代码搜索：Raspberry Pi
找到约 10,000 项符合「Raspberry Pi」的源代码
代码结果 10,000
www.eeworm.com/read/211604/15176746
m xggszb.m
%
%产生服从高斯分布的随机数
%
%clear;
%clc;
N=1024;
u1=rand(1,N); %标准正态分布的随机数
u2=rand(1,N); %
for i=1:N
x(i)=sqrt((-2)*log2(u1(i)))*cos(2*pi*u2(i));
end
%
%求滤波器系数
%
f=20;
f0=512;
t=1/f0;
www.eeworm.com/read/211004/15188828
m sin102.m
clear all
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Yule-Walker法估计（自相关法）库函数%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
fs=500;
nfft=1024;
t=0:1/fs:1;
f1=60;
f2=130;
f3=210;
xn=sin(2*pi*f1*t
www.eeworm.com/read/210913/15189971
m regression3.m
%SVC的三维回归！！！
clc
clear all
global p1;
p1=102.111;
e=0.005;
C=9000;
t=linspace(8*pi,20*pi,501);
%画出三维的曲线
figure(1);
subplot(221);plot3(1.5*t.*sin(t),1.5*t.*cos(t),t,'-r');title('空间三维曲线');
Bo
www.eeworm.com/read/210736/15192637
m pm2_wv_doa_test.m
clear;
clear all;
format short;
c=3*10.^8;
L=8;
N=256*1;
fm1=0.25;
fm2=0.32;
Km1=0.025;
Km2=0.033;
f0=0.32;
lamta=c/f0;
len=lamta/2;
P1=20*pi/180;
P2=40*pi/180;
snr=10;
Amp=sqrt(2*10^(
www.eeworm.com/read/210159/15205776
m ip_07_07.m
% MATLAB script for Illustrative Problem 7.7.
echo on
Tb=1;
f1=1000/Tb;
f2=f1+1/Tb;
phi=pi/4;
N=5000; % number of samples
t=0:Tb/(N-1):Tb;
u1=cos(2*pi*f1*t);
u2=cos(2*pi*f2*t);
% Assu
www.eeworm.com/read/209401/15220655
m source2.m
% SOURCE SIGNAL 2
%-------------------------------------------------------------------------------
% s = source signal 2
%----------------------------------------------------------------------------
www.eeworm.com/read/208657/15239781
m x2t.m
function T=x2t(x,str)
% T=x2t(x,str)
%
% Converts a generalized position vector x, which contains
% position and orientation vectors of B with respect to A,
% into transformation matrix T betw
www.eeworm.com/read/208637/15241773
c 1cpdv.c
#include "math.h"
void cpdv(pr,pi,m,qr,qi,n,sr,si,k,rr,ri,l)
int m,n,k,l;
double pr[],pi[],qr[],qi[],sr[],si[],rr[],ri[];
{ int i,j,mm,ll;
double a,b,c,d,u,v;
void cmul(doubl
www.eeworm.com/read/207935/15256533
txt sinwn.txt
#include"math.h"
#include"gauss.c"
void sinwn(a,f,ph,m,fs,snr,seed,x,n)
int m,n;
long seed;
double fs,snr,a[],f[],ph[],x[];
{int i,k;
double z,PI,nsr;
PI=4.0*atan(1.0;
z=snr/10.0;
z=pow(10.0
www.eeworm.com/read/206816/15289196
m drawearth.m
%time是参数
%利用这个参数，可以绘制一个看起来是旋转的地球
function DrawEarth(time)
r=6400; %地球半径
j1=[0:pi/6:pi/2];
w1=[-pi/2:pi/6:pi/2];
L1=length(w1);
L2=length(j1);
for n=1:L1
z=ones(L2,1);
z=z*r*sin(w1(n