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22355981wsn_matlab2
wsn wit mat lab code
- 2010-12-08 20:58:33下载
- 积分：1
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ARIMAtest
对于时间序列模型建立ARMA预测模型，可对未来值进行预报。(For the time series model, the ARMA prediction model is established, and the future value can be forecasted.)
- 2017-07-12 16:48:59下载
- 积分：1
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fast_WT_1D
本程序实现不用对数据进行2抽取和2插值的快速小波分解,只需对滤波器进行2插值即 可，所以不要求数据的长度为2的幂次方，共分解了3次，然后进行了重构！！！(This procedure achieved without the data extraction and 2 Fast 2 interpolation wavelet decomposition, only 2 of the interpolation filter that can be, so the data length does not require a power of 2, a total break down three times, then refactored! ! !)
- 2011-10-20 16:50:13下载
- 积分：1
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FSK_signals_demod
fsk modulation with matlab
- 2012-05-27 18:46:07下载
- 积分：1
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three-algorithms-about-histogram
本代码实现了三种直方图相关的算法，分别为直方图均衡化，直方图规定化，对比受限自适应直方图均衡化。三种方法放在3个不同的文件夹下，直接运行test文件，就可以得到运行结果(This code implements three histogram algorithm, respectively histogram equalization, histogram provides contrast limited adaptive histogram equalization. Three methods on three different folder, run directly test file, you can get the results)
- 2021-02-12 11:39:52下载
- 积分：1
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fbmc
一个FBMC/OQAM的仿真程序，可以实现调制与解调，得到原型滤波器和OQAM调制解调信号图(A FBMC/OQAM simulation program, you can achieve modulation and demodulation, the prototype filter and OQAM modulation and demodulation signal)
- 2021-03-09 21:39:27下载
- 积分：1
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rcs_frequency
雷达rcs是雷达的重要参数，为此编写了雷达rcs随频率的变化关系。方便新入门的了解rcs在不同情况下的特征(Radar rcs important parameters of the radar, for the preparation of the Radar rcs variation with frequency. Understanding rcs facilitate new entry characteristics in different cases)
- 2020-11-04 19:19:52下载
- 积分：1
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code
说明： matlab环境下的iris 和 sonar的数据聚类。包括GA、k-means、FCM（只有iris）。(Iris and sonar data clustering in Matlab environment. It includes GA, K-means and FCM(only for iris).)
- 2020-07-15 10:50:28下载
- 积分：1
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Demo3_TransferLearningwithDeepNetworkDesigner
说明： 使用Matlab深度学习工具箱-迁移学习(Transfer Learning with Deep Network Designer)
- 2021-01-02 18:09:13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
