代码搜索：Raspberry Pi
找到约 10,000 项符合「Raspberry Pi」的源代码
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www.eeworm.com/read/174605/9580311
cpp p1-4.cpp
#include //包含iostream.h头文件
void main()
{
//输出字符常量、变量和字符串
char c1='A';
cout
www.eeworm.com/read/366428/9815528
m q729.m
%《MATLAB在电子信息课程中的应用》第七章例7.29程序q729
% 用脉冲响应不变法和双线性变换法将模拟滤波器离散化
% 电子工业出版社出版 陈怀琛 吴大正 高西全合著 2001年10月
%脉冲响应不变法实现模拟到数字滤波器的转换
clear;close all
b=1000;a=[1,1000];
w=[0:1000*2*pi]; % 设定模拟频率
[hf,w]
www.eeworm.com/read/365864/9842836
m calculatewavelet.m
function W=CalculateWavelet(width,height);
%计算小波矩阵W；将结果保存在文件wavele.dat中；
%计算小波变换矩阵簇；
d=2*pi;
x0=width/2+i*height/2;
for v=1:5
for u=1:8
k1=pi*pow2(-(v+1)/2)*cos(pi*(u-1)/8);
www.eeworm.com/read/365527/9858287
m exa5_31.m
clear
n=12;
m=moviein(n);
t=0:2/n*pi:4*pi;
x=0:1/n*pi:4*pi;
nj=length(x);
for i=1:n
for j=1:nj
y(j)=sin(x(j)-t(i));
end
plot(x,y)
axis([0,4*pi,-1.5,1.5]);
m(i)=
www.eeworm.com/read/365527/9858321
asv exa5_31.asv
n=12;
m=moviein(n);
t=0:2/n*pi:4*pi;
x=0:1/n*pi:4*pi;
nj=length(x);
for i=1:n
for j=1:nj
y(j)=sin(x(j)-t(i));
end
plot(x,y)
axis([0,4*pi,-1.5,1.5]);
m(i)=getfram
www.eeworm.com/read/169467/9859958
m cic.m
clear all;
N3=16;a=0.03;b=0.4;N1 = 16-b;N2=16+b;n1 =2 ;n2 =N1/n1;n3=2;n4=N2/n3;n5=2;n6=N3/n5;k=1;l=2;% Define the filter parameters
w = linspace(0,pi,1024); % Set the frequency in radians
% Calcula
www.eeworm.com/read/365319/9869861
m wrap.m
function wrapped = wrap(imatrix);
% WRAP(MATRIX) wrap (phase values) to principal interval.
% MAT = WRAP(MATRIX); wrap to principal interval
% return MAT wrapped MATRIX to principal interval [-pi
www.eeworm.com/read/364547/9903751
m dqfenjie.m
clear all;
hold off;
N=800;
fs=100;
x=1:1:N;%总共取4个周期的波形，800点
for a=1:1:length(x)
t(a)=20/fs*(x(a)-1)/1000
if a600
f(a)=sin(2
www.eeworm.com/read/168396/9918151
m gngauss.m
function gsrv=gngauss(sgma);
u=rand;
z=sgma*(sqrt(2*log(1/(1-u))));
u=rand;
gsrv=z*cos(2*pi*u);
www.eeworm.com/read/364103/9921968
m modifiedmusic.m
%修正的Music算法,前向空间平滑技术
%多源，未知源个数,信号源相干
%产生信号源
clear;
clc;
N=7;%信号源个数为N
%我们这样设计信号源，频率分别为f0,2f0,…，(N-1)f0,相位为随机初始相位
f0=1000;
fs=10000;%采样频率
t=0:1/fs:0.1-1/fs;
L=length(t);
for ii=1:5
s(ii,