我该如何求解先验方程?

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我想在matlab中求解给定的方程，以找到beta的值，但会出错.我在下面发布代码和错误.

I want to solve the given equation in matlab to find the value of beta but getting error. I am posting the code and error below.

谢谢.

n1=1.77; n2=1.45; d=1e-6; lambda = 1e-6; ko = 2*pi/lambda; A=(ko*n1)^2; B=(ko*n2)^2; syms beta; s = 'sqrt(A-beta^2)*(d/2)*tan(sqrt(A-beta^2)*d/2)=sqrt(beta^2-B)*(d/2)'; solve (s);

错误:

Error using solve>processString (line 337) ' sqrt(A-beta^2)*(d/2)*tan(sqrt(A-beta^2)*d/2)=sqrt(beta^2-B)*(d/2) ' is not a valid expression or equation. Error in solve>getEqns (line 267) eqns = processString(eqns, v, vc); Error in solve (line 150) [eqns,vars,options] = getEqns(varargin{:}); Error in transcendetal (line 9) solve (s);

推荐答案

我确认R2013a的以下工作:

I confirm the following works on R2013a:

syms beta A B d n1=1.77; n2=1.45; d=1e-6; lambda = 1e-6; ko = 2*pi/lambda; A=(ko*n1)^2; B=(ko*n2)^2; solve ( sqrt(A-beta^2)*(d/2)*tan(sqrt(A-beta^2)*d/2)==sqrt(beta^2-B)*(d/2))

但是

Warning: Explicit solution could not be found. > In solve at 179 ans = [ empty sym ]

这很奇怪，因为在R2010a中，您使用上面提供的旧"语法，我得到了

This is odd, because in R2010a with the 'old' syntax you give above, I get

ans = 0 -(log((A + B - 2*beta^2 + 2*(A - beta^2)^(1/2)*(beta^2 - B)^(1/2)*i)/(A - B))*i)/(A - beta^2)^(1/2)

请注意，最后一个解决方案是复数值，这可能是您在更高版本中必须启用的某些选项.我仍然更喜欢白板方法:p

Note that this last solution is complex-valued, which might be some option you have to enable in later versions...I don't know the symbolic math toolbox very well; I still prefer the whiteboard method :p