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采用旋转双圆楔形棱镜的光束2维扫描系统结构如图 1所示。该系统主要由两个完全相同(楔角为ε,折射率为n)的圆楔形棱镜Π1和Π2及其传动电机M1和M2组成,其中Π1和Π2共轴且垂直于轴线z的两个面Σ1和Σ2, 背对放置,Π1和Π2分别由电机M1和M2控制, 绕轴线z独立转动。入射光束1沿轴线z依次经过Σ1′、Σ1、Σ2和Σ2′ 4个折射面,最后从Π2的倾斜面Σ2′出射。经过分析可知,随着Π1的旋转角Φ1和Π2的旋转角Φ2的变化,出射光束2的天顶角γ和方位角Φ可在一定空间范围内任意变化。
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旋转双圆楔形棱镜的光束指向正向解析问题为:由楔形棱镜Π1和Π2绕转轴的旋转角Φ1、Φ2,确定出射光束天顶角γ与方位角Φ。如图 2a所示,设定两个倾斜面Σ1′和Σ2′平行时为系统初态,此时Π1和Π2的对称面重合,设该面为零旋转角度面,Φ1=Φ2=0°, 如图 2b所示,利用折射定律可确定光束经由Π1和Π2的完整传播轨迹。对图 2b仔细分析后可得光束在Π1和Π2中传播轨迹的简单模型,如图 2c所示。图 2c中NN′为共轴线,NP为Σ2′面的法线,当Π2绕NN′旋转360°时,NP也将绕NN′旋转360°,此时就会形成一个以NP为母线的圆锥面, AN和BN分别为Σ2′面的入射光束和折射光束。
图 2 旋转双圆楔形棱镜中光束传播轨迹及其简化模型
Figure 2. Beam propagation trajectory and its simplified model in double rotating round wedge prisms
在图 2b中,假设Π1绕轴线NN′转动了角度Φ1,入射光束1沿NN′轴线从Σ1′面入射,入射角γi, 1=ε。根据折射定律及几何关系,经过Σ1′面折射光束3的偏折角γp可表示为:
$ \gamma_{\mathrm{p}}=\gamma_{\mathrm{i}, 1}-\gamma_{\mathrm{r}, 1}=\varepsilon-\arcsin [(\sin \varepsilon) / n] $
(1) 式中,γr, 1为折射角。从图中可知,此时光束的传播平面与设定的零旋转角度面的夹角为Φ1。随后,折射光束3从Σ1面出射,经过Π1和Π2间的平行空气隙后,再入射至Σ2面。在实际工程设计加工过程中,Σ1和Σ2两个面无法做到严格平行,假定该平行误差很小,可忽略其对出射光束角度的影响。显然,折射光束3在经过平行空气隙前后方向未改变,只产生一个与隙厚成正比的平移。假设Σ1和Σ2的间距很小,即隙厚很小,可忽略其所产生的光束平移,进入Π2面的光束仍可用折射光束3表示,即图 2c中的AN。
如图 2c所示,假定Π2绕轴线NN′转动了角度Φ2,折射光束3传播至Σ2′面的入射角和折射角分别为γi, 2和γr, 2,显然有sinγr, 2=nsinγi, 2。图中令$\overline{N N^{\prime}}$= l,BP =a,AP =c,且有∠N′NP=ε,∠N′NA=γp,∠N′NB=γ,∠PNB =γr, 2,∠PNA=γi, 2。在△AN′P和△ANP中采用余弦定理可得:
$ \begin{gathered} c^2=(l \tan \varepsilon)^2+\left(l \tan \gamma_{\mathrm{p}}\right)^2- \\ 2 l^2 \tan \varepsilon \tan \gamma_{\mathrm{p}} \cdot \cos \left({\mathit{\Phi}}_2-{\mathit{\Phi}}_1\right) \end{gathered} $
(2) $ \begin{gathered} c^2=(l / \cos \varepsilon)^2+\left(l / \cos \gamma_{\mathrm{p}}\right)^2- \\ 2 l^2 /\left(\cos \varepsilon \cdot \cos \gamma_{\mathrm{p}}\right) \cdot \cos \gamma_{\mathrm{i}, 2} \end{gathered} $
(3) 联立上述两式得:
$ \cos \gamma_{\mathrm{i}, 2}=\cos \gamma_{\mathrm{p}} \cos \varepsilon+\sin \gamma_{\mathrm{p}} \sin \varepsilon \cos \left({\mathit{\Phi}}_2-{\mathit{\Phi}}_1\right) $
(4) 在△BN′P和△BNP中采用余弦定理得:
$ \begin{gathered} a^2=(l \tan \varepsilon)^2+(l \tan \gamma)^2- \\ 2 l^2 \tan \varepsilon \tan \gamma \cdot \cos \left({\mathit{\Phi}}_2-{\mathit{\Phi}}\right) \end{gathered} $
(5) $a^2=(l / \cos \varepsilon)^2+(l / \cos \gamma)^2-\\ 2 l^2 /(\cos \varepsilon \cos \gamma) \cdot \cos \gamma_{\mathrm{r}, 2} $
(6) 联立上述两式得:
$ \cos \gamma_{\mathrm{r}, 2}=\cos \gamma \cos \varepsilon+\sin \gamma \sin \varepsilon \cos \left({\mathit{\Phi}}_2-{\mathit{\Phi}}\right) $
(7) 在△BNP和△ANP中分别采用正弦定理得:
$ a / \sin \gamma_{\mathrm{r}, 2}=(l / \cos \gamma) / \sin \angle A P N $
(8) $ c / \sin \gamma_{\mathrm{i}, 2}=\left(l / \cos \gamma_{\mathrm{p}}\right) / \sin \angle A P N $
(9) 联立上述两式得:
$ \frac{a}{c}=\frac{\sin \gamma_{\mathrm{r}, 2}}{\sin \gamma_{\mathrm{i}, 2}} \frac{\cos \gamma_{\mathrm{p}}}{\cos \gamma}=\frac{n \cos \gamma_{\mathrm{p}}}{\cos \gamma} $
(10) 在△BNP和△BNA中分别采用正弦定理得:
$ a / \sin \gamma_{\mathrm{r}, 2}=(l / \cos \varepsilon) / \sin \angle A B N $
(11) $ \begin{gathered} (a-c) / \sin \left(\gamma_{\mathrm{r}, 2}-\gamma_{\mathrm{i}, 2}\right)= \\ \left(l / \cos \gamma_{\mathrm{p}}\right) / \sin \angle A B N \end{gathered} $
(12) 联立上述两式得:
$ \frac{a}{c}=\frac{\sin \gamma_{\mathrm{r}, 2} \cos \gamma_{\mathrm{p}}}{\sin \gamma_{\mathrm{r}, 2} \cos \gamma_{\mathrm{p}}-\sin \left(\gamma_{\mathrm{r}, 2}-\gamma_{\mathrm{i}, 2}\right) \cos \varepsilon} $
(13) 联立(10)式、(13)式并化简可得:
$ \cos \gamma=n \cos \gamma_{\mathrm{p}}+\left(\cos \gamma_{\mathrm{r}, 2}-n \cos \gamma_{\mathrm{i}, 2}\right) \cos \varepsilon $
(14) 根据sinγr, 2=nsinγi, 2且γr, 2≤90°,可得:
$ \cos \gamma_{\mathrm{r}, 2}=\sqrt{1-n^2+n^2 \cos ^2 \gamma_{\mathrm{i}, 2}} $
(15) 将上式代入(14)式、(7)式并整理得到如下两式:
$ \begin{gathered} \gamma=\arccos \left[n \cos \gamma_{\mathrm{p}}+\right. \\ \left.\left(\sqrt{1-n^2+n^2 \cos ^2 \gamma_{\mathrm{i}, 2}}-n \cos \gamma_{\mathrm{i}, 2}\right) \cos \varepsilon\right] \end{gathered} $
(16) $ {\mathit{\Phi}}=\left\{\begin{array}{l} {\mathit{\Phi}}_2-\varphi, \left(0 \leqslant {\mathit{\Phi}}_2-{\mathit{\Phi}}_1 \leqslant \pi\right) \\ {\mathit{\Phi}}_2+\varphi-2 \pi, \left(\pi <{\mathit{\Phi}}_2-{\mathit{\Phi}}_1 \leqslant 2 \pi\right) \\ {\mathit{\Phi}}_2+\varphi, \left(0 \leqslant {\mathit{\Phi}}_1-{\mathit{\Phi}}_2 \leqslant \pi\right) \\ {\mathit{\Phi}}_2-\varphi+2 \pi, \left(\pi<{\mathit{\Phi}}_1-{\mathit{\Phi}}_2 \leqslant 2 \pi\right) \end{array}\right. $
(17) 其中:
$ \varphi=\arccos \left(\frac{\sqrt{1-n^2+n^2 \cos ^2 \gamma_{\mathrm{i}, 2}}}{\sin \gamma \sin \varepsilon}-\cot \gamma \cot \varepsilon\right) $
(18) 将(1)式、(4)式代入(16)式,然后再将(1)式、(4)式和(16)式代入(17)式,可先后求得γ(Φ1, Φ2)和Φ(Φ1, Φ2)两个函数关系式。
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旋转双圆楔形棱镜的光束指向反向解析问题为:由出射光束天顶角γ和方位角Φ,确定双圆楔形棱镜Π1和Π2的旋转角度。为简便起见,设定Φ2-Φ1在[0, π]内变化,则由(16)式得:
$ \cos \gamma_{\mathrm{i}, 2}=\frac{n \cos \varepsilon\left(n^{-2}-1\right)}{2\left(\cos \gamma-n \cos \gamma_{\mathrm{p}}\right)}-\frac{\cos \gamma-n \cos \gamma_{\mathrm{p}}}{2 n \cos \varepsilon} $
(19) 根据(17)式得:
$ \;\;\;\;\;\;\;\;\;\;\;{\mathit{\Phi}}_2=\\ {\mathit{\Phi}}+\arccos \left(\frac{\sqrt{1-n^2+n^2 \cos ^2 \gamma_{\mathrm{i}, 2}}}{\sin \gamma \sin \varepsilon}-\cot \gamma \cot \varepsilon\right) $
(20) 联立(4)式、(19)式可得:
$ \begin{gathered} {\mathit{\Phi}}_2-{\mathit{\Phi}}_1=\arccos \left[\frac{n \cot \varepsilon\left(n^{-2}-1\right)}{2\left(\cos \gamma-n \cos \gamma_{\mathrm{p}}\right) \sin \gamma_{\mathrm{p}}}-\right. \\ \left.\frac{\cos \gamma-n \cos \gamma_{\mathrm{p}}}{n \sin \gamma_{\mathrm{p}} \sin (2 \varepsilon)}-\cot \gamma_{\mathrm{p}} \cot \varepsilon\right] \end{gathered} $
(21) 将(1)式、(19)式代入(20)式,然后再将(1)式、(20)式代入(21)式,可求得Φ2(γ, Φ)和Φ1(γ, Φ)两个函数关系式。
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由(4)式可发现:
$ \cos \left(\gamma_p+\varepsilon\right) \leqslant \cos \gamma_{\mathrm{i}, 2} \leqslant \cos \left(\gamma_{\mathrm{p}}-\varepsilon\right) $
(22) 当Φ2-Φ1=π时,γi, 2取最大值γp+ε。此时,为确保折射光束3在Σ2′面折射后仍能保持正向传播,须满足如下条件:
$ \sin \left(\gamma_{\mathrm{p}}+\varepsilon\right) \leqslant 1 / n $
(23) 当(23)式取“=”时,折射光束3将在Σ2′面发生临界全反射,此时出射光束2的天顶角γ最大,且γmax= π/2-ε。进一步分析可知,当折射率n和楔角ε的值满足临界全反射条件时,γ的取值范围最大,为[0, γmax]。在临界条件下,将(1)式代入(23)式可得:
$ 4 n^2 \sin ^3 \varepsilon+4\left(n^2-1\right) \sin ^2 \varepsilon-3 \sin \varepsilon-1=0 $
(24) 由(24)式可绘出满足临界条件时,折射率n和棱镜楔角ε的变化关系曲线,如图 3所示。显然,楔角ε随着折射率n的增大而逐渐减小。
图 3 临界全反射条件下,棱镜楔角随折射率的变化曲线
Figure 3. Curve of prism wedge angle with refractive index under critical condition of total internal reflection
综上分析可知,为获得一个尽可能大的出射光束径向扫描范围,楔角ε和折射率n的取值在满足(23)式的同时,楔角ε应尽可能小;满足(23)式的楔角ε和折射率n近似成反比例变化关系。因此,为获得大的出射光束扫描范围,双楔形棱镜应尽量选择折射率高的材料制作。本文中以直探式测风激光雷达为例,直探式系统常用的3个工作波长分别为355 nm(高层风场探测,以大气分子为探测目标)、532 nm(兼顾高低层风场探测,以大气分子和气溶胶为探测目标)和1064 nm(低层风场探测,以大气气溶胶粒子为探测目标)。假定便携式多普勒激光雷达的发射波长为532 nm,选取肖特公司的LASF35镧系光学玻璃作为楔形棱镜的制作材料,LASF35在532 nm波长的折射率n达到了2.03。由图 3可得,当n=2.03时,楔角ε=19.5°。此时,出射光束2在理想条件下的天顶角变化范围为0°~70.5°,即仰角变化范围为19.5°~90°。
需要指出的是,上述结论是假定入射光束严格平行的理想条件下得出的。然而,实际入射光束具有一定发散角和口径大小,需要考虑大偏转角时光束压缩效应的影响。容易得出,偏转角越大,压缩效应就越明显。初略估算双楔形棱镜的压缩比(入射光斑与出射光斑的垂直截面积之比)RCR为:
$ \begin{gathered} R_{\mathrm{CR}}=\frac{\cos \gamma_{\mathrm{i}, 1}}{\cos \gamma_{\mathrm{r}, 1}} \frac{\cos \gamma_{\mathrm{i}, 2}}{\cos \gamma_{\mathrm{r}, 2}}= \\ \frac{\cos \varepsilon}{\sqrt{1-\sin ^2 \varepsilon / n^2}} \frac{\cos \gamma_{\mathrm{i}, 2}}{\sqrt{1-n^2+n^2 \cos ^2 \gamma_{\mathrm{i}, 2}}} \end{gathered} $
(25) 且有:
$ \Delta \gamma_{\mathrm{r}, 2}=\Delta \gamma_{\mathrm{i}, 1} R_{\mathrm{CR}} $
(26) 式中,Δγi, 1和Δγr,2分别为入射光束发散角和出射光束在折射面上的发散角。RCR越大,表明压缩效应越显著。压缩效应将使出射光束在某些方向上的发散角增大,远场能量分布区域变宽。扫描式测风激光雷达系统一般为收发同置,压缩效应将导致发射和接收视场均增大,一方面造成系统探测远距离区域的空间分辨率降低;另一方面导致系统接收到的天空背景噪声增大,白天探测信噪比有所下降,风速探测误差增大。因此,为避免光束压缩效应对激光雷达探测性能造成大的影响,需合理限定RCR的最大值。
将(19)式代入(25)式后可得RCR与γ的关系,如图 4所示。由图 4可见:随着光束天顶角γ增大,RCR经历了从缓慢增大到迅速增大的过程。实际中,首先构建RCR与具体激光雷达探测性能的关系;然后根据探测性能要求确定RCR的阈值;最后由图 4确定出射光束实际能取得的最大天顶角γmax。例如:假定Δγi, 1=0.5 mrad,若要求Δγr, 2不能超过2 mrad,则RCR不能超过4,由图可得γmax=58.7°;若要求Δγr, 2不能超过3 mrad,则RCR不能超过6,由图可得γmax=62.6°。关于压缩比增大对测风激光雷达探测性能影响问题,后续将深入进行定量仿真研究。
测风激光雷达双棱镜2维扫描系统的光学设计
Optical design of biprism 2-D scanning system for wind LiDAR
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摘要: 传统的测风激光雷达双反射镜式2维扫描系统体积较大、结构相对复杂, 不利于系统小型一体化集成。基于旋转双圆楔形棱镜, 研究了新型2维光学扫描系统; 分析了系统的工作原理, 推导出了双圆楔形棱镜的旋转角与出射光束方位角及天顶角之间的简单正反向函数关系式, 对楔形棱镜的折射率和楔角进行了优化选取和设计。结果表明, 当工作波长为532 nm、楔形棱镜材料折射率为2.03时, 最优设计楔角为19.5°; 出射光束最大天顶角不仅取决于楔形棱镜折射率和楔角, 还受光束压缩效应的制约。该系统结构紧凑、便于集成, 能实现出射光束大范围和快速高精度的扫描, 也能实现测风激光雷达以平面位置显示、距离高度显示等光束扫描模式工作。Abstract: 2-D scanning system with double mirrors of a traditional wind light detection and ranging(LiDAR) was large in volume and complex in structure, which was not conducive to the small-scale integration of the system. A new 2-D optical scanning system based on rotating double circular wedge prism was studied. The working principle of the system was analyzed, and the simple forward and inverse functional relationship between the rotation angle of the double circular wedge prism and the azimuth and zenith angle of the outgoing beam was derived. The refractive index and wedge angle of the wedge prism were optimized and designed. The results show that when the working wavelength is 532 nm and the refractive index of wedge prism material is 2.03, the optimal wedge angle is 19.5°. The maximum zenith angle of the outgoing beam depends not only on the refractive index and wedge angle of the wedge prism, but also on the beam compression effect. The system is compact and easy to integrate, and large-scale, fast and high-precision scanning of the outgoing beam can be realized. The wind LiDAR can also work in beam scanning modes such as plane position display and distance height display.
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