# Performance Assessment of the Medium Frequency R-Mode Baltic Testbed at Sea near Rostock
## Appendix A  
The section shows the derivation of the formulas used in the paper. As is well-known, the sum of two sinusoidal signals with the circular frequency \(\omega=2\pi f\) is also a sinusoidal signal with the same frequency. By using the trigonometric addition formulas, we define the two sinusoids, as follows  
\[s_{1}(t)=A_{1}\cos(\omega t+\theta_{1})=A_{1}[\cos(\omega t)\cos(\theta_{1})- \sin(\omega t)\sin(\theta_{1})] \tag{1}\]  
and  
\[s_{2}(t)=A_{2}\cos(\omega t+\theta_{2})=A_{2}[\cos(\omega t)\cos(\theta_{2})- \sin(\omega t)\sin(\theta_{2})]. \tag{2}\]  
Then the sum is given as  
\[s_{3}(t)=s_{1}+s_{2}=\cos(\omega t)[A_{1}\cos(\theta_{1})+A_{2}\cos(\theta_{2} )]-\sin(\omega t)[A_{1}\sin(\theta_{1})+A_{2}\sin(\theta_{2})]. \tag{3}\]  
Our goal is to find amplitude \(A_{3}\) and phase \(\theta_{3}\) of \(s_{3}\). We define, as follows  
\[A_{3}\cos(\theta_{3})=A_{1}\cos(\theta_{1})+A_{2}\cos(\theta_{2}) \tag{4}\]  
and  
\[A_{3}\sin(\theta_{3})=A_{1}\sin(\theta_{1})+A_{2}\sin(\theta_{2}). \tag{5}\]Substituting Equations (A4) and (A5) in Equation (A3) and by knowing the trigonometric addition formula, we obtain  
\[s_{3}=A_{3}[\cos(\omega t)\cos(\theta_{3})-\sin(\omega t)\sin(\theta_{3})]=A_{3} \cos(\omega t+\theta_{3}).\] (A6)  
To derive the formula which shows the dependency of \(A_{3}\) on amplitude and phase of \(s_{1}\) and \(s_{2}\), we square and sum Equations (A4) and (A5). After some passages, we obtain  
\[A_{3}=\sqrt{A_{1}^{2}+A_{2}^{2}+2A_{1}A_{2}\cos(\theta_{1}-\theta_{2})}.\] (A7)  
To find \(\theta_{3}\), we divide Equation (A5) by (A4), obtaining  
\[\tan(\theta_{3})=\frac{A_{1}\sin(\theta_{1})+A_{2}\sin(\theta_{2})}{A_{1}\cos (\theta_{1})+A_{2}\cos(\theta_{2})}.\] (A8)  
By inverting Equation (A8), we can write \(\theta_{3}\), as follows  
\[\theta_{3}=\tan^{-1}\biggl{(}\frac{A_{1}\sin(\theta_{1})+A_{2}\sin(\theta_{2} )}{A_{1}\cos(\theta_{1})+A_{2}\cos(\theta_{2})}\biggr{)}.\] (A9)  
We are now interested in the maximum phase error caused by the signal \(s_{2}\) when it is summed to \(s_{1}\). For simplicity, we assume \(\theta_{1}=0\). Therefore, the phase error is directly represented by \(\theta_{3}\). To find the maximum, we compute the first derivative of Equation (A9) and find the zeros. We first define the ratio between \(A_{1}\) and \(A_{2}\) as \(\rho\) and then, remembering that \(\theta_{1}=0\), we write Equation (A9), as follows  
\[\theta_{3}=\tan^{-1}\biggl{(}\frac{\sin(\theta_{2})}{\rho+\cos(\theta_{2})} \biggr{)}.\] (A10)  
We also assume \(\rho>1\), which means \(A_{1}>A_{2}\). Then, we compute the derivative with respect to \(\theta_{2}\)  
\[ \begin{split}\frac{d\theta_{3}}{d\theta_{2}}&=\frac {1}{1+\Bigl{(}\frac{\sin\theta_{2}}{\rho+\cos\theta_{2}}\Bigr{)}^{2}}\biggl{[} \frac{\cos\theta_{2}(\rho+\cos\theta_{2}+\sin^{2}\theta_{2})}{(\rho+\cos\theta _{2})^{2}}\biggr{]}\\ &=\frac{(\rho+\cos\theta_{2})^{2}}{(\rho+\cos\theta_{2})^{2}+ \sin^{2}\theta_{2}}\frac{\cos\theta_{2}(\rho+\cos\theta_{2}+\sin^{2}\theta_{2 })}{(\rho+\cos\theta_{2})^{2}}\\ &=\frac{\rho\cos\theta_{2}+\cos^{2}\theta_{2}+\sin^{2}\theta_{2}} {\rho^{2}+2\rho\cos\theta_{2}+\cos^{2}\theta_{2}+\sin^{2}\theta_{2}}\\ &=\frac{1+\rho\cos\theta_{2}}{\rho^{2}+2\rho\cos\theta_{2}+1} \end{split} \] (A11)  
from which we find  
\[\cos\theta_{2}=-1/\rho\longrightarrow\theta_{2}=\arccos(-1/\rho).\] (A12)  
Substituting Equation (A12) in (A10), we obtain  
\[\theta_{3}=\tan^{-1}\biggl{(}\frac{\sin[\arccos(-1/\rho)]}{\rho+\cos[\arccos(- 1/\rho)]}\biggr{)}=\tan^{-1}\left(\frac{\sqrt{1-\frac{1}{\rho^{2}}}}{\rho- \frac{1}{\rho}}\right)=\tan^{-1}\biggl{(}\frac{1}{\sqrt{\rho^{2}-1}}\biggr{)}.\] (A13)  
By observing Equation (A13), it can be noted that the maximum error depends purely on the ratio \(\rho\). This formula cannot be applied if \(0<\rho\leq 1\). In such a case, we consider the four-quadrant arctangent instead of the normal arctangent operator.