Question

matlab 解一元非线性方程 参数传递疑问

在对一组信号做处理时，需要把之前解得的方差、均值等结果，赋值到一个非线性方程中求解，方程如下：
log(x/a)-(b/a)^2/(c/d)^2*log(x/d)=0
其中，a、b、c、d为之前处理得到的方差与平均值，x为未知量
我的问题是，如何实现MATLAB中动态的参数传递？用什么函数能解决呢？
现在我的做法是把a、b、c、d的值复制黏贴到solve中求解，这么做太笨了。。。
望赐教，不盛感激

Answer 1

clear all;
close all;
warning off all;

a = 1; %%此处a,b,c,d换成你自己的数值
b = 3;
c = 5;
d = 7;
r = solve('log(x/a)-(b/a)^2/(c/d)^2*log(x/d)=0','x');
r_temp = subs(r,'a',a); %%利用subs函数将符号表达式中的指定符号替换成数据
r_temp = subs(r_temp,'b',b);
r_temp = subs(r_temp,'c',c);
r_temp = subs(r_temp,'d',d);
result = r_temp %%result为最终结果

Answer 2

function main
a=2;
b=3;
c=4;
d=5;
%%%%%%你求出这些参数的代码
x=fun(a,b,c,d)
end

function x=fun(a,b,c,d)
syms x
y=log(x/a)-(b/a)^2/(c/d)^2*log(x/d);
x1=solve(y);
x=double(x1);
end

Answer 3

代码是这样的：
[a,b,c,d]=solve('(a+300)*(a+300)+b*b=242500','(c-300)*(c-300)+d*d=242500','(a-c)*(a-c)+(b-d)*(b-d)=1000000','(b-d)/(a-c)=0.140540834')
结果有四组：
1，a,b,c,d=
-645.1911329 351.2023374 345.0769359 490.3754376
2，
-345.0769359 -490.3754376 645.1911329 -351.2023374
3，
495.13403441871544929862197087467+54.764098493321742310644330274791*i
69.586550138990974453113046837450-625.76458366341173499493975505703*i
-495.13403441871544929862197087467+54.764098493321742310644330274791*i
-69.586550138990974453113046837450-625.76458366341173499493975505703*i

4，
495.13403441871544929862197087467-54.764098493321742310644330274791*i
69.586550138990974453113046837450+625.76458366341173499493975505703*i
-495.13403441871544929862197087467-54.764098493321742310644330274791*i
-69.586550138990974453113046837450+625.76458366341173499493975505703*i