MatlabGUI

采用四边形网格划分,利用matlab实现二维拓扑优化问题!(Using quadrilateral mesh, using matlab to achieve two-dimensional topology optimization problem!)

龙贝格算法的matlab源代码，包含图像，收敛速度分析(Romberg algorithm matlab source code, including images, convergence rate analysis)

弦割法求解非线性方程 迭代法 间接求法(Indirect method for finding chord secant method for solving nonlinear equations iterative method)

用MATLAB实现信道化，可以根据自己的需求改变信道数，有仿真图(Using MATLAB channelization, you can change the number of channels according to their needs, there are simulation map)

卫星光网络接入控制中微波与激光中继卫星卫星资源调度相关代码(Satellite optical network access control in microwave and laser relay satellite satellite resources scheduling related code)

This paper examines the asymptotic (large sample) performance of a family of non-data aided feedforward (NDA FF) nonlinear least-squares (NLS) type carrier frequency estimators for burst-mode phase shift keying (PSK) modulations transmitted through AWGN and flat Ricean-fading channels. The asymptotic performance of these estimators is established in closed-form expression and compared with the modified Cram`er-Rao bound (MCRB). A best linear unbiased estimator (BLUE), which exhibits the lowest asymptotic variance within the family of NDA FF NLS-type estimators, is also proposed.

相控阵雷达处理MVDR 自适应算法 没有考虑宽带问题(MVDR adaptive algorithm)

Reed Solomon WITH SNR

近代平差，含有近代经典平差函数模型，适用于数据处理中的近代平差应用(Modern adjustment, containing a model of modern classical adjustment function for data processing applications in modern Adjustment)

Converts a floating point value to IEEE 754 32 bit binary representation. Sign Bit: 1 Exp: 2 to 9 Frac: 10:32 Single-precision floating-point format is a computer number format that occupies 4 bytes (32 bits) in computer memory and represents a wide dynamic range of values by using a floating point. In IEEE 754-2008 the 32-bit base 2 format is officially referred to as binary32. It was called single in IEEE 754-1985. In older computers, other floating-point formats of 4 bytes were used. For example, the 32-bit floating point bit pattern 1 01111101 00101100000000000000000 is interpreted as the following: (− 1)1 × (1 + 0.34375) × 2(125–127) = -1.34375× 2− 2 = -0.3359375 ​