# Performance Analysis of Interferometer Algorithm under Phase Measurement Error
## 5 Equivalent Derivation of the MSE of Azimuth and the MSE of Elevation  
In this section, another way to obtain (18) and (19) is described. The theoretical phase value on the \(m\)-th sensor for an incident signal from azimuth, \(\theta\), and elevation, \(\psi\), can be written as  
\[\phi_{m}=\frac{2\pi r}{\lambda}\cos\left(\theta-\frac{2\pi m}{M}\right)\cos(\psi) \tag{20}\]  
Explicit expressions of \(\frac{\partial\phi_{m}}{\partial\theta}\) and \(\frac{\partial\phi_{m}}{\partial\psi}\) can be obtained from partial differentiations of (20) with respect to \(\theta\) and \(\psi\) :  
\[\frac{\partial\phi_{m}}{\partial\theta}=-\frac{2\pi r}{\lambda}\sin\left( \theta-\frac{2\pi m}{M}\right)\cos(\psi) \tag{21}\]  
\[\frac{\partial\phi_{m}}{\partial\psi}=-\frac{2\pi r}{\lambda}\cos\left( \theta-\frac{2\pi m}{M}\right)\sin(\psi) \tag{22}\]  
Perturbation in \(\phi_{m}\) due to small change in \(\theta\) and \(\psi\) can be formulated as  
\[\Delta\phi_{m}=\frac{\partial\phi_{m}}{\partial\theta}\Delta\theta+\frac{ \partial\phi_{m}}{\partial\psi}\Delta\psi \tag{23}\]  
Note that (23) is equivalent to the first-order Taylor series expansion. (23) can be written in matrix form as follows :\[\left[ \begin{array}{c}\Delta\phi_{1}\\ \Delta\phi_{2}\\ \vdots\\ \Delta\phi_{M}\end{array} \right]=\left[ \begin{array}{cc}\frac{\partial\phi_{1} }{\partial\theta}&\frac{\partial\phi_{1}}{\partial\phi}\\ \frac{\partial\phi_{2}}{\partial\theta}&\frac{\partial\phi_{2}}{\partial\phi} \\ \vdots&\vdots\end{array} \right]\left[ \begin{array}{c}\Delta\theta\\ \Delta\psi\end{array} \right] \tag{24}\]  
Let \(\mathbf{A}\) and \(\mathbf{b}\) defined as  
\[\mathbf{A}\equiv\left[ \begin{array}{cc}\frac{\partial\phi_{0}}{\partial \theta}&\frac{\partial\phi_{0}}{\partial\phi}\\ \frac{\partial\phi_{1}}{\partial\theta}&\frac{\partial\phi_{1}}{\partial\phi} \\ \vdots&\vdots\\ \frac{\partial\phi_{M-1}}{\partial\theta}&\frac{\partial\phi_{M-1}}{\partial \phi}\end{array} \right] \tag{25}\]  
\[\mathbf{b}=\left[ \begin{array}{c}\Delta\phi_{0}\\ \Delta\phi_{1}\\ \vdots\\ \Delta\phi_{M-1}\end{array} \right] \tag{26}\]  
By substituting (21) and (22) in (25), \(\mathbf{A}\) can be explicitly written as  
\[\mathbf{A}\equiv-\frac{2\pi r}{\lambda}\left[ \begin{array}{cc}\sin(\theta) \cos(\psi)&\cos(\theta)\sin(\psi)\\ \sin(\theta-\frac{2\pi}{M})\cos(\psi)&\cos(\theta-\frac{2\pi}{M})\sin(\psi) \\ \vdots&\vdots\\ \sin(\theta-\frac{2\pi(M-1)}{M})\cos(\psi)&\cos(\theta-\frac{2\pi(M-1)}{M}) \sin(\psi)\end{array} \right]. \tag{27}\]  
In terms of \(\mathbf{A}\) and \(\mathbf{b}\), (24) can be expressed as  
\[\mathbf{b}=\mathbf{A}\left[ \begin{array}{c}\Delta\theta\\ \Delta\psi\end{array} \right]. \tag{28}\]  
\(\mathbf{A}\) defined in (25) is not a square matrix. Therefore, it is not invertible. Since (28) is an overdetermined system, the least squares approach can be adopted to get \(\left[ \begin{array}{cc}\Delta\theta&\Delta\psi\end{array} \right]^{T}\) from (28):  
\[\left[ \begin{array}{c}\Delta\theta\\ \Delta\psi\end{array} \right]=\left(\mathbf{A}^{T}\mathbf{A}\right)^{-1}\left( \mathbf{A}^{T}\mathbf{b}\right). \tag{29}\]  
By tranposing (29), \(\left[ \begin{array}{cc}\Delta\theta&\Delta\psi\end{array} \right]\) can be written as  
\[\left[ \begin{array}{cc}\Delta\theta&\Delta\psi\end{array} \right]=\mathbf{b} ^{T}\mathbf{A}\left(\mathbf{A}\mathbf{A}^{T}\right)^{-1}. \tag{30}\]  
From (29) and (30), we have  
\[\left[ \begin{array}{c}\Delta\theta\\ \Delta\psi\end{array} \right][\Delta\theta\Delta\psi]=\left[ \begin{array}{cc }\Delta\theta^{2}&\Delta\theta\Delta\psi\\ \Delta\psi\Delta\theta&\Delta\psi^{2}\end{array} \right]=\left(\mathbf{A}^{T} \mathbf{A}\right)^{-1}\mathbf{A}^{T}\mathbf{b}\mathbf{b}^{T}\mathbf{A}\left( \mathbf{A}\mathbf{A}^{T}\right)^{-1}. \tag{31}\]  
An additive noise on each sensor location is spatially white. From uncorrelatedness of \(\Delta\phi_{m}\), we have  
\[E[\Delta\phi_{m}\ \Delta\phi_{n}]=\left\{ \begin{array}{cc}E\left[\Delta\phi_{m}^{2} \right]=\sigma_{\phi}^{2}&m=n\\ E[\Delta\phi_{m}]E[\Delta\phi_{n}]=0&m\
eq n\end{array}