CA-matlab

说明：  关于ARNOLD算法的内容,对了解ARNOLD有参考意义(guanyu arnold suanfa de neirong)

Tutoriel MATLAB, The integrated using Matlab is very effective, learn to use the basics.

一个kalmen filter 的介绍，就是matlab中如何使用kalmen filter(A paper about the introduction of the kalmen filter)

本书简绍了matlab编写有限元程序的整个流程(The goal of this document is to give a very brief overview and direction in the writing of nite element code using Matlab. It is assumed that the reader has a basic familiarity with the theory of the nite element method, and our attention will be mostly on the implementation. An example nite element code for analyzing static linear elastic problems written in Matlab is presented to illustrate how to program the nite element method.)

牛顿法求解非线性方程Matlab编程实现(Newton method for solving nonlinear equations Matlab programming )

说明：  图像复原--数字图像处理--matlab程序(Image Recovery- Digital image processing- matlab program)

大学数值分析算法，重要参数的提取，采用的是脉冲对消法，基于kaiser窗的双谱线插值FFT谐波分析，可以动态调节运行环境的参数，三相光伏逆变并网的仿真，相关分析过程的matlab方法，结合PCA的尺度不变特征变换(SIFT)算法。( University of numerical analysis algorithms, Extract important parameters, It uses a pulse of consumer law, Dual-line interpolation FFT harmonic analysis kaiser windows, Can dynamically adjust the parameters of the operating environment, Three-phase photovoltaic inverter and network simulation, Correlation analysis process matlab method, Combined with PCA scale invariant feature transform (SIFT) algorithm.)

induction machine closed loor v/f control simulink model

软件-调用matlab,通过vc++软件编程软件-调用matlab。(Software- call matlab)

％File: c9_estimatepi.m ％有5个pi的估计，每一个都是基于500次重复随机试验，所得的pi的五个估计值用以下向量表示 ％pi的估计值=[3.0960 3.0720 2.9920 3.1600 3.0480] ％如果对5个结果进平均，则pi的估计值=3.0736,这样的结果等价于2500次的试验结果。( File: c9_estimatepi.m have 5 pi estimates are based on a randomized trial to repeat 500 times the income of the five estimated value of pi with the following vector, said the estimated value of pi = [3.0960 3.0720 2.9920 3.1600 3.0480] if the results into an average of 5, then the estimated value of pi = 3.0736, so the result is equivalent to 2500 times the test results.)