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hafumanbianma
输入各个字符及对应出现的次数 ; 建立赫夫曼树并对各个字符进行编码 ;然后 对 输入 的 二进制 串进行 译码(Enter the various characters and the corresponding number of emerging establish Huffman tree and the various characters coding then enter strings of binary decoding)
- 2007-12-16 18:50:01下载
- 积分:1
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hamming
汉明码编码译码源码,谢谢大家分享,~~~~(Hamming code encoding decoding source code, thank you for sharing)
- 2010-03-12 15:00:04下载
- 积分:1
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rvm
说明: 这是一个完善的RVM算法。其性能良好,代码清晰。能够很好的用于分类。(This is a perfect RVM algorithm. Its good performance, the code clear. Can be good for classification.)
- 2011-04-09 11:31:07下载
- 积分:1
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xinhao2
信号处理源程序,振动、模态分析经常用的算法,省去每次编写(Source signal processing, vibration, modal analysis method often used, eliminating the preparation time)
- 2011-05-10 23:04:35下载
- 积分:1
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Burg
说明: 用MATLAB编的 谱估计里的伯格算法 效果很好(Made use of MATLAB in the Burg spectral estimation algorithm works well)
- 2008-09-03 20:56:09下载
- 积分:1
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gsc1
generalized sidelobe cancellation implented frome "Adaptive filter theory 3th edition, page 320-335"
author: ب ه م ن ی ا ن د ا ن ش گ ا ه ش ی ر ا ز
- 2013-07-17 13:02:03下载
- 积分:1
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fdct_wrapping_matlab
采用中心旋转法来进行曲波变换,带有几个demos。方便的进行curvelet变换(Center of rotation method used to carry out curvelet transform, with a few demos. Convenient for curvelet transform)
- 2010-06-10 18:51:18下载
- 积分:1
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DFT
1)编写DFT子函数(计算并画出幅频特性),以后可调用。
2)N=32,M=32,128,256时,分析频率分辨率、频谱的光滑程度。
3)N=64,256,512时,分析频率分辨率和频谱的光滑程度。
(1) the preparation of DFT Functions (calculated and drawn amplitude-frequency characteristics), the future can be called. 2) N = 32, M = 32,128,256, the analysis of the frequency resolution, the degree of smooth spectrum. 3) N = 64,256,512, the analysis of the frequency spectrum resolution and the degree of smoothness.)
- 2009-04-08 22:47:25下载
- 积分:1
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shiyan1
Matlab通信系统的建模仿真
实验一 模拟通信系统的建模仿真(matlab)
- 2009-12-22 12:16:01下载
- 积分:1
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new1dfdtd
增加了声子晶体周期条件:
if (mod(ii,19)<9)
u(ii,3)=2*u(ii,2)-u(ii,1)+c*(u(ii+1,2)-2*u(ii,2)+u(ii-1,2))
else
u(ii,3)=2*u(ii,2)-u(ii,1)+0.5*c*(u(ii+1,2)-2*u(ii,2)+u(ii-1,2))
end(if (mod(ii,19)<9)
u(ii,3)=2*u(ii,2)-u(ii,1)+c*(u(ii+1,2)-2*u(ii,2)+u(ii-1,2))
else
u(ii,3)=2*u(ii,2)-u(ii,1)+0.5*c*(u(ii+1,2)-2*u(ii,2)+u(ii-1,2))
end)
- 2015-03-05 02:15:46下载
- 积分:1