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OFDM 系统的误码率的MATLAB仿真代码 可用欢迎下载(OFDM BER )

s1.m: 低频波检测。能从高频信号中检测出低频信号。并将信号输出以70倍因子downsample. s2.m: 将downsampling滤波器输出信号，用单级速率和多相upsampling。 s3.m: 自动识别车牌位置。此程序能从一张图片当中，自动识别出车牌的位置。 s4.m: 用Hough变换识别出一张带有水池的图片的水池分界线。 s5.m: 用原始的静态多副图片（例如：恐龙图片）构造出三维旋转立体图片。(signal processing)

利用PWM脉宽控制的半桥电路实现的DCDC变换器(The converter that use the PWM to control the voltage of the output。 )

root music 算法估计精度更加优秀(root music The estimation accuracy of the algorithm is better)

在数字信号的基带传输过程中，信息传输速率在满足香农公式的条件下还极大地受到码间干扰的限制(It is also limited by intersymbol interference when transmission speed of code accord with Shannon expressions in the course of baseband transmission of digital signal.)

通信系统中，服从标准分布、正态分布、泊松分布和瑞利分布的随机数的生成(the genneration of the uniform distribution, Gaussian distribution, Rayleigh distribution and Possion distribution, in communication system)

lms的matlab算法，dsp源程序，包含效果图(lms algorithm matlab, dsp source code, including renderings)

In order to reduce the volume and cost of the Z-source network, we need to eliminate the low-frequency current ripple by using a constant shoot-through duty ratio

PCM的编码端仿真程序,主要包括抽样量化\编码三部分.用程序语言编写(PCM terminal emulation program code, including sample quantization encoding three parts. With the preparation of programming language)

数值解法 例1 用改进的欧拉方法求一阶方程初值问题的 数值解 例2 数学摆的运动方程与近似方程数值解比较 例3 四阶龙格库塔公式求数值解 Matlab实现求数值解 例1 求初值问题y =x*sin(x)-y,y(0)=0的数值 解.先编辑一个定义方程的ODEF函数文件 ODEfun1.m. 例2 求方程组初值问题的数值解,按指定步长 0.05划分节点. 例3 按指定的精度求高阶方程初值问题的数值解.(Numerical Solution of the law a method of using the improved Euler' s equation of order to a first-order initial value problem of the numerical solution of cases of two mathematical pendulum equations of motion compared with the approximate numerical solutions of equations of three cases of fourth-order Runge-Kutta numerical solution of the formula Matlab order to achieve the numerical solution of demand cases seeking an initial value problem y ' = x* sin (x)-y, y (0) = 0 the numerical solution. first, the definition of an equation editor ODEF function file ODEfun1.m. cases of two demand equations initial value problem of numerical solution, according to the specified division of the node step 0.05. Example 3 according to the specified precision of seeking higher-order equation of the numerical solution of initial value problem.)