feijiruanjiakifajishu

matlab实现样条插值法，方便快捷实现各种拟合(matlab implementation spline interpolation method, to achieve a variety of quick and easy fitting)

迷你MATLAB 支持基础运算 矩阵运算 数学统计函数作图 线性方程组和一般简单方程求解(Mini MATLAB matrix operations support basic computing functions mapping mathematical statistics and general linear equations to solve simple equations)

mfcc的matlab显示方法，可以直接运行(mfcc s showing by matlab)

基于比例积分微分控制的永磁同步电机直接转矩控制仿真(Permanent Magnet Synchronous Motor Direct Torque Control Simulation of PID control)

another code from the sample code category

matlab语音工具箱的源代码语音工具箱的源代码语音工具箱的源代码语音工具箱的源代码(matlab voiceboxvoiceboxvoicebox)

采用四参数生成法自动生成多孔介质(二维)，优化了出现独立单个节点、实体内部出现孔洞的问题(Four-parameter generation method automatically generates a porous medium, optimize the emergence of an independent single node, hole problems arise internal entities(2D))

将matlab的m文件生成可供labview调用的dll(The matlab m file is generated for LabVIEW call DLL)

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.)

解决新版本MATLAB与VC++混合编程问题。经过验证能够使用(Solve the problem of the new version MATLAB and VC++ for hybrid programming. Proven ability to use)