## 3 Hybrid FE-SEA Modeling Parameter Identification Algorithm
### Derivation of the Energy Response of Random Subsystems of Three-Coupled Systems
#### 3.2.1 Random Subsystem Energy Response of Three-Coupled Systems
The two terms are unrelated, and the right-hand side of the equation is expressed as [20,21]:  
\[P_{in,dir}^{ext}=\frac{\omega}{2}\sum\limits_{rs}\Big{[}\text{Im}\Big{\{}D_{ dir,js}^{(j)}\Big{\}}\Big{(}D_{tot}^{-1}S_{ff}^{ext}D_{tot}^{-1H}\Big{)}_{rs} \Big{]}=\frac{\omega}{2}\sum\limits_{rs}\Big{[}\text{Im}\Big{\{}D_{dir,rs}^{(1) }\Big{\}}\Big{(}D_{tot}^{-1}S_{ff}^{ext}D_{tot}^{-1H}\Big{)}_{rs}\Big{]} \tag{39}\]  
\[P_{in,k\to j}=\frac{2E_{k}}{\pi n_{k}}\sum\limits_{rs}\Big{[}\text{Im}\Big{\{}D_ {dir,rs}^{(j)}\Big{\}}\Big{(}D_{tot}^{-1}\Big{\{}D_{dir}^{(k)}\Big{\}}D_{tot}^{- 1H}\Big{)}_{rs}\Big{]}=\frac{2E_{k}}{\pi n_{k}}\sum\limits_{rs}\Big{[}\text{Im} \Big{\{}D_{dir,rs}^{(1)}\Big{\}}\Big{(}D_{tot}^{-1}\Big{\{}D_{dir}^{(2)} \Big{\}}D_{tot}^{-1H}\Big{)}_{rs}\Big{]} \tag{40}\]  
Figure 3: FE-SEA subsystem power balance. Substitute Equation (39) and Equation (40) into Equation (38); then:  
\[P_{in,dir}^{(j)}=P_{in,dir}^{ext}+\sum_{k=1}^{N}P_{in,k\to j}=P_{in,dir}^{ext}+ \sum_{k=1}^{2}P_{in,2\to 1} \tag{41}\]  
The power dissipation (\(P_{out,rev}^{(j)}\)) of the subsystem (\(j=1\)) consists of two parts. Part of the lost power is transmitted to the deterministic subsystem (d) through the reverberation field of the random subsystem (\(j\)). The other part of power loss (\(\sum\limits_{k=1}^{N}p_{tran,j\to k}\)) is transferred to other random subsystems, and it can be concluded that:  
\[P_{out,rev}^{(j)}=P_{tran,j\to d}+\sum_{k=1}^{N}P_{tran,j\to k}=P_{tran,1\to 3}+\sum_{k=1}^{2}P_{tran,1\to 2} \tag{42}\]  
where the first part of the right side of the equation can be further expressed as:  
\[P_{tran,1\to 3}=E\Big{[}\frac{\omega}{2}\text{Im}\Big{\{}q^{(1),H}D_{d}q^{(1)} \Big{\}}\Big{]}=\frac{\omega}{2}\text{Im}\Bigg{\{}\sum_{r}\sum_{s}E\Big{[}q_{ r}^{(1)*}D_{drs}q_{s}^{(1)}\Big{]}\Bigg{\}}=\frac{\omega}{2}\sum_{rs}\Big{[} \text{Im}\{D_{drs}\}S_{q^{(1)}q(1),rs}\Big{]} \tag{43}\]  
Through Equations (35) and (36), it can be concluded that:  
\[S_{q(1)q(1)}=E\bigg{[}\Big{(}D_{tot}^{-1}f_{rev}^{(1)}\Big{)}\Big{(}D_{tot}^{ -1}f_{rev}^{(1)}\Big{)}^{H}\bigg{]}=D_{tot}^{-1}\text{E}\Big{[}f_{rev}^{(1)H} \Big{]}D_{tot}^{-1H}=D_{tot}^{-1}\bigg{(}\frac{4E_{2}}{\omega\pi n_{2}}\bigg{)} \text{Im}\Big{\{}D_{dir}^{(1)}\Big{\}}D_{tot}^{-1H} \tag{44}\]  
Substituting Equation (44) into Equation (43), the transmitted power of the random subsystem (\(j=1\)) can be deduced:  
\[P_{tran,1\to 3}=\frac{2E_{1}}{\pi n_{1}}\sum_{rs}\text{Im}\{D_{drs}\}\Big{(}D_{ tot}^{-1}\text{Im}\Big{\{}D_{dir}^{(1)}\Big{\}}D_{tot}^{-1H}\Big{)}_{rs} \tag{45}\]  
By comparing Equation (41), the second expression on the right of Equation (42) can be deduced as follows:  
\[P_{tran,1\to 2}=\frac{2E_{1}}{\pi n_{1}}\sum_{rs}\text{Im}\Big{\{}D_{dirrs}^{(2)} \Big{\}}\Big{(}D_{tot}^{-1}\text{Im}\Big{\{}D_{dir}^{(1)}\Big{\}}D_{tot}^{-1H }\Big{)}_{rs} \tag{46}\]  
Substitute Equation (45) and Equation (46) into Equation (43) to obtain the power loss of the subsystem (\(j=1\)) in the reverberation field.