BPSK_CONVOLUTION_SOFT

时频分析工具——径向高斯核时频分布。为了在时频分布平面内不出现交叉项，则在模糊函数域内应有效去除互分量而仅保留自分量。设计一个与信号相匹配的核函数可以达到上述目的。基于信号的径向高斯核时频分布是一种比较理想的时频分布(Optimal radially Gaussian kernel time-frequency representation)

8 puzzle 程序, 采用 matlab 编程(8 puzzle)

非常好的工具箱，包括了各种著名的算法，可以试一试(Very good kit, including a variety of well-known algorithm, can try)

it is finite differeance time domain code for antenna

matlab命令大全电子书，matlab入门必备(Matlab order Daquan e-books, Matlab entry required)

%求解n乘n线性系统

说明：  这里有2组图，“1000.0.05”这一组是运行结果比较好的一组图，但是当时修改了时间间隔而没有改坐标说明，所以db1000.0.05.fig这幅图的坐标说明后来做了人工改正（应该是“time(0～250ms分成5000组,步长0.05ms”）。(Reuters, "1000.0.05" This group is running a relatively good results Reuters. But then amended the time interval and not changed coordinates Note So db1000.0.05.fig site map coordinates statement later became the artificial correction (should be "ti me (0 ~ 250ms group into 5000, 0.05 Step ").)

椭圆带通滤波器设计，根据椭圆原型滤波器，进行双线性变换和低通转带通，获得所需滤波器(Elliptic bandpass filter design based on elliptic prototype filter, the bilinear transformation turn bandpass and low-pass, to obtain the desired filter)

结合通信/雷达中的阵列测向应用，进行方位估计(Combining communications and radar array detection applications, azimuth estimate)

猴子摘桃海滩上有一堆桃子，五只猴子来分。第一只猴子把这堆桃子凭据分为五份，多了一个，这只 猴子把多的一个扔入海中，拿走了一份。第二只猴子把剩下的桃子又平均分成五份，又多了 一个，它同样把多的一个扔入海中，拿走了一份，第三、第四、第五只猴子都是这样做的， 问海滩上原来最少有多少个桃子？-Peach Monkeys on the beach there is a pile of peaches, five monkeys for points. The first monkey peach credentials this heap is divided into five copies, more than one, the monkey to more than one thrown into the sea, took a. The second monkey the rest of the peaches and what the average is divided into five, and there are one, and it is likewise more than one thrown into the sea, took a third, fourth, fifth monkeys are actually doing that , the question on the beach for at least the number of the original peach?