-
ofdm
说明： OFDM全系统仿真代码，包括卷积编码，调制解调，四倍过采样(OFDM system-wide simulation code, including the convolution coding, modulation and demodulation, four times oversampling)
- 2011-02-24 09:40:58下载
- 积分：1
-
Intel_Architecture_Software_Developer_Manual
Intel Architecture Software Developer s Manual(Intel Architecture Software Developer's M Profile)
- 2007-05-12 18:11:58下载
- 积分：1
-
runonly
单实例运行技术，通过检测互斥对象来实现.Createmutex(run only)
- 2014-09-12 18:15:07下载
- 积分：1
-
siebiu_v79
Gabor小波变换与PCA的人脸识别代码，基于人工神经网络的常用数字信号调制，相参脉冲串复调制信号。( Gabor wavelet transform and PCA face recognition code, The commonly used digital signal modulation based on artificial neural network, Complex modulation coherent pulse train signal.)
- 2017-01-16 21:05:07下载
- 积分：1
-
CaculateNetvalue
计算A加权后的噪声净噪声值并将结果保存到excel文档(Calculate the Netnoise after A-filter and save the result into excel file.)
- 2015-01-04 10:40:45下载
- 积分：1
-
OFDM_code
本代码是关于OFDM通信系统的仿真源程序，其中信道编码，QPSK调制，QPSK解调，FFT，IFFT，viterbi编译码，误比特统计，滤波等模块。(The code is on the OFDM communication system simulation source, which channel coding, QPSK modulation, QPSK demodulation, FFT, IFFT, viterbi decoding, bit error statistics, the filtering module.)
- 2010-07-26 16:49:23下载
- 积分：1
-
program_20_1
说明： ％ 本程序实现小波在矩阵方程求解中的应用( This program in the matrix equation Wavelet Applications)
- 2008-12-03 22:38:08下载
- 积分：1
-
vectorcontrol3
异步电机的矢量控制，详细的介绍了异步电机矢量控制系统建立的过程，以及仿真结果(Induction motor vector control, a detailed description of the induction motor vector control system established process and simulation results)
- 2021-01-14 17:18:47下载
- 积分：1
-
matlab_queue_system
自己写的用Matlab模拟一个缓冲区大小（包括正在服务的那个）为10的随机排队系统。
（1）到达过程是的泊松过程（到达速率为Mu），服务时间服从独立指数分布（均值为1/Lamda）。
对Mu//Lamda=0.2，Mu//Lamda=0.8和Mu//Lamda=1.1三种情况进行仿真，求出队列中接受服务用户为n的概率P(n), n=0,...,10。并与理论结果进行比较。
（2）假设到达过程为均匀过程（到达速率为Mu），服务时间服从独立瑞利分布（均值为1/Lamda）。(yourself writing a simulation using Matlab buffer size (including services that are) for the following 10 Machine queuing system. (1) arrival process is the Poisson process (arrival rate of Mu), submit to independent service index distribution (mean of 1/Lamda). Right Mu// Lamda = 0.2, Mu// Lamda = 0.8 and the Mu// = 1.1 Lamda three of the simulation, Queue calculated users receive services n the probability P (n), n = 0 ,..., 10. And with the theoretical results. (2) It is assumed that the arrival process for the uniform process (arrival rate of Mu), submit to independent service Rayleigh distribution (mean of 1/Lamda).)
- 2007-05-16 09:14:40下载
- 积分：1
-
g.721
Codec G.721 hecho en MATBAL
- 2011-07-23 07:10:52下载
- 积分：1
