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momitor
监视程序,能监视系统有没有打开扫雷程序,如果打开就关闭,其窗口自动隐藏,按ALT+F10调出窗口,(Monitoring program, can monitor the system has no open mine program, if the opening is closed, its windows auto-hide, press ALT+ F10 transferred out of the window,)
- 2009-10-03 03:34:46下载
- 积分:1
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LMD
局域均值分解的程序,此为经验模态分解方法的另一种改进,此方法能够减小端点效应的影响,但是模态混叠问题还是没有得到完全抑制(Local mean decomposition process, this is the empirical mode decomposition method is another improvement, this method can reduce the effect of the endpoint, but the modal aliasing problem has still not been completely suppressed)
- 2020-12-28 10:49:02下载
- 积分:1
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P
说明: WLS的电力系统状态估计,能够迅速完成节点迭代(WLS state estimation of power system that can quickly complete node iterator)
- 2015-05-21 17:03:42下载
- 积分:1
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df
说明: 分布式环境的计算,课件的使用,以及各种资源。。。(The research paper of “Segmentation methods of fruit image based on color difference” is mainly about four segmentation methods in fruit-harvesting robot vision, which respectively are dynamic threshold segmentation method, extended Otsu method, improved Otsu combined with genetic arithmetic and adaptive segmentation method based on LVQ network. It is written in a good structure. )
- 2009-11-10 22:53:26下载
- 积分:1
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P_w
描述不同风速下风力机转速和功率之间的关系曲线(Describe the relationship between the speed and power of the wind turbine in different wind speeds curve)
- 2021-04-21 20:08:49下载
- 积分:1
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R(D)
这个程序可用来画R(D)函数,学习信息论的一般要做这个。(This procedure can be used to draw R (D) function, learning the general information theory to do this.)
- 2010-05-11 17:15:41下载
- 积分:1
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computer1
利用matlab进行商品销售的计算机仿真算法
根据售货情况,计算进货情况(Merchandising using matlab simulation algorithm based on the sales situation, calculate the purchase situation)
- 2013-09-06 01:01:53下载
- 积分:1
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kmanes2
k means clustering algorithm new 2 cluster
- 2013-11-04 05:00:43下载
- 积分:1
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GraphSeg
this program for efficient graph based image segmentation
- 2010-08-24 12:07:09下载
- 积分:1
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LTE_System_Level_1.4_r570
To solve this instance of the decision problem we must determine whether there is a truth value (TRUE or FALSE) we can assign to each of the variables (x1 through x4) such that the entire expression is TRUE. In this instance, there is such an assignment (x1 = TRUE,x2 = TRUE, x3=TRUE, x4=TRUE), so the answer to this instance is YES. This is one of many possible assignments, with for instance, any set of assignments including x1 = TRUE being sufficient. If there were no such assignment(s), the answer would be NO.
- 2012-04-23 22:00:12下载
- 积分:1