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Description: 这是关于LDPC信道编码模块设计的程序
打开源程序，先运行gengrate_h.m程序，陆续将码长设置为756bit，列重设置为3，行重设置为9。在Workspace中同时将H、A、B、C、D、E、Hget、Fget、g、Tget这是个变量选择另存为encode_in.mat 格式。再运行main_encode.m进行编码，主程序运行后，在当前目录下，自动生成编码结果文件“encode—out.mat”，这将作为下一次扩频调制仿真实验的的输入信号。最后分别查看Workspace中的变量s（编码前数据）和xyuan（编码后数据）的波形。
对比后，可以看出编码前的数据码片长度为504bit，编码后的码片数据长度为756bit。编码效率=编码前码片长度/编码后码片长度=2/3。-This is about the LDPC channel coding module design process
Open source, first run gengrate_h.m program, phasing out the code length is set to 756bit, column re-set to 3, line weight is set to 9. In the Workspace in the same time, H, A, B, C, D, E, Hget, Fget, g, Tget This is a variable select Save As encode_in.mat format. Then run main_encode.m encoding, the main program running in the current directory, the results of automatically generated code file "encode-out.mat", which will serve as the next simulation of the spread spectrum modulation input signal. Finally, the variables were View Workspace in s (before encoding data) and xyuan (encoded data) waveform.
After comparison, we can see the data before encoding chip length of 504bit, encoded data length of chip 756bit. Coded before coding efficiency = chip length/length of the encoded chip = 2/3.
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Size: 15360 |
Author: 吴健 |
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Description: （1）、设计伪随机码电路：产生八位伪随机序列（如M序列、Gold序列等）；
（2）、了解D/A的工作原理及使用方法，将伪随机序列输入D/A中（如DAC0808），观察其模拟信号的特性；
（3）、分析信号源的特点，使用EWB软件进行仿真；
-(1), pseudo-random code circuit design: eight pseudo-random sequence generation (such as M sequences, Gold sequences, etc.) (2) to understand the D/A and use the working principle, the pseudo-random sequence input D/A in (such as the DAC0808), to observe the characteristics of the analog signal (3), analysis of the characteristics of the signal source, using EWB software simulation
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Size: 233472 |
Author: 张阳 |
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Description: 双击可执行程序RQSZ.exe，在出现的提示信息中输入大写字母“D”,可显示系统当前日期；输入大写字母“T”,可显示系统当前时间；输入大写字母
“Q”，可结束程序。-a assembly program that can show the data and time on your computer by input "D" or "T" to choose date or time to show on the screen.
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Size: 2048 |
Author: xiao 5 |
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Description: D触发器设计图形输入法,设计软件quartus-Input D flip-flop design graphics, design software quartus
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Size: 4096 |
Author: wangchenlin2000 |
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Description: 本课程设计主要是使学生增进对伪随机序列的认识，加深对通信原理理论方面的理解，使学生了解如何产生伪随机序列以及D/A的工作原理及使用方法，并将伪随机序列输入D/A转换器，观察其模拟信号特性。-This course is designed to make the students improve the understanding of the pseudo random sequence, to deepen the understanding of communication principle theory, is to make the students know how to generate pseudo random sequence and D/A working principle and method of use, and will pseudo random sequence input D/A converter, observe the analog signal characteristics.
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Size: 404480 |
Author: 刘志友 |
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Description: 汇编语言 输入成绩并排序 汇编语言 输入成绩并排序 汇编语言 输入成绩并排序-Assembly language input results and sequencingAssembly language input results and sequencingAssembly language input results and sequencing
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Size: 1024 |
Author: skyfish |
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Description: 输入一个从星期日到星期六中一天的英文字符串，判断它是一个星期的第几天，如果该字符串不存在，输出-1。-Input a from Sunday to Saturday in a day of English string, and judge it is a week s first day, if the string is not exist, output-1.
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Size: 10240 |
Author: qiyanyan |
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Description: 输入法注入源码，能通过输入法注入任意D-Input into the source code
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Size: 106496 |
Author: woaini |
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Description: You will no be able to see the output:
Welcome to ICS 103: Programming in C
To overcome this we can stop the output window in two ways:
Use an input statement such as getch(), getchar(), or scanf before the return statement
Use the Operating System call system("PAUSE") before the return statement
The first method is better than the second as it results in a more efficient program.
-You will no be able to see the output:
Welcome to ICS 103: Programming in C
To overcome this we can stop the output window in two ways:
Use an input statement such as getch(), getchar(), or scanf before the return statement
Use the Operating System call system("PAUSE") before the return statement
The first method is better than the second as it results in a more efficient program.
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Size: 11264 |
Author: alaa |
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Description: Function to implement Linear Discriminant Analysis
Usage:-
[Xnew,V,mX,D] = lda(X,index,mX)
[V,D] = lda(X,index)
where:-
X = concatenated grand profile vectors (GPV)
i = class index of each vector
Xnew = new feature vectors
P = eigen vectors
mX = mean of GPV (optional in input)
D = eigen values- Function to implement Linear Discriminant Analysis
Usage:-
[Xnew,V,mX,D] = lda(X,index,mX)
[V,D] = lda(X,index)
where:-
X = concatenated grand profile vectors (GPV)
i = class index of each vector
Xnew = new feature vectors
P = eigen vectors
mX = mean of GPV (optional in input)
D = eigen values
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Size: 1024 |
Author: Praveen |
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Description: 设计一个D型触发器，输入CK（时钟信号, ↑表示上升沿时刻），D（数据），Clear端（“0”时清零），输出Q-Design of a D-type flip-flop, the input CK (clock signal, ↑ indicates rising time), D (data), Clear end (" 0" is cleared), the output Q
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Size: 3072 |
Author: 许光达 |
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Description: 编写程序，登录幼儿园200个小朋友的数据：姓名、性别、年龄、身高、体重、出生日期，分别按年龄排序后输出。
要求：
（1）登录数据用函数input()
（2）按身高排序用函数sort()
（3）输出排序结果用函数output()，输出格式为：
幼儿园小朋友一览（依身高排序）
===================================================
姓名 性别 年龄 身高 出生日期-Write a program, log K 200 children' s data: name, gender, age, height, weight, date of birth, respectively, after sorting the output by age. Requirements: (a) the login data using the function input () (2) Sort by height using the function sort () (3) Output sort results with the function output (), output format: kindergartners list (sorted by height) === ================================================ Name Sex Age Height Date of birth
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Size: 1024 |
Author: yean |
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Description: 第四届华为杯程序设计竞赛D题目。问题输入
输入包括多个测试用例，首先给出测试用例数N，接着给出N个测试用例，每一个测试用例包括1行，首先给出正整数序列数M，接着依次给出M个正整数，1≤N≤100000，2≤M≤500000，正整数大小不超过1000。
问题输出
针对每一个测试用例，输出一行，给出对一系列正整数进行任意次变换后求得的最小之和。
样例输入
2
4 2 4 8 16
3 3 2 2
样例输出
8
3
-Problem input
Inputs include multiple test cases, test cases given first number N, then given N test cases, including a line for each test case, the first number given sequence of positive integers M, then turn gives M a positive integer, a ≤ N ≤ 100000,2 ≤ M ≤ 500000, positive integer size does not exceed 1000.
Problem Output
For each test case, output a line, given after a series of positive integers transform any number of times to obtain the minimum sum.
Sample Input
2
424816
3322
Sample Output
8
3
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Size: 6144 |
Author: 赵博伟 |
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Description: 1.从键盘上输入某班学生的某科目成绩.输入按学生的学号由小到大的顺序输入.
2.统计检查每个学生的名次.
3.将统计结果在屏幕上显示.
4.为便于观察,输入学生数目不宜太多,以不超过一屏为宜.输出应便于阅读.尽可
能考虑美观.
5.输入要有检错手段.
3.对输入要有检错措施,以防止非法字符输入,并有适当的提示.
4.将整个程序分解为若干模块,分别用子程序实现.在报告中要给出模块层次图.-1 Enter a class of students in a course result from the keyboard. Enter the order number from small to big school student input.
2. Statistical examination of each student ranking.
3 The statistical results displayed on the screen.
(4) To facilitate the observation, enter the number of students should not be too much to not more than one screen is appropriate. Output should be easy to read. Feel
Can be considered beautiful.
5 Enter to have error detection methods.
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Size: 1024 |
Author: 王程序 |
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Description: 设计一个稀疏矩阵计算器，要求能够：⑴输入并建立稀疏矩阵；⑵输出稀疏矩阵；⑶执行两个矩阵相加；⑷执行两个矩阵相乘；⑸求一个矩阵的转置矩阵；⑹求一个矩阵的逆矩阵（选做）。-Design of a sparse matrix calculator, requires the ability to: ⑴ input and the establishment sparse matrix ⑵ output sparse matrix ⑶ perform two matrices are added ⑷ perform two matrix multiplication ⑸ seeking a matrix transpose matrix ⑹ seeking a inverse matrix (Optional).
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Size: 3072 |
Author: David |
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Description: 1.从键盘上输入某班学生的某科目成绩.输入按学生的学号由小到大的顺序输入.
2.统计检查每个学生的名次.
3.将统计结果在屏幕上显示.-1. Enter the keyboard of a subject in a class of students score. Enter the order number small to big school student input. 2. Check each student' s ranking statistics. 3. The statistical results are displayed on the screen.
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Size: 1024 |
Author: 尹航 |
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Description: 稀疏矩阵的压缩存储：
实现稀疏矩阵压缩存储，并实现矩阵转置和求和。
输入矩阵时，首先需要输入非零元素的个数，然后分别输入矩阵的 行号，列号和值。
输完2个矩阵后，自动进行计算第一个矩阵的转置以及两个矩阵的和。
例如：输入如下：
100 90 5 //矩阵的行数为100，列数为90，共5个非零元素。
1 10 100 //a(1,10)=100
50 60 200//a(50,60)=200
50 80 100//a(50,80)=100
60 60 200//a(60,60)=200
99 89 10//a(99,89)=10
100 90 4 //矩阵b的行数为100，列数为90，共4个非零元素。
1 1 10 //b(1,1)=10
50 60 -200//b(50,60)=-200
50 80 100 //b(50,80)=100
70 70 10 //b(70,70)=10-Compressed storage sparse matrix: Implementing sparse matrix compression storage and realization matrix transpose and summation. When you enter the matrix, you first need to enter the number of non-zero elements of the matrix are input and line number, column number and value. Losers 2 matrix, the automatic calculation of a matrix transpose and matrix and the two. Example: Enter the following: Number of lines 100 905 // matrix is 100, the number of columns is 90, a total of five non-zero elements. 1 10 100 // a (1,10) = 100 50 60 200 // a (50,60) = 200 50 80 100 // a (50,80) = 100 60 60 200 // a (60,60) = 200 99 89 10 // a (99,89) = 10 100 90 4 // matrix b is the number of lines is 100, the number of columns is 90, a total of four non-zero elements. 1 1 10 // b (1,1) = 10 50 60-200 // b (50,60) =- 200 50 80 100 // b (50,80) = 100 70 70 10 // b (70, 70) = 10
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Size: 688128 |
Author: 刘忠威 |
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Description: You are given N numbers: d[0], d[1], ..., d[N - 1].
Output a string of bits with N bits (0/1) in which the i-th bit is 1 if there is any number before d[i] which equals d[i], otherwise output 0.
Then, output a string of bits with N bits (0/1) in which the i-th bit is 1 if there is any number after d[i] which equals d[i], otherwise output 0.
Input Format
N // count of the numbers in the input.
d[0] // 1st number
d[1] // 2nd number
...
d[n - 1] // last number
Output Format
The first line contains a string of n bits, where the i-th bit (as read left to right) indicates if there is any number before b[i] which is equal to b[i].
The second line contains a string of n bits, where the i-th bit (as read left to right) indicates if there is any number after b[i] which is equal to b[i].-You are given N numbers: d[0], d[1], ..., d[N - 1].
Output a string of bits with N bits (0/1) in which the i-th bit is 1 if there is any number before d[i] which equals d[i], otherwise output 0.
Then, output a string of bits with N bits (0/1) in which the i-th bit is 1 if there is any number after d[i] which equals d[i], otherwise output 0.
Input Format
N // count of the numbers in the input.
d[0] // 1st number
d[1] // 2nd number
...
d[n - 1] // last number
Output Format
The first line contains a string of n bits, where the i-th bit (as read left to right) indicates if there is any number before b[i] which is equal to b[i].
The second line contains a string of n bits, where the i-th bit (as read left to right) indicates if there is any number after b[i] which is equal to b[i].
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Size: 2048 |
Author: Syed Nauman |
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Description: 将3D文件导入OpenGL中去，方便显示-import in the OpenGL 3 d files, convenient display
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Size: 22214656 |
Author: Mr.kf |
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Description: 基本的D触发器，可实现基本的保持功能。输入到输出不变。(The basic D flip flops enable basic retention functions. Input to output remain unchanged.)
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Size: 3072 |
Author: 紫芩
|
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