# Experimental Study on Partially Coherent Optical Coherent Detection
## Theoretical Analysis
### Polarization States and Mixing Efficiency
\[E_{0}=-\frac{1}{2}ik_{1}E_{s}\sin(2\theta)e^{i(\varphi+\varphi(t)+\tau)}+\frac {1}{2}k_{2}E_{s}(1+i\cos(2\theta))e^{i(\varphi+\varphi(t)+\tau)}+\frac{\sqrt{2} }{2}k_{3}E_{Lo}e^{i(\psi+2\tau+\frac{\pi}{4})} \tag{40}\]  
\[E_{90}=\frac{1}{2}k_{1}E_{s}(1-i\cos(2\theta))e^{i(\varphi(t)+2\tau)}-\frac{1} {2}ik_{2}E_{s}\sin(2\theta)e^{i(\varphi(t)+2\tau)}+\frac{\sqrt{2}}{2}k_{4}E_{ Lo}e^{i(\psi+2\tau-\frac{\pi}{4})} \tag{41}\]  
\[E_{180}=-\frac{1}{2}ik_{1}E_{s}\sin(2\theta)e^{i(2\varphi+\varphi(t))}-\frac{ 1}{2}k_{2}E_{s}(1+i\cos(2\theta))e^{i(\varphi+\varphi(t)+\tau)}+\frac{\sqrt{2} }{2}k_{3}E_{Lo}e^{i(\psi+\rho+\tau+\frac{\pi}{4})} \tag{42}\]  
\[E_{270}=\frac{1}{2}k_{1}E_{s}(1-i\cos(2\theta))e^{i(\varphi(t)+\tau+\rho)}- \frac{1}{2}ik_{2}E_{s}\sin(2\theta)e^{i(\varphi+\varphi(t)+\tau)}-\frac{\sqrt{ 2}}{2}k_{4}E_{Lo}e^{i(\psi+2\rho-\frac{\pi}{4})} \tag{43}\]  
In path \(I\) and path \(Q\), the phase information is replaced by parameters \(x\) and \(y\), respectively; where \(x=\varphi(t)-\psi+\rho-\tau-\frac{\pi}{4}\) and \(y=\varphi(t)-\psi-\rho+\tau+\frac{\pi}{4}\), and the photocurrent of the four signal light and local oscillator light output beams following coherent mixing can be calculated using the Euler formula, as follows:  
\[ \begin{array}{ll}I_{0}&=\frac{1}{4}\Big{\{}[(k_{1}\sin(2\theta)-k_{2}\cos( 2\theta))^{2}+k_{2}^{2}]|E_{s}|^{2}+2k_{3}|E_{Lo}|^{2}\\ &+2\sqrt{2}[k_{1}k_{3}\sin(2\theta)\sin x+k_{2}k_{3}\cos x+k_{2}k_{3}\cos(2 \theta)\sin(-x)]|E_{s}E_{Lo}|\Big{\}}\end{array}  \tag{44}\]  
\[ \begin{array}{ll}I_{90}&=\frac{1}{4}\Big{\{}[(k_{1}\cos(2\theta)+k_{2}\sin( 2\theta))^{2}+k_{1}^{2}]|E_{s}|^{2}+2k_{4}|E_{Lo}|^{2}\\ &+2\sqrt{2}[k_{2}k_{4}\sin(2\theta)\sin y+k_{1}k_{4}\cos y+k_{1}k_{4}\cos(2 \theta)\sin y]|E_{s}E_{Lo}|\Big{\}}\end{array}  \tag{45}\]  
\[ \begin{array}{ll}I_{180}&=\frac{1}{4}\Big{\{}[(k_{1}\sin(2\theta)-k_{2}\cos (2\theta))^{2}+k_{2}^{2}]|E_{s}|^{2}+2k_{3}|E_{Lo}|^{2}\\ &+2\sqrt{2}[k_{2}k_{3}\cos(2\theta)\sin x-k_{2}k_{3}\cos x+k_{1}k_{3}\sin(2 \theta)\sin(-x)]|E_{s}E_{Lo}|\Big{\}}\end{array}  \tag{46}\]  
\[ \begin{array}{ll}I_{270}&=\frac{1}{4}\Big{\{}[(k_{1}\cos(2\theta)+k_{2}\sin (2\theta))^{2}+k_{1}^{2}]|E_{s}|^{2}+2k_{4}|E_{Lo}|^{2}\\ &+2\sqrt{2}[k_{1}k_{4}\cos(2\theta)\sin(-y)+k_{2}k_{4}\sin(2\theta)\sin(-y)-k _{1}k_{4}\cos y]|E_{s}E_{Lo}|\Big{\}}\end{array}  \tag{47}\]  
Subtracting the photocurrent of the 0\({}^{\circ}\) and 180\({}^{\circ}\) branches obtains the DC signal \(I_{I}^{\prime}\). The photocurrents of the 90\({}^{\circ}\) and 270\({}^{\circ}\) branches can be subtracted to obtain an orthogonal signal \(I_{Q}{}^{\prime}\). Then, \(I_{I}^{\prime}\) and \(I_{Q}{}^{\prime}\) can be represented as  
\[I_{I}^{\prime}=I_{0}-I_{180}=\sqrt{2}[k_{2}k_{3}\cos x+k_{3}\sin x(k_{1}\sin 2 \theta-k_{2}\cos 2\theta)]|E_{s}E_{LO}| \tag{48}\]  
\[I_{Q}^{\prime}=I_{90}-I_{270}=\sqrt{2}[k_{1}k_{4}\cos y+k_{4}\sin y(k_{1}\sin 2 \theta+k_{2}\cos 2\theta)]|E_{s}E_{LO}| \tag{49}\]  
When ignoring external limiting factors such as channel conditions, encoding methods, and the randomness of atmospheric transmission, and only considering the impact of the polarization state changes of the signal light on mixing efficiency, assuming that the linearly polarized EGSM light enters the mixer in an ideal state, with a polarization vector angle of 45\({}^{\circ}\), the linearly polarized EGSM light is uniformly distributed in both horizontal and vertical directions. Therefore, substituting \(k_{1}=k_{2}=k_{3}=k_{4}=\frac{\sqrt{2}}{2}\) yields  
\[ \begin{array}{ll}I_{I}^{\prime}&=I_{0}-I_{180}=\sqrt{2}[\frac{1}{2}\cos( \varphi(t)-\psi)\\ &+\frac{\sqrt{2}}{2}\sin(\varphi(t)-\psi)\Big{(}\frac{\sqrt{2}}{2}\sin(2 \theta)-\frac{\sqrt{2}}{2}\cos(2\theta)\Big{)}]|E_{s}E_{LO}|\end{array}  \tag{50}\]\[ \begin{array}{ll}I^{\prime}_{Q}&=I_{90}-I_{270}=\sqrt{2}[\frac{1}{2}\cos(\varphi(t) -\psi)\\ &+\frac{\sqrt{2}}{2}\sin(\varphi(t)-\psi)\Big{(}\frac{\sqrt{2}}{2}\cos(2\theta)+ \frac{\sqrt{2}}{2}\sin(2\theta)\Big{)}||E_{s}E_{LO}|\end{array}