radio

自己写的关于方程求根的matlab程序，仅供参考(Roots of equations to write about the matlab program, for reference only)

FDTD 时域有界差分 MATLAB程序语言(Bounded difference time domain FDTD MATLAB programming language)

实现了高斯随机变量的生成 包涵了C++程序 和matlab程序(Gaussian varible generation)

参数变化对最小相位系统的影响，动态矩阵算法（DMC）仿真实例。(Effects of parameter changes on the minimum-phase system。Dynamic matrix algorithm (DMC) simulation.)

数学形态学对一维信号处理程序,供大家学习参考(Mathematical morphology to one dimensional signal handler)

CC2530的定时器实验，使用IAR编写。(CC2530 key experiments, using IAR preparation procedures.)

某市有一码头，每次仅容一辆船停泊装卸货，由于经常有船等候进港，部分人提出要扩建码头。经过调查历史资料发现，码头平均每月停船24艘，每艘船的停泊时间为24±20小时，相邻两艘船的到达时间间隔为30±15小时，如果一艘船因有船在港而等候1小时，其消耗成本为1000元。经预算，扩建码头大约需要1350万元，故市长决策如下：如果未来五年内停泊船只因等候的成本消耗总和超过扩建码头花费则扩建码头，否则，不予扩建。因此，希望你能够帮助市长做出决策。此问题已知到达的大概时间和大概停泊时间，对于此问题用概率统计的方法来做比较复杂，可用程序随机产生到达时间和停泊时间来模拟未来五年内船的停泊，多次模拟预测停泊情况，以做出决策(。-Problem in a city with a terminal expansion dock, each only allow a boat moored loading and unloading, as often there are waiting boat into port, some people proposed to expand pier. Terminal stopping 24 per month, per vessel berthing time of 24 ± 20 hours, the arrival of two adjacent vessels interval 30 ± 15 hours, if a ship due to boat in the harbor while waiting for one hour, and its consumption cost of 1,000 yuan. The budget of about 13.5 million yuan expansion dock, so the mayor decision as follows: If a ship is moored next five years, the cost of waiting more than the sum of consumption spending is expanding marina pier extension, otherwise, not expanded.)

固高控制卡程序,点到点运动,双轴插补,测试程序(GuGao Control Card Codes)

matlab dsp中函数downsample和upsamplec代码(matlab dsp functions in downsample and upsamplec code)

D算法求最短路径利用matlab解决 输出S矩阵和行向量(Dijkstra algorithm)