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基于 飞思卡尔单片机 MODBUS汇编源程序
可以
基于 飞思卡尔单片机 MODBUS汇编源程序
可以-FREESCALE
- 2023-03-06 18:00:03下载
- 积分:1
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,片选接到p1.5
复位端 p1.4
中断端接到 int0
82c250工作在斜率模式下 ,RS 接51k电阻到地。
2,SJA10...
,片选接到p1.5
复位端 p1.4
中断端接到 int0
82c250工作在斜率模式下 ,RS 接51k电阻到地。
2,SJA1000与82C250的接口,逻辑上为SJA1000输出0(显性位),表现为低点平输出给82C250,则82C250输出
差分的低电平。SJA1000输出1(隐性位),表现为高电平或悬空输出给82C250,则82C250输出悬空状态。
3,周立功的中SJA_BCANCONF.ASM中对输出控制寄存器 的TX0的输出 极性的位的定义不对。
还有里面定义有接收错误标志码,不知道何时用。-, The election received a P1.5-chip client p1.4 interrupt reset int0 82c250 client received in the slope mode, RS then 51k resistor to ground. 2, SJA1000 with the 82C250 interface, logic for SJA1000 Output 0 (dominant bit), showing low level output to the 82C250, while the low-level differential output 82C250. SJA1000 Output 1 (recessive), and the performance of the output is high or left vacant to the 82C250, the vacant state 82C250 output. 3, Zhou Ligong in the SJA_BCANCONF.ASM of output control register TX0 output polarity bit the definition of wrong. Inside the definition also has to receive error flag code, I do not know when to use.
- 2022-06-28 07:07:40下载
- 积分:1
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霍夫曼编码,根据给定权值创建赫夫曼编码,显示每一步如何进行及所得编码...
霍夫曼编码,根据给定权值创建赫夫曼编码,显示每一步如何进行及所得编码-Huffman coding, according to the weight given to create Huffman codes, showing how each step and the resultant code
- 2022-11-17 16:05:03下载
- 积分:1
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单周期完成版
写一个单周期处理器运行一段mips指令,并包含mips指令转汇编码的程序(Write a single cycle processor to run a section of MIPS instruction)
- 2020-07-03 04:20:02下载
- 积分:1
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myNotepad
使用RadAsm做的一个有基本功能的记事本程序(Use RadAsm to make a basic functional Notepad)
- 2011-07-01 04:24:34下载
- 积分:1
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进制转换,将数字以任意进制(2,8,10,16)输入,选择另一个进制(2,8,10,16)输出。...
进制转换,将数字以任意进制(2,8,10,16)输入,选择另一个进制(2,8,10,16)输出。-Hexadecimal conversion, digital to arbitrary M-ary (2,8,10,16) input, select another band (2,8,10,16) output.
- 2023-06-23 03:25:02下载
- 积分:1
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在C语言中使用汇编语言设计边框范例 .zip
在C语言中使用汇编语言设计边框范例 .zip-in the C language used assembly language frame design paradigm. Zip
- 2022-03-12 07:28:18下载
- 积分:1
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Dining philosophers problem is presented and solved Dijkatra typical synchroniza...
哲学家就餐问题是由Dijkatra提出并解决的典型同步问题。该问题描述的是有五个哲学家共用一张圆桌,分别坐在周围的五张椅子上,圆桌上有五个碗和五只筷子,他们的生活方式是交替的进行思考和就餐。平时,一个哲学家进行思考,饥饿时便试图取用其左右最靠近他的筷子,只有在他拿到两只筷子时才能进餐。进餐完,放下筷子继续思考。-Dining philosophers problem is presented and solved Dijkatra typical synchronization problems. Description of the problem is that there are five philosophers share a round-table were sitting around the five chair, round table has five bowl and five chopsticks, their way of life is the turn of thinking and eating. Normally, a philosopher to think, hunger when trying to access their nearest him about the chopsticks, only if he can get two chopsticks when eating. End meals, to lay down their chopsticks to continue thinking.
- 2023-06-13 12:15:03下载
- 积分:1
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51汇编 三线驱动5片74HC595级联输出并由5片74HC165级联读回,proteus仿真
51汇编 三线驱动5片74HC595级联输出并由5片74HC165级联读回,有proteus仿真
- 2022-01-28 07:18:27下载
- 积分:1
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27个源代码,其中包括数字方阵,图形,游戏
27个源代码,其中包括数字方阵,图形,游戏-27 source code, including digital matrix, graphics, games
- 2022-02-04 04:50:13下载
- 积分:1