## 2 Theory
### Single Direction Algorithm
Hence, the amplitude and angle of each eccentric disc's unbalance can be calculated based on Equations (8) and (10), in which the unbalance responses in the \(y\) direction are required as inputs. According to Equations (4) and (5), the matrix \(H_{2}\) in Equations (8) and (10) can be calculated using \(m+n\) unbalance responses as inputs. The unbalance response of the eccentric disc whose unbalance is to be identified must be measured according to Equation (7). Hence, the total number of input unbalance responses is equal to \(m+n+1\). To identify all the eccentric discs' unbalances simultaneously, the unbalance responses of all the discs should be included. Similarly, Equation (11) can be obtained according to Equation (2) in the \(x\) direction. \[\frac{V(q)\cdot EI}{L^{2}}=\left[ \begin{array}{c}G_{v}(q,q_{1d})\\ \vdots\\ G_{v}(q,q_{md})\\ G_{v}(q,q_{1b})\\ \vdots\\ G_{v}(q,q_{(m+n)b})\end{array} \right]^{T}\left[ \begin{array}{c}\pi\cdot w^{2 }\cdot m_{u1}\cdot e_{1}\cdot(\cos\alpha_{1}+i\cdot\sin\alpha_{1})+w^{2}\cdot m _{1d}L\cdot V_{1d}\\ \vdots\\ \pi\cdot w^{2}\cdot m_{mu}\cdot e_{m}\cdot(\cos\alpha_{m}+i\cdot\sin\alpha_{m} )+w^{2}\cdot m_{md}L\cdot V_{md}\\ -L\cdot k_{1s\cdot xy}U_{1b}-L\cdot k_{1s\cdot xx}V_{1b}\\ \vdots\\ -L\cdot k_{ns\cdot xy}U_{nb}-L\cdot k_{ns\cdot xx}V_{nb}\end{array} \right]_{m+n} \tag{11}\]  
\(m+n\) equations can be obtained based on Equation (11) using the unbalance responses as inputs and considering rotor unbalances and bearing coefficients as unknown variables. Equation (12) can be obtained by writing the equations in matrix form. \[\frac{EI}{L^{2}}\cdot\left[ \begin{array}{c}V(q_{1s})\\ \vdots\\ V(q_{ms})\\ \vdots\\ V(q_{ms})\end{array} \right]=H_{3}\cdot\left[ \begin{array}{c}\pi\cdot w^{2} \cdot m_{u1}\cdot e_{1}\cdot(\cos\alpha_{1}+i\cdot\sin\alpha_{1})+w^{2}\cdot m _{1d}L\cdot V_{1d}\\ \vdots\\ \pi\cdot w^{2}\cdot m_{mu}\cdot e_{m}\cdot(\cos\alpha_{m}+i\cdot\sin\alpha_{m} )+w^{2}\cdot m_{md}L\cdot V_{md}\\ -L\cdot k_{1s\cdot xy}U_{1b}-L\cdot k_{1s\cdot xx}V_{1b}\\ \vdots\\ -L\cdot k_{ns\cdot xy}U_{nb}-L\cdot k_{ns\cdot xx}V_{nb}\end{array} \right]_{m+n} \tag{12}\]  
where \(V(q_{1s}),\ldots,V(q_{ms})\) and \(V\Big{(}q_{(m+1)s}\Big{)},\ldots,V\Big{(}q_{(m+n)s}\Big{)}\) are the measured dimensionless-less unbalance responses in the frequency domain in the \(y\) direction of locations, excluding the discs and bearings; the number of \(V(q_{1s}),\ldots,V(q_{ms})\) is \(m\) and the number of \(V\Big{(}q_{(m+1)s}\Big{)},\ldots,V(q_{ns})\) is \(n\); \(H_{3}\) is a \((m+n)\times(m+n)\) matrix and can be calculated using [PERSON]'s functions \(G_{v}(q,q_{i})\) as long as the locations of the measuring points, the bearings, and the discs are known; \(H_{3}\) is as follows:\[H_{3}=\left[ \begin{array}{cccccc}G_{v}(q_{1s},q_{1d})&\cdots&G_{v}(q_{1s},q_{md})&G _{v}(q_{1s},q_{1b})&\cdots&G_{v}(q_{1s},q_{nb})\\ G_{v}(q_{ms},q_{1d})&G_{v}(q_{ms},q_{md})&G_{v}(q_{ms},q_{1b})&G_{v}(q_{ms},q_{ nb})\\ G_{v}\Big{(}q_{(m+1)s},q_{1d}\Big{)}&G_{v}\Big{(}q_{(m+1)s},q_{md}\Big{)}&G_{v} \Big{(}q_{(m+1)s},q_{1b}\Big{)}&G_{v}\Big{(}q_{(m+1)s},q_{nb}\Big{)}\\ G_{v}(q_{ns},q_{1d})&G_{v}(q_{ns},q_{md})&G_{v}(q_{ns},q_{1b})&G_{v}(q_{ns},q_{ nb})\end{array} \right].\]  
Hence, Equation (13) is obtained. \[\left[ \begin{array}{cccc}\pi\cdot w^{2}\cdot m_{u1}\cdot e_{1}\cdot(\cos \alpha_{1}+i\cdot\sin\alpha_{1})+w^{2}\cdot m_{1d}L\cdot V_{1d}\\ \vdots\\ \pi\cdot w^{2}\cdot m_{mu}\cdot e_{m}\cdot(\cos\alpha_{m}+i\cdot\sin\alpha_{m })+w^{2}\cdot m_{md}L\cdot V_{md}\\ -L\cdot k_{1s\cdot xy}U_{1b}-L\cdot k_{1s\cdot xx}V_{1b}\\ \vdots\\ -L\cdot k_{ns\cdot xy}U_{nb}-L\cdot k_{ns\cdot xx}V_{nb}\end{array} \right]_{m +n}=\frac{EI}{L^{2}}\cdot H_{3}{}^{-1}\left[ \begin{array}{c}V(q_{1s})\\ \vdots\\ V(q_{ms})\\ \vdots\\ V(q_{ns})\end{array} \right] \tag{13}\]  
Define \(H_{4}\) as following. \[H_{4}=\frac{EI}{L^{2}}\cdot H_{3}{}^{-1}\left[ \begin{array}{c}V(q_{1s})\\ \vdots\\ V(q_{ms})\\ V\Big{(}q_{(m+1)s}\Big{)}\\ \vdots\\ V(q_{ns})\end{array} \right].\]  
According to Equations (13) and (14) can be obtained. \[\left\{