2 Analysis

One might expect that with one equation and two unknowns there are two possible solutions for the phases. However, previous work [2,3,4,5,6,7,8] has not attempted to explicitly determine these solutions, relying instead on tracking loops or other estimators. In this section, the phase solutions are derived analytically. From Eq. (1),

$$\Vert r\Vert^2 = \Vert Ae^{j\theta} + Be^{j\phi}\Vert^2 = A^2 + B^2 + 2AB\cos(\phi-\theta),$$

and thus

$$\cos(\phi-\theta) = {\Vert r\Vert^2 - A^2 - B^2 \over 2AB}.$$ (2)

From $r = e^{j\theta} (A + Be^{j(\phi-\theta)})$, it follows that

$$e^{j\theta} = {r \over A + Be^{j(\phi-\theta)}} = {r (A + Be^{-j(\phi-\theta)}) \over A^2 + B^2 + 2AB\cos(\phi-\theta)}$$

$$= {r (A + Be^{-j(\phi-\theta)}) \over \Vert r\Vert^2}.$$

Hence, $\theta = \arg[r (A + Be^{-j(\phi-\theta)})] = \arg[r (A + B\cos(\phi-\theta) - jB\sin(\phi-\theta))].$ By the symmetry of Eq. (1), it also immediately follows that $\phi = \arg[r (B + A\cos(\phi-\theta) + jA\sin(\phi-\theta))].$ By letting $D \triangleq (\Vert r\Vert^2 - A^2 - B^2)/(2AB) = \cos(\phi-\theta)$, the solutions are given by

$$\theta = \arg\left[r(A+BD + j sB\sqrt{1-D^2})\right]$$ (3)

$$\phi = \arg\left[r(B+AD - jsA\sqrt{1-D^2})\right],$$

where $s\in \{-1,+1\}$. The phases have now been determined exactly as a function of $A$, $B$, and $r$, to within one of two possibilities.