本代码为编码开关代码，编码开关也就是数字音响中的 360度旋转的数字音量以及显示器上用的（单键飞梭开 关）等类似鼠标滚轮的手动计数输入设备。 我使用的编码开关为5个引脚的，其中2个引脚为按下 转轮开关（也就相当于鼠标中键）。另外3个引脚用来 检测旋转方向以及旋转步数的检测端。引脚分别为a,b,c b接地a，c分别接到P2.0和P2.1口并分别接两个10K上拉 电阻，并且a，c需要分别对地接一个104的电容，否则 因为编码开关的触点抖动会引起轻微误动作。本程序不 使用定时器，不占用中断，不使用延时代码，并对每个 细分步数进行判断，避免一切误动作，性能超级稳定。 我使用的编码器是APLS的EC11B可以参照附件的时序图 编码器控制流水灯最能说明问题，下面是以一段流水 灯来演示。
上传时间： 2017-07-03
上传用户：gaojiao1999
【问题描述】 在一个N*N的点阵中，如N=4,你现在站在（1，1），出口在（4，4）。你可以通过上、下、左、右四种移动方法，在迷宫内行走，但是同一个位置不可以访问两次，亦不可以越界。表格最上面的一行加黑数字A[1..4]分别表示迷宫第I列中需要访问并仅可以访问的格子数。右边一行加下划线数字B[1..4]则表示迷宫第I行需要访问并仅可以访问的格子数。如图中带括号红色数字就是一条符合条件的路线。 给定N，A[1..N] B[1..N]。输出一条符合条件的路线，若无解，输出NO ANSWER。（使用U，D，L，R分别表示上、下、左、右。） 2 2 1 2 （4，4） 1 （2，3） （3，3） （4，3） 3 （1，2） （2，2） 2 （1，1） 1 【输入格式】 第一行是数m (n < 6 )。第二行有n个数，表示a[1]..a[n]。第三行有n个数，表示b[1]..b[n]。 【输出格式】 仅有一行。若有解则输出一条可行路线，否则输出“NO ANSWER”。
标签： 点阵
上传时间： 2014-06-21
上传用户：llandlu
userial is an Free project building an USB to I2C/SPI/GPIO bridge, using the Atmel AT90USB647 chip. Hardware and Software are released under an Open Source licence. It supports the following interfaces: * USB interface (serial emulation) * JTAG * I2C (TWI) * SPI * 8 General purpose digital I/O * 4 Analog to Digital converters (currently no firmware support)
标签： USB building userial project
上传时间： 2013-12-25
上传用户：小鹏
A large body of computer-aided techniques has been developed in recent years to assist in the process of modeling, analyzing, and designing communication systems . These computer-aided techniques fall into two categories: formula-based approaches, where the computer is used to evaluate complex formulas, and simulation-based approaches, where the computer is used to simulate the waveforms or signals that flow through the system. The second approach, which involves “waveform”-level simulation (and often incorporates analytical techniques), is the subject of this book. Since performance evaluation and trade off studies are the central issues in the analysis and design of communication systems, we will focus on the use of simulation for evaluating the performance of analog and digital communication systems with the emphasis on digitalcommunication systems.
标签： computer-aided techniques developed assist
上传时间： 2014-01-01
上传用户：541657925
实验源代码 //Warshall.cpp #include<stdio.h> void warshall(int k,int n) { int i , j, t; int temp[20][20]; for(int a=0;a<k;a++) { printf("请输入矩阵第%d 行元素:",a); for(int b=0;b<n;b++) { scanf ("%d",&temp[a][b]); } } for(i=0;i<k;i++){ for( j=0;j<k;j++){ if(temp[ j][i]==1) { for(t=0;t<n;t++) { temp[ j][t]=temp[i][t]||temp[ j][t]; } } } } printf("可传递闭包关系矩阵是:\n"); for(i=0;i<k;i++) { for( j=0;j<n;j++) { printf("%d", temp[i][ j]); } printf("\n"); } } void main() { printf("利用 Warshall 算法求二元关系的可传递闭包\n"); void warshall(int,int); int k , n; printf("请输入矩阵的行数 i: "); scanf("%d",&k); 四川大学实验报告 printf("请输入矩阵的列数 j: "); scanf("%d",&n); warshall(k,n); }
上传时间： 2016-06-27
上传用户：梁雪文以
#include "iostream" using namespace std; class Matrix { private: double** A; //矩阵A double *b; //向量b public: int size; Matrix(int ); ~Matrix(); friend double* Dooli(Matrix& ); void Input(); void Disp(); }; Matrix::Matrix(int x) { size=x; //为向量b分配空间并初始化为0 b=new double [x]; for(int j=0;j<x;j++) b[j]=0; //为向量A分配空间并初始化为0 A=new double* [x]; for(int i=0;i<x;i++) A[i]=new double [x]; for(int m=0;m<x;m++) for(int n=0;n<x;n++) A[m][n]=0; } Matrix::~Matrix() { cout<<"正在析构中~~~~"<<endl; delete b; for(int i=0;i<size;i++) delete A[i]; delete A; } void Matrix::Disp() { for(int i=0;i<size;i++) { for(int j=0;j<size;j++) cout<<A[i][j]<<" "; cout<<endl; } } void Matrix::Input() { cout<<"请输入A:"<<endl; for(int i=0;i<size;i++) for(int j=0;j<size;j++){ cout<<"第"<<i+1<<"行"<<"第"<<j+1<<"列:"<<endl; cin>>A[i][j]; } cout<<"请输入b:"<<endl; for(int j=0;j<size;j++){ cout<<"第"<<j+1<<"个:"<<endl; cin>>b[j]; } } double* Dooli(Matrix& A) { double *Xn=new double [A.size]; Matrix L(A.size),U(A.size); //分别求得U，L的第一行与第一列 for(int i=0;i<A.size;i++) U.A[0][i]=A.A[0][i]; for(int j=1;j<A.size;j++) L.A[j][0]=A.A[j][0]/U.A[0][0]; //分别求得U，L的第r行，第r列 double temp1=0,temp2=0; for(int r=1;r<A.size;r++){ //U for(int i=r;i<A.size;i++){ for(int k=0;k<r-1;k++) temp1=temp1+L.A[r][k]*U.A[k][i]; U.A[r][i]=A.A[r][i]-temp1; } //L for(int i=r+1;i<A.size;i++){ for(int k=0;k<r-1;k++) temp2=temp2+L.A[i][k]*U.A[k][r]; L.A[i][r]=(A.A[i][r]-temp2)/U.A[r][r]; } } cout<<"计算U得:"<<endl; U.Disp(); cout<<"计算L的:"<<endl; L.Disp(); double *Y=new double [A.size]; Y[0]=A.b[0]; for(int i=1;i<A.size;i++ ){ double temp3=0; for(int k=0;k<i-1;k++) temp3=temp3+L.A[i][k]*Y[k]; Y[i]=A.b[i]-temp3; } Xn[A.size-1]=Y[A.size-1]/U.A[A.size-1][A.size-1]; for(int i=A.size-1;i>=0;i--){ double temp4=0; for(int k=i+1;k<A.size;k++) temp4=temp4+U.A[i][k]*Xn[k]; Xn[i]=(Y[i]-temp4)/U.A[i][i]; } return Xn; } int main() { Matrix B(4); B.Input(); double *X; X=Dooli(B); cout<<"~~~~解得:"<<endl; for(int i=0;i<B.size;i++) cout<<"X["<<i<<"]:"<<X[i]<<" "; cout<<endl<<"呵呵呵呵呵"; return 0; }
标签： 道理特分解法
上传时间： 2018-05-20
上传用户：Aa123456789
16-, 14-, 12-Bit, Six-Channel, Simultaneous Sampling ANALOG-TO-DIGITAL CONVERTERS
上传时间： 2018-06-07
上传用户：nj精灵
From the transition of analog to digital communication along with seamless mobility and high computing power of small handheld devices, the wireless communications industry has seen tremendous changes leading to the integration of several telecommunication networks, devices and services over last 30 years. The rate of this progress and growth has increased particularly in the past decade because people no longer use their devices and networks for voice only, but demand bundle contents such as data download/streaming, HDTV, HD video , 3D video conferencing with higher efficiency, seamless connectivity, intelligence, reliability and better user experience. Although the challenges facing service providers and telecommunication companies differ by product, region, market size, and their areas of concentration but time to market, efficient utilization of their assets and revenue expansion, have impacted significantly how to manage and conduct their business while maintaining sufficient margin.
标签： Convergence Networks Beyond 4G of
上传时间： 2020-05-26
上传用户：shancjb
The continued reduction of integrated circuit feature sizes and commensurate improvements in device performance are fueling the progress to higher functionality and new application areas. For example, over the last 15 years, the performance of microprocessors has increased 1000 times. Analog circuit performance has also improved, albeit at a slower pace. For example, over the same period the speed/resolution figure-of-merit of analog-to-digital converters improved by only a factor 10.
标签： Digitally Assisted Pipeline ADCs
上传时间： 2020-05-27
上传用户：shancjb
Mobile multimedia communication is increasingly in demand because of the basic need to communi- cate at any time, anywhere, using any technology. In addition, to voice communication, people have a desire to access a range of other services that comprise multimedia elements—text, image, animation, high fidelity audio and video using mobile communication networks. To meet these demands, mobile communication technologies has evolved from analog to digital, and the networks have passed through a number of generations from first generation (1G) to fourth generation (4G).
标签： Communications Multimedia Concepts Mobile
上传时间： 2020-05-30
上传用户：shancjb
