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viterbi
说明: 实现了卷积码的viterbi译码,软判决和硬判决均可实现,是从英文教学书上抄下来的。(Achieved viterbi decoding of convolutional codes, soft decision and hard decision can be realized, from the English teaching books, copied down.)
- 2008-11-03 09:18:45下载
- 积分:1
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ldpc_decode
基于对数的置信度传播算法的LDPC译码程序,ldpc_decode.m
(Based on the number of belief propagation algorithm for LDPC decoding process, ldpc_decode.m)
- 2011-07-31 17:26:43下载
- 积分:1
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nwse
caculate newy-west standard deviation
- 2009-10-18 10:08:11下载
- 积分:1
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MATLAB_DSP_dev_integeration
用matlab针对TI的TMS32050xx系列DSP进行开发的硕士论文,比一般的文章讲得详细(master dissertation about using matlab simulink to program TMS32050XX series DSP)
- 2010-12-28 22:07:00下载
- 积分:1
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brovey_method
Image fusion is the process of combining two images and obtaining the resultant image which is having more information than the two input images.
- 2011-04-28 16:45:05下载
- 积分:1
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pufenxi
小波谱分析mallat算法经典程序,用的也是一维信号(small spectral analysis algorithm Mallat classic procedures, as well as the one-dimensional signal)
- 2006-10-17 19:50:52下载
- 积分:1
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OFDMsourceCode
用matlab仿真实现了25 kHz 带宽内同时传送总共48 kbit/s 的两路语音(每路速率9. 6
kbit/s) 和两路数据信息(速率分别为19. 2 kbit/s 和9. 6 kbit/s) , 仿真结果表明达到了较好的误码性能。并比较了4 种数字调制方式下O FDM 系统的误码性能。(using Matlab Simulation bandwidth of 25 kHz simultaneous transmission within a total of 48 kbit/s 2 Road Voice (rate of 9 per road. 6 kbit/s) and two road data information (rate of 19. 2 kbit/s and 9. 6 kbit/s), the simulation results indicate achieve a better BER. And comparison of four types of digital modulation mode O FDM system BER.)
- 2007-03-14 13:35:41下载
- 积分:1
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MIMOLS
MIMO阵列的最小二乘算法,LS算法。编写并验证可用。(LS for MIMO array。)
- 2015-03-18 21:18:15下载
- 积分:1
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0
说明: 给定n种物品和一个背包,物品i的重量是Wi,价值是vi,被包容量是C,应该如何选择装入被包的物品,使得装入背包中物品的总价值最大?对于每种物品i,只有两种选择,装入或者不装入.
输入:第一行两个正整数n(物品数)和C(容量)
第二行n个正整数n,表示物品价值
第三行n个正整数n,表示物品重量
输出:第一行表示总价值
第二行n个数(0/1),0表示物品不装入,1表示装入
例如,
输入:
5 10
6 3 5 4 6
2 2 6 5 4
输出
15
1 1 0 0 1
(Species of a given n items and a backpack, the weight of item i is Wi, the value of vi, was packet capacity is C, should be how to choose a package of items to be loaded, the backpack load in the largest total value of goods? For each item i, there are only two options, loaded or not loaded. Input: the first line of the two positive integer n (the number of items) and C (capacity) of the second line of n positive integer n, that the value of the third line of n positive integer n, that the weight of output: the total value of the first line of the second line, said n number of (0/1), 0 that the items do not load, that load 1 for example, enter: 5,106,354,622 6 5 4 output 15 1 1 0 0 1)
- 2009-07-01 23:07:01下载
- 积分:1
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Numerical
These sources help to solve problems on numerical methods
- 2012-09-05 08:53:23下载
- 积分:1