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MATLAB-statement
说明： matlab基础知识，PPT文件，内部资料简单，对于深入学习的人不是很合适(matlab basic knowledge, PPT files, internal information is simple, the depth of learning are not very suitable)
- 2008-11-22 15:22:11下载
- 积分：1
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m1
说明： 交通流问题的matlab模拟，元胞自动机的思想实现(Matlab simulation, cellular automata traffic flow problems thought to achieve)
- 2014-12-21 21:46:07下载
- 积分：1
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code
根据双尺度差分方程构造小波函数 初始函数 矩形波/三角波(The dual-scale construction of wavelet function differential equation initial wave function rectangular/triangular wave)
- 2015-01-02 19:06:50下载
- 积分：1
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lad2src.tar
code for shell command implementation(code for shell peogramming)
- 2013-11-23 15:42:59下载
- 积分：1
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m
说明： matlab圆柱投影的代码，辛苦管理员了，初学者，求教啊(Matlab code of cylindrical projection, hard administrator, beginners, for advice
)
- 2015-10-10 19:57:20下载
- 积分：1
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powclip
COFDM系统在不同调制方式（BPSK/QPSK/16PSK）下的PPCR和BER的比较。不含噪声和多径(COFDM modulation system in different ways (BPSK/QPSK/16PSK) under the PPC R and BER comparison. Without noise and multipath)
- 2007-04-05 17:09:53下载
- 积分：1
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ScieCompLec6Q1
solve boundary value problem of
Cousera课程Scientific Computing的Lecture 6对应的Quiz中Question 1的求解程序。
求解常微分方程的边值问题。
y =-4y (y of Theory: y=0.5*sin(2x))
x∈[0,1]
BC: y(0) = 0 y (1) = cos(2).
(solve boundary value problem of Cousera course Scientific Computing' s Lecture 6 corresponds Quiz Question 1 in the solution process. Solving boundary value problems for ordinary differential equations. y' ' =-4y (y of Theory: y = 0.5* sin (2x)) x ∈ [0,1] BC: y (0) = 0 y ' (1) = cos (2).)
- 2013-05-22 12:47:54下载
- 积分：1
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matlab11
matlab的使用与实现，对于刚入门matlab的会员来说绝对有用啊！(matlab use and realize, for just a member of matlab entry is absolutely useful ah!)
- 2008-02-29 15:52:01下载
- 积分：1
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ditujiasupso
说明： 一种带有梯度加速的粒子群算法，可以实现多种优化工作的需要哦。(With a gradient to accelerate the particle swarm algorithm, can achieve the needs of a variety of optimization Oh.)
- 2008-10-06 09:22:45下载
- 积分：1
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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
