firfilter

最大流最小截问题，用ｍａｔｌａｂ编写的，ｒａｒ中有说明，较清晰(Maximum flow minimum cut problem, written with matlab, rar have stated more clearly)

说明：  《MATLAB 7.0编程基础》第6章 （源码实例）主要讲解MATLAB图形基础("MATLAB 7.0 programming foundation" Chapter 6 (source code examples) mainly based on MATLAB graphics)

通过迭代法计算方同轴线的电势分布，进而求得内导体电流，最终计算出同轴线的阻抗(Iterative method to calculate the square coaxial electric potential distribution, and thus obtain the inner conductor current, the final calculation of the impedance of the coaxial)

　k-means 算法接受参数 k ；然后将事先输入的n个数据对象划分为 k个聚类以便使得所获得的聚类满足：同一聚类中的对象相似度较高；而不同聚类中的对象相似度较小。聚类相似度是利用各聚类中对象的均值所获得一个“中心对象”（引力中心）来进行计算的。(K-means algorithm accept parameter k Then will the n of prior input data object is divided into k clustering to make won clustering meet: the same clustering the object in the similarity is higher And different clustering the object in the smaller similarity. Clustering similarity is using the cluster of the object in the mean gained a "center object" (the center of gravity) to calculate. )

利用函数MadHG生成规则LDPC码的校验矩阵H，其行重为6,列重为3,行数为列数一半(行数越大越好)，H中任意两列没有围长为4的圈，并得到H对应的生成矩阵G，可以保证mod(G*H ,2)=0。使用方法为:[H,G] = MadHG(m,n,x),x= 1(得到的G左半部分为单位阵) or 2(G右半部分为单位阵),(use function MadHG Generation rules LDPC check matrix H, re-6, series of three, out for a few half of the number (A bigger the better), H 2 arbitrary no girth of the four circles, H counterparts and with the formation of matrix G, we can be assured of mod (G* H, 2) = 0. To be used : [H, G] = MadHG (m, n, x), x = 1 (the G part of the Left Front units) or 2 (G right half of the units RUF),)

example of matlab code to calculate a Fitzhug Nagumo system of equation

自适应噪声对消器，消除语音中的噪声，算是探讨性质的小程序吧(Adaptive noise cancellation, and voice in the elimination of noise)

一些边缘检测的matlab程序，希望对大家有帮助(Some of the MATLAB edge detection program)

MODELING OF STATIC SERIES VOLTAGE REGULATOR IN DISTRIBUTION SYSTEM FOR LOSS REDUCTION

实现多输入单输出的预测功能，预测精度高，程序操作简单，易于修改应用。(The prediction function of multi input and single output is realized. The prediction accuracy is high, the program operation is simple, and it is easy to modify the application.)