gaot_toolbox

采用辛普生计算方法进行数值计算，并对结果进行画图(simpson)

The recognition of noise as a source of annoyance began in antiquity, but therelationship, sometimes subtle, that may exist between noise and money seems to be a development of more recent times. For example, the manager of a large wind tunnel once told one of the authors that in the evening he liked to hear, from the back porch of his home, the steady hum of his machine 2 km away, for to him the hum meant money. However, to his neighbours it meant only annoyance and he eventually had to do without his evening pleasure.

说明：  利用互信息法求得混沌时间序列相空间重构最优化时延的Matlab程序(use of mutual information obtained chaotic time series reconstruction phase space optimization procedures Matlab Delay)

MATLAB编程经典教程，中文译本，简单易懂(MATLAB programming classic handbook, Chinese translation, easy to understand)

QuantLloyd using matlab

在线字典的学习算法，此方法可以对大量的特征进行学习，完整的代码。(Online dictionary learning algorithm, this method can be for a lot of the characteristics of the study, the complete code.)

Trans. Funcion de transferencia. Filtro pasa bajas. MATLAB. Trans. Funcion de transferencia. Filtro pasa bajas. MATLAB. Trans. Funcion de transferencia. Filtro pasa bajas. MATLAB. Trans. Funcion de transferencia. Filtro pasa bajas. MATLAB.

基于matlab简单图像处理系统中用图像平均的方法消除噪声(Matlab eliminate noise using a simple image processing system based on the method of averaging the image)

波束形成中DOA估计MUSIC算法 matlab程序(MUSIC)

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 ​