SMC_I

雷达系统分析与设计使用MATLAB第二版 Pdf(Radar Systems Analysis and Design Using MATLAB Second Edition Pdf)

算法交易案例，包含止盈止损等，代码比较简单。(Algorithmic trading case)

根据对真空蒸发镀膜的分析，在40×40的范围内分别蒸 发：1000、5000、10000、20000、30000个粒子。  利用2维数组记录每个点的粒子数。  初始粒的横纵坐标为0～40的均匀分布。  能量为（15，3）的正态分布。  在基体表面自由行走时损失1的能量。  若该点有粒子，即与其他粒子相遇时损失3的能量。  由低处向高处扩散，每扩散一层，损失4的能量，由高处 向低处扩散，损失3的能量。  当粒子的能量小于等于0时停留下来，该点的粒子数加1。  临界核的尺寸为10个粒子。  根据每个点粒子个数的不同，以不同的灰度显示这些点。 点越亮，则表示该点的膜越厚。  以图形方式显示粒子初始能量的正态分布情况(Based on the analysis of the vacuum evaporation coating, respectively, in the range of from 4040 evaporated: 1000,5000,10000,20000,30000 particles.  each point using 2-dimensional array to record the number of particles. The  initial grains transverse and longitudinal coordinates of a uniform distribution of 0 to 40. The  energy as a normal distribution (15,3).  1 energy losses in the substrate surface to walk.  point particles that 3 energy loss in the encounter with the other particles.  by the lower diffusion tops every diffusion layer, a loss of 4 energy diffusion from high to low, the energy loss of 3.  When the energy of the particles is less than or equal to 0:00 stay down, add 1 to the number of particles that point.  critical nucleus size of 10 particles.  these points, depending on the number of particles of each point is displayed in a different shade. Point, the brighter, the thicker the point of the fi)

Matlab符号数学工具箱，主要对比较复杂的部分做一些介绍(Matlab Symbolic Math Toolbox, the main part of some more complex description)

shannon fano algorithma source matlab

主要讲述了c的一些应用程序及其在具体地方的应用(C described some of the major applications and applications in specific areas)

MATLAB编写的一个贪吃蛇游戏 休闲娱乐(MATLAB to prepare a recreational game Snake)

说明：  matalab编程精通matalab6.5.1一书的所有例题程序(matalab progeaming master matalab6.5.1all of the examples programs)

说明：  ofdm的matlab编程,绝对是好东西,简单明了,对于ofdm学习很有帮助(The OFDM matlab programming is definitely good things, simple and clear, very helpful for OFDM)

自己写的用Matlab模拟一个缓冲区大小（包括正在服务的那个）为10的随机排队系统。 （1）到达过程是的泊松过程（到达速率为Mu），服务时间服从独立指数分布（均值为1/Lamda）。 对Mu//Lamda=0.2，Mu//Lamda=0.8和Mu//Lamda=1.1三种情况进行仿真，求出队列中接受服务用户为n的概率P(n), n=0,...,10。并与理论结果进行比较。 （2）假设到达过程为均匀过程（到达速率为Mu），服务时间服从独立瑞利分布（均值为1/Lamda）。(yourself writing a simulation using Matlab buffer size (including services that are) for the following 10 Machine queuing system. (1) arrival process is the Poisson process (arrival rate of Mu), submit to independent service index distribution (mean of 1/Lamda). Right Mu// Lamda = 0.2, Mu// Lamda = 0.8 and the Mu// = 1.1 Lamda three of the simulation, Queue calculated users receive services n the probability P (n), n = 0 ,..., 10. And with the theoretical results. (2) It is assumed that the arrival process for the uniform process (arrival rate of Mu), submit to independent service Rayleigh distribution (mean of 1/Lamda).)