| 查看: 1604 | 回复: 0 | ||
[求助]
超越方程求解的实现
|
|
程序大致如下,主要是想通过求解超越方程,将解得的结果用于循环。但是在超越方程部分编程出现问题,想请大家帮忙完成注释部分的超越方程的编写。谢谢!【公式稍微有点长】 clear;clc L=1e-3; C=40e-6; E=48; T=0.0002; R=5; T1=0.0001;T2=0.0001;e=0; % R=13;%注意该值的正负性I=6; for m=1:300 K=0.1+0.001*(m-1); i(1)=0; v(1)=0; a1=(((- 4*C*R^2 + L)/L)^(1/2) - 1)/(2*C*R); a2=-(((- 4*C*R^2 + L)/L)^(1/2) + 1)/(2*C*R); for n=1:1000 %%%%%%%%%%%%%%%%%%%%%%%% %在该处求解超越方程,解得的结果T1,从而可以得到T2=T-T1然后再进行下面的循环计算. %超越方程为1.5-24*K-K*X1-3*T1;其中的X1 如下: %X1=(E*(L*((a2*exp(T1*a1))/(a1 - a2) - (a1*exp(T1*a2))/(a1 - a2) +(exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))/(C*R) + 1) - (L*(exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2)))/(C*R)))/L - v(n)*((a2*exp(T1*a1))/(a1 - a2) - (a1*exp(T1*a2))/(a1 - a2) + (exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))/(C*R)) + (i(n)*(exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2)))/C; %%%%%%%%%%%%%%%%%%%%%%%% i(n+1)= i(n)*(((a1*exp(T1*a2))/(a1 - a2) - (a2*exp(T1*a1))/(a1 - a2))*((a1*exp(T2*a2))/(a1 - a2) - (a2*exp(T2*a1))/(a1 - a2)) - ((exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))*(exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2)))/(C*L)) - v(n)*((((a1*exp(T2*a2))/(a1 - a2) - (a2*exp(T2*a1))/(a1 - a2))*(exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2)))/L - ((exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2))*((a2*exp(T1*a1))/(a1 - a2) - (a1*exp(T1*a2))/(a1 - a2) + (exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))/(C*R)))/L) + (E*(L*((((a1*exp(T2*a2))/(a1 - a2) - (a2*exp(T2*a1))/(a1 - a2))*(exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2)))/L - ((exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2))*((a2*exp(T1*a1))/(a1 - a2) - (a1*exp(T1*a2))/(a1 - a2) + (exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))/(C*R) + 1))/L) + (L*(((a1*exp(T2*a2))/(a1 - a2) - (a2*exp(T2*a1))/(a1 - a2))*((a2*exp(T1*a1))/(a1 - a2) - (a1*exp(T1*a2))/(a1 - a2) + 1) + ((exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))*(exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2)))/(C*L)))/R))/L; v(n+1)=v(n)*(((a2*exp(T1*a1))/(a1 - a2) - (a1*exp(T1*a2))/(a1 - a2) + (exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))/(C*R))*((a2*exp(T2*a1))/(a1 - a2) - (a1*exp(T2*a2))/(a1 - a2) + (exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2))/(C*R)) - ((exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))*(exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2)))/(C*L)) + i(n)*((((a1*exp(T1*a2))/(a1 - a2) - (a2*exp(T1*a1))/(a1 - a2))*(exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2)))/C - ((exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))*((a2*exp(T2*a1))/(a1 - a2) - (a1*exp(T2*a2))/(a1 - a2) + (exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2))/(C*R)))/C) - (E*(L*(((a2*exp(T2*a1))/(a1 - a2) - (a1*exp(T2*a2))/(a1 - a2) + (exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2))/(C*R))*((a2*exp(T1*a1))/(a1 - a2) - (a1*exp(T1*a2))/(a1 - a2) + (exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))/(C*R) + 1) - ((exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))*(exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2)))/(C*L)) - (L*(((exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2))*((a2*exp(T1*a1))/(a1 - a2) - (a1*exp(T1*a2))/(a1 - a2) + 1))/C + ((exp(T1*a1)/(a1 - a2) - exp(T1*a2)/(a1 - a2))*((a2*exp(T2*a1))/(a1 - a2) - (a1*exp(T2*a2))/(a1 - a2) + (exp(T2*a1)/(a1 - a2) - exp(T2*a2)/(a1 - a2))/(C*R)))/C))/R))/L; end end |
» 猜你喜欢
三甲基碘化亚砜的氧化反应
已经有4人回复
请问下大家为什么这个铃木偶联几乎不反应呢
已经有5人回复
请问有评职称,把科研教学业绩算分排序的高校吗
已经有5人回复
孩子确诊有中度注意力缺陷
已经有12人回复
2025冷门绝学什么时候出结果
已经有3人回复
天津工业大学郑柳春团队欢迎化学化工、高分子化学或有机合成方向的博士生和硕士生加入
已经有4人回复
康复大学泰山学者周祺惠团队招收博士研究生
已经有6人回复
AI论文写作工具:是科研加速器还是学术作弊器?
已经有3人回复
论文投稿,期刊推荐
已经有4人回复
硕士和导师闹得不愉快
已经有13人回复














回复此楼