VHDL編寫的4選一數(shù)據(jù)選擇器 entity mux41a is port(a,b:in std_logic; s1,s2,s3,s4:in std_logic; y: out std_logic); end entity mux41a; architecture one of mux41a is signal ab:std_logic_vector(1 downto 0);
標(biāo)簽： VHDL 數(shù)據(jù)選擇器
上傳時(shí)間： 2020-05-15
上傳用戶：cdga
Many times I have been asked to explain “ briefl y ” how SDH, SONET, and the OTN “ exactly ” work. The questions came mainly from new colleagues, stu- dents, and users of these technologies, personally or via the usenet newsgroup comp.dcom.sdh - sonet. I could have referred them to the standards documents, but to provide a more consistent and clear answer I decided to write this pocket guide. The objective of this book is that it can be used both as an introduction as well as a reference guide to these technologies and their spe- cifi c standards documents.
標(biāo)簽： Generation Transport Optical ComSoc Guide
上傳時(shí)間： 2020-05-27
上傳用戶：shancjb
function y=lagr(x0,y0,x) %x0,y0為節(jié)點(diǎn) %x是插值點(diǎn) n=length(x0); m=length(x); for i=1:m z=x(i); s=0.0; for k=1:n p=1.0; for j=1:n if j~=k p=p*(z-x0(j))/(x0(k)-x0(j)); end end s=p*y0(k)+s; end y(i)=s; end
標(biāo)簽： lagr
上傳時(shí)間： 2020-06-09
上傳用戶：shiyc2020
My association with the theory of controls in continuous time started during my studies at the Indian Institute of Technology, Kharagpur, India, in 1974 as an undergraduate student in the Controls and Power program. The initial introduction by Professors Kesavamurthy, Y. P. Singh, and Rajagopalan laid the foundation for a good basic understanding of the subject matter. This pursuit and further advanced study in the field of digital controls continued during my days as a graduate student in the Electrical and Systems Engineering Department at the University of Connecticut in Storrs, from 1983 to 1988.
標(biāo)簽： Applications Digital Control
上傳時(shí)間： 2020-06-10
上傳用戶：shancjb
#include<stdio.h> #define TREEMAX 100 typedef struct BT { char data; BT *lchild; BT *rchild; }BT; BT *CreateTree(); void Preorder(BT *T); void Postorder(BT *T); void Inorder(BT *T); void Leafnum(BT *T); void Nodenum(BT *T); int TreeDepth(BT *T); int count=0; void main() { BT *T=NULL; char ch1,ch2,a; ch1='y'; while(ch1=='y'||ch1=='y') { printf("\n"); printf("\n\t\t 二叉樹子系統(tǒng)"); printf("\n\t\t*****************************************"); printf("\n\t\t 1---------建二叉樹 "); printf("\n\t\t 2---------先序遍歷 "); printf("\n\t\t 3---------中序遍歷 "); printf("\n\t\t 4---------后序遍歷 "); printf("\n\t\t 5---------求葉子數(shù) "); printf("\n\t\t 6---------求結(jié)點(diǎn)數(shù) "); printf("\n\t\t 7---------求樹深度 "); printf("\n\t\t 0---------返 回 "); printf("\n\t\t*****************************************"); printf("\n\t\t 請(qǐng)選擇菜單號(hào) （0--7）"); scanf("%c",&ch2); getchar(); printf("\n"); switch(ch2) { case'1': printf("\n\t\t請(qǐng)按先序序列輸入二叉樹的結(jié)點(diǎn):\n"); printf("\n\t\t說明：輸入結(jié)點(diǎn)（‘0’代表后繼結(jié)點(diǎn)為空）后按回車。\n"); printf("\n\t\t請(qǐng)輸入根結(jié)點(diǎn):"); T=CreateTree(); printf("\n\t\t二叉樹成功建立！\n");break; case'2': printf("\n\t\t該二叉樹的先序遍歷序列為:"); Preorder(T);break; case'3': printf("\n\t\t該二叉樹的中序遍歷序列為:"); Inorder(T);break; case'4': printf("\n\t\t該二叉樹的后序遍歷序列為:"); Postorder(T);break; case'5': count=0;Leafnum(T); printf("\n\t\t該二叉樹有%d個(gè)葉子。\n",count);break; case'6': count=0;Nodenum(T); printf("\n\t\t該二叉樹總共有%d個(gè)結(jié)點(diǎn)。\n",count);break; case'7': printf("\n\t\t該樹的深度為:%d",TreeDepth(T)); break; case'0': ch1='n';break; default: printf("\n\t\t***請(qǐng)注意：輸入有誤！***"); } if(ch2!='0') { printf("\n\n\t\t按【Enter】鍵繼續(xù)，按任意鍵返回主菜單！\n"); a=getchar(); if(a!='\xA') { getchar(); ch1='n'; } } } } BT *CreateTree() { BT *t; char x; scanf("%c",&x); getchar(); if(x=='0') t=NULL; else { t=new BT; t->data=x; printf("\n\t\t請(qǐng)輸入%c結(jié)點(diǎn)的左子結(jié)點(diǎn):",t->data); t->lchild=CreateTree(); printf("\n\t\t請(qǐng)輸入%c結(jié)點(diǎn)的右子結(jié)點(diǎn):",t->data); t->rchild=CreateTree(); } return t; } void Preorder(BT *T) { if(T) { printf("%3c",T->data); Preorder(T->lchild); Preorder(T->rchild); } } void Inorder(BT *T) { if(T) { Inorder(T->lchild); printf("%3c",T->data); Inorder(T->rchild); } } void Postorder(BT *T) { if(T) { Postorder(T->lchild); Postorder(T->rchild); printf("%3c",T->data); } } void Leafnum(BT *T) { if(T) { if(T->lchild==NULL&&T->rchild==NULL) count++; Leafnum(T->lchild); Leafnum(T->rchild); } } void Nodenum(BT *T) { if(T) { count++; Nodenum(T->lchild); Nodenum(T->rchild); } } int TreeDepth(BT *T) { int ldep,rdep; if(T==NULL) return 0; else { ldep=TreeDepth(T->lchild); rdep=TreeDepth(T->rchild); if(ldep>rdep) return ldep+1; else return rdep+1; } }
上傳時(shí)間： 2020-06-11
上傳用戶：ccccy
#include <stdio.h> #include <stdlib.h> #define SMAX 100 typedef struct SPNode { int i,j,v; }SPNode; struct sparmatrix { int rows,cols,terms; SPNode data [SMAX]; }; sparmatrix CreateSparmatrix() { sparmatrix A; printf("\n\t\t請(qǐng)輸入稀疏矩陣的行數(shù)，列數(shù)和非零元素個(gè)數(shù)（用逗號(hào)隔開）:"); scanf("%d,%d,%d",&A.cols,&A.terms); for(int n=0;n<=A.terms-1;n++) { printf("\n\t\t輸入非零元素值（格式：行號(hào)，列號(hào)，值）:"); scanf("%d,%d,%d",&A.data[n].i,&A.data[n].j,&A.data[n].v); } return A; } void ShowSparmatrix(sparmatrix A) { int k; printf("\n\t\t"); for(int x=0;x<=A.rows-1;x++) { for(int y=0;y<=A.cols-1;y++) { k=0; for(int n=0;n<=A.terms-1;n++) { if((A.data[n].i-1==x)&&(A.data[n].j-1==y)) { printf("%8d",A.data[n].v); k=1; } } if(k==0) printf("%8d",k); } printf("\n\t\t"); } } void sumsparmatrix(sparmatrix A) { SPNode *p; p=(SPNode*)malloc(sizeof(SPNode)); p->v=0; int k; k=0; printf("\n\t\t"); for(int x=0;x<=A.rows-1;x++) { for(int y=0;y<=A.cols-1;y++) { for(int n=0;n<=A.terms;n++) { if((A.data[n].i==x)&&(A.data[n].j==y)&&(x==y)) { p->v=p->v+A.data[n].v; k=1; } } } printf("\n\t\t"); } if(k==1) printf("\n\t\t對(duì)角線元素的和：:%d\n",p->v); else printf("\n\t\t對(duì)角線元素的和為：:0"); } int main() { int ch=1,choice; struct sparmatrix A; A.terms=0; while(ch) { printf("\n"); printf("\n\t\t 稀疏矩陣的三元組系統(tǒng) "); printf("\n\t\t*********************************"); printf("\n\t\t 1------------創(chuàng)建 "); printf("\n\t\t 2------------顯示 "); printf("\n\t\t 3------------求對(duì)角線元素和"); printf("\n\t\t 4------------返回 "); printf("\n\t\t*********************************"); printf("\n\t\t請(qǐng)選擇菜單號(hào)（0-3）:"); scanf("%d",&choice); switch(choice) { case 1: A=CreateSparmatrix(); break; case 2: ShowSparmatrix(A); break; case 3: SumSparmatrix(A); break; default: system("cls"); printf("\n\t\t輸入錯(cuò)誤！請(qǐng)重新輸入！\n"); break; } if (choice==1||choice==2||choice==3) { printf("\n\t\t"); system("pause"); system("cls"); } else system("cls"); } }
上傳時(shí)間： 2020-06-11
上傳用戶：ccccy
某疾病發(fā)生率y‰和年齡段x（每五年為一段，例如0~5歲為第一段..）
標(biāo)簽： C代碼
上傳時(shí)間： 2020-06-14
上傳用戶：ll12346
第一種邊界條件下的三次樣條插值問題（高斯消元法） 題目 計(jì)算方法上機(jī)實(shí)習(xí)題目（二） ——第一種邊界條件下的三次樣條插值問題（高斯消元法） 已知直升飛機(jī)旋轉(zhuǎn)機(jī)翼外形曲線輪廓線上的某些型值點(diǎn)（見表）及端點(diǎn)處的一階導(dǎo)數(shù)值 y' (x ) = 1.86548, y' (x ) = -0.046115
標(biāo)簽： 計(jì)算方法上機(jī)實(shí)習(xí)題目
上傳時(shí)間： 2020-06-27
上傳用戶：zg250
x=[1,2,0,-1,3,2];h=[1,-1,1]; y1=x*h(1); y2=x*h(2); y3=x*h(3); Y1=[0,0,y1]; Y2=[0,y2,0]; Y3=[y3,0,0]; y=Y1+Y2+Y3; L=-2:1:5; figure(1); subplot(211);stem(L,y,'*'); xlabel('L');ylabel('y');title('(1)'); X=x.';X=X'; r1=X*y(1);r2=X*y(2);r3=X*y(3);r4=X*y(4); r5=X*y(5);r6=X*y(6);r7=X*y(7);r8=X*y(8); R1=[0,0,0,0,0,0,0,r1];R2=[0,0,0,0,0,0,r2,0]; R3=[0,0,0,0,0,r3,0,0];R4=[0,0,0,0,r4,0,0,0]; R5=[0,0,0,r5,0,0,0,0];R6=[0,0,r6,0,0,0,0,0]; R7=[0,r7,0,0,0,0,0,0];R8=[r8,0,0,0,0,0,0,0]; R=R1+R2+R3+R4+R5+R6+R7+R8; n=-7:5; subplot(212);stem(n,R);title('(2)');
標(biāo)簽： ketang
上傳時(shí)間： 2020-11-10
上傳用戶：
本設(shè)計(jì)方案中我采用多路復(fù)用器，2-4譯碼器，LED燈和或門等器件來完成設(shè)計(jì)。用2個(gè)74x151多路復(fù)用器擴(kuò)展為16-2多路復(fù)用器,題目中的地址代碼A、B、C、D4個(gè)輸入端作為擴(kuò)展的多路復(fù)用器的地址端，DO-D8作為數(shù)據(jù)端。開箱鑰匙孔信號(hào)E作為2-4decoder的使能端。設(shè)計(jì)開鎖的正確代碼為0101，當(dāng)用鑰匙開鎖(即2-4decoder的使能端有效〉時(shí)，如果正確輸入開鎖密碼:0101，則輸出Y為邏輯高電平，Y’為邏輯低電平，鎖被打開，而LED燈不會(huì)亮（即不會(huì)報(bào)警);如果輸入的密碼錯(cuò)誤或者鑰匙孔信號(hào)無效，則輸出Y為邏輯低電平，Y’為邏輯高電平，鎖無法打開，邏輯高電平Y(jié)’驅(qū)動(dòng)LED燈亮，產(chǎn)生報(bào)警效果。2.設(shè)計(jì)原理圖:(以下電路圖為用QuartusI蹤合后截屏所得) 本設(shè)計(jì)方案中我采用多路復(fù)用器，2-4譯碼器，LED燈和或門等器件來完成設(shè)計(jì)。用2個(gè)74x151多路復(fù)用器擴(kuò)展為16-2多路復(fù)用器,題目中的地址代碼A、B、C、D4個(gè)輸入端作為擴(kuò)展的多路復(fù)用器的地址端，DO-D8作為數(shù)據(jù)端。開箱鑰匙孔信號(hào)E作為2-4decoder的使能端。設(shè)計(jì)開鎖的正確代碼為0101，當(dāng)用鑰匙開鎖(即2-4decoder的使能端有效〉時(shí)，如果正確輸入開鎖密碼:0101，則輸出Y為邏輯高電平，Y’為邏輯低電平，鎖被打開，而LED燈不會(huì)亮（即不會(huì)報(bào)警);如果輸入的密碼錯(cuò)誤或者鑰匙孔信號(hào)無效，則輸出Y為邏輯低電平，Y’為邏輯高電平，鎖無法打開，邏輯高電平Y(jié)’驅(qū)動(dòng)LED燈亮，產(chǎn)生報(bào)警效果。2.設(shè)計(jì)原理圖:(以下電路圖為用QuartusI蹤合后截屏所得)
標(biāo)簽： 密碼鎖
上傳時(shí)間： 2021-04-26
上傳用戶：情可傾想
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