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LeastSquareP2
基于普通最小二乘法的ARMA模型谐波频率估计
(Ordinary least squares method based on the ARMA model of harmonic frequency estimation)
- 2009-01-03 19:33:54下载
- 积分：1
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1
说明： 2003年，ITU-R制订完成了下一代移动通信系统的纲领性文件M.1645，其中明确要求，下一代移动通信系统支持低速用户100Mb/s，高速用户1Gb/s的传输速率。由于频谱资源紧张，如何在占用有限的频带资源的条件下，实现大范围网络覆盖，支持更高速率的无线数据(bu cuo deyuan ma hen hao hao de xue xi azi x)
- 2009-03-14 09:48:11下载
- 积分：1
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Radar_Systems_Analysis_and_Design_Using_MatLab
Radar systems analysis with Matlab. Theoretical texbook + matlab codes in the text.(Radar systems analysis with Matlab. Theoretical texbook+ matlab codes in the text.)
- 2009-06-15 20:23:42下载
- 积分：1
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phased-array
相控阵系统学习，包含线阵和矩形面阵的仿真，非常有参考价值。(Simulation learning of Phased array system learning,includeing linear and rectangular array，which is very valuable.)
- 2014-09-15 13:57:56下载
- 积分：1
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CH3-Source-Seeking
《Extremum seeking control and
applications》
第三章实例代码(Extremum seeking control and
applications
code of the third chapter )
- 2014-11-03 13:05:57下载
- 积分：1
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CEEMD_V
ceemdan是对EMD EEMD的改进算法，此程序包中有子程序和测试例子，可以运行(his algorithm was presented at ICASSP 2011, Prague, Czech Republic
Plese, if you use this code in your work, please cite the paper where the
algorithm was first presented.
If you use this code, please cite:
M.E.TORRES, M.A. COLOMINAS, G. SCHLOTTHAUER, P. FLANDRIN,
"A complete Ensemble Empirical Mode decomposition with adaptive noise,"
IEEE Int. Conf. on Acoust., Speech and Signal Proc. ICASSP-11, pp. 4144-4147, Prague (CZ))
- 2014-06-28 09:35:47下载
- 积分：1
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vechicle_parking_control_system
automatic vehicle parking control system controls the door of parking area. and also it limits the number of vehicles automatically.
- 2010-09-08 19:10:34下载
- 积分：1
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s_3
this function is getting the background by mean function for rael time .
the function calc the mean of every pixel .
- 2009-12-15 18:15:27下载
- 积分：1
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caeser1
Decryption algorithm of caeser which generate all possible combinations
- 2011-10-17 08:56:09下载
- 积分：1
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NR_power-Flow
x0 = ones(2,2) Make a starting guess at the solution
options = optimset( Display , off ) Turn off Display
[x,Fval,exitflag] = fsolve(@myfun,x0,options)
The solution is
x =
-0.1291 0.8602
1.2903 1.1612
Fval =
1.0e-009 *
-0.1619 0.0776
0.1161 -0.0469
exitflag =
1
and the residual is close to zero.
sum(sum(Fval.*Fval))
ans =
4.7915e-020
- 2014-02-22 08:37:54下载
- 积分：1
