本期实战一个用Newton–Raphson Method求解非线性方程组根的例子。 使用Newton–Raphson方法,求解下述非线性弹簧的两个节点位移。使用1×10-5的收敛误差和初始估计u0= [0,0]T 。同时,估计收敛速度。 考虑两个串联的非线性弹簧,如Fig. 1所示。两个弹簧的刚度取决于弹簧的伸长率,其中 k1 =50+500u [N/m]和k2...
下面是解决非线性弹簧问题的基于改进的Newton–Raphson方法的MATLAB程序。 tol=1.0e-5;iter=0;u=[0.3;0.6];uold=u;c=0;f=[0;100];P=[300*u(1)^2+400*u(1)*u(2)-200*u(2)^2+150*u(1)-100*u(2)200*u(1)^2-400*u(1)*u(2)+200*u(2)^2-100*u(1)+100*u(2)];R=f-P;conv...
MATLAB function [F, dF] = equation(x) F = x^2 - 2;示例方程为x^2 - 2 = 0 dF = 2*x;方程的导数为2*x end 在这个示例中,我们定义了一个名为equation的函数,该函数接收一个变量x作为输入,并返回方程F(x)和其导数F'(x)的值。请注意,这只是一个示例,实际问题中的方程和导数可能更加复杂。
I downloaded power system test case data from the University of Washington Power Systems Test Case Archive (UW Power Systems Test Case Archive). I have formatted the data for use in my code verification process. This is for educational purposes only. Please note that the formatted data ...
MATLAB Online で開く I want to approximate the zeros of the following nonlinear system, using N-R method: f(x,y)=x+y^9/3+x^{243}/9+y^{2187}/27=0; g(x,y)=y+x^{27}/3+y^{243}/9+x^{6561}/27=0. I have tried with the following Ma...
Why is this code running so long? I'm not getting the expected value, which is just over 5, any help is appreciated. %% Problem 3b. Newton-Raphson Method (Needs Work) clear all clc R=15/2; v=500; h=8; i=1; while (1)
MATLAB Online에서 열기 Here is my code and my output is a function and not a numerical value as I expected. Can anyone debug this code? 테마복사 syms f y x df f=@(y) exp(y)-(sin(pi*y/3)); df=@(y) exp(y)-((pi*cos(pi*y/3))/3); x(1)=-3.0; error...
Newton-Raphson Numerical Method (https://www.mathworks.com/matlabcentral/fileexchange/61683-newton-raphson-numerical-method), MATLAB Central File Exchange. Retrieved April 12, 2025. MATLAB Release Compatibility Created with R2015a Compatible with any release Platform Compatibility Windows macOS Linux...
牛顿-拉夫逊法潮流计算matlab程序(NewtonRaphsonmethodof powerflowcalculationmatlabprogram) 主程序”powerflow_nr。” [bus_res功能,s_res]=powerflow_nr_2%牛顿-拉夫逊法解潮 流方程的主程序 【巴士线]=opdf_;%打开数据文件的子程序,返回总线(节点数据) ...
Depending on the starting guess value the Newton Raphson method yields one of the three roots. For example, if we start with value xg = − 0.5 the method yields the root x = − 0.093361 after just 5 iterations. A MATLAB code has been presented below for this choice of the starting ...