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/179152/9368169
m exp2_4.m
close all
clc
clear
%定义时间范围
t=[0:pi/20:9*pi];
figure(1) %选择图像
plot(t,sin(t),'r:*')
grid on %在所画出的图形坐标中添加栅格，注意用在plot之后
grid off %删除栅格
figure(2)
plot(t,cos(t))
grid on
grid off
www.eeworm.com/read/179061/9375596
m chap3_11f.m
function [y]=func(x1,x2,x3)
for l1=1:1:3
gs1=-[(x1+pi/6-(l1-1)*pi/6)/(pi/12)]^2;
u1(l1)=exp(gs1);
end
for l2=1:1:3
gs2=-[(x2+pi/6-(l2-1)*pi/6)/(pi/12)]^2;
u2(l2)=exp(gs2);
end
www.eeworm.com/read/374795/9384144
txt p1-4.txt
#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/374795/9384192
txt p1-14.txt
#include
const double PI=3.1416; //声明常量(const变量)PI为3.1416
main()
{
//声明3个变量
double r=3,l,s;
//计算圆的周长
l=2*PI*r;
cout
www.eeworm.com/read/178406/9399670
m testana1.m
%
% simply supported Euler beam
L=1; E=2.15e11; rho=7800; d=50e-3; nu=0.3;
r=d/2;
ind=1:4;
I=pi*d^4/64; EI=E*I;
A=pi*(r^2);
lam=(ind*pi).^2/L^2*sqrt(EI/rho/A)/2/pi;
fprintf(
www.eeworm.com/read/178406/9399695
m testanax.m
%
% simply supported Euler beam
global L A Gs E Iy rho nn
% use Genta page 87 eq. 2-40 to compute simply supported Timoshenko beam
% eigen frequencies
dw=[];
%
L=1; E=2.15e11; rho=7800; d=50e-3
www.eeworm.com/read/178406/9399715
m testana.m
%
% simply supported Euler beam
global L A Gs E Iy rho
L=1; E=2.15e11; rho=7800; d=50e-3; nu=0.3;
r=d/2;
ind=1:5;
I=pi*d^4/64; EI=E*I;
A=pi*(r^2);
lam=(ind*pi).^2/L^2*sqrt(EI/rh
www.eeworm.com/read/178406/9399732
m testanax2.m
%
% simply supported Euler beam
global L A Gs E Iy rho nn
% use Genta page 87 eq. 2-40 to compute simply supported Timoshenko beam
% eigen frequencies
dw=[];
%
L=1; E=2.15e11; rho=7800; d=50e-3
www.eeworm.com/read/374387/9407763
m ep2_p5.m
% Ep2_p5: > 参数方程的曲面图形表面积
% Designed by FGH
m= 260;
R= 30;
[v,u]= meshgrid(0:pi/2/m:pi/2,0:2*pi/m:2*pi);
x= R*sin(v).*cos(u).*(1+0.1*abs(sin(6*u)));
y= R*sin(v).*sin(u).*(1+0.1*abs(s
www.eeworm.com/read/178117/9417976
m c9_estimatepi.m
%File: c9_estimate<mark>pi</mark>.m
%有5个<mark>pi</mark>的估计，每一个都是基于500次重复随机试验，所得的<mark>pi</mark>的五个估计值用以下向量表示
%<mark>pi</mark>的估计值=[3.0960 3.0720 2.9920 3.1600 3.0480]
%如果对5个结果进平均，则<mark>pi</mark>的估计值=3.0736,这样的结果等价于2500次的试验结果。
%用来产生该结果的MATLAB程序如下
m=input('Ente ...