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EpEsQ

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数值模拟三维受激布里渊散射过程，考虑了泵浦光的能量损耗等等(three dimention numerical stimulation of the stimulated Brullouin Scattering ,considering the depletion of the pump energy)

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EpEsQ.m,4454,2014-08-09

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MATLABxuexi
不错的matlab教程，大家可以看看，对matlab的学习有帮助的。呵呵！(Matlab good tutorial, we will look at the matlab learning to be helpful. Ha ha!)

2007-09-09 09:48:35下载

积分：1
OFDM
matlab已调好，并加上了注释。希望对大家有帮助！(matlab)

2009-04-02 21:21:59下载

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FuzzyMath
说明：用PALKING模糊方法进行边缘检测，还有3个自适应改进方法(PALKING fuzzy method with edge detection, there are three adaptive Improvement)

2011-03-07 20:20:21下载

积分：1
C_point
采用轮廓跟踪法，得到两条曲线的坐标集，再通过求质心确定两个基准点 (This function is used to get two basis points between fingers )

2010-06-02 13:55:25下载

积分：1
MATLAB7.8
这本书是老师推荐给我的，非常好，我用这个学习matlab然后毕设时候很顺利。(This book is the teacher recommended to me, very good learning matlab and then completed the set up very well when I use this.)

2012-06-13 10:55:34下载

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Hamiltion
程序理论详见内附的pdf：《Hamilton图的充要条件及圈的矩阵算法》主程序为 [K]=getPk(A,k) A表示图的邻接矩阵，k表示k阶长路和圈返回K表示该图的所有k阶长路和圈其他函数介绍： Ppro定义路的乘法 Psum定义路的加法 MatPro定义路矩阵的乘法 isaPorC判断是否为路或圈 P2subP提取路的和式中的所有路 getP1从邻接矩阵得到1阶路矩阵(get a k-path from a graph)

2014-09-13 07:05:38下载

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labrosa-coversongid
Beat Tracking and Music Matching - an easy-to-use beat tracker in Matlab, plus code for describing music audio as per-beat chroma features, which is an effective representation for matching songs that have the same melodic/harmonic content despite changes in tempo or instrumentation.

2015-03-30 22:22:34下载

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LMF
调频连续波仿真程序。完成调频连续波的调制仿真。(Frequency Modulated Continuous Wave Simulation)

2013-12-26 10:19:47下载

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深度学习中的DBN学习，内含有matlab的m文件，学习深度学习不可错过的东西！

2023-04-05 13:05:04下载

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anquanguohe
先定义允许状态S={(x,y)|x=0或3,y=0,1,2,3 x=y=1,2}, 允许决策D={(u,v)|u+v=1,2},求dk<-D(k=0,1,2,…,n),使状态Sk<-S按照状态转移律S(k+1)=S(k)+(-1)^k*dk,由初始状态 S1=[3,3]经有限步到达状态S(n+1)=[0,0].此为求解多步决策问题( First define to allow the state S = ((x, y) | x = 0 or 3, y = 0,1,2,3 x = y = 1,2), to allow the decision-making D = ((u, v) | u+ v = 1,2), seek dk < -D (k = 0,1,2, ..., n), so that state Sk < -S state transition in accordance with law S (k+1) = S (k )+ (-1) ^ k* dk, from the initial state S1 = [3,3] by the finite-step to reach the status S (n+1) = [0,0]. This is a multi-step decision-making problem solving)

2010-02-17 12:02:50下载

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EpEsQ

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