登录
首页 » Others » beamfor

beamfor

于 2012-10-30 发布 文件大小:3KB
0 198
下载积分: 1 下载次数: 6

代码说明:

  The return is the beampattern of the arrays in dB

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

  • An-example-of-programming-interface
    一个实例搞定MATLAB界面编程一个实例搞定MATLAB界面编程(An example to get MATLAB programming interface)
    2013-08-23 16:13:07下载
    积分:1
  • LK3D
    3D Lucas Kanade Optical Flow
    2013-04-09 04:13:48下载
    积分:1
  • fuzzypid2
    一个用simulink做的模糊PID自适应系统,FIS等都比较简单,适合初学者。通过与PID的比较曲线可以看出此系统更加优越。(A simulink do with self-adaptive fuzzy PID system, FIS are relatively simple, suitable for beginners. Comparison with the PID system curve can be seen even more superior.)
    2009-04-12 19:15:58下载
    积分:1
  • 01
    说明:  matlab中增减采样以及样条差值的程序实例,为读者提供初等帮助。(matlab to increase or decrease the difference between sampling and spline examples of procedures that provide readers with elementary help.)
    2011-12-01 22:07:35下载
    积分:1
  • SVM
    说明:  SVM支持向量机,利用matlab语言实现,很简单,以word的形式给出-(SVM support vector machine, using matlab language is very simple to word given in the form-)
    2010-04-17 18:38:33下载
    积分:1
  • hinf_m
    H-infinity Control Example, robust control
    2021-02-01 10:40:00下载
    积分:1
  • (Matlab-Code)Direct_pro_loudspeaker
    動圈式揚聲器的非線性頻響諧波失真模擬,代碼包含非線性響應模擬、微分方程式數值求解、(Feedback control)參數非線性補償代碼。(Nonlinear response of harmonic distortion dynamic speaker simulation, simulation code contains non-linear response, the numerical solution of differential equations, (Feedback control) parameter nonlinear compensation code.)
    2015-03-25 13:13:35下载
    积分:1
  • statistic-signal
    统计信号检测与估值的课件资料 浅显易懂包括如何用统计的方法检测出信号频谱的存在并估计出来(Courseware statistics signal detection and valuation easy to understand)
    2015-04-17 11:11:35下载
    积分:1
  • Modalresponse
    模态叠加法计算结构响应,function函数程序(Mode superposition method to calculate the structural response, function function program)
    2020-07-03 20:20:02下载
    积分:1
  • patternrecognition
    高频系数融合均采用基于像素点绝对值取大的规则,这里要注意的是,比较的对象是两幅原始图像的像素点的绝对值,而融合后图像的像素点的取值是两幅原始图像像素点绝对值大的那一点的值,也就是说,要取绝对值大的那个点的实际值
    2010-10-23 20:50:34下载
    积分:1
  • 696518资源总数
  • 106164会员总数
  • 18今日下载