FPGA三国志，不得不看的FPGA发展历史

来自《VC6.0可视化编程》的源码，适用于初学者。 这是第一章，关于multi windows 的代码-from "VC6.0 visual programming," the source, applicable to beginners. This is the first chapter of the multi windows code

LINUX内核完全注释文档,详细有代码注释及说明,此为上部分-LINUX kernel entirely Notes documents, detailed notes and descriptions of code, this is partly

不错的文章，建议-good articles, look at the recommendations

随机中点位移法在三维地形插值显示的适用性分析.pdf-Random midpoint displacement method in three-dimensional terrain interpolation shows the applicability of analysis. Pdf

redhat9.0的很好的基础操作资料-redhat9.0 a good foundation operation information

Understanding the Linux Kernel, 3rd Edition -Understanding the Linux Kernel. 3rd Edition

：介绍了扩展卡尔曼滤波算法和无迹变换(unscented transformation，UT)算法，并对扩展卡尔曼滤波算法(EKF)和无 迹卡尔曼滤波算法(UKF)进行比较，阐明了UKF优于EKF。在此基础上，提出了一种基于Unscented变换(uT)的高斯和滤 波算法，该算法首先通过合并准则得到适当个数的混合高斯模型，逼近系统中非高斯噪声的概率密度-: Introduction of the extended Kalman filter algorithm and unscented transform (unscented transformation, UT) algorithm, the extended Kalman filter algorithm (EKF) and unscented Kalman filter (UKF) for comparison to clarify the UKF is superior to EKF. On this basis, we propose an approach based on Unscented Transform (uT) and the Gaussian filtering algorithm, which first of all, by merging the appropriate number of criteria to be a mixture of Gaussian model, which was close to the system of the Central African Gaussian noise probability density

一个很好的javabeans的开发书，详细地介绍了javabeans的开发的过程实现了逐步提高，通熟易懂，是初学者很容易的上手，也可以作为高手提高的教材。-a good Javabeans development, a detailed introduction of the Javabeans development process of achieving a progressively higher-cooked understandable, is very easy to beginners and thus can be used as a master to improve teaching.

vc2005 入门经典 电子书 高清晰 第八章 -vc2005 e-book HD eighth

假设检验例题 7.1 设总体 ，其中参数 ， 为未知，试指出下面统计假设中哪些是简单假设，哪些是复合假设： （1） ； （2） ； （3） ； （4） ； （5） . 解：（1）是简单假设，其余位复合假设 -Hypothesis testing for Example 7.1 overall, including parameters, is unknown, pointed out that the following statistical hypothesis test which is a simple assumption, which is a compound hypothesis: (1) (2) (3) (4) (5). Solution : (1) is a simple assumption, the remaining bit composite hypothesis