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PID
基于二进制编码遗传算法的ＰＩＤ整定的仿真程序(Binary-coded genetic algorithm based on the PID tuning the simulation program)
- 2009-03-09 20:38:51下载
- 积分：1
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matlab-wl
一些使用的MATLAB例子的程序代码，如利立体透视，弹出菜单等(some of MATLAB code examples, such as the three-dimensional perspective Lee, pop-up menus, etc.)
- 2006-07-10 12:41:26下载
- 积分：1
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navimatlabcode
matlab code for inertial navigation
- 2009-09-26 16:56:42下载
- 积分：1
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boshutu
标准线阵的频率波数响应、波束图
对于N=11的标准线阵：
1、画出频率响应波束图，并标出可视区间；
2、画出极坐标下的波束图；
3、分析栅瓣问题。(Standard linear frequency response, the beam pattern)
- 2012-05-06 12:35:02下载
- 积分：1
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code_multihop_last-1
The objective of this project is to investigate the possibility of applying a dynamic channel allocation to cMCN. The proposed method is a multihop dynamic channel assignment (MDCA) scheme that works by assigning channels based on interference information in the surrounding cells. A channel searching strategy, sequential channel searching (SCS) is developed for the MDCA scheme. This strategy is responsible for the channel updation which includes channel assignment for a new request and channel release after the completion of the communication process. Through computer simulation, we show that the proposed MDCA scheme using the SCS strategy improves the system capacity significantly.
- 2013-07-13 01:28:59下载
- 积分：1
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gst
广义S变换，用于时频分析，是S的改进形式(Generalized S-transform, used in time-frequency analysis, is an improved form of S)
- 2019-04-02 12:29:17下载
- 积分：1
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stft_test
matlab短时傅里叶变换例程，绘出等高线图形(matlab short time Fourier transform routines, draw contour graph)
- 2010-07-26 11:06:14下载
- 积分：1
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ExperimentReport3
说明： 《信号与系统》计算机练习－利用matlab
西安交通大学出版社
实验三("Signals and systems" computer exercises-using Matlab Xi'an Jiaotong University Press Experiment 3)
- 2006-04-13 19:01:06下载
- 积分：1
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Untitled
说明： 面阵波束形成，由两条垂直线阵组成的面阵波束接收天线(Planar array beam forming, by the vertical line array consisting of two side beam receiving antenna array)
- 2011-04-09 21:32:24下载
- 积分：1
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we
说明： 七单元天线阵DOA估计
clear clc
d=1 天线阵元的间距
lma=2 信号中心波长
q1=1*pi/4 q2=1*pi/3 q3=1*pi/6 q4=3*pi/4 四输入信号的方向
A1=[exp(-2*pi*j*d*[0:6]*cos(q1)/lma)] 求阵因子
A2=[exp(-2*pi*j*d*[0:6]*cos(q2)/lma)]
A3=[exp(-2*pi*j*d*[0:6]*cos(q3)/lma)]
A4=[exp(-2*pi*j*d*[0:6]*cos(q4)/lma)]
A=[A1,A2,A3,A4] 得出A矩阵
n=1:1900
v1=.015 四信号的频率
v2=.05
v3=.02
v4=.035
d=[1.3*cos(v1*n) 1*sin(v2(we)
- 2010-05-06 15:26:50下载
- 积分：1
