畢業(yè)論文——發(fā)動(dòng)機(jī)自動(dòng)熄火的診斷分析2

1、課程設(shè)計(jì)題目：基于單片機(jī)音樂(lè)演奏曲title: instrumental music based on single chip姓 名： 學(xué) 號(hào)： 系 別： 專 業(yè)： 年 級(jí)： 指導(dǎo)教師： 2012年5 月 25 日基于單片機(jī)音樂(lè)演奏曲摘要單片機(jī)具有強(qiáng)大的控制功能和靈活的編程實(shí)現(xiàn)特性，它已經(jīng)溶入現(xiàn)代人們的生活中，成為不可替代的一部分。本設(shè)計(jì)以at89c2051為核心，主要由電源電路、復(fù)位電路、音頻放大電路、時(shí)鐘電路和數(shù)碼管電路和蜂鳴器電路構(gòu)成單片機(jī)奏樂(lè)附加時(shí)鐘的一個(gè)小系統(tǒng)。電路中i/o口采用分時(shí)復(fù)用的借口技術(shù)，使at89c2051單片機(jī)的引腳資源得以充分利用，本系統(tǒng)的電路簡(jiǎn)單，實(shí)現(xiàn)的功能強(qiáng)大，所

2、用芯片比較便宜，性價(jià)比較高。關(guān)鍵詞：at89c2051，數(shù)碼管，單片機(jī)奏樂(lè)，分時(shí)復(fù)用目 錄摘要i1.引言32.系統(tǒng)整體結(jié)構(gòu)42.1系統(tǒng)總設(shè)計(jì)42.2實(shí)現(xiàn)的功能42.3主要芯片介紹42.3.1 at89c2051芯片介紹43.系統(tǒng)硬件設(shè)計(jì)53.1鍵盤輸入模塊53.2時(shí)鐘模塊53.3顯示模塊63.4復(fù)位電路63.5蜂鳴器電路設(shè)計(jì)74.系統(tǒng)軟件設(shè)計(jì)74.1系統(tǒng)主程序流程圖系74.2 部分子程序流程圖8結(jié)束語(yǔ)10參考文獻(xiàn)10致謝11附錄11 - 2 -1.引言隨著社會(huì)的發(fā)展、科技的進(jìn)步以及人們生活水平的逐步提高，各種方便于生活的自動(dòng)控制系統(tǒng)開始進(jìn)入了人們的生活，以單片機(jī)為核心的各種系統(tǒng)也越來(lái)越多。同時(shí)

3、也標(biāo)志了自動(dòng)控制領(lǐng)域成為了數(shù)字化時(shí)代的一員。它實(shí)用性強(qiáng)，功能齊全，技術(shù)先進(jìn)，使人們相信這是科技進(jìn)步的成果。它更讓人類懂得，數(shù)字時(shí)代的發(fā)展將改變?nèi)祟惖纳?，將加快科學(xué)技術(shù)的發(fā)展。本次設(shè)計(jì)為單片機(jī)奏樂(lè)器,硬件部分它以單片機(jī)at89c2051為核心,由功放電路、數(shù)碼管等組成。當(dāng)接上電源按下開關(guān)時(shí)，就能聽到優(yōu)美的旋律。當(dāng)然這些音樂(lè)都是通過(guò)軟件編程實(shí)現(xiàn)的，把它存儲(chǔ)在存儲(chǔ)器里，根據(jù)存儲(chǔ)容量大小決定存儲(chǔ)音樂(lè)的數(shù)目。 2 2.系統(tǒng)整體結(jié)構(gòu)2.1系統(tǒng)總設(shè)計(jì)本設(shè)計(jì)主要由單片機(jī)at89c2051、獨(dú)立鍵盤、復(fù)位電路、數(shù)碼管顯示電路和時(shí)鐘電路組成，用于實(shí)現(xiàn)單片機(jī)奏樂(lè)功能。系統(tǒng)設(shè)計(jì)總框圖如圖1所示：系統(tǒng)設(shè)計(jì)總框圖12.

4、2實(shí)現(xiàn)的功能設(shè)有四個(gè)按鍵，其中兩個(gè)能實(shí)現(xiàn)多首音樂(lè)的選擇，另一個(gè)能實(shí)現(xiàn)播放功能和暫停的功能，最后一個(gè)能實(shí)現(xiàn)復(fù)位和關(guān)閉歌曲的功能。設(shè)有一位數(shù)碼管，每選擇一首曲子能在數(shù)碼管上顯示當(dāng)前的曲目（只是數(shù)字）。2.3主要芯片介紹2.3.1 at89c2051芯片介紹 at89c2051芯片的引腳圖2：at89c2051的20 個(gè)引腳的功能為：vccgnd：電源電壓，接地引腳xtal1、xtal2：反向震蕩放大器內(nèi)部時(shí)鐘工作電路的輸入、反向震蕩放大器內(nèi)部時(shí)鐘工作電路的輸出引腳。rst：復(fù)位信號(hào)輸入引腳。當(dāng)rst編為高電平并保持2個(gè)機(jī)器周期，所有i/o引腳復(fù)位至“1”電平。p1.0p1.7：p1口的8位雙向i/

5、o引腳。p3.0p3.5與p3.7：為7個(gè)帶內(nèi)部上拉的雙向i/o口 3.系統(tǒng)硬件設(shè)計(jì)3.1鍵盤輸入模塊設(shè)計(jì)所用到的按鍵數(shù)量較少，所以可以采用獨(dú)立按鍵式鍵盤。獨(dú)立式按鍵盤每一個(gè)按鍵與一個(gè)i/o口相連。共有3個(gè)按鍵，sw2用于播放和暫停，sw1用于上一曲，sw3用于下一曲。如圖所示： 3.2時(shí)鐘模塊單片機(jī)工作的時(shí)間基準(zhǔn)是由時(shí)鐘電路提供的。在單片機(jī)的xtal1與xtal2兩個(gè)引腳間，接一只晶振及兩只電容就構(gòu)成了單片機(jī)的時(shí)鐘電路，如下圖所示。電容器c5和c6對(duì)振蕩頻率有微調(diào)的作用，通常的取值是30pf；石英晶體選12mhz。 3.3顯示模塊顯示部分是使用一位數(shù)碼管，采用共陽(yáng)接法，a-g分別接p1.0-

6、p1.6口。3.4復(fù)位電路單片機(jī)的rst引腳為主機(jī)提供一個(gè)外部信號(hào)輸入端口。信號(hào)時(shí)高電平有效，高電平有效的持續(xù)時(shí)間應(yīng)為2個(gè)機(jī)器周期以上。復(fù)位以后，單片機(jī)內(nèi)各個(gè)部件恢復(fù)到初始狀態(tài)，單片機(jī)從rom的0000h開始執(zhí)行程序。單片機(jī)的復(fù)位方式有上電復(fù)位和手工復(fù)位兩種，在這邊我采用的按鍵復(fù)位方式，只要vcc上升時(shí)間不超過(guò)1ms，它們都能很好的工作。該單元的電路圖：3.5蜂鳴器電路設(shè)計(jì)設(shè)計(jì)要求定時(shí)時(shí)間到的時(shí)候來(lái)回置低電平，來(lái)驅(qū)動(dòng)蜂鳴器發(fā)出聲音。4.系統(tǒng)軟件設(shè)計(jì)4.1系統(tǒng)主程序流程圖系4.2 部分子程序流程圖 獨(dú)立按鍵掃描 音樂(lè)播放模塊 結(jié)束語(yǔ)本次設(shè)計(jì)總體上達(dá)到了預(yù)期的要求，運(yùn)用我們所學(xué)的單片機(jī)知識(shí)完成了以

7、單片機(jī)at89c2051為核心的單片機(jī)奏樂(lè)器，經(jīng)過(guò)這段時(shí)間有關(guān)于單片機(jī)奏樂(lè)器的課程設(shè)計(jì)，使我對(duì)單片機(jī)的應(yīng)用有了更深的了解。在設(shè)計(jì)過(guò)程中，碰到了許多問(wèn)題。當(dāng)然也收獲了很多，動(dòng)手能力得到了很大的提高，學(xué)會(huì)了查閱資料和對(duì)各種資料的融會(huì)變通。參考文獻(xiàn)1 趙建領(lǐng)著.51單片機(jī)開發(fā)與應(yīng)用技術(shù)詳解.電子工業(yè)出版社,2009.01. 2 倪云峰主編.單片機(jī)原理與應(yīng)用.西安電子科技大學(xué)出版社,2009.06.10附錄1附錄一 源程序#include sbit fmq=p37;sbit sw0=p34;sbit sw1=p33;sbit sw2=p35;unsigned char flag=0;unsigned

8、char data l;unsigned int sta=1;unsigned int data j;unsigned char data i;unsigned char data k;unsigned char count;void delay(unsigned int t); /延時(shí)函數(shù)聲明void play();void xuanz();void keyscan(); /按鍵掃描unsigned char *pmmusic;unsigned char code led=0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90;unsigned c

9、har code music30=0xff,0xff,0xfb,0x90,0xfc,0x0c,0xfc,0x44,0xfc,0xac, 0xfd,0x09,0xfd,0x34,0xfd,0x82,0xfd,0xc8,0xfe,0x06, 0xfe,0x22,0xfa,0x15,0xfb,0x04,0xfa,0x67,0xfe,0x85;/音調(diào)數(shù)組unsigned char code mmusic=0x35,0x31,0x34,0x34,0x24,0x35,0x32,0x32,0x24,0x38,0x44,0x58,0x48,0x34,0x31,0x34,0x34,0x24,0x38,0x34,

10、0x7f,0x32,0x34,0x32,0x34,0x24,0x38,0x34,0x24,0x38,0x43,0x58,0x48,0x34,0x32,0x34,0x38,0x24,0x38,0x34,0x7f,0x16,0x4c,0x74,0x78,0x64,0x54,0x48,0x54,0x64,0x58,0x44,0x34,0x24,0x38,0x24,0x14,0x12,0x21,0x14,0x78,0x68,0x3f,0x4c,0x74,0x78,0x64,0x52,0x42,0x48,0x53,0x64,0x58,0x44,0x34,0x24,0x38,0x24,0x24,0x38,

11、0x44,0x58,0x48,0x3c,0xff;unsigned char code mmusic1= 0xc2,0xb2,0x36,0x52,0x42,0x32,0x11,0xc1,0xb2,0xc6, 0x22,0xc4,0x12,0xc2,0x76,0xa2,0x84,0x72,0x51,0x41, 0x32,0x4c,0x41,0x36,0x74,0x82,0x74,0x82,0x72,0x52, 0x41,0x38,0x32,0x42,0x14,0xc2,0x72,0x52,0x42,0x42, 0x3c,0xc2,0xb2,0x36,0x52,0x42,0x32,0x11,0xc

12、1,0xb2, 0xc6,0x12,0xc4,0x12,0xc2,0x76,0xa2,0x82,0x72,0x51, 0x41,0x32,0x4c,0x42,0x32,0x76,0x82,0x74,0x82,0x72, 0x53,0x41,0x38,0x32,0x42,0x16,0xc2,0x72,0x52,0x41, 0x53,0x12,0xcc,0xc2,0xb2,0x16,0xc2,0x72,0x52,0x42, 0x41,0x3c,0x32,0x32,0x26,0x22,0x24,0x42,0x32,0x22,0xff; unsigned char code mmusic2= 0x34

13、,0x31,0x31,0x34,0x44,0x52,0x62,0x52,0x42,0x32, 0x34,0x04,0x74,0x74,0x62,0x62,0x64,0x3c,0x04,0x64, 0x62,0x52,0x42,0x32,0x34,0x33,0x31,0x32,0x72,0x76, 0x72,0x83,0x81,0x82,0x82,0x82,0x74,0x72,0x7c,0x04, 0x63,0x61,0x64,0x64,0x64,0x72,0x82,0x72,0x74,0x72, 0x62,0x52,0x42,0x32,0x42,0x44,0x42,0x42,0x52,0x62

14、, 0x52,0x5c,0x04,0x64,0x62,0x62,0x64,0x72,0x82,0x72, 0x74,0x72,0x62,0x52,0x42,0x32,0x42,0x46,0x53,0x41, 0x42,0x32,0x3c,0x04,0x44,0x48,0x02,0x32,0x3f,0x44, 0x48,0x01,0x34,0x3f,0x44,0x0c,0xff;/主程序void main() tmod=0x11;/初始化 ie=0x8a;th1=(65536-5000)/256;tl1=(65536-5000)%256;tr1=1;tr0=0;while(1)p1=ledsta

15、; if(tr0=1)flag=0;xuanz(); play(); void xuanz()if(sta=1)pmmusic=mmusic; else if(sta=2)pmmusic=mmusic1; else if(sta=3)pmmusic=mmusic2; /*else if(sta=4)pmmusic=mmusic3; else if(sta=5)pmmusic=mmusic4; else if(sta=6)pmmusic=mmusic5; else if(sta=7)pmmusic=mmusic6; else if(sta=8)pmmusic=mmusic7;*/ else pm

16、music=mmusic1;void play()if(*(pmmusic+j)!=0xff) k=*(pmmusic+j)&0x0f; l=*(pmmusic+j)4; th0=music2*l; tl0=music2*l+1; if(flag=1)goto next; if (music2*l=0xff)&(music2*l+1=0xff) tr0=0; for(i=k;i0;-i) delay(23350); j+; elsenext:j=0;tr0=0;void timer0(void) interrupt 1 th0=music2*l; tl0=music2*l+1; fmq=!fmq; void timer1(void) interrupt 3th1=(65536-5000)/256;tl1=(65536-5000)%256;keyscan();void keyscan() /獨(dú)立按鍵掃描if(sw0=0) count+; if(count=40)count=0; if(sw0=0) tr0=tr0;fmq=1; if(sw1=0) count+; if(count=40)