## 5 Conclusions
Appendix A Phase Difference between the Phase at the \(m^{\text{th}}\) Sensor Location and the Phase at the Origin for an Incident Wave from \(\theta=\theta^{(0)}\) and \(\psi=\psi^{(0)}\) with a Wavelength of \(\lambda_{RF}\)  
In this Appendix, the sensor location vector for a uniform circular array of a radius \(r\) with \(M\) sensors is given by:  
\[\mathbf{z}_{m}=\left[r\cos\left(\frac{2\pi m}{M}\right),\ r\sin\left(\frac{2 \pi m}{M}\right),\ \ 0\right]^{T}. \tag{1}\]  
The direction vector for an incident signal from \((\theta^{(0)},\psi^{(0)})\) can be written as:  
\[\kappa=[\cos(\psi^{(0)})\cos(\theta^{(0)}),\ \ \cos(\psi^{(0)})\sin(\theta^{(0)} ),\ \ \sin(\psi^{(0)})]^{T}. \tag{2}\]  
The time delay between the \(m^{\text{th}}\) sensor location and the origin can be written as:  
\[\tau(m)=\kappa^{T}\mathbf{z}_{m}/c \tag{3}\]  
The phase difference between the \(m^{\text{th}}\) sensor location and the origin is given by the product of the angular frequency and time delay:  
\[\phi_{m}=\omega\times\text{Timedelay}=\omega\frac{\kappa^{T}\mathbf{z}_{m}}{ c}=2\pi\frac{\kappa^{T}\mathbf{z}_{m}}{\lambda} \tag{4}\]  
By substituting (1) and (2) in (4), we have:  
\[ \begin{array}{c}\phi_{m}=\omega\times\text{Timedelay}=\omega\frac{\kappa^{T }\mathbf{z}_{m}}{c}=2\pi\frac{\kappa^{T}\mathbf{z}_{m}}{\lambda}\\ \\ =2\pi\frac{\tau}{\lambda}\left[\cos\left(\frac{2\pi m}{M}\right)\cos\left( \theta^{(0)}\right)\cos(\psi^{(0)})+\sin\left(\frac{2\pi m}{M}\right)\sin\left( \theta^{(0)}\right)\cos(\psi^{(0)})\right]\\ \\ =2\pi\frac{\tau}{\lambda}\left[\cos\left(\theta^{(0)}-\frac{2\pi m}{M}\right) \cos(\psi^{(0)})\right]\end{array}  \tag{5}\]  
**Appendix B. Proof of \(\sum\limits_{m=0}^{M-1}\cos^{2}(a_{m})=\sum\limits_{m=0}^{M-1}\sin^{2}(a_{m})= \frac{M}{2}\), \(\sum\limits_{m=0}^{M-1}\sin(a_{m})\cos(a_{m})=0\)**  
If \(\sum\limits_{m=0}^{M-1}\exp\left(j\left(2\theta^{(0)}-\frac{4\pi mt}{M}\right)\right)\) can be shown to be identically zero, \(\sum\limits_{m=0}^{M-1}\left(\cos\left(2\theta^{(0)}-\frac{4\pi mt}{M}\right) +j\sin\left(2\theta^{(0)}-\frac{4\pi mt}{M}\right)\right)\) is also equal to zero via the Euler formula. Based on the summation of a geometric series, \(\sum\limits_{m=0}^{M-1}\exp\left(j\left(2\theta^{(0)}-\frac{4\pi mt}{M}\right)\right)\) is equal to zero:  
\[ \begin{array}{l}\sum\limits_{m=0}^{M-1}\exp\left(j\cdot\left(2\theta^{(0)}- \frac{4\pi mt}{M}\right)\right)=\exp\left(j\cdot 2\theta^{(0)}\right)\left(1+ \exp\left(-j\cdot\frac{4\pi}{M}\right)+\exp\left(-j\cdot\frac{8\pi}{M}\right) +\cdots+\exp\left(-j\cdot\frac{4\pi(M-1)}{M}\right)\right)\\ =\exp\left(j\cdot 2\theta^{(0)}\right)\left(\frac{1-\exp\left(-j\cdot\frac{4\pi }{M}M\right)}{1-\exp\left(-j\cdot\frac{4\pi}{M}\right)}\right)=0.\end{array} \] (A6)  
From (A6), since \(\sum\limits_{m=0}^{M-1}\exp j(2a_{m})\) is identically zero, we have:  
\[\sum\limits_{m=0}^{M-1}\left(\cos\left(2a_{m}\right)+j\sin\left(2a_{m}\right) \right)=0\] (A7)  
\[\sum\limits_{m=0}^{M-1}\cos\left(2a_{m}\right)+j\sum\limits_{m=0}^{M-1}\sin \left(2a_{m}\right)=0\] (A8)  
Note that \(\cos\left(2a_{m}\right)\) and \(\sin\left(2a_{m}\right)\) are real-valued for all \(m\). Since the real number is closed under addition, \(\sum\limits_{m=0}^{M-1}\cos\left(2a_{m}\right)\) is also real. Similarly, \(\sum\limits_{m=0}^{M-1}\sin\left(2a_{m}\right)\) is also real. From (A8), both \(\sum\limits_{m=0}^{M-1}\cos\left(2a_{m}\right)\) and \(\sum\limits_{m=0}^{M-1}\sin\left(2a_{m}\right)\) should be equal to zero since \(\sum\limits_{m=0}^{M-1}\cos\left(2a_{m}\right)\) and \(\sum\limits_{m=0}^{M-1}\sin\left(2a_{m}\right)\) are real-valued:  
\[ \begin{array}{l}\sum\limits_{m=0}^{M-1}\cos\left(2a_{m}\right)=0\\ \sum\limits_{m=0}^{M-1}\sin\left(2a_{m}\right)=0.\end{array}