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FSK
基于matlab的加性高斯白噪声信道的FSK调制(Matlab based on the additive white Gaussian noise channel FSK modulation)
- 2009-04-25 15:36:42下载
- 积分:1
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table
说明: 含有多种table表格样式,满足界面开发素材需求(table)
- 2010-03-26 12:15:12下载
- 积分:1
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LSVM
经典局域支持向量机(LSVM)算法,常用于预测领域。(Classic local support vector machine (LSVM) algorithm, commonly used in forecasting.)
- 2011-07-12 09:58:45下载
- 积分:1
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07051919v1
Optimal Watermark Embedding and Detection Strategies Under Limited Detection Resources
关于数字水印的最新外文资料(2007年)(Optimal Watermark Embedding and Detectio Under n Strategies Limited Resource Detection s digital watermark on the latest information on foreign language (2007))
- 2007-06-23 11:23:16下载
- 积分:1
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Airplane.Stability
飞行稳定性与自动控制,尼尔森的书,中文版,希望对大家有所帮助(Flight stability and automatic control, Nelson' s book, the Chinese version, we hope to help)
- 2013-04-09 15:34:06下载
- 积分:1
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mainNewBlockMatrix
阻塞矩阵多频信号主瓣干扰抑制-画自适应波束形成图f=6、7、8、9MHz,目标角度10,干扰11,20,-20强度分别为20db,40db,20db,20db-(Blocking matrix multi-frequency signal main lobe interference suppression- painting adaptive beamforming Figure f = 6, 7, 8, 9 MHz, target angle, interference 11,20,-20 strength to 20db, 40db, 20db, 20db-)
- 2012-11-01 10:58:14下载
- 积分:1
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State_of_the-art_MOS_Comparison
阵列信号处理信源数估计算法,用来估计信源数目(Matlab code of state-of-the-art 1-D model order selction (also known as source enumeration, estimating the number of source ignals impinging on an array of sensor) methods. The main function "main.m" compares the performance of them.)
- 2013-11-26 15:56:59下载
- 积分:1
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polar matlab
极化码在高斯信道下CA——SCL译码算法(CA - SCL decoding algorithm for polarization code in Gauss channel)
- 2017-07-11 09:42:55下载
- 积分:1
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YTY_GA_Final
To find the largest fitness value and its location
寻找最大适应性及相应的位置!Population N=50;crossover bits=n/2 (half of bits of an individual) with random locations,mutation bits = 4 种群数 N=50,交换位数= n/2, 即个体位数的一半,且位置随机;
变异位数统一取为4;
Nc=20,28,36,44,individuals for crossover(交换的个数)。Nm=1,5,10,15, individuals for mutation(变异的个数)。(To find the largest fitness value and its location to find the location of the maximum adaptability and the corresponding! Population N = 50 crossover bits = n/2 (half of bits of an individual) with random locations, mutation bits = 4 population size N = 50, exchange digits = n/2, ie half of the median individual, and random position variation is taken as unity median 4 Nc = 20,28,36,44, individuals for crossover (exchange number). Nm = 1,5,10,15, individuals for mutation (the number of variations.))
- 2010-12-27 21:23:27下载
- 积分:1
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Polar_code
polar code simulation result
- 2020-11-16 14:59:40下载
- 积分:1