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exam4

于 2013-07-18 发布 文件大小:33KB
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  一个时滞微分方程的例子及matlab程序代码(These are some delay differential equations matlab code.)

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exam4.m,1832,2010-10-12

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  • shenjingwangluo
    T=[1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1] 输入向量的最大值和最小值 threshold=[0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1] net=newff(threshold,[31 3],{ tansig , logsig }, trainlm ) 训练次数为1000,训练目标为0.01,学习速率为0.1 net.trainParam.epochs=1000 net.trainParam.goal=0.01 LP.lr=0.1 net = train(net,P,T) 测试数据,和训练数据不一致 P_test=[0.2101 0.0950 0.1298 0.1359 0.2601 0.1001 0.0753 0.0890 0.0389 0.1451 0.0128 0.1590 0.2452 0.0512 0.1319 0.2593 0.1800 0.0711 0.2801 0.1501 0.1298 0.1001 0.1891 0.2531 0.0875 0.0058 0.1803 0.0992 0.0802 0.1002 (T = [1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1] ' of the maximum and minimum input vector threshold = [0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1] net = newff (threshold, [31 3], {' tansig' , ' logsig' }, ' trainlm' ) training times for the 1000 target of 0.01 training, learning rate of 0.1 net.trainParam.epochs = 1000 net. trainParam.goal = 0.01 LP.lr = 0.1 net = train (net, P, T) test data, and training data inconsistencies P_test = [0.2101 0.0950 0.1298 0.1359 0.2601 0.1001 0.0753 0.0890 0.0389 0.1451 0.0128 0.1590 0.2452 0.0512 0.1319 0.2593 0.1800 0.0711 0.2801 0.1501 0.1298 0.1001 0.1891 0.2531 0.0875 0.0058 0.1803 0.0992 0.0802 0.1002 )
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