attachments

cement, modeling cemented sand

在一种变步长LMS 算法的基础上，引进动量因式，提出了一种新的改进LMS 的算法。新算法整体性能优 于变步长LMS 算法以及LMS 算法。通过理论分析，比较了新的算法和变步长LMS 算法以及LMS 算法的收敛性 和稳态性，提出了一种设想以提高新算法的稳态性。仿真试验证明了新算法的优越性以及设想的在仿真条件下的正 确性。(In a variable step size LMS algorithm, based on the introduction of momentum factor, a new improvement of LMS algorithms. The overall performance of the new algorithm is superior to variable step size LMS algorithm and LMS algorithm. Through theoretical analysis, comparing the new algorithm and variable step size LMS algorithm and LMS algorithm convergence and steady-state, and proposes a new algorithm is envisaged to enhance the steady-state nature. Simulation tests proved the superiority of the new algorithm, as well as envisaged under the conditions in the simulation is correct.)

《MatLab工程数学应用》是很多研究生做工程时必备的一本参考书。很好用。(" MatLab Engineering Mathematics Applications" is a lot of graduate students to do projects must have a reference book. It just works.)

说明：  Ar气在CO2激光作用下的电离matlab程序代码。(co2 laser induced Ar ionzation matlab codes.)

获取对象骨架的方法，不同于matlab自带的bwmorph方法，但速度较慢(Access to the object skeleton methods, different from the matlab built-in bwmorph, but slower)

gmsk的生成程序，毕设做的，希望对大家有一定帮助(generator gmsk completed the set to do, we hope to have some help)

hvdc converter circuit used in the 12 pulse

书的中文名是matlab宝典，详细介绍了matlab7.适合matlab工作者查阅。(The Chinese name is written matlab canon, detailing matlab7. Matlab workers for inspection.)

说明：  matlab 功能工具 matlab 功能工具(matlab functions matlab function tool tool)

实验一：三次样条插值（P56，例6） 一、实验目的： 1) 掌握三次样条插值的运用 2) 了解拉格朗日插值在高次上的误差 二、实验环境：Matlab6.5 三、实验内容： 1) 给定函数f(x)=1/(1+x2)，-5<=x<=5，节点xk=-5+k，(k=0,1,2…10)，用三次样条插值求S10(x)，S10’(-5)=f’(-5)，S10’(5)=f’(5)。 2）作原函数f(x)，拉格朗日插值函数Ln(x)，三次样条差值函数Sn(x)。画出三个函数的图像，比较它们的区别。 (Experiment I: cubic spline interpolation (P56, cases of 6) First, the experiment was: 1) cubic spline interpolation to master the use of 2) understanding of Lagrange interpolation in high-error second, experimental environment: Matlab6 .5 c, experimental elements: 1) a given function f (x) = 1/(1+ x2),-5 <= x <= 5, node xk =- 5+ k, (k = 0,1,2 ... 10), using cubic spline interpolation for S10 (x), S10 (-5) = f (-5), S10 (5) = f (5). 2) for the original function f (x), the Lagrange interpolation function Ln (x), cubic spline difference function Sn (x). Draw the three functions of images, compare the difference between them.)