登录
首页 » matlab » DSTBC

DSTBC

于 2016-06-30 发布 文件大小:6KB
0 211
下载积分: 1 下载次数: 27

代码说明:

  差分空时编码,解码译码都有,无信道,含新版本matlab缺少的旧版函数dmodce(Differential space-time encoding, decoding decoding has no channel)

文件列表:

apkconst.m,2332,2016-06-27
dmodce.m,9532,2016-06-27
STBCDecode.m,1286,2016-06-29
STBCEncode.m,876,2016-06-29

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

  • ParameterEstimation
    说明:  用matlab解决数理统计中的对参数的区间估计的问题(Using matlab to solve mathematical statistics in interval estimation of the parameters of the problem)
    2010-04-24 21:00:17下载
    积分:1
  • zernike
    zernike矩的源代码,非常不错,用matlab编的,值得下载(zernike moments of the source code, very good, with matlab code, it is worth downloading)
    2014-09-20 14:28:11下载
    积分:1
  • newton_interpolation.m
    newton_interpolation berechnet das Newton-Interpolationspolynom
    2009-10-25 01:01:39下载
    积分:1
  • New-WinRAR-ZIP-archive
    MATLAB PROGRAM FOR BISECTION METHOD
    2012-07-23 12:17:51下载
    积分:1
  • rgb2raw
    实现rgb图像到raw格式的image的转化。(it can use to change the rgb image into raw image.)
    2010-12-01 13:47:54下载
    积分:1
  • linearequationsthe-iterative-method
    第12章 解线性方程组的迭代法.zip。是从书中光盘里直接拷贝的代码,放在这里,与大家共享,共同交流,希望对大家有帮助。(Chapter 12 solution of linear equations of the iterative method)
    2012-03-27 11:03:40下载
    积分:1
  • gui.tar
    ejemplo de desarrollo de guis en matlab
    2010-05-15 19:36:56下载
    积分:1
  • narayani__ref
    A multiplexing system with NT transmit and NR equals NT receive antennas is designed. The main challenge fo systems lies in the non-orthogonality of the transmission channel. This can be solved by turbo processing. Turbo channel encoder is used for converting the bits in its suitable form. Then different M-ary QAM is compared.
    2013-11-26 13:19:04下载
    积分:1
  • matlab-Ant-colony-algorithm
    蚁群算法的源代码 并带有详细注释 强烈推荐!(Ant colony algorithm source code with detailed notes and highly recommended!)
    2013-04-17 22:28:32下载
    积分:1
  • viterbi-matlab
    卷积编码器g1=111,g2=101 x为输入的待译码序列 x=[1,0,1,0,0,1,0,0,0,1,0,1,1,1] x=[1,1,0,1,0,1,1,0,0,1] x=[1,1,1,0,0,0,1,0,1,1] a=size(x) 输入序列的长度 s=a(2)/2 译码后的m序列长度为x的一半 m=zeros(1,s) 最终结果存放 ma=zeros(1,s+1) 存放Fa路径的 mb=zeros(1,s+1) 存放Fb路径的 mc=zeros(1,s+1) 存放Fc路径的 md=zeros(1,s+1) 存放Fd路径的(viterbi decoder)
    2015-04-20 17:06:18下载
    积分:1
  • 696516资源总数
  • 106914会员总数
  • 0今日下载