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设发射信号为单色非偏振光,其琼斯矩阵为[10]:E =[Ex Ey]。起偏器、偏振分束镜反射、透射,半波片正向入射、反向入射的琼斯矩阵分别为[10]:
$ \left\{ \begin{array}{l} {\mathit{\boldsymbol{J}}_{\rm{p}}} = \left[ {\begin{array}{*{20}{c}} 0&0\\ 0&1 \end{array}} \right]\\ {\mathit{\boldsymbol{J}}_{{\rm{PBS, r}}}} = \left[ {\begin{array}{*{20}{c}} 0&0\\ 0&1 \end{array}} \right]\\ {\mathit{\boldsymbol{J}}_{{\rm{PBS, t}}}} = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&0 \end{array}} \right]\\ {\mathit{\boldsymbol{J}}_{{\rm{h + }}}} = {\rm{i}}\left[ {\begin{array}{*{20}{c}} {1/\sqrt 2 }&{1/\sqrt 2 }\\ {1/\sqrt 2 }&{ - 1/\sqrt 2 } \end{array}} \right]\\ {\mathit{\boldsymbol{J}}_{{\rm{h}} - }} = {\rm{i}}\left[ {\begin{array}{*{20}{c}} {1/\sqrt 2 }&{ - 1/\sqrt 2 }\\ { - 1/\sqrt 2 }&{ - 1/\sqrt 2 } \end{array}} \right] \end{array} \right. $
(1) 发射光经过起偏器后可以被偏振分束镜反射,通过光学天线出射,不被探测器接收。同时,接收光信号到达探测器的光矢量可以表示为:
$ {\mathit{\boldsymbol{E}}_{\rm{r}}} = {\mathit{\boldsymbol{J}}_{{\rm{PBS, t}}}}{\mathit{\boldsymbol{J}}_{{\rm{h - }}}}{\mathit{\boldsymbol{J}}_{{\rm{h + }}}}{\mathit{\boldsymbol{J}}_{{\rm{PBS, r}}}}{\mathit{\boldsymbol{J}}_{\rm{p}}}\mathit{\boldsymbol{E}} = \left[ {\begin{array}{*{20}{c}} {{E_{\rm{y}}}}&0 \end{array}} \right] $
(2) 结果表明,接收信号光可以完全透过偏振分光片到达信号探测器,实现收发隔离。
由于偏振分束镜、半波片、扩束镜等表面存在的反射光有可能进入探测器,利用琼斯矩阵分析半波片及透镜的前后表面反射光。其中反射面和透镜的琼斯矩阵为[10]:
$ \left\{ \begin{array}{l} {\mathit{\boldsymbol{J}}_{\rm{r}}} = \left[ {\begin{array}{*{20}{c}} 1&0\\ 0&{ - 1} \end{array}} \right]\\ {\rm{ }}{\mathit{\boldsymbol{J}}_{{\rm{lens}}}} = \left[ {\begin{array}{*{20}{c}} 1&0\\ { - 1/f}&1 \end{array}} \right] \end{array} \right. $
(3) 式中, f为扩束镜焦距。则偏振分束镜后表面及半波片前表面的反射光到达探测器的琼斯矩阵为:E1=JPBS, tJrJPBS, rJpE=[0 0]。半波片后表面及扩束镜前表面反射光到达探测器的琼斯矩阵为:E2=JPBS, tJh-·JrJh+JPBS, rJpE=[0 0]。扩束镜后表面反射光到达探测器的琼斯矩阵为:E3=JPBS, tJh-JlensJrJlensJh+JPBS, r·JpE=[0 0]。由此可知,在理想的偏振分束情况下,偏振片及透镜前后表面反射光均不会进入信号探测器。扩束镜组其它镜片的反射光到达探测器的琼斯矩阵与第1片扩束镜后表面反射光到达探测器的琼斯矩阵一致,也不会进入信号探测器。但是由于发射光束偏振度、分束镜偏振性能及其造成的影响,难以实现绝对收发隔离。
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假设在收发端调节半波片快轴与信号偏振方向的夹角误差为θ,则半波片正向入射、反向入射的琼斯矩阵分别为:
$ \left\{ \begin{array}{l} {J_{{\rm{h + }}}}\prime = {\rm{i}}\left[ {\begin{array}{*{20}{c}} {{\rm{cos}}\left( {\frac{{\rm{ \mathsf{ π} }}}{4} + 4\theta } \right)}&{{\rm{sin}}\left( {\frac{{\rm{ \mathsf{ π} }}}{4} + 4\theta } \right)}\\ {{\rm{sin}}\left( {\frac{{\rm{ \mathsf{ π} }}}{4} + 4\theta } \right)}&{ - {\rm{cos}}\left( {\frac{{\rm{ \mathsf{ π} }}}{4} + 4\theta } \right)} \end{array}} \right]\\ {J_{{\rm{h}} - }}\prime = {\rm{i}}\left[ {\begin{array}{*{20}{c}} {{\rm{cos}}\left( {\frac{{\rm{ \mathsf{ π} }}}{4} + 4\theta } \right)}&{ - {\rm{sin}}\left( {\frac{{\rm{ \mathsf{ π} }}}{4} + 4\theta } \right)}\\ { - {\rm{sin}}\left( {\frac{{\rm{ \mathsf{ π} }}}{4} + 4\theta } \right)}&{ - {\rm{cos}}\left( {\frac{{\rm{ \mathsf{ π} }}}{4} + 4\theta } \right)} \end{array}} \right] \end{array} \right. $
(4) 此时,信号探测器接收到的光矢量变为:Er′=JPBS, tJh-′Jh+′JPBS, rJpE=Ey[cos(4θ) 0]。若存在角度偏差时,偏振分束镜后表面及半波片前表面反射光到达探测器的琼斯矩阵为:E1′=JPBS, tJrJPBS, rJpE=[0 0]。半波片后表面及透镜前表面反射光到达探测器的琼斯矩阵可以表示为:E2′=JPBS, tJh-′JrJh+′·JPBS, rJpE=[0 0]。透镜后表面反射光到达探测器的琼斯矩阵可以表示为:E3′=JPBS, tJh-′JlensJrJlensJh+′·JPBS, rJpE=[0 0]。由此可知,半波片快轴与信号偏振方向夹角误差不会影响偏振片、透镜反射信号光到达信号探测器的光强,即收发隔离性能不会受到影响。
同时,存在调节误差与不存在调节误差时探测器接收光强不同,它们的比值为cos2(4θ)。由此可知,半波片快轴与信号偏振方向的夹角为22.5°时,探测器接收到的信号最强,当两者方向夹角偏离22.5°、并且偏离角为θ时,信号探测器接收到的光强按cos2(4θ)减小。因此,存在调节误差时会影响接收到的信号光光强,但是不会影响光学系统的偏振隔离度。
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为了进一步提高系统隔离度,对偏振分束器的前表面进行楔角处理,由此可以达到更高的收发隔离度[11-12]。收发隔离度定义为在接收光敏面上,发射光功率中的后向反射、散射及光路系统后向散射光多次散射等后向接收光功率Pb和传输功率Pt的比值[3, 13-15],即:R=10lg(Pt/Pb)。为了得到系统收发隔离度,建立模型对所设计系统进行分析。
模拟光路模型如图 2a所示,信号发射光源波长采用1550nm,发射信号总功率为1W。图 2中各个元件表示意义与图 1相同,并假设各个透镜、起偏器、半波片透射率为99.5%,反射率为0.5%,直径为20mm。偏振分束镜对p光反射率为99.9%,透射率为0.1%,对s光透射率为99.9%,发射率为0.1%,镜筒为完美吸收体,探测器直径为100μm。图 2a中小图所示为偏振分束镜前表面的楔角θ。图 2b为系统模拟时的光路图。由于接收信号可以实现自动跟踪,在改变偏振分束镜角度的过程中可以保证接收信号光斑始终位于探测器中心,回波信号则会因为偏振分束镜角度的增大逐渐偏离探测器中心,从而在保证接收信号功率不变的同时增加系统收发隔离度。
图 3所示为探测器上接收回波光强分布在不同参量,尤其在不同偏振分束镜楔楔角时的情况。图 3a中楔角为0mrad,接收光功率为1.8×10-5W,隔离度为47.3dB;图 3b中楔角为3.5mrad,接收光功率为3.01× 10-7W,隔离度为65.21dB;图 3c中楔角为3.98mrad,接收光功率为1.07× 10-9W,隔离度为89.70dB;图 3d中楔角为3.99mrad,接收光功率为5.06×10-13W,隔离度为122.96dB。本文中探测器的探测尺寸为0.1mm,因此它存在一定的视场范围,由于透镜前聚光镜的焦距为34.92mm,可得探测器全视场为:FFOV=arctan(0.1/34.92)=2.9mrad。随着楔角度的增加,探测器接收到的信号回波功率越来越小,偏振隔离度越来越大,在倾角大约为3.99mrad时超出探测器视场。可以看出, 在某一楔角度时,系统可以达到100dB的隔离度。对于不同的光学系统,该楔角与探测器光敏面尺寸及聚光镜的焦距有关。
在试验过程中, 由于遮光罩不可能完全遮挡所有外界杂散光,加上探测器自身噪声的影响、各种器件的制造公差影响,以及探测器的灵敏度影响等因素, 还不能达到如此高的隔离度,只有不断提升实验环境及器件来达到预期。
一种收发隔离光学系统的设计
Design of a transmitting-receiving isolation optical system
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摘要: 为了实现高隔离度的收发隔离,采用一种偏振收发隔离光学系统,利用琼斯矩阵进行了理论分析,验证了偏振隔离在收发同轴光学系统中的可行性。采取空间隔离来抑制回波进入接收系统,研究了模拟系统中偏振分束镜前表面楔角大小对收发隔离度的影响。结果表明,在达到一定楔角时,可取得100dB收发隔离度。这一结果对收发同轴光学系统的设计具有一定指导意义。Abstract: In order to achieve high-degree isolation in a transceiver system, an isolation optical system was put forward based on polarization and analyzed by means of Jones matrix. The feasibility of polarization isolation in a coaxial optical transceiver was verified. The space separation was used to suppress the echo into the receiving system. The influence of the wedge angle of polarization beam splitter on transmitting-receiving isolation was studied. The results show that, the transmitting-receiving isolation of 100dB is achieved with a certain wedge angle. The results will be of guiding significance for the design of transmitting-receiving coaxial optical systems.
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Key words:
- optical design /
- transmitting-receiving isolation /
- polarization beam splitter /
- isolation degree /
- echo
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