登录
首页 » Proteus » 蜂鸣器控制

蜂鸣器控制

于 2020-07-01 发布 文件大小:82KB
0 297
下载积分: 1 下载次数: 10

代码说明:

  LPC2138单片机控制蜂鸣器播放音乐, 蜂鸣器电路图(LPC2138 microcontroller control buzzer play music, Buzzer circuit)

文件列表:

蜂鸣器控制\GPIO1.axf
蜂鸣器控制\GPIO1.hex
蜂鸣器控制\GPIO1.htm
蜂鸣器控制\GPIO1.lnp
蜂鸣器控制\GPIO1.map
蜂鸣器控制\GPIO1.Opt
蜂鸣器控制\GPIO1.plg
蜂鸣器控制\GPIO1.sct
蜂鸣器控制\GPIO1.tra
蜂鸣器控制\GPIO1.Uv2
蜂鸣器控制\GPIO1_Opt.Bak
蜂鸣器控制\GPIO1_Target 1.dep
蜂鸣器控制\GPIO1_Uv2.Bak
蜂鸣器控制\L4_1.DSN
蜂鸣器控制\L4_1.PWI
蜂鸣器控制\Last Loaded L4_1.DBK
蜂鸣器控制\main.c
蜂鸣器控制\main.crf
蜂鸣器控制\main.d
蜂鸣器控制\main.o
蜂鸣器控制\main.__i
蜂鸣器控制\Startup.lst
蜂鸣器控制\Startup.o
蜂鸣器控制\Startup.s
蜂鸣器控制

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • 直插式电阻电容封装与尺寸图解(20190725092446)
    说明:  直插式电阻电容封装与尺寸图解(20190725092446)(Direct-plug Resistance Capacitor Packaging and Dimension Diagram (20190725092446))
    2020-06-15 22:22:59下载
    积分:1
  • 遗传算法&雷达手册
    遗传算法 本部分主要为了了解遗传算法的应用,选择一个复杂的二维函数来进行遗传算法优化。函数显示为y=10*sin(5*x)+7*abs(x-5)+10, 另附上雷达手册(genetic algorithm This part is mainly to understand the application of genetic algorithm, select a complex two-dimensional function to optimize genetic algorithm. The function is shown as y=10*sin (5*x) +7*abs (X-5) +10,)
    2017-10-23 19:35:22下载
    积分:1
  • nilm-eval-master
    NILM-Eval是一个MATLAB框架,允许在不同场景中评估非侵入式负载监控算法,以全面了解其性能.NILM-Eval可以轻松评估多个数据集,住户,数据粒度,时间段和特定算法参数的算法。通过将这些参数封装在配置中。NILM评价仪进一步允许用户轻松地重复其他人执行的实验,评估新数据集上的算法,以及微调配置以提高新算法的性能设置(Nilm-eval is a MATLAB framework that allows evaluation of non-invasive load monitoring algorithms in different scenarios to fully understand their performance. Nilm-eval can easily evaluate algorithms for multiple data sets, households, data granularity, time periods and specific algorithm parameters.By encapsulating these parameters in the configuration.NILM evaluator further allows users to easily repeat experiments performed by others, evaluate algorithms on new data sets, and fine-tune configurations to improve performance Settings for new algorithms)
    2021-01-27 19:38:41下载
    积分:1
  • 01 糊涂塔克学习笔记【创享电子科技版权所有】
    总结了labview学习中遇到的问题以及解决方案 适用于初学者(please unzip and read it)
    2018-12-30 16:01:49下载
    积分:1
  • PAT
    Given N rational numbers in the form "numerator/denominator", you are supposed to calculate their sum. 输入描述: Each input file contains one test case. Each case starts with a positive integer N (<=100), followed in the next line N rational numbers "a1/b1 a2/b2 ..." where all the numerators and denominators are in the range of "long int". If there is a negative number, then the sign must appear in front of the numerator. 输出描述: For each test case, output the sum in the simplest form "integer numerator/denominator" where "integer" is the integer part of the sum, "numerator" < "denominator", and the numerator and the denominator have no common factor. You must output only the fractional part if the integer part is 0.(Input Description: Each input file contains one test case. Each case starts with a positive integer N (<=100), followed in the next line N rational numbers "a1/b1 a2/b2 ..." where all the numerators and denominators are in the range of "long int". If there is a negative number, then the sign must appear in front of the numerator. Output Description: For each test case, output the sum in the simplest form "integer numerator/denominator" where "integer" is the integer part of the sum, "numerator" < "denominator", and the numerator and the denominator have no common factor. You must output only the fractional part if the integer part is 0.)
    2019-03-02 14:33:27下载
    积分:1
  • 用LABVIEW开发的贪吃蛇小游戏,娱乐的同时学习编程思想和方法。...
    用LABVIEW开发的贪吃蛇小游戏,娱乐的同时学习编程思想和方法。-Developed using LABVIEW Snake game and entertainment programming at the same time learning the ideas and methods.
    2022-01-25 19:12:19下载
    积分:1
  • Practice1_14
    字符串交叉输入,就是将一个字符串交叉的输入到了另一个字符串当中,比如: char a[4]="ABCD" char b[4]="abcd" 则char c[8]="AaBbCcDd" (Cross-string input, is to cross a string input to another string of them, such as: char a [4] = "ABCD" char b [4] = "abcd" While char c [8] = "AaBbCcDd" )
    2009-02-19 18:26:20下载
    积分:1
  • C++ example 100
    C++ example, 适合初学者。常用的实现方法。对于自学C++而苦于缺少示例模仿的朋友们提供了大量的实例。教材实验设计完全按照C++标准规范,给自学的朋友们提供了好的典范
    2022-02-26 21:07:12下载
    积分:1
  • 专家系统 :农夫过河问题的源代码 农夫过河问题.CLP
    专家系统 :农夫过河问题的源代码 农夫过河问题.CLP -expert system : the farmer across the river source code farmer river issues. CLP
    2022-12-23 05:00:03下载
    积分:1
  • DIV+CSS网页模板2
    CSS和DIV布局学习,提供一些模板,参照例子进行。(A template designed by a web page)
    2017-08-03 18:31:59下载
    积分:1
  • 696516资源总数
  • 106658会员总数
  • 16今日下载