登录
首页 » matlab » 多普勒频率估计 Doppler

多普勒频率估计 Doppler

于 2014-07-20 发布 文件大小:42KB
0 389
下载积分: 1 下载次数: 62

代码说明:

  结合通信/雷达中的阵列测向应用,建立多个信号存在条件下的信号模型,进行多普勒频率估计,仿真实现并分析其性能(Combined communication/radar array detection applications, the signal model to establish the presence of multiple signals, the Doppler frequency estimation, simulation and analyze its performance)

文件列表:

Doppler
.......\Doppler.m,676,2014-04-15
.......\实验三 多普勒频率估计.docx,51860,2014-07-20

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

  • minmax
    minmax filtering in matlab
    2011-01-15 22:33:37下载
    积分:1
  • MATLAB_-Programming-Style-Guide
    对于matlab编程的一些风格的指南,感觉还是比较重要,特别对于初学者,分享一下啦(For some matlab programming style guide, I feel it is quite important, especially for beginners, to share it)
    2013-07-24 10:22:37下载
    积分:1
  • simulink
    simulink使用详细介绍;smulink参数设置教程( )
    2013-09-03 19:14:00下载
    积分:1
  • mianyi
    这是一个免疫克隆算法程序,求解了一个非线性的函数问题的最优值。(This is an immune clone algorithm procedure for solving a nonlinear problem of the optimal value function.)
    2009-03-29 13:26:52下载
    积分:1
  • growcut
    growcut image segmentation gui
    2009-05-01 22:53:22下载
    积分:1
  • statecom
    this file contains a matlab simulation of a smulin system that represents the power compensation
    2014-12-31 22:11:13下载
    积分:1
  • jhkhgkghkjgh
    是数模方面的一个很好资料,它为数模参赛者提供一个好的引导.(few die of a good information, it will provide participants with the number die a good guide.)
    2006-07-21 10:34:12下载
    积分:1
  • ieee754_32
    Converts a floating point value to IEEE 754 32 bit binary representation. Sign Bit: 1 Exp: 2 to 9 Frac: 10:32 Single-precision floating-point format is a computer number format that occupies 4 bytes (32 bits) in computer memory and represents a wide dynamic range of values by using a floating point. In IEEE 754-2008 the 32-bit base 2 format is officially referred to as binary32. It was called single in IEEE 754-1985. In older computers, other floating-point formats of 4 bytes were used. For example, the 32-bit floating point bit pattern 1 01111101 00101100000000000000000 is interpreted as the following: (− 1)1 × (1 + 0.34375) × 2(125–127) = -1.34375× 2− 2 = -0.3359375 ​
    2013-09-07 22:20:32下载
    积分:1
  • ist
    说明:  解决基追踪降噪(Basis Pursuit De-Noising, BPDN)问题(Solve the problem of base pursuit de noising (bpdn))
    2019-12-30 23:19:13下载
    积分:1
  • GaussJordan
    matlab编写的适用于数值分析求解线性方程组的程序(matlab numerical analysis prepared for the procedure for solving linear equations)
    2009-12-07 21:11:36下载
    积分:1
  • 696516资源总数
  • 106442会员总数
  • 11今日下载