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MatlabtutorialNotes
Tutorial on Matlab: Basic Matlab Commands and Syntax (very useful), Enjoy
- 2010-10-15 23:54:12下载
- 积分:1
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wavread
说明: 基于RlS(递归式最小均方)自适应滤波算法的C程序的仿真代码(Adaptive filtering algorithm based RlS C program simulation code)
- 2013-12-19 10:52:56下载
- 积分:1
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现代永磁同步电机控制原理及MATLAB仿真
说明: 书籍以及对应的例程,都可以使用 非常好用,而且每个章节也会有对应的模型,是有关于PMSM模型的,可以学习到 滑模控制、FOC控制以及弱磁等控制巴拉巴拉的(Books and corresponding routines can be used very easily. Moreover, there will be corresponding models in each chapter, which is about PMSM model. You can learn sliding mode control, FOC control and flux weakening control of Balabala)
- 2021-04-27 21:08:45下载
- 积分:1
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Matlab
MAtLAB实用教程,有大量的例题,适合初学者快速上手(MATLAB practical tutorials, there are a large number of Example, suitable for beginners and running quickly)
- 2009-09-20 23:43:21下载
- 积分:1
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zxg
实现白噪声、正弦信号和带有白噪声的正弦信号的自相关函数的分析。(Achieve white noise, white noise with a sinusoidal signal and a sinusoidal signal of the autocorrelation function.)
- 2013-08-30 09:46:14下载
- 积分:1
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Flight-Vehicle-Design
基于分布式视觉的飞行器位姿估计与路径规划(Document of design)
- 2014-12-08 08:48:57下载
- 积分:1
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Project_uGUI_map_file
dfydrtyrt df d erzftg zer ter ter
- 2019-01-09 18:49:18下载
- 积分:1
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HarmonicsANDapfft
电力系统谐波监测相关的文章其中的fft和apfft之间的对比很有参考意义(Fft and apfft contrast between the power system harmonic monitoring related article which is very useful)
- 2014-12-11 11:20:01下载
- 积分:1
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Untitled
设有一个随机信号 服从AR(4)过程,它是一宽带过程,参数如下:
我们通过观测方程 来测量该信号, 是方差为1的高斯白噪声,用LMS算法和RLS算法通过观测方程来估计原信号。并用Matlab对此问题进行仿真。
(There is a random signal obedience AR (4) process, which is a broadband process parameters are as follows: We are measured by observing the signal equation is a Gaussian white noise variance, with the LMS algorithm and RLS algorithm to estimate the observation equation the original signal. This issue with Matlab simulation.)
- 2020-11-19 22:09:37下载
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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