Noise-cancelling

非常有用的学习资料，用它学习matlab，从入门到精通，面面俱到(Very useful learning materials, use it to learn matlab, from entry to the master, everything)

自适应阵列波束形成算法研究.kdh 硕士毕业论文(Adaptive array beamforming algorithm. Kdh Master' s Thesis)

双DQ锁相PLL算法的研究，具有一定参考价值。(Research Documentation double DQ phase-locked PLL algorithm has some reference value.)

用鼠标画线，运行程序后，可以在axis界面上画任意曲线，且每个点的地址会返回到matlab主窗口(draw a line with a mouse)

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.)

pagerank的一个简单改进，用来识别核心节点。(A simple improvement of PageRank is used to identify core nodes.)

register file in vhdl and alu

说明：  applied mathematics in integrated navigation systems第三版的随书光盘(applied mathematics in integrated navigation systems with the third edition of the book CD-ROM)

说明：  这是一篇关于移动机器人路径规划文献。很好的(This is a literature on mobile robot path planning . Good)

主成分分析法将图像分为低秩部分和稀疏部分(Principal Component Analysis (PCA) divides images into low rank and sparse parts)