simple_SR

说明：  这是我找到的一个运用神经网络的实例，希望对大家有用。(This is what I found one instance of the use of neural networks and hope for all of us.)

本程序是关于MATALB图象处理应用例子，用MATLAB编程把图象石块的总个数计算出来。(This program is an example on the MATALB image processing applications, using MATLAB programming to the total number of images derived rocks.)

将idNO类中序列相差2以内的子序列进行连接，用于交流留的数据处理(Conecting the subsequences which are next to others (less than 2). This is used for data processing of transpotation flow)

基于matlab 的动态阈值 的白平衡算法，实现分区处理；(Dynamic threshold of the white balance algorithm based on matlab)

2D DCT in matlab. inputs are original data and block size. ex) block size = [8 8]

北京航空航天大学《惯性导航原理》第二次大作业源程序(Beijing University of Aeronautics and Astronautics " principle of inertial navigation" second source operations)

关于cuda计算,我认为很好。NVIDIA_CUDA_Programming_Guide_2.3.pdf(Calculated on the cuda, I think very well. NVIDIA_CUDA_Programming_Guide_2.3.pdf)

used to simple calculator

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).)

denoise In BayesShrink[5] we determine the threshold for each subband assuming a Generalized Gaussian Distribution(GGD) . The GGD is given by GG¾ X ¯ (x) = C(¾ X ¯ )exp¡ [® (¾ X ¯ )jxj]¯ (6) ¡ 1 < x < 1 ¯ > 0, where ® (¾ X ¯ ) = ¾ ¡ 1 X [ ¡ (3=¯ ) ¡ (1=¯ ) ]1=2 and C(¾ X ¯ ) = ¯ ¢ ® (¾ X ¯ ) 2¡ ( 1 ¯ ) and ¡ (t) = R1 0 e¡ uut¡ 1du. The parameter ¾ X is the standard deviation and ¯ is the shape parameter It has been observed[5] that with a shape parameter ¯ ranging from 0.5 to 1, we can describe the the distribution of coefficients in a subband for a large set of natural images.Assuming such a distribution for the wavelet coefficients, we empirically estimate ¯ and ¾ X for each subband and try to find the threshold T which minimizes the Bayesian Risk, i.e, the expected value of the mean square error.