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liziqunann
说明: 5篇有关粒子群优化神经网络的文章,caj格式的(Particle swarm optimization of neural networks related to the article, caj format)
- 2011-04-17 11:18:52下载
- 积分:1
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Exponential-growth-model
由1790-1980年间美国每隔十年的人口记录表为基础进行分析。(Exponential growth model of population
)
- 2011-05-26 00:24:00下载
- 积分:1
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fun_smpccon_smpcsim
模型预测控制仿真实例,采用无约束的模型预测控制策略(model predictive control design and simulation)
- 2012-05-10 14:45:14下载
- 积分:1
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LET
toolbox for Lyapunov exponent compute
- 2011-10-10 15:18:23下载
- 积分:1
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1
说明: 精通MATLAB综合辅导与指南例程
里面包含所有的例程(Comprehensive counseling MATLAB routines)
- 2009-05-14 03:14:43下载
- 积分:1
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ZTrans_InvTrans
说明: 对输入的一有理Z变换表达式(输入分子和分母系数序列),分别画出:
(1)零极点图
(2)z在单位圆上时z变换所对应的DTFT的幅度谱和相位谱
(3)逆z变换所得序列,在各种不同的收敛域下(包括了右边序列,左边序列和双边序列的情况)
(the importation of a rational expression Z-transform (input coefficient of the numerator and the denominator sequence), respectively paint : (1) The pole-zero plan (2) z in the unit circle at z transform the X corresponding to the amplitude spectrum and phase spectrum (3) Inverse z transform sequences derived from the convergence of various jurisdictions (including the right sequence, the sequence of the left and bilateral series))
- 2006-05-04 21:27:38下载
- 积分:1
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Perceptron
shabake asabiPerceptron.m
- 2012-05-28 01:53:29下载
- 积分:1
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ct1
变声器的matlab实现,能够变声,将男人的声音变成女人的声音等等(The the of the matlab of voice-changer to achieve, be able to voice changed turned into the, will be the the voice of the a man' s to the the voice of the a woman' s, and and so on on)
- 2013-04-09 19:24:28下载
- 积分:1
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AdaptiveequalizerofLMSalgorithmanditssimulation
自适应均衡器的LMS算法实现及其仿真 :自适应均衡器已广泛应用于通信、雷达、声纳、控制和生物医学工程等许多领域,为克服多径衰落和信道失真
引起的码间干扰,实时跟踪移动通信信道的时变特性,笔者设计了一个基于LMS 算法的自适应线性均衡器,并通过改变步
长因子Δ来分析其收敛速度和均方误差特性.(Adaptive equalizer of LMS algorithm and its simulation)
- 2009-05-25 00:38:03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