登录
首页 » matlab » evolutionary_game_WSLS

evolutionary_game_WSLS

于 2009-01-31 发布 文件大小:2KB
0 197
下载积分: 1 下载次数: 133

代码说明:

  wsls在一个无标度网络上的进化博弈,迭代了20代,代数可更改(wsls in a scale-free network on the evolutionary game, iterative 20 generation, algebra can change)

文件列表:

evolutionary_game_WSLS.m

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • thewholeloop
    rls自适应滤波算法实现,仿真结果是误码率曲线的比较(rls adaptive filtering algorithm, simulation results of bit error rate curve is relatively)
    2009-06-23 21:07:59下载
    积分:1
  • K-PNN
    K-PNN Algorithm a type of k nearest neighbor algorithm
    2010-02-04 16:35:14下载
    积分:1
  • AuxParticle
    辅助粒子滤波matlab代码,包括和粒子滤波的比较(Auxiliary particle filter matlab code, including comparison of the particle filter)
    2013-05-10 20:45:33下载
    积分:1
  • denoise_fft_angle_Analysis
    计算时间序列的快速傅里叶变换并画出时域和频域图,之前进行小波降噪。(Calculate the time series and fast Fourier transform time-domain and frequency domain plot diagram, before wavelet decomposition.)
    2010-09-25 21:47:10下载
    积分:1
  • Speaker-recognition-system
    matlab实现语者识别系统,可以实现录音保存播放功能,在后和库中的声音进行对比,检测出这是谁的声音。(matlab implementation speaker recognition system, can save the recording playback, the voice of the library and after comparison, to detect who is this voice.)
    2011-04-26 09:28:31下载
    积分:1
  • friction-model-Simulink
    friction model simulink
    2011-07-01 13:32:55下载
    积分:1
  • New-Microsoft-Word-Document
    Thresholding of i/p image
    2013-12-17 18:57:07下载
    积分:1
  • IIT_Filter
    MATLAB下通过impulse invariant transformation方法设计数字滤波器,包含低通和带通滤波器代码。(Using MATLAB m-file to design LP and BP digital filter by impulse invariant transformation method.)
    2010-06-27 03:32:53下载
    积分:1
  • optical-filter
    sevenlayer optical filter comparision the analytical method with reflections
    2011-10-01 17:40:58下载
    积分:1
  • we
    说明:  七单元天线阵DOA估计 clear clc d=1 天线阵元的间距 lma=2 信号中心波长 q1=1*pi/4 q2=1*pi/3 q3=1*pi/6 q4=3*pi/4 四输入信号的方向 A1=[exp(-2*pi*j*d*[0:6]*cos(q1)/lma)] 求阵因子 A2=[exp(-2*pi*j*d*[0:6]*cos(q2)/lma)] A3=[exp(-2*pi*j*d*[0:6]*cos(q3)/lma)] A4=[exp(-2*pi*j*d*[0:6]*cos(q4)/lma)] A=[A1,A2,A3,A4] 得出A矩阵 n=1:1900 v1=.015 四信号的频率 v2=.05 v3=.02 v4=.035 d=[1.3*cos(v1*n) 1*sin(v2(we)
    2010-05-06 15:26:50下载
    积分:1
  • 696516资源总数
  • 106914会员总数
  • 0今日下载