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此代码是本人一年前所编写 对JAVA初学者可能会又帮助 在数据库连接上 效率较高...
此代码是本人一年前所编写 对JAVA初学者可能会又帮助 在数据库连接上 效率较高-year ago, I prepared for Java beginners may also help connect the database more efficient
- 2023-08-01 16:30:04下载
- 积分:1
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基于可编程跨导运算放大器POTA低通有源滤波器设计。该文给出了基于可编程跨导运算放大器(POTA)的有源滤波器设计方法,能在一定范围内实现跨导值程控调节, 提高...
基于可编程跨导运算放大器POTA低通有源滤波器设计。该文给出了基于可编程跨导运算放大器(POTA)的有源滤波器设计方法,能在一定范围内实现跨导值程控调节, 提高调节精度和准确度。并且利用Jacobi 法求解相似对角形矩阵,避免解高次方程的难题。所设计的有源滤波电路不易受分布电容的影响,稳定性好、灵敏度低。本文给出了设计实例,显示了该方法的优点。-Based on programmable transconductance operational amplifier POTA low-pass active filter design. In this paper, based on the programmable transconductance amplifier (POTA) of the active filter design method that can realize in a certain range transconductance value of program-controlled adjustment, improve precision and accuracy regulation. Jacobi method and use a similar matrix of angular, high-equation solution to avoid problems. Designed active filter circuit less susceptible to the effects of distributed capacitance, good stability, low sensitivity. In this paper, design examples, showing the advantages of the method.
- 2022-10-29 13:30:03下载
- 积分:1
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绘制B样条曲线,可以试一下
绘制B样条曲线,可以试一下-mapping of B-spline curves, Try
- 2022-07-22 18:36:17下载
- 积分:1
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CT仿真源代码
基于Shepp_Logan编写的
CT仿真源代码
基于Shepp_Logan编写的-CT-based simulation of the source code written Shepp_Logan
- 2022-09-05 10:10:03下载
- 积分:1
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Neusoft embedded programming training, quality is very good, well
东软嵌入式编程培训,质量非常不错,著名公司的质料,对编程非常有帮助-Neusoft embedded programming training, quality is very good, well-known companies of the material, on the programming very helpful
- 2022-10-10 10:40:04下载
- 积分:1
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数根
数也有根吗?有!数根就是把一个数的各个数位上的值相加,直到得到一个一位数为止。请你编写一个程序来计算任意数N(0≤N≤1000000000000)的...
数根
数也有根吗?有!数根就是把一个数的各个数位上的值相加,直到得到一个一位数为止。请你编写一个程序来计算任意数N(0≤N≤1000000000000)的根。
输入:
输入有多个案例。每一个案例为一行,每行只有一个整数N。当输入的N是-1时,输入结束。
输出:
每一个案例输出一行。每行只有一个数字,就是N对应的数根。
Sample input
39
-1
Sample output
3-Number of root number also has roots do? There is! The number of roots is to a number of different values on the digital sum until they have been a number of up to one. Please write a program to calculate an arbitrary number N (0 ≤ N ≤ 1000000000000) roots. Input: Enter a number of cases. Each case for the line, each line is only an integer N. When the input of N is-1, the input end. Output: the output line of each case. There is only one number per line, that is, N corresponds to the number of roots. Sample input 39-1 Sample output 3
- 2023-09-05 21:25:03下载
- 积分:1
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一个关于画图的简单的程序,大家随便看看,自己编的
一个关于画图的简单的程序,大家随便看看,自己编的-A simple drawing program, we just look at the own
- 2022-10-15 18:30:02下载
- 积分:1
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Introduction to Algorithms MIT course lecture notes, very complete, is a compute...
MIT算法导论课程的讲义,十分齐全,是计算机专业的学生学习算法设计的经典入门材料-Introduction to Algorithms MIT course lecture notes, very complete, is a computer science students to learn the classic algorithm design entry materials
- 2022-08-08 10:27:05下载
- 积分:1
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用Access增强NET功能报表,超使用的,用Access增强NET功能报表,超使用的...
用Access增强NET功能报表,超使用的,用Access增强NET功能报表,超使用的-Access
- 2023-09-07 17:25:02下载
- 积分:1
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短路潮流计算软件,可以用来完成短路潮流的计算
短路潮流计算软件,可以用来完成短路潮流的计算-power flow
- 2023-05-06 06:40:03下载
- 积分:1