-
frequeceestimation
基于均匀线阵，用music对信号进行多频率估计(frequece of estimaton )
- 2009-06-27 11:02:08下载
- 积分：1
-
smith_pid
带有smith预估器的PID控制 matlab仿真程序(The smith predictor PID control matlab simulation program)
- 2013-05-11 12:45:51下载
- 积分：1
-
Iain-Haslam
国外经典LBM D2Q9 D3Q19模型的matlab程序，亲测可用。可用于渗流场和热传递模拟。自己根据需要修改模型。引用时记得注明出处。版权问题概不负责。(Foreign LBM D2Q9 D3Q19 model matlab procedures, pro-test available. It can be used for seepage and heat transfer simulation. Make changes to the model as needed. Remember when referring to attribution. We are not responsible for copyright issues.)
- 2021-03-18 10:29:20下载
- 积分：1
-
ParallelNotch
说明： 用matlab对并行notch滤波器进行仿真，此为源代码(With notch filter matlab parallel simulation, this is the source code)
- 2009-07-24 22:07:59下载
- 积分：1
-
matlab-chapter7
matlab应用技术chapter7 ppt(matlab application technology chapter7 ppt)
- 2015-02-02 20:03:59下载
- 积分：1
-
MATLAB-DUO-HUANG-FANGZ
matlab环境下的多路串口数据实时绘图显示，GUI界面,这个程序是matlab软件编写，为大家在学习中提供了帮助(multiplexed serial data real-time graphics display matlab environment, GUI interface, this program is written in matlab software, provide help in learning)
- 2013-12-18 19:22:03下载
- 积分：1
-
Matlab-program-of-ant-colony
这是一个用蚁群算法解决最短路径问题的通用matlab程序，供大家参考使用(This is an ant algorithm to solve the shortest path problem common matlab program, for reference use)
- 2013-05-20 10:58:10下载
- 积分：1
-
Spline-way
利用样条差值法，根据已知坐标，绘制整条曲线。 数学建模实例，很有用的。(Using spline difference method, based on the known coordinates, drawing the whole curve. Mathematical modeling examples, very useful.)
- 2011-05-12 17:41:28下载
- 积分：1
-
diedai
用雅克比迭代法解椭圆型方程，这是椭圆型方程的差分解法。见于孙志忠编的《偏微分方程数值解法》的p41例题(
Solutions of Elliptic Equations with Jacques iterative method, which is the solution of elliptic differential equations. Seen SUN Zhi-zhong series "Numerical Solution of Partial Differential Equations" examples of p41)
- 2013-12-08 12:33:50下载
- 积分：1
-
NR_power-Flow
x0 = ones(2,2) Make a starting guess at the solution
options = optimset( Display , off ) Turn off Display
[x,Fval,exitflag] = fsolve(@myfun,x0,options)
The solution is
x =
-0.1291 0.8602
1.2903 1.1612
Fval =
1.0e-009 *
-0.1619 0.0776
0.1161 -0.0469
exitflag =
1
and the residual is close to zero.
sum(sum(Fval.*Fval))
ans =
4.7915e-020
- 2014-02-22 08:37:54下载
- 积分：1
