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    [原创]在框架结构确定的情况下,基于matlab的消四种像差的三反系统初始结构的求解 [复制链接]

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    离线songshaoman
     
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    只看楼主 倒序阅读 楼主  发表于: 2020-05-25
    %无中间像,焦距输入为负数 ;AjY-w  
    function sjr=nfdre(~) jgz}  
    O@Ro_sPG(  
    %系统焦距及各镜间距输入,间距取负正负 "$lE~d">  
    ;$\?o  
    f=input('f:'); n.323tNY  
    d1=input('d1:'); OIqisQ7ZB  
    d2=input('d2:'); 0|D^_1W`R  
    d3=input('d3:'); d"P\ =`+  
    sv g`s,g  
    A=f^2/(d3*d2)-f/d1; vz[-8m:f  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); Tx+Bkfj  
    C=d3/d2-f/d1; w,,QXJe{Z_  
    b,k%n_&n  
    a1=(-B+sqrt(B^2-4*A*C))/(2*A);%α1 2$fFl,v!z  
    a2=d3/(a1*f);%α2 =Fd!wkB'{  
    b2=a1*(1-a2)*f/d2;%β2 4Z& i\#Q  
    b1=(1-a1)*f/(d1*b2);%β1 5Dhpcgq<<  
    }:b6WN;c  
    |2YkZ nJn  
    %曲率半径 O]XdPH20  
    6c?;-5.  
    R1=2*f/(b1*b2) w6PKr^  
    R2=2*a1*f/(b2*(1+b1)) YV6@SXy  
    R3=2*a1*a2*f/(1+b2)  L$Uy  
    &V$qIvN$  
    A1=b2^3*(a1-1)*(1+b1)^3; _4#8o\  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; {x-iBg9#l2  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; sy ]k  
    Q?Y\WD  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); =)2sehU/  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); tiTJ.uz6  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); <3 b|Sk:T  
    7jj.maK  
    CB=[C1 B1;C2 B2]; <;O -N=  
    AB=[A1 B1;A2 B2]; HB>&}z0  
    AC=[A1 C1;A2 C2]; d~G, *  
    \A{ [2  
    %非球面系数 ,t2yw  
    k2=-(det(CB)/det(AB)); @scy v@5)F  
    k3=-(det(AC)/det(AB)); zK Y 9 'y  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 k^JV37;bl  
    k2=k2 >nqDUGnEo>  
    k3=k3 ^AI5SjOUx  
    dGNg[  
    end H*gX90{!2  
    w%3R[Kdzk  
    %有中间像,焦距输入为正数 rt _k }  
    2 SJ N;A~}  
    function sjr=yfdre(~) _3pME9l  
    RF#S=X6  
    f=input('f:'); (3_m[N\F  
    d1=input('d1:'); 'l&),]|$)  
    d2=input('d2:'); wWaJ%z>3y  
    d3=input('d3:'); JKXb$  
    RTL A*  
    A=f^2/(d3*d2)-f/d1; [!<W{ ($5  
    B=f/d1-f/d2+f/d1+f/d3-d3*f/(d3*d2); RRasX;zK  
    C=d3/d2-f/d1; pZ IDGy=~  
    k)a3j{{  
    a1=(-B-sqrt(B^2-4*A*C))/(2*A); iB1+4wa  
    a2=d3/(a1*f); &>YdX$8x  
    b2=a1*(1-a2)*f/d2; q#6K'=AC  
    b1=(1-a1)*f/(d1*b2); >U.f`24  
    >k)zd-  
    %曲率半径 BJL*Dih m[  
    )J"*[[e  
    R1=2*f/(b1*b2) /Cl=;^)  
    R2=2*a1*f/(b2*(1+b1)) ag7(nn0!  
    R3=2*a1*a2*f/(1+b2) Y\e8oIYu7  
    H[u[3  
    A1=b2^3*(a1-1)*(1+b1)^3; 0wA?.~ L  
    B1=-(a2*(a1-1)+b1*(1-a2))*(1+b2)^3; Q&@~<!t  
    C1=(a1-1)*b2^3*(1+b1)*(1-b1)^2-(a2*(a1-1)+b1*(1-a2))*(1+b2)*(1-b2)^2-2*b1*b2; 9iwSE(},  
    =8p *Ijs  
    A2=b2*(a1-1)^2*(1+b1)^3/(4*a1*b1^2); [8h~:.d`  
    B2=-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)^3/(4*a1*a2*b1^2*b2^2); '1Z3MjX  
    C2=b2*(a1-1)^2*(1+b1)*(1-b1)^2/(4*a1*b1^2)-(a2*(a1-1)+b1*(1-a2))^2*(1+b2)*(1-b2)^2/(4*a1*a2*b1^2*b2^2)-b2*(a1-1)*(1-b1)*(1+b1)/(a1*b1)-(a2*(a1-1)+b1*(1-a2))*(1-b2)*(1+b2)/(a1*a2*b1*b2)-b1*b2+b2*(1+b1)/a1-(1+b2)/(a1*a2); X`+8r O[  
    f\zu7,GU  
    CB=[C1 B1;C2 B2]; 8 etNS~^  
    AB=[A1 B1;A2 B2]; }[2|86,G;  
    AC=[A1 C1;A2 C2]; Tp.t.Qic  
    G(*7hs  
    %二次系数 'A#bBn,|  
    %"C%pA  
    k2=-(det(CB)/det(AB)); 9P)28\4  
    k3=-(det(AC)/det(AB)); KJLC2,  
    k1=(k2*a1*b2^3*(1+b1)^3-k3*a1*a2*(1+b2)^3+a1*b2^3*(1+b1)*(1-b1)^2-a1*a2*(1+b2)*(1-b2)^2)/(b1^3*b2^3)-1 rDEd MT  
    k2=k2 [3~mil3rO  
    k3=k3 ;LhNz()b  
    U~}cib5W5  
    end
     
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    离线doushan
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    只看该作者 1楼 发表于: 2023-03-01
    谢谢分享,学习一下 /yd<+on^