登录
首页 » matlab » 光通信中3*3耦合器相位解调的仿真 phase-demodulation

光通信中3*3耦合器相位解调的仿真 phase-demodulation

于 2014-08-28 发布 文件大小:2KB
0 321
下载积分: 1 下载次数: 25

代码说明:

  光通信中3*3耦合器相位解调的仿真,包括对称解调与非对称解调两种。(Optical Communication 3* 3 coupler phase demodulation of the simulation, including demodulation and demodulation of two symmetric asymmetric.)

文件列表:

phase demodulation1.m,2180,2013-11-15
phase demodulation2.m,1649,2014-01-12

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

  • 8puzzle
    这是一个非常好的程序,主要用来解决模糊变结构控制问题,是我导师给我的,现在和大家分享。(fsekjl dfsdfj psdfsdf sdfklsdjfopsd)
    2009-02-19 10:57:18下载
    积分:1
  • CUM3X
    说明:  计算信号的高阶统计量,用于信号估计和分析( Use, duplication, or disclosure by the Government is subject to restrictions as set forth in subparagraph (c) (1) (ii) of the Rights in Technical Data and Computer Software clause of DFARS)
    2010-04-10 17:22:35下载
    积分:1
  • MTDataRazor
    读取大地电磁测深时间序列文件.ts,由V5-2000采集(read the mt .ts files acqured by v5-2000 system.)
    2014-12-19 17:01:29下载
    积分:1
  • exp
    基于matlab的gui代码实现下拉菜单(matlab gui menu)
    2011-12-04 20:31:57下载
    积分:1
  • ESmart
    智能天线波束赋型的研究,使智能天线波束赋型角度更加准确。(Beamforming smart antenna research, so that smart antenna beamforming point of view more accurate.)
    2009-11-15 11:20:13下载
    积分:1
  • Structures.tar
    Structures Theory in Spanish.
    2010-01-06 23:37:59下载
    积分:1
  • SOM
    说明:  基于自组织特征映射(SOM)的系统辨识算法实现(Based on self-organizing feature map (SOM) of the system identification algorithm)
    2008-10-14 23:16:44下载
    积分:1
  • music
    用matlab实现非相干信号music算法的DOA估计(Using matlab to achieve non-coherent signal music algorithm for DOA estimation)
    2012-05-10 14:49:31下载
    积分:1
  • upf_demos
    UKF PF 等滤波算法代码,比较全,适合与初学者仿真实验(UKF PF and other filtering algorithm code, more whole, suitable for beginners simulation with)
    2011-08-19 16:46:04下载
    积分:1
  • EvaluateReversePolishNotation
    Evaluate the value of an arithmetic expression in Reverse Polish Notation. Valid operators are +, -, *, /. Each operand may be an integer or another expression. Some examples: [ 2 , 1 , + , 3 , * ] -> ((2 + 1) * 3) -> 9 [ 4 , 13 , 5 , / , + ] -> (4 + (13 / 5)) -> 6 分析:这一题就是写程序计算逆波兰式的结果,遍历表达式,碰到操作数入栈,碰到操作符就从栈顶取出两个操作数,再将计算后的结果入栈,最后栈中剩余的唯一操作数就是计算结果。(. Evaluate the value of an arithmetic expression in Reverse Polish Notation Valid operators are+,-,*,/Each operand may be an integer or another expression Some examples:.. [" 2" , " 1" , " +" , " 3 " ," * " ]-> ((2+ 1)* 3)-> 9 [" 4 " ," 13 " ," 5 " ," /" ," + " ]-> (4+ (13/After this problem is to write a program to calculate the inverse Polish results, iterate expression operand stack encounter, encounter operator to fetch two operands the stack, and then calculated as follows:> 6 Analysis- 5)) The results of the stack, the only remaining operand stack is the final calculations.)
    2015-03-15 23:06:09下载
    积分:1
  • 696516资源总数
  • 106914会员总数
  • 0今日下载