the sampling rate is 200 Hz. As described before, the excitation force should be fitted by the TLSF according to Equation (13). Defining the frequency of LSF \(f_{LSF}\) as:  
\[f_{LSF}=f_{s}/2l \tag{21}\]  
Figure 1: Continuous beam model. Figure 2: Excitations and corresponding responses. where \(f_{s}\) is the sampling rate. \(l\) is the length of each TLSF. During the total measuring time \(T\), there will be m TLSFs:  
\[T\times f_{s}=l\times m \tag{22}\]  
\(f_{LSF}\) is required to not be smaller than the main range of frequency of the unknown force in order to simulate all its details. In forced vibration, the frequency of unknown excitation closes to the frequency of the structural responses. Using the fast Fourier transformation of the response, the main frequency of the structural response is determined below 5 Hz. Hence \(f_{LSF}\) is defined as 5 Hz in this case, \(l\) and \(m\) are calculated as 20 and 50, respectively. Therefore, the dimension of the influence matrix \(\mathbf{R}\) is 1000 \(\times\) 100, which is 10 times smaller than the dimension of impulse response matrix \(I\) (\(1000\times 1000\)). It is worth noting that the physical meaning of influence matrix \(\mathbf{R}\) is the structural response due to LSF, which is usually obtained by finite element simulation. In this case, the time step that should be calculated is 1000. By introducing TLSF, defining the truncated value \(\mu\) as 90% according to the damping effect, the effective length \(\delta\) is calculated as 64. In this way, the calculation time step can be reduced to 64, and the affective dimension of TLSF matrix \(\mathbf{\Lambda}\) can be decreased to 64 \(\times\) 100. In addition, because the TLSF smooth the impulse response, the singularity of \(\mathbf{\Lambda}\) is small. Figure 3 show the L-curve of \(\mathbf{\Lambda}\)_and_\(\mathbf{I}\) with respect to the displacement response with noise. It is obvious that with the improvement of TLSF, no typical corner is found in the L-curve, even with 5% white noise in the measurement. This proves that TLSF can significantly enhance the condition of the matrix without a regularization procedure. The determined results are presented in Figure 4. The relative percentage error (RPE) is defined in Equation (23), the results are listed in Table 1.  
\[\mathrm{RPE}=\frac{\sum_{i}\left|F_{identified}-F_{true}\right|}{F_{true}} \times 100\% \tag{23}\] \begin{table}
\begin{tabular}{c c c} \hline \hline
**RPE (\%)** & **Harmonic Force** & **Square Wave Force** \\ \hline via D1 response & \(1.5\) & \(6.72\) \\ via D1 response with 5\% noise & \(5.73\) & \(9.42\) \\ via D2 response & \(-48.1\) & \(13.41\) \\ via D2 response with 5\% noise & \(-49.1\) & \(22.85\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: RPE of fixed periodic load identification. Figure 3: L-curves of TLSF response matrix and Impulse response matrix with respect to D1 with 5% noise. (**a**) L-curve of \(\mathbf{\Lambda}\); (**b**) L-curve of \(\mathbf{I}\).