# Performance Analysis of Interferometer Algorithm under Phase Measurement Error
## Appendix A Derivation of Partial Derivative of Cost Function with Respect to Elevation Angle  
The result of differentiating the cost function with respect to the elevation angle and the results of linear approximation can be expressed as follows:  
\[ \begin{array}{l}\frac{\partial J(\theta,\varphi)}{\partial\varphi}\left| \begin{array}{l}\theta=\theta=\theta^{(0)}+\Delta\theta\\ \psi=\psi^{(0)}+\Delta\psi\end{array}\right.\\ \end{array}  \tag{18}\]  
\[ \begin{array}{l}=\sum\limits_{m=0}^{M-1}\left[\frac{2\pi\lambda}{\lambda} \cos(\theta^{(0)}+\Delta\theta-\frac{2\pi m}{M})\cos(\psi^{(0)}+\Delta\psi)- \hat{\phi}_{m}\right]\cos(\theta^{(0)}+\Delta\theta-\frac{2\pi m}{M})\sin( \psi^{(0)}+\Delta\psi)\\ =\sum\limits_{m=0}^{M-1}\left[\begin{array}{l}\frac{2\pi\lambda}{\lambda} \cos^{2}(a_{m}+\Delta\theta)\left(\cos(\psi^{(0)})\cos(\Delta\psi)-\sin(\psi^ {(0)})\sin(\Delta\psi)\right)\left(\sin(\psi^{(0)})\cos(\Delta\psi)+\cos(\psi ^{(0)})\sin(\Delta\psi)\right)\\ -\hat{\phi}_{m}\cos(a_{m}+\Delta\theta)\sin(\psi^{(0)}+\Delta\psi)\end{array} \right]=0.\end{array}  \tag{19}\]  
Using \(\cos(\Delta\theta)\approx 1\) and \(\sin(\Delta\psi)\approx\Delta\psi\), (19) is simplified to  
\[ \begin{array}{l}\frac{\partial J(\theta,\psi)}{\partial\psi}\left| \begin{array}{l}\theta=\hat{\theta}=\theta^{(0)}+\Delta\theta\\ \psi=\psi=\psi^{(0)}+\Delta\psi\end{array}\right.\end{array}  \tag{20}\]  
Based on the approximations of \((\Delta\psi)^{2}\approx 0\), (20) can be reduced to  
\[ \begin{array}{l}\frac{\partial J(\theta,\psi)}{\partial\psi}\left| \begin{array}{l}\theta=\theta=\theta^{(0)}+\Delta\theta\\ \psi=\psi=\psi^{(0)}+\Delta\psi\end{array}\right.\end{array}  \tag{21}\]Appendix B Proof of \(\sum\limits_{m=0}^{M-1}\cos^{2}(a_{m})=\sum\limits_{m=0}^{M-1}\sin^{2}(a_{m})= \frac{M}{2}\)  
If \(\sum\limits_{m=0}^{M-1}\exp\left(j\left(2\theta-\frac{4\pi m}{M}\right)\right)\) can be shown to be identically zero, \(\sum\limits_{m=0}^{M-1}\left(\cos\left(2\theta-\frac{4\pi m}{M}\right)+j\sin \left(2\theta-\frac{4\pi m}{M}\right)\right)\) is also equal to zero via the Euler formula. Based on the summation of a geometric series, \(\sum\limits_{m=0}^{M-1}\exp\left(j\left(2\theta-\frac{4\pi m}{M}\right)\right)\) is equal to zero:  
\[ \begin{array}{l}\sum\limits_{m=0}^{M-1}\exp\left(j\cdot\left(2\theta-\frac{4 \pi m}{M}\right)\right)=\exp(j\cdot 2\theta)\left(\exp\left(-j\cdot\frac{4\pi}{M} \right)+\exp\left(-j\cdot\frac{8\pi}{M}\right)+\cdots+\exp\left(-j\cdot\frac{4 \pi M}{M}\right)\right)\\ =\exp(j\cdot 2\theta)\left(\frac{1-\exp\left(-j\cdot\frac{4\pi}{M}\cdot M \right)}{1-\exp\left(-j\cdot\frac{4\pi}{M}\right)}\right)=0.\end{array}  \tag{10}\]  
The number of sensor elements, \(M\), can not be one since there should be at least two sensor elements. Since the common ratio of a geometric sequence, \(\exp\left[-j\frac{4\pi}{M}\right]\), should not be equal to one, \(M\) cannot be two. Therefore, \(M\) should be greater than two. Note that \(M\) denotes the number of sensors. Therefore, the derivation in this paper holds when the number of sensors is at least three. Note that the number of sensors in the interferometer algorithm is usually greater than two since adopting only two sensors results in performance degradation due to phase ambiguity and vulnerability to an additive noise. Apart from the phase ambiguity, there is another reason why there should be at least three sensors. Array with only two sensors can not simultaneously estimate the azimuth and the elevation since there exist infinitely many different directions, which result in the same phase difference between two antenna elements. Therefore, it is impossible to uniquely determine both the azimuth and the elevation from the phase difference between two antenna elements. This problem occurs regardless of the distance between two sensor elements. To obviate this problem, at least three antennas should be used.