HPW

本代码为牛顿下山法求解非线性方程组。其调用格式为[r,m]=mulDNewton(F,x0,eps) 其中F:非线性方程组，x0：初始解，eps：解的精度；r：求得的一组解，n：迭代步数。(The code for Newton' s method for solving nonlinear equations. Its call format [r, m] = mulDNewton (F, x0, eps) where F: non-linear equations, x0: initial solution, eps: solution accuracy r: obtain a set of solution, n : the number of iterative steps.)

Matlab Image Fusion Toolbox for image fusion algorithm based on bandelet transform. This toolbox contains Matlab files that implement the image fusion algorithms described in paper Xiaobo Qu, Jingwen Yan, Guofu Xie, et al. A Novel Image Fusion Algorithm Based on Bandelet Transform. Chinese Optics Letters,Vol.5,No.10,pp:569-572, 2007

关于matlab解决0-1背包问题的修改，原先下载的matlab代码中有错误(Solve 0-1 knapsack problem matlab modifications, there is an error in the original download matlab code)

ZEMAX是美国 Radiant Zemax 公司所发展出的光学设计软件，可做光学组件设计与照明系统的照度分析，也可建立反射，折射，绕射等光学模型，并结合优化，公差等分析功能，是套可以运算sequential及Non-Sequential的软件。(ZEMAX is a wonderful too for designing light .)

Wind turbine output power vs. rotational speed, with wind speed as parameter

在matlab/simulink目前还没有更好地实现小波包模块，simuwave包括了各种各样的小波包子模块，可以自己实现小波分解。( At present, matlab/simulink is still not better to implement the wavelet packet module, simuwave includes a variety of wavelet packet module, you can realize the wavelet decomposition. )

1.对染噪doppler信号进行小波包3层分解：分解层次j=1,2时，都是信号的概貌；当j=3时，反映概貌的已几乎不含噪声分量，而其它噪声分量的幅值已很小。 2.对加噪Blocks信号进行不同阈值及不同阈值的使用方式降噪。 (1. On noise pollution doppler signal decomposition of wavelet packet Layer 3: decomposition level j = 1,2, are the signal profiles when j = 3 to reflect the general picture has been almost non-noise component, while other noise components amplitude has been very small. 2. Blocks of noise signals of different thresholds and different thresholds for the use of noise reduction methods.)

用MATLAB编程 主要讲解小波变换的一些知识包括源代码，适用于研究小波变换的人(Mainly on the use of MATLAB programming knowledge, including some of the wavelet transform source code, for the people of wavelet transform)

engineering statistic

题目：有1、2、3、4个数字，能组成多少个互不相同且无重复数字的三位数？都是多少？ 1.程序分析：可填在百位、十位、个位的数字都是1、2、3、4。组成所有的排列后再去 　　　　　　掉不满足条件的排列。 (Title: There are numbers 1,2,3,4, how many can be composed of distinct three-digit numbers without repetition? Is how much? 1. Program Analysis: fill in the hundred, ten, a bit of the numbers are 1,2,3,4. Composed of all the permutations and then remove the arrangement does not meet the conditions.)