## 2 Derivation of the Azimuth and the Elevation Error under Frequency Mismatch
### Partial Derivative of the Cost Function with Respect to the Azimuth  
The partial derivative of the cost function, \(J(\theta,\psi)\), in (2) with respect to the azimuth, \(\theta\), is given by:  
\[\left. \begin{array}{l}\frac{\partial J(\theta,\psi)}{\partial\theta}=\sum \limits_{m=0}^{M-1}\left[\begin{array}{c}\sin\left(\frac{2\pi r}{\lambda}\cos (\theta-\frac{2\pi m}{M})\cos(\psi)-\frac{2\pi r}{\lambda_{RF}}\cos(\theta^{(0 )}-\frac{2\pi m}{M})\cos(\psi^{(0)})\right)\\ \cdot\frac{2\pi r}{\lambda}\sin(\theta-\frac{2\pi m}{M})\cos(\psi)\end{array} \right]\end{array} \right] \tag{5}\]  
The first necessary condition for \(\theta=\theta^{(0)}+\Delta\theta,\psi=\psi^{(0)}+\Delta\psi\) to be a critical point of the cost function \(J(\theta,\psi)\) is that the partial derivative of the cost function, \(J(\theta,\psi)\), in (2) with respect to the azimuth, \(\theta\), should be identically zero at \(\theta=\theta^{(0)}+\Delta\theta,\psi=\psi^{(0)}+\Delta\psi\):  
\[\left.\begin{array}{l}\frac{\partial J(\theta,\psi)}{\partial\theta}\Big{|} \begin{array}{l}\theta=\theta^{(0)}+\Delta\theta\\ \psi=\psi^{(0)}+\Delta\psi\end{array} =\sum\limits_{m=0}^{M-1}\left[ \begin{array} []{c}\sin\left(\frac{2\pi r}{\lambda}\cos(\theta^{(0)}+\Delta\theta-\frac{2 \pi m}{M})\cos(\psi^{(0)}+\Delta\psi)-\frac{2\pi m}{M}\cos(\theta^{(0)}-\frac {2\pi m}{M})\cos(\psi^{(0)})\right)\\ \cdot\frac{2\pi r}{\lambda}\sin(\theta^{(0)}+\Delta\theta-\frac{2\pi m}{M}) \cos(\psi^{(0)}+\Delta\psi)\end{array} \right]=0. \tag{6}\]  
Since \(\sin(\phi)\) is approximately equal to \(\phi\) for small \(\phi\), (6) can be written as:  
\[\left.\begin{array}{l}\frac{\partial J(\theta,\psi)}{\partial\psi}\Big{|} \begin{array}{l}\theta=\theta^{(0)}+\Delta\theta\\ \psi=\psi^{(0)}+\Delta\psi\end{array} =\sum\limits_{m=0}^{M-1}\left[ \begin{array} []{c}\left(\frac{2\pi r}{\lambda}\cos(\theta^{(0)}+\Delta\theta-\frac{2\pi m }{M})\cos(\psi^{(0)}+\Delta\psi)-\frac{2\pi m}{M}\cos(\psi^{(0)}-\frac{2\pi m }{M})\cos(\psi^{(0)})\right)\\ \cdot\frac{2\pi r}{\lambda}\sin(\theta^{(0)}+\Delta\theta-\frac{2\pi m}{M}) \cos(\psi^{(0)}+\Delta\psi)\end{array} \right]=0. \tag{7}\]  
Let \(a_{m}\) be defined as \(\theta^{(0)}-\frac{2\pi m}{M}\). Dividing both sides of (7) by \(\frac{(2\pi r)^{2}}{\lambda}\) yields:  
\[\sum\limits_{m=0}^{M-1}\left[\left( \begin{array}{c}\frac{1}{\lambda}\cos(a_ {m}+\Delta\theta)\cos(\psi^{(0)}+\Delta\psi)\left(\sin(a_{m}+\Delta\theta) \cos(\psi^{(0)}+\Delta\psi)\right)\\ -\frac{1}{\lambda_{RF}}\cos(a_{m})\cos(\psi^{(0)})\cdot\left(\sin(a_{m}+\Delta \theta)\cos(\psi^{(0)}+\Delta\psi)\right)\end{array} \right)\right]=0. \tag{8}\]  
From the double angle formula for the sine function, (8) can be written as:  
\[\sum\limits_{m=0}^{M-1}\left[\left( \begin{array}{c}\frac{1}{\lambda}\left( \frac{1}{2}\sin\left(2a_{m}+2\Delta\theta\right)\cos^{2}(\psi^{(0)}+\Delta\psi )\right)\\ -\frac{1}{\lambda_{RF}}\cos(a_{m})\cos(\psi^{(0)})\cdot\left(\sin(a_{m}+\Delta \theta)\cos(\psi^{(0)}+\Delta\psi)\right)\end{array} \right)\right]=0. \tag{9}\]  
Applying the angle sum identity for the sine function results in:\[\sum\limits_{m=0}^{M-1}\left[\left( \begin{array}{c}\frac{1}{\lambda}\left( \frac{1}{2}\left(\sin\left(2a_{m}\right)\cos\left(2\Delta\theta\right)+\cos \left(2a_{m}\right)\sin\left(2\Delta\theta\right)\right)\cos^{2}(\psi^{(0)}+ \Delta\psi)\right)\\ -\frac{1}{\lambda_{RF}}\cos(a_{m})\cos(\psi^{(0)})\cdot\left(\sin(a_{m}+ \Delta\theta)\cos(\psi^{(0)}+\Delta\psi)\right)\end{array} \right)\right]=0.