搜索资源列表
matlab-exp
- 用matlab解决一些实际的问题,如交通,销售,动物繁殖等-using Matlab solve some practical problems, such as the transport, sale, breeding, etc.
movie_cartoon
- ％电影动画： ％1.首先调用moviein函数对内存初始化.创建一个足够大的矩阵来容纳一系列指定的图形（帧） ％2.调用getframe函数生成每一帧.该函数返回一个矢量，利用这个矢量创建一个电影动画矩阵 ％3.调用movie函数按照指定速度进行指定次数的播放 ％例子2：演示如何实现快速傅立叶变换（exp(j*2*pi/n)）的可视化过程- Movie animation: 1. First transfers the moviein function to the me
nonlinear(exp)_FLSSVM
- 基于模糊的最小支持响亮机,用来分类效果比一些常用的要好很多.-based on fuzzy loud support of the smallest machines, used classification results than some frequently used a lot nicer.
matlab(GA)2
- matlab遗传算法工具箱函数及实例讲解2 【问题】在－5<=Xi<=5,i=1,2区间内，求解 f(x1,x2)=-20*exp(-0.2*sqrt(0.5*(x1.^2+x2.^2)))-exp(0.5*(cos(2*pi*x1)+cos(2*pi*x2)))+22.71282的最小值。
finalreport
- matlab实现的多个基础程序和报告并有流程图（1） 绘出正弦信号波形及频谱。 （2） 单极性归零（RZ）波形及其功率谱，占空比为50％。 （3） 升余弦滚降波形的眼图及其功率谱。滚降系数为0.5。发送码元取值为0、2。 （4） 最佳基带系统的Pe~Eb\\No曲线，升余弦滚降系数a=0.5，取样值的偏差是Ts/4。 （5） Pe~Eb\\No，升余弦滚降系数a=0.5，取样时间无偏差，但信道是多径信道，C(f)=abs(1-0.5*exp(-j*2*pi
radon_ambiguity
- ％radon transform clear all ％ N=800 n=1:N fs=200 t=n/fs x1=exp(j*2*pi*(5*t+0.5*5*t.^2)) x2=exp(j*2*pi*(5*t+0.5*15*t.^2)) x=x1+x2 ％N=length(x) ％ ambifunb(x ) ％*****************************************RAT na
s.t
- 有约束2a+b^2+tan(c)=15 目标函数a*exp(b/x)+c=y 的最小二乘解-Binding 2a+ B ^ 2+ Tan (c) = 15 the objective function a* exp (b/x)+ C = y the least squares solution
MOMantena
- matlab求解半波天线,矩量法解双位方程，求直线天线表面电流分布 -matlab--antenau=u+I1(1,m).*sin(thita(1,n)).*exp(j*k0*z(1,m).*cos(thita(1,n))).*L/(N*lamda)
math_exp
- matlab常用算法的实现，还包括PPT讲解，很详细-matlab algorithm used, but also on the PPT, a very detailed
MyGUI
- 这是用MATlab进行用户界面设计的程程序这是界面的功能是：对连续时间函数w(t)=exp(-abs(t))在【-5，5】区间内进行采样，利用样条插值重构该连续函数，绘出重构图形并输出重构误差。-It is used MATlab process of user interface design process which is the interface functions are: a continuous time function w (t) = exp (-abs (t)) in th
centered_difference_numerical_diff
- Implementation of the centered difference method for numerical differentiation. Evaluation of the derivative of the following functions (numerically) f(x) = 0.3x^2 + x*exp(sin(2x)), for − 6 <= x <= 6-Implementation of the centered d
steepest_method_with_const_step
- NUMERICAL OPTIMIZATION: This steepest descent method with constant step length to find the minima of f(x, y) = xy exp(− 2x^2 − y^2 + 0.3y) Graphical represxentation in 5 ways of solution, simple and clear explained.-NUMERICAL OPTIMIZ
exp
- 一个将32路64kbps信息流合成一个基群的复用模块和相应的解复用模块，要能观察合成或分解后信息流的一致性，演示的最小时间刻度为时隙（一次采样，即1/8000秒）。-64kbps would be a 32-way flow of information into one group based reuse modules and the corresponding de-multiplexing module, to be able to observe the post-synthesis
sig
- approximate the integral of exp(x)/(1+exp(x) )* N(m,ss)
two_ray_model
- Modelo de dos rayos de Andrea Goldsmith (figura 2.5, según expresión 2.12) f = 0.9 frecuencia en GHz landa = 0.3/f en m R = -1 coeficiente de reflexión en tierra ht = 10 altura del transmisor en m hr = 2 altura del receptor en m Gt
EXP
- M文件编写的BP神经网络例子，有学习和识别的功能-M document prepared by the example of BP neural network, learn and identify the function
HVEDITTT.ZIP
- To design a filter with the following specifications: a) a low pass filter with cut off 2pi/K b) use DFT modulation eq. Hk(z)=H(z) exp(-j*2*pi*(k/K)). The outputs should be shifted in the frequency domian-To design a filter with the following spe
yuanma
- 求解方程-u"(x)+u(x)=exp(x)*(sin(x)-2*cos(x))两点边值问题-Solve the equation-u " (x)+ u (x) = exp (x)* (sin (x)-2* cos (x)) Boundary Value Problems
we
- 七单元天线阵DOA估计 clear clc d=1 天线阵元的间距 lma=2 信号中心波长 q1=1*pi/4 q2=1*pi/3 q3=1*pi/6 q4=3*pi/4 四输入信号的方向 A1=[exp(-2*pi*j*d*[0:6]*cos(q1)/lma)] 求阵因子 A2=[exp(-2*pi*j*d*[0:6]*cos(q2)/lma)] A3=[exp(-2*pi*j*d*[0:6]*cos(q3)/lma)] A4=[exp(-2*
DSP-exp
- 这是中国科学技术大学的dsp实验内容，包括FFT，卷积，滤波器设计等内容，可供学习MATLAB以及深入理解dsp原理之用-This is dsp experiment of USTC, including FFT, convolution, filter design, etc., for learning MATLAB and in-depth understanding of the principle of dsp.