-
anolog_circuit
遗传编程实现模拟电路设计，模拟电路在诸如通信等高速应用领域方面十分重要(automated analog circuit design__ using genetic programming)
- 2010-06-03 10:36:48下载
- 积分：1
-
Sudoku
SUDOKU CONTROL VIA MATLAB APPROACH
- 2014-08-18 00:33:37下载
- 积分：1
-
Factorize
matlab code for factorization
- 2010-03-10 14:39:44下载
- 积分：1
-
QPSKcode
qpsk干扰仿真，很不错的源代码，欢迎大家下载，(qpsk interference simulation, a very good source, all are welcome to download, Oh)
- 2007-01-02 21:40:48下载
- 积分：1
-
crc_10g
为了解决! 以太网接入系统中大规模并行编码器的设计问题! 提出了矩阵法
代入法流水线法等三种设计方法以此为基础!给出了!以太网接入系统中 编码器的实现方
案.(In order to solve! " # Ethernet access system, $ $ massively parallel encoder design! Proposed a matrix method" substitution method " , three line method as a basis for design#! Shows!" Ethernet access system# $ $ encoder implementations)
- 2011-10-18 17:37:53下载
- 积分：1
-
peridogramse
本程序在MATLAB开发环境下，用周期图谱估计的方法对信号的频谱进行了估计，估计结构为两个频率分量。(The procedure developed in the MATLAB environment, with the cycle map estimation method to estimate the signal s spectrum, estimated structure for the two frequency components)
- 2010-07-20 19:07:02下载
- 积分：1
-
rayleigh
Rayleigh fading channel based on
the Filtered Gaussian Noise method and
r based on the Sum of Sinusoids method(simulation for Rayleigh fading channel
-Plot the channel output for fm
T = 0.01, 0.1 and 0.5 (t/T =
0 ~ 300)
– Plot the channel output autocorrelation for fm
T = 0.01,
0.1 and 1.0 (fm
= 0 ~ 10))
- 2013-01-14 22:51:33下载
- 积分：1
-
PCA
Principal component analysis (PCA)/
Matlab Code/
If You have Any Question Mail To ME:
ebiabdollahian@ce.sharif.edu
- 2013-03-07 22:33:28下载
- 积分：1
-
Solve-Modular-Linear-Equations
解模块线性方程组, solve modular linear equations, 用到了extended-euclid算法求最大公约数...(solve modular linear equations, use extended-euclid to compute the greatest common divisor..)
- 2012-04-25 01:03:27下载
- 积分：1
-
codeerr
bpsk的调制与解调的过程，并采用卷积码，讨论其误码性能随着其功率、频偏、相偏而变化.(bpsk modulation and demodulation process and using convolutional codes, to discuss its error performance with its power, frequency offset, phase offset changes.)
- 2014-02-14 00:19:41下载
- 积分：1
