FFT2

通信系统中接收机采用训练序列进行CFO估计的MATLAB仿真程序(The receiver uses the training sequence in the communication system CFO estimation MATLAB simulation program)

newton-raphon 法解方程程序(newton-raphon )

again numerical computing using matlab environment

用平行截面法讨论由曲面z=x^2-y^2构成的马鞍面形状.m(code)

basic full converter lab circuit

函数用于图象质量评价， 计算梯度均值，标准差，峰值(Function for image quality evaluation, and to calculate the gradient of the mean, standard deviation, peak)

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.)

说明：  通过修改inp文件，用于对abaqus的模型进行自动创建裂纹的程序(by modifying inp. file, used for create a crack in abaqus model)

通过PCM调制实现信源编码,通过抽样,调制通过MATLAB软件实现(Through the realization of PCM modulation source coding, through sampling, modulation through the MATLAB software)

列文森-杜宾算法，用于语音的线性预测控制中，也可以用于谱包络。(Levenson-Durbin algorithm for voice linear predictive control, but also can be used for spectral envelope.)