Avgfilt

主要介绍 matlab微分方程解法 里面的例子较多(mathematical modelling in matlab)

说明：  演化算法经典教材 入门者必备 很好的 非常深(Teaching beginners classical evolutionary algorithm must be very deep well)

这是CDMA信道的matlab仿真-AWGN多径衰落信道(This is the CDMA channel Matlab simulation-AWGN over Fading Channels)

OFDM多载波调制方式的DFT变换，以及OFDM波形的仿真(OFDM multi-carrier modulation of the DFT transform, and the OFDM waveform simulation)

形态学操作，图像的膨胀和腐蚀操作，使用matlab进行实现。(Morphological operations, image dilation and erosion operations, using matlab to achieve.)

说明：  运用高斯积分法计算一维二维三维函数的数值积分(Gaussian integration method using two-dimensional three-dimensional function of one-dimensional numerical integration)

This MATLAB Code plot BER Vs Eb/No for M_ary PSK with different values of M. The user can choose the desired BER and the code will tell what is the required Eb/No to achieve this BER for different values of M. In addition, a plot for shannon limit is available.

newton-raphson search method optimization methodsd and algorithms

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.)

Drop fall algorithm used to segmentate digits