登录
首页 » matlab » 抗积分饱和PID控制算法及仿真抗积分饱PIDsimulation2

抗积分饱和PID控制算法及仿真抗积分饱PIDsimulation2

于 2020-07-07 发布 文件大小:1KB
0 252
下载积分: 1 下载次数: 26

代码说明:

  抗积分饱和PID控制算法及仿真抗积分饱和PID控制算法及仿真抗积分饱和PID控制算法及仿真抗积分饱和PID控制算法及仿真(Anti-integral saturation PID control algorithm and simulation)

文件列表:

CHAP1_6.M

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论


0 个回复

  • console1.6
    从多媒体文件中搜索给定的视频帧号,这些帧有一个start_code标志位,本算法是融合顺序和预测相结合的方法,算法的执行时间以及内存空间都比较少。(search the presibed viedo frame in multimedia files,with the startcode flag.Use the predict algorithm and sequence algorithm. The algorithm demand small time and little memory space.)
    2011-06-17 15:00:59下载
    积分:1
  • e00compr-0.5.tar
    GIS相关工具,可以读写压缩的e00文件(GIS tools that can read and write file compression e00)
    2005-02-02 19:40:08下载
    积分:1
  • 9019633
    The physical layer of WCDMA is explained in detail. The downlink physical channel structure, uplink physical channel structure and timing relationship between channels are described in detail.
    2019-01-04 19:01:44下载
    积分:1
  • versinfo
    class for retrival of executable file and product version information from VERSIONINFO resource in C++
    2012-10-19 00:30:05下载
    积分:1
  • bldcbcheng
    用来仿真直流无刷电机控制系统,该系统运行良好,修改参数可以模拟任意电机的运行(used for simulation brushless DC motor control system, which is operating well, Parameter changes can simulate the operation of any motor)
    2007-05-28 08:57:11下载
    积分:1
  • medo
    设X[ 0 : n - 1]和Y[ 0 : n – 1 ]为两个数组,每个数组中含有n个已排好序的数。找出X和Y的2n个数的中位数。  编程任务 利用分治策略试设计一个O (log n)时间的算法求出这2n个数的中位数。 数据输入 由文件input.txt提供输入数据。文件的第1行中有1个正整数n(n<=200),表示每个数组有n个数。接下来的两行分别是X,Y数组的元素。结果输出 程序运行结束时,将计算出的中位数输出到文件output.txt中(Let X [0: n- 1] and Y [0: n- 1] for the two arrays, each array containing the n number has been sorted. 2n X and Y to identify the number of digits.  programming tasks using the divide and conquer strategy try to design an O (log n) time algorithm to calculate this median number 2n. Data input by the input data provided input.txt file. The first line in the file has a positive integer n (n < = 200), that there are n numbers of each array. The next two lines are the X, Y array elements. The end result is output program runs, the calculated median output to file output.txt)
    2021-03-22 16:29:16下载
    积分:1
  • Chap11
    c plusplus beginner structures example source codes
    2012-06-11 01:46:02下载
    积分:1
  • Hufferman
    使用哈弗曼编码技术进行数据压缩和解压缩,缺点在于只能用于文本(Havermann coding techniques using data compression and decompression, disadvantage is that the text can only be used)
    2010-01-03 19:23:10下载
    积分:1
  • Fibonacci
    Gives the user s desired number of Fibonacci Sequence
    2010-11-24 21:51:35下载
    积分:1
  • Dry-eye
    this is CLIPS code for eye disease
    2011-08-18 11:04:38下载
    积分:1
  • 696518资源总数
  • 104881会员总数
  • 38今日下载