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PID_controller_with_integration_sturation
说明： 该算法连续系统的抗积分饱和的PID算法,用MATlab开发，能够管直接运行,可明显减弱几分饱和现象对系统稳定性的影响，当控制量与设定值偏差较大时，取消积分作用，以免积分作用过大引起系统的稳定性降低。(The algorithm of the anti-continuous system saturated PID integral algorithm, developed using Matlab and can directly run the tube can be significantly weakened somewhat saturated phenomenon on system stability, when the control volume and settings deviation is larger, integral role in the abolition of so as not to cause too large integral role in the stability of the system lower.)
- 2008-09-06 13:20:50下载
- 积分：1
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Perceptron
说明： matlab语言编写的iris数据的单层感知器分类(matlab data written in a single layer perceptron iris classification)
- 2011-04-02 11:16:29下载
- 积分：1
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coherence_of_injection1and2
相关系数计算的模拟,用于衡量两个数列的相关程度(coherence coefficient calculation)
- 2010-05-17 23:04:49下载
- 积分：1
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robotbz
本程序模拟一个机器人躲避障碍的仿真.其中有不完善的地方
迫于精力有限,具体过程还有点粗糙,读者可以加以修正
(a simulation of the robot avoid obstacles to the simulation. Which forced the inadequacies in the energy are limited, There is a process in concrete rough, and readers can be amended)
- 2006-10-21 21:32:48下载
- 积分：1
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jili
本代码是解决结构动力学中经典的问题的源程序并且采用了gui设计，一定程度上实现了人工智能的作用(This code is to solve the classic problem in structural dynamics of the source and using the gui design, implemented to some extent the role of artificial intelligence)
- 2011-06-04 00:08:30下载
- 积分：1
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FTF
FTF自适应算法
求：自适应滤波器的两个权值ω0 (n)和ω1(n)分别收敛于0.2和0.7的收敛轨迹，并与LMS算法进行对比。(FTF adaptive algorithm
Requirements: two weights ω0 is (n) of the adaptive filter and ω1 (n) converges to 0.2 and 0.7, convergence trajectory, and compared with the LMS algorithm.)
- 2020-12-10 23:39:18下载
- 积分：1
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masslessfourbar
基于SimMechanics的刚体系统建模masslessfourbar案例(SimMechanics modeling)
- 2015-03-29 22:46:18下载
- 积分：1
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案例27
说明： 运用遗传算法对BP神经网络输入自变量进行降维(Dimension Reduction of Input Independent Variables of BP Neural Network Using Genetic Algorithms)
- 2019-04-24 20:10:53下载
- 积分：1
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nntest6
function nntest6(action,flag)
NNTEST6 View and Control Neural network.
See also SIMUFF.
Koos j. den Oudsten, 1-20-99
koos@phil.uu.nl
Copyright (c) 1998-99 by KoosSoft vof
$Revision: 0.4 $ $Date: 1999/01/27 22:29:28 $
global Xx w1 b1 w2 b2 hidden mlp
if nargin < 1
action = start
end
On recursive calls get all necessary handles and data.
if ~strcmp(action, start )
childList = allchild(0)
nn_fig = childList(find(childList == gcf))
ud = gco
popupvalue = get(ud.popup, Value )
- 2010-03-04 17:12:54下载
- 积分：1
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shuxuejianmo_1
说明： 某厂生产甲乙两种口味的饮料,每百箱甲饮料需用原料6千克,工人10名,可获利10万元 每百箱乙饮料需用原料5千克,工人20名,可获利9万元.今工厂共有原料60千克,工人150名,又由于其他条件所限甲饮料产量不超过8百箱.问如何安排生产计划,即两种饮料各生产多少使获利最大.进一步讨论:
1)若投资0.8万元可增加原料1千克,问应否作这项投资.
2)若每百箱甲饮料获利可增加1万元,问应否改变生产计划.
(Factory production of beverages and B in two flavors, each box 100 need a drink six kilograms of raw materials, workers 10 to 100,000 yuan profit B beverage cartons per cent five kilograms of raw materials needed, workers 20, the profitability of 90,000 yuan. this plant a total of 60 kilograms of raw materials, 150 workers, and because other conditions do not yield a beverage cartons more than 8 per cent. ask what arrangements the production of projects, namely the production of two the number of drinks to make the biggest profit. further discussion : 1) if the investment 08,000 yuan to increase one kilograms of raw materials and asked whether we should make this investment. 2) If a beverage cartons per cent profit increase 10,000 yuan and asked whether or not to change the production plan.)
- 2009-07-24 22:57:35下载
- 积分：1
