MATLAB用for循环解个方程syms n i qH=1.8;lanmuda=1.4;n1=1.55;n2=1.67;n3=1.50;k0=2*pi/lanmuda;for q=0:2 i=q; Si=solve(sqrt(n2*n2-n^2)*k0*H-atan(sqrt(n^2-n1*n1)/sqrt(n2*n2-n^2))-atan(sqrt(n^2-n3*n3)/sqrt(n2*n2-n^2))-q*pi==0,n); end主要是
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![MATLAB用for循环解个方程syms n i qH=1.8;lanmuda=1.4;n1=1.55;n2=1.67;n3=1.50;k0=2*pi/lanmuda;for q=0:2 i=q; Si=solve(sqrt(n2*n2-n^2)*k0*H-atan(sqrt(n^2-n1*n1)/sqrt(n2*n2-n^2))-atan(sqrt(n^2-n3*n3)/sqrt(n2*n2-n^2))-q*pi==0,n); end主要是](/uploads/image/z/2476977-33-7.jpg?t=MATLAB%E7%94%A8for%E5%BE%AA%E7%8E%AF%E8%A7%A3%E4%B8%AA%E6%96%B9%E7%A8%8Bsyms+n+i+qH%3D1.8%3Blanmuda%3D1.4%3Bn1%3D1.55%3Bn2%3D1.67%3Bn3%3D1.50%3Bk0%3D2%2Api%2Flanmuda%3Bfor+q%3D0%3A2++++i%3Dq%3B++++Si%3Dsolve%28sqrt%28n2%2An2-n%5E2%29%2Ak0%2AH-atan%28sqrt%28n%5E2-n1%2An1%29%2Fsqrt%28n2%2An2-n%5E2%29%29-atan%28sqrt%28n%5E2-n3%2An3%29%2Fsqrt%28n2%2An2-n%5E2%29%29-q%2Api%3D%3D0%2Cn%29%3B++end%E4%B8%BB%E8%A6%81%E6%98%AF)
MATLAB用for循环解个方程syms n i qH=1.8;lanmuda=1.4;n1=1.55;n2=1.67;n3=1.50;k0=2*pi/lanmuda;for q=0:2 i=q; Si=solve(sqrt(n2*n2-n^2)*k0*H-atan(sqrt(n^2-n1*n1)/sqrt(n2*n2-n^2))-atan(sqrt(n^2-n3*n3)/sqrt(n2*n2-n^2))-q*pi==0,n); end主要是
MATLAB用for循环解个方程
syms n i q
H=1.8;lanmuda=1.4;n1=1.55;n2=1.67;n3=1.50;
k0=2*pi/lanmuda;
for q=0:2
i=q;
Si=solve(sqrt(n2*n2-n^2)*k0*H-atan(sqrt(n^2-n1*n1)/sqrt(n2*n2-n^2))-atan(sqrt(n^2-n3*n3)/sqrt(n2*n2-n^2))-q*pi==0,n);
end
主要是是想求在q=0,1,2,3.下的不同的n值,循环程序不对,求帮忙修改.
MATLAB用for循环解个方程syms n i qH=1.8;lanmuda=1.4;n1=1.55;n2=1.67;n3=1.50;k0=2*pi/lanmuda;for q=0:2 i=q; Si=solve(sqrt(n2*n2-n^2)*k0*H-atan(sqrt(n^2-n1*n1)/sqrt(n2*n2-n^2))-atan(sqrt(n^2-n3*n3)/sqrt(n2*n2-n^2))-q*pi==0,n); end主要是
把“solve”那行里最后的==0去掉即可.