## 2 Theory
### Two Orthogonal Direction Algorithms  
According to Equations (6) and (14), Equation (18) can be obtained. \[\left\{ \begin{array}{c}\frac{H_{2}(1,1)-\pi\cdot w^{2}\cdot m_{1u}e_{1} \cdot(\sin\alpha_{1}-i\cos\alpha_{1})}{H_{4}(1,1)-\pi\cdot w^{2}\cdot m_{1u}e_ {1}\cdot(\cos\alpha_{1}+i\sin\alpha_{1})}=\frac{U_{1d}}{V_{1d}}=C_{1d}\\ \vdots\\ \frac{H_{2}(m,1)-\pi\cdot w^{2}\cdot m_{md}e_{m}\cdot(\sin\alpha_{m}-i\cos \alpha_{m})}{H_{4}(m,1)-\pi\cdot w^{2}\cdot m_{md}e_{m}\cdot(\cos\alpha_{m}+i \sin\alpha_{m})}=\frac{U_{md}}{V_{md}}=C_{md}\end{array} \right. \tag{18}\]  
\[\left\{ \begin{array}{c}cx_{1}=m_{1u}e_{1}\cdot\sin\alpha_{1}\\ cy_{1}=m_{1u}e_{1}\cdot\cos\alpha_{1}\\ \vdots\end{array} \right.\]  
According to Equations (18) and (19) is obtained. \[\left\{ \begin{array}{c}c\\ H_{2}(1,1)-\pi\cdot w^{2}\cdot(cx_{1}-i\cdot cy_{1})=C_{1d}\cdot H_{4}(1,1)-C_ {1d}\cdot\pi\cdot w^{2}\cdot m_{u1}\cdot e_{1}\cdot(cy_{1}+i\cdot cx_{1})\\ \vdots\\ H_{2}(m,1)-\pi\cdot w^{2}\cdot(cx_{m}-i\cdot cy_{m})=C_{md}\cdot H_{4}(m,1)-C_ {md}\cdot\pi\cdot w^{2}\cdot(cy_{m}+i\cdot cx_{m})\end{array} \right. \tag{19}\]  
In Equation (19), \(H_{2}(1,1)\cdots H_{2}(m,1),H_{4}(1,1)\cdots H_{4}(m,1),C_{1d}\cdots C_{md}\) are complex numbers and the others are real numbers. According to Equations (19) and (20) can be obtained. \[\left\{\begin{array}{c}\pi\cdot w^{2}\cdot\bigg{[} \begin{array}{c}-(1+iC _{1d})&rC_{1d}\\ rC_{1d}&(1+iC_{1d})\end{array} \bigg{]}\cdot\bigg{[} \begin{array}{c}cx_{1}\\ cy_{1}\end{array} \bigg{]}=\bigg{[} \begin{array}{c}r(C_{1d}H_{4}(1,1))-rH_{2 }(1,1)\\ i(C_{1d}H_{4}(1,1))-iH_{2}(1,1)\end{array} \bigg{]}\\ \vdots\\ \pi\cdot w^{2}\cdot\bigg{[} \begin{array}{c}-(1+iC_{md})&rC_{md}\\ rC_{md}&(1+iC_{md})\end{array} \bigg{]}\cdot\bigg{[} \begin{array}{c}cx_{m} \\ cy_{m}\end{array} \bigg{]}=\bigg{[} \begin{array}{c}r(C_{md}H_{4}(m,1))-rH_{2 }(m,1)\\ i(C_{md}H_{4}(m,1))-iH_{2}(m,1)\end{array} \bigg{]}\\ \text{where }rH_{2}(1,1)\cdots rH_{2}(m,1),rH_{4}(1,1)\cdots rH_{4}(m,1),rC_{1d} \cdots rC_{md}\) are the real parts of the above-mentioned complex number; and \(iH_{2}(1,1)\cdots iH_{2}(m,1),iH_{4}(1,1)\cdots iH_{4}(m,1),\)\(iC_{1d}\cdots iC_{md}\) are the imaginary parts of the above-mentioned complex number. Hence, Equation (21) is obtained according to Equation (20). \[\left\{ \begin{array}{c}\left[\begin{array}{c}cx_{1}\\ cy_{1}\end{array}\right]=\frac{1}{\pi\cdot w^{2}}\cdot\bigg{[}\begin{array}{ c}-(1+iC_{1d})&rC_{1d}\\ rC_{1d}&(1+iC_{1d})\end{array}\bigg{]}^{-1}\bigg{[}\begin{array}{c}r(C_{1d} H_{4}(1,1))-rH_{2}(1,1)\\ i(C_{1d}H_{4}(1,1))-iH_{2}(1,1)\end{array}\bigg{]}\\ \vdots\\ \left[\begin{array}{c}cx_{m}\\ cy_{m}\end{array}\right]=\frac{1}{\pi\cdot w^{2}}\cdot\bigg{[}\begin{array}{ c}-(1+iC_{md})&rC_{md}\\ rC_{md}&(1+iC_{md})\end{array}\bigg{]}^{-1}\bigg{[}\begin{array}{c}r(C_{md} H_{4}(m,1))-rH_{2}(m,1)\\ i(C_{md}H_{4}(m,1))-iH_{2}(m,1)\end{array}\bigg{]}\end{array} \right. \tag{21}\]  
The amplitude and angle of the rotor unbalance are:  
\[\left\{\begin{array}{c}\left\{ \begin{array}{c}m_{1u}e_{1}=\text{abs}(cy_{1 }+i\cdot cx_{1})\\ \alpha_{1}=\text{angle}(cy_{1}+i\cdot cx_{1})\\ \vdots\\ \left\{\begin{array}{c}m_{mu}e_{m}=\text{abs}(cy_{m}+i\cdot cx_{m})\\ \alpha_{m}=\text{angle}(cy_{m}+i\cdot cx_{m})\end{array}\right.\end{array} \right. \tag{22}\]where \(\text{abs}()\) is the function used to obtain the module of a complex number and \(\text{angle}()\) is the function used to obtain the angle of a complex number. Therefore, the amplitude and angle of each eccentric disc's unbalance can be calculated based on Equations (21) and (22). There should be \(m+n\) unbalance responses in both the \(x\) and \(y\) directions to obtain the matrix \(H_{2}\) and \(H_{4}\). If an eccentric disc's unbalance is to be identified, its unbalance responses in both the \(x\) and \(y\) directions are required according to Equation (18). Hence, there should be \(m+n+1\) measured unbalance responses in the two orthogonal directions.