下面这个函数大家都不会陌生,计算zernike函数值的,并根据此可以还原出图像来, B}p.fE
我输入10阶的n、m,r,theta为38025*1向量,最后得到的z是29525*10阶的矩阵, !{uV-c-5,
这个,跟我们用zygo干涉仪直接拟合出的36项zernike系数,有何关系呢? hN1[*cF
那些系数是通过对29525*10阶的矩阵每列的值算出来的嘛? e2;=OoBK
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function z = zernfun(n,m,r,theta,nflag) _J"J[$
%ZERNFUN Zernike functions of order N and frequency M on the unit circle. QiRx2Z*\
% Z = ZERNFUN(N,M,R,THETA) returns the Zernike functions of order N }gX4dv
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% and angular frequency M, evaluated at positions (R,THETA) on the yFFNzw{
% unit circle. N is a vector of positive integers (including 0), and c No)LF
% M is a vector with the same number of elements as N. Each element {Y]3t9!\
% k of M must be a positive integer, with possible values M(k) = -N(k) #&{)`+!"
% to +N(k) in steps of 2. R is a vector of numbers between 0 and 1, m]=G73jzO
% and THETA is a vector of angles. R and THETA must have the same B]7QOf"
% length. The output Z is a matrix with one column for every (N,M) P8CIKoKCV
% pair, and one row for every (R,THETA) pair. waV4~BdL
% n1+J{EPH
% Z = ZERNFUN(N,M,R,THETA,'norm') returns the normalized Zernike 9@Z++J.^y
% functions. The normalization factor sqrt((2-delta(m,0))*(n+1)/pi), L`^v"W()
% with delta(m,0) the Kronecker delta, is chosen so that the integral )s 1
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% of (r * [Znm(r,theta)]^2) over the unit circle (from r=0 to r=1, q> #P|
% and theta=0 to theta=2*pi) is unity. For the non-normalized 3i}$ ~rz]U
% polynomials, max(Znm(r=1,theta))=1 for all [n,m]. )MM(HS
% ZhoB/TgdL
% The Zernike functions are an orthogonal basis on the unit circle. 1&kf