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例如RS232通讯RS232
RS232 Example for RS232 communication
- 2022-12-18 14:00:03下载
- 积分：1
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自动化代码
资源描述工厂产品测试自动化代码，包括机械手控制代码
- 2023-03-10 23:25:06下载
- 积分：1
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TLC2543介绍中国，用汇编操作，TLC2543是TI SW…
TLC2543的中文简介,附有汇编操作,TLC2543是德州仪器公司生产的12位开关电容型逐次逼近模数转换器，它具有三个控制输入端，采用简单的3线SPI串行接口可方便地与微机进行连接，是12位数据采集系统的最佳选择器件之一。本文介绍了该芯片的功能、时序，并给出了8051单片机的接口电路。-TLC2543 briefed the Chinese, with the compilation operation, TLC2543 is by TI Switched-Capacitor 12-successive approximation ADC. it has three control input, a simple three-wire SPI serial interface can be easily connected with the computer, yes 12 Data Acquisition System devices to choose the best one. This paper introduces the chip functional, timing, and gives the 8051 interface circuit.
- 2022-08-03 17:02:33下载
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单项链表的实现
单项链表的实现-individual achievement Chain
- 2023-02-05 07:35:03下载
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1个高级的剪贴板程序VC
1个高级的剪贴板程序VC-a senior clipboard procedures VC
- 2022-02-18 15:00:25下载
- 积分：1
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student performance management system is a C language series of the procedures l...
学生成绩管理系统是用C语言编的 程序不大，希望对初学者有用-student performance management system is a C language series of the procedures little hope useful for beginners
- 2022-02-02 01:44:52下载
- 积分：1
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哲学家进餐问题，进程是独立参与分配资源的最小单位，在有线程的OS中，线程是运行的最小单位，
课堂所述进程之间的同步与互斥，实际上是属于不同进程的线程间的同步...
哲学家进餐问题，进程是独立参与分配资源的最小单位，在有线程的OS中，线程是运行的最小单位，
课堂所述进程之间的同步与互斥，实际上是属于不同进程的线程间的同步与互斥，当
然，属于同一进程的不同线程一样存在同步与互斥，其控制同步与互斥的原理跟进程
是一样的-dining philosophers problem, the process is independent participation in the allocation of resources as the smallest unit, in the OS thread, thread is the smallest unit of operation, the process described in the classroom between the synchronization and mutual exclusion. actually belong to different threads in the process of synchronization and mutual exclusion, of course, the same process there is a different thread synchronization and mutual exclusion, control and coordinate the process with the principle of the same
- 2022-08-22 22:48:11下载
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详细地介绍了世界上先进的表面处理工艺
很实用
详细地介绍了世界上先进的表面处理工艺
很实用-detailed introduction of the world"s advanced surface treatment technology is practical
- 2023-02-06 01:05:04下载
- 积分：1
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问题描述:
一杯沸水冷却，圆柱体模型，底面半径0.05m，高0.1m，周围温度20度，初始水温100度
方程是四维输运方程（常数a^2=k/(c*...
问题描述:
一杯沸水冷却，圆柱体模型，底面半径0.05m，高0.1m，周围温度20度，初始水温100度
方程是四维输运方程（常数a^2=k/(c*p)，k是热传导系数0.6006焦/（米*秒*度））
初始条件：t=0时水等于100度
边界条件：1.上下壁都是自由冷却，第三类边界条件，周围温度保持在20度（H=k/h,h取1）
2.杯壁绝热，第二类边界条件
图形显示格式,取过圆柱轴的截面温度变化将其做成动画.
-Problem description : a cup of boiling water cooling cylinder model, the bottom radius of 0.05m, 0.1m high. temperature around 20 degrees, initial temperature of 100 degrees equation is four-dimensional equations (a constant ^ 2 = k/(c* p), k is the thermal conductivity coefficient of 0.6006 Coke/(m** seconds degrees)) initial conditions : t = 0 to 100 degree water boundary conditions : 1. upper and lower walls are free cooling, the third boundary conditions, maintain the temperature around 20 degrees (H = k/h, h for 1) 2. Beibi insulation, The second category boundary conditions graphics format, cylindrical shaft for the cross section of its temperature changes caused animation.
- 2022-05-12 17:55:03下载
- 积分：1
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在线电影院的简单的web管理系统采用的是php编码
在线电影院的简单的web管理系统采用的是php编码-online cinema simple web management system using the php code
- 2022-08-12 14:34:32下载
- 积分：1
