improve-canny

用来实现三维阻抗及光学断层成像重建的matlab程序(used 3D impedance and optical tomography reconstruction procedures Matlab)

非负矩阵应用于电能质量中的程序，用于谐波检测，构造和分解基本谐波(Non-negative matrix used in the power quality program for harmonic detection, the basic harmonic structure and decomposition)

hspice toolbox interface between hspice and matlab

通过输入关参数，求相得四阶累积量。方便的解决了求四阶累积量的计算复杂问题。(Through the input parameters, phase fourth order cumulants. Convenient to solve the fourth-order cumulant of calculating the complex problems. )

基于MATLAB的并网逆变器双闭环控制策略仿真(Double closed loop control)

我们的目的是将一个分数拆分成若干个分子是1分母是正整数的分数之和的形式。例如2/3可以拆分成1/2+1/6的形式。这里不考虑顺序，比如1/6+1/2和1/2+1/6相同。 输入p, q, a, n, 其中p/q为分数，对这个分数进行拆分，要求拆分的分数个数不能多于n个，所有分母的乘积不能大于a。求有多少种拆分方法? 输入：最多200组数据。在每行中都有四个数分别表示p, q, a, n，之间以空格隔开，并且满足p,q<=800，a<=12000，n<=7，最后以四个0做为输入的结束。 输入示例: 2 3 120 3 2 3 300 3 0 0 0 0 输出示例: 4 7 (Our aim is to split a fraction into a number of elements is a denominator is a positive integer fraction of the sum of form. For example, 2/3 can be split into 1/2+1/6 form. Not considered here, the order, such as 1/6+1/2 and 1/2+1/6 of the same. Input p, q, a, n, where p/q as a fraction of this split scores require no more than the number of split points n-all the product of the denominator can not be larger than a. How many kinds of demand split method? Input: Up to 200 sets of data. In each line has four numbers indicate the p, q, a, n, separated by a space between, and meet the p, q < = 800, a < = 12000, n < = 7, the end, four 0 as the input end. Input Example: 2,312,032,330,030,000 sample output: 47)

机械设计及理论 转子动力学中对轴承的研究(Mechanical Design and Theory Rotor Dynamics)

virtools用户手册，讲述了vt的基本构成，图形学原理，基本的使用方法，两个实例制作过程(vt user guide ,please download)

说明：  基本源代码程序，针对Artificial neural network人工神经元网络书中的源代码实例(Artificial neural network)

说明：   改进的牛顿法求解： 方程的构造方法:给出[0,1]区间上的随机数(服从均匀分布)作为方程的根p*. 设你的班级数为a3,学号的后两位数分别为a2与a1,从而得到你的三次方程 例如:你的31班的12号,则你的方程是21x3+60x2+2x+a0=0的形式. 方程中的系数a0由你得到的根p*来确定. (improve Newton's method : Construction of the equation : given interval [0,1] on the random number (subject to uniform distribution) as the root equation p*. set up your classes at a3, after learning of the double-digit for a1 and a2, so you get the three equations for example : Your 31 classes on the 12th, then your equation is 21x3 60x2 2x a0 = 0 forms. the equation coefficients a0 you get from the root of p* to determine.)