ISAR

Cooperative relaying with Decode and Forward scheme code rate = 1/4 C.C for single-relay case

说明：  N-cut聚类算法,详细给出了谱聚类算法的基本实现，并详细给出了注释说明，编写两个函数(Normalized cut Clustering)

该函数是用限定记忆法进行最小二乘法辨识。得到系统的参数。(This function is limited memory method of least squares identification method. Get the system parameters.)

this file is used to simulation of dam break problem in MATLAB codes

产生均匀分布的随机白噪信号,并观察数据分布的直方图产生均匀分布的随机白噪信号,并观察数据分布的直方图(Generate uniformly distributed random white noise signal, and to observe the distribution of the histogram data generated uniformly distributed random white noise signal, and to observe the data distribution histogram)

说明：  A new technique of free-space simulation has been developed for solving unbounded electromagnetic problems with the finite-difference time-domain method.

精通Matlab综合辅导与指南-源程序，跟书籍配套的(Proficient in Matlab comprehensive counseling and guidance- source)

说明：  DPCM压缩编码算法的子函数，需要在另外编写程序才可用(DPCM compression algorithm of the subroutine, you need to be used in other programming)

Algorithm Given are P training pairs {X1,d1,X2,d2....Xp,dp}, where Xi is (n*1) di is (n*1) No of Categories=R. i=1,2,...P Yi= Augmented input pattern( obtained by appending 1 to the input vector) i=1,2,…P In the following, k denotes the training step and p denotes the step counter within the training cycle Step 1: c>0 , Emin is chosen, Step 2: Weights are initialized at w at small values, w is (n+1)*1. Counters and error are initialized. k=1,p=1,E=0 Step 3: The training cycle begins here. Input is presented and output computed: Y=Yp, d=dp Oi=f(wtY) for i=1,2,….R Step 4: Weights are updated: wi=wi+1/2c(di-oi)Y for i=1,2,…..R. Step 5: Cycle error is computed: E=1/2(di-oi)2+E for i=1,2,…..R. Step 6: If p<P then p=p+1,k=k+1, and go to Step 3: Otherwise go to Step 7. Step 7: The training cycle is completed. For E=0,terminate the training session. Outputs weights and k. If E>0,then E=0 ,p=1, and enter the new training cycle by going to step 3.

改进的量子遗传算法，求解最值问题，绝对没有密码(The improved quantum genetic algorithm to solve the most value problem There is absolutely no password)