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calculate-FFT

于 2013-12-04 发布 文件大小:279KB
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下载积分: 1 下载次数: 46

代码说明:

  基于FFT平面相控阵天线方向图快速计算 (Figure quickly calculate FFT-based planar phased array antenna direction)

文件列表:

calculate FFT.pdf,316840,2013-12-04

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  • mianyi
    免疫算法的matlab代码,应该比较好用,谢谢支持!(immune algorithm Matlab code, it should be more convenient, thank you support!)
    2007-03-20 12:20:00下载
    积分:1
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    2009-12-17 11:05:46下载
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    VC与MATLAB之间的转换,通过设计可以在VC环境打开matlab代码(Conversion between VC and MATLAB open the matlab code in VC environment through design)
    2013-03-14 10:36:03下载
    积分:1
  • 用VC调用matlab的dll文件
    用VC调用matlab的dll文件(VC call Matlab dll file)
    2005-01-18 12:47:03下载
    积分:1
  • Selective
    数据链路层滑动窗口协议的设计与实现源码 信道模型为8000bps 全双工卫星信道,信道传播时延270 毫秒,信道误码率为10-5,信道提供字节流传输服务,网络层分组长度在240~256字节范围。(Sliding window data link layer protocol design and implementation of source-channel model 8000bps full-duplex satellite channel, the channel propagation delay 270 ms, the channel bit error rate of 10-5, the channel provides a byte stream transfer service, network layer packet length in the range of 240 to 256 bytes.)
    2013-12-08 11:45:53下载
    积分:1
  • bumeixia20100203
    说明:  通过此程序实现对微装配的调用微动平台的实现(Achieve through this process called micro-assembly platform for the realization of micro-)
    2010-04-08 21:49:04下载
    积分:1
  • M200042316412a
    matlab的雷达相关程序源码,包含MTIMMTDCFAR,动目标自适应处理 可直接使用。 (Matlab radar program source code contains MTIMMTDCFAR moving target adaptive processing can be used directly.)
    2012-07-19 20:25:52下载
    积分:1
  • LMS2
    说明:  LMS算法的MATlab仿真分析,结果表明LMS算法有较好的噪声消除效果!(LMS algorithm MATlab simulation results show that the LMS algorithm has better noise reduction effect!)
    2010-03-31 10:59:38下载
    积分:1
  • farsight070825DSP
    了一些资料,都没有说这个仿真平台是什么,是matlab还是直接在VC++的编译环境编程?还是有专门的仿真环境呢? 3、如果用实物验证的话,我觉得有两种方式:一是用摄像头来采集整个地图信息,智能车的位置也有摄像头来反馈;二是,整个地图保存在内存中,小车根据传感器信息自我定位,然后和保存的地图比较。(You can download the current points to 1, the source download get points one day, you can download up to 20 times. [View your upload log] [continue to upload source to increase points] You have recently downloaded the 240 source code, you should contribute something new other friends 【Source】 it? Joined the group】 【QQ, blowing boasting a chat. You can also create a QQ group you join the union,)
    2010-08-08 11:44:31下载
    积分:1
  • 0
    说明:  给定n种物品和一个背包,物品i的重量是Wi,价值是vi,被包容量是C,应该如何选择装入被包的物品,使得装入背包中物品的总价值最大?对于每种物品i,只有两种选择,装入或者不装入. 输入:第一行两个正整数n(物品数)和C(容量) 第二行n个正整数n,表示物品价值 第三行n个正整数n,表示物品重量 输出:第一行表示总价值 第二行n个数(0/1),0表示物品不装入,1表示装入 例如, 输入: 5 10 6 3 5 4 6 2 2 6 5 4 输出 15 1 1 0 0 1 (Species of a given n items and a backpack, the weight of item i is Wi, the value of vi, was packet capacity is C, should be how to choose a package of items to be loaded, the backpack load in the largest total value of goods? For each item i, there are only two options, loaded or not loaded. Input: the first line of the two positive integer n (the number of items) and C (capacity) of the second line of n positive integer n, that the value of the third line of n positive integer n, that the weight of output: the total value of the first line of the second line, said n number of (0/1), 0 that the items do not load, that load 1 for example, enter: 5,106,354,622 6 5 4 output 15 1 1 0 0 1)
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    积分:1
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