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aco
蚁群算法实现tsp问题的解，适合初学者使用(Ant colony optimization tsp solution of the problem, for beginners)
- 2010-06-28 16:55:01下载
- 积分：1
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CarBacker
基于matlab的倒车模糊控制器设计，避免 死区 问题(Matlab fuzzy controller based on the reverse design, to avoid the ‘dead zone’ problem)
- 2021-03-25 13:49:13下载
- 积分：1
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music
经典的music算法（包含多个），用于无线通信的来波方向估计。(Classical music algorithm (contains more than one) for wireless communications, estimated DOA.)
- 2008-06-24 16:15:02下载
- 积分：1
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IIR
使用iir方法设计的数字滤波器，通过验证，功各位使用。(Designed using iir digital filter, through verification, power you use.)
- 2010-05-21 21:14:24下载
- 积分：1
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P92_ODE_EulerMethod
Euler method used for solving ordinary differential equation.
- 2014-10-15 17:48:59下载
- 积分：1
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MLAMBDA1.0
GPS，rtk模糊度解算，和文章“A modified LAMBDA method for integer least-squares estimation”匹配的程序，配合学习(GPS, RTK, code for A modified LAMBDA method for integer least-squares estimation )
- 2016-06-10 03:30:19下载
- 积分：1
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多材料热接触的传热分析
基于ANSYS的多材料热接触的传热分析的命令流(Command Flow for Heat Transfer Analysis of Multi-material Thermal Contact Based on ANSYS)
- 2020-06-16 19:40:01下载
- 积分：1
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QAM
编写一个含QAM传输的发送和接收模块
1)分别采用复基带仿真和实带通仿真两种形式
a) 实带通仿真时, 载波频率为符号率的4倍
b)发端采用滚降系数0.5的根号升余弦滤波器
2)画出Eb/N0=15dB时的接收波形(收滤波前),取100个采样
3)画出发端输出的眼图和收端过匹配滤波后的眼图(画眼图时不加噪声)
统计误符号率和误比特率与Eb/N0的关系,画出曲线, 与理论计算的曲线相对比
(Prepare to send of one with QAM transmission and receiver module 1), respectively, the complex baseband simulation and the real band-pass simulation in two forms a) the real band-pass simulation, the carrier frequency used for the symbol rate of 4 times b) the originator roll-off factor of 0.5 root raised cosine filter, 2) draw Eb/N0 = 15dB when the received waveform (closing filter), take 100 samples) to draw the originator of the output eye diagram and income side of the eye diagram after the matched filter ( draw the eye diagram without noise) statistics symbol error rate and bit error rate and Eb/N0, draw a curve calculated with the theoretical curve compared)
- 2012-05-29 16:51:47下载
- 积分：1
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LDPCcode
利用函数MadHG生成规则LDPC码的校验矩阵H，其行重为6,列重为3,行数为列数一半(行数越大越好)，H中任意两列没有围长为4的圈，并得到H对应的生成矩阵G，可以保证mod(G*H ,2)=0。使用方法为:[H,G] = MadHG(m,n,x),x= 1(得到的G左半部分为单位阵) or 2(G右半部分为单位阵),(use function MadHG Generation rules LDPC check matrix H, re-6, series of three, out for a few half of the number (A bigger the better), H 2 arbitrary no girth of the four circles, H counterparts and with the formation of matrix G, we can be assured of mod (G* H, 2) = 0. To be used : [H, G] = MadHG (m, n, x), x = 1 (the G part of the Left Front units) or 2 (G right half of the units RUF),)
- 2007-05-12 10:49:48下载
- 积分：1
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Moravec
morave code for matlab
- 2013-01-27 14:35:57下载
- 积分：1
