## 2 Derivation of the Azimuth and the Elevation Error under Frequency Mismatch
### Partial Derivative of the Cost Function with Respect to Elevation
\tag{22}\]  
By Appendix B, due to \(\sum\limits_{m=0}^{M-1}\cos\left(2a_{m}\right)=\sum\limits_{m=0}^{M-1}\sin \left(2a_{m}\right)=0\) and \(\sum\limits_{m=0}^{M-1}\sin\left(a_{m}\right)\cos\left(a_{m}\right)=0\), (22) is simplified to:  
\[\sin(\psi^{(0)}+\Delta\psi)\frac{1}{\lambda}\frac{M}{2}\cos(\psi^{(0)}+\Delta \psi)-\sin(\psi^{(0)}+\Delta\psi)\frac{1}{\lambda_{RF}}\cos(\psi^{(0)})\cos \left(\Delta\theta\right)\sum\limits_{m=0}^{M-1}\cos^{2}(a_{m})=0 \tag{23}\]  
Substituting (A8) in (23) yields:  
\[\sin(\psi^{(0)}+\Delta\psi)\frac{1}{\lambda}\frac{M}{2}\cos(\psi^{(0)}+\Delta \psi)-\sin(\psi^{(0)}+\Delta\psi)\frac{1}{\lambda_{RF}}\frac{M}{2}\cos(\psi^{ (0)})\cos\left(\Delta\theta\right)=0. \tag{24}\]  
Since \(\frac{M}{2}\) is not zero, by dividing both sides of (24) by \(\frac{M}{2}\), we have:  
\[\sin(\psi^{(0)}+\Delta\psi)\left(\frac{1}{\lambda}\cos(\psi^{(0)}+\Delta\psi)- \frac{1}{\lambda_{RF}}\cos(\psi^{(0)})\cos\left(\Delta\theta\right)\right)=0. \tag{25}\]  
2.1 First Condition for the Partial Derivative with Respect to the Azimuth at \(\theta=\theta^{(0)}+\Delta\theta,\psi=\psi^{(0)}+\Delta\psi\) to be Zero: \(\sin(\Delta\theta)=0\)  
Note that, for (15) to be true, \(\Delta\theta\) should be integer multiples of \(\pi\). \(\Delta\theta\) is an azimuth estimation error. It cannot be nonzero integer multiples of \(\pi\) since the absolute value of the azimuth error should be much smaller than \(\pi=180^{\circ}\). Therefore, from (15), we have:  
\[\Delta\theta=0. \tag{26}\]  
Note that \(\Delta\theta\) in (26) is independent of \(\theta^{(0)},\psi^{(0)},\lambda\), and \(\lambda_{RF}\), where \(\theta^{(0)}\) and \(\psi^{(0)}\) denote the azimuth and the elevation of an incident signal, respectively, and \(\lambda_{RF}\) and \(\lambda\) denote the wavelength of the incident signal and the nominal wavelength for the calculation of an array manifold. Therefore, the azimuth estimation error, \(\Delta\theta\), is zero irrespective of in which direction the signal is incident and how different the frequency of the incident signal is from the nominal frequency used for the calculation of an array manifold. From (26), the azimuth error is approximately equal to zero even when the array manifold frequency is different from the carrier frequency of the signal source in the absence of many uncertainties except the small frequency offset between the nominal frequency in the array manifold and the frequency of the incident signal. By substituting (26) in (25), we have:  
\[\sin(\psi^{(0)}+\Delta\psi)\left(\frac{1}{\lambda}\cos(\psi^{(0)}+\Delta\psi)- \frac{1}{\lambda_{RF}}\cos(\psi^{(0)})\right)=0. \tag{27}\]  
Applying angle addition and the subtraction theorem to (27) yields:  
\[\left(\sin\left(\psi^{(0)}\right)\cos\left(\Delta\psi\right)+\cos\left(\psi^{( 0)}\right)\sin\left(\Delta\psi\right)\right)\left(\frac{1}{4}\left(\cos\left( \psi^{(0)}\right)\cos\left(\Delta\psi\right)-\sin\left(\psi^{(0)}\right)\sin \left(\Delta\psi\right)\right)-\frac{1}{\lambda_{RF}}\cos(\psi^{(0)})\right)=0. \tag{28}\]  
Since (28) is not a polynomial equation in \(\Delta\psi\), the Taylor approximation is used to make (28) a polynomial equation in \(\Delta\psi\). First, applying the approximations of \(\cos(\Delta\psi)\approx 1,\sin(\Delta\psi)\approx\Delta\psi\) in (28) and a little manipulation yield:  
\[\left(\sin\left(\psi^{(0)}\right)+\cos\left(\psi^{(0)}\right)\Delta\psi \right)\left(\frac{1}{\lambda}\left(\cos\left(\psi^{(0)}\right)-\sin\left( \psi^{(0)}\right)\Delta\psi\right)-\frac{1}{\lambda_{RF}}\cos(\psi^{(0)}) \right)=0. \tag{29}\]  
With the distributive law, (29) can be rewritten as:  
\[ \begin{array}{c}\left(\Delta\psi\right)^{2}\left(\frac{1}{\lambda}\sin \left(\psi^{(0)}\right)\cos\left(\psi^{(0)}\right)\right)\\ +\Delta\psi\left(\frac{1}{\lambda}\cos^{2}\left(\psi^{(0)}\right)-\frac{1}{ \lambda}\sin^{2}\left(\psi^{(0)}\right)-\frac{1}{\lambda_{RF}}\cos^{2}\left( \psi^{(0)}\right)\right)\\ +\left(\frac{1}{\lambda}-\frac{1}{\lambda_{RF}}\right)\sin\left(\psi^{(0)} \right)\cos\left(\psi^{(0)}\right)=0.\end{array}