51单片机实验-蜂鸣器播放音乐

51单片机实验-蜂鸣器播放音乐

题目：编程实现利用单片机使蜂鸣器播放“茉莉花”的乐曲。
学习单片机时写的，利用单片机控制蜂鸣器播放乐曲，开发板上的蜂鸣器可以用三极管驱动，需要根据乐谱来设置当前节拍的发声时间。中断函数来控制蜂鸣器的发声。

#include 
sbit Buzz = P1^6; //声明绑定蜂鸣器，接线时只需将此IO口与蜂鸣器（扬声器）信号输入端相连即可
unsigned int code NoteFrequ[]={   
	523,587,659,698,784,880,988,    //中音对应的1-7  
	1047,1175,1319,1397,1568,1760,1976,	//高音对应的1-7
	262,294,330,349,392,440,494};  //低音对应的1-7

unsigned int code NoteReload[]={ //中音1-7和高音1-7对应的定时器重载值
	65536 - (11059200/12) /(523*2),//中音1-7  
	65536 - (11059200/12) /(587*2),  
	65536 - (11059200/12) /(659*2),
	65536 - (11059200/12) /(698*2),
	65536 - (11059200/12) /(784*2),  
	65536 - (11059200/12) /(880*2),
	65536 - (11059200/12) /(988*2),
	65536 - (11059200/12) /(1047*2),//高音1-7  
	65536 - (11059200/12) /(1175*2),  
	65536 - (11059200/12) /(1319*2),  
	65536 - (11059200/12) /(1397*2),
	65536 - (11059200/12) /(1568*2),
	65536 - (11059200/12) /(1760*2),
	65536 - (11059200/12) /(1976*2),
	65536 - (11059200/12) /(262*2), //低音1-7
	65536 - (11059200/12) /(294*2),
	65536 - (11059200/12) /(330*2),
	65536 - (11059200/12) /(349*2),
	65536 - (11059200/12) /(392*2),
	65536 - (11059200/12) /(440*2),
	65536 - (11059200/12) /(494*2)};

bit enable = 1; //发声使能表标识
bit tmrflay = 0; //定时器中中断完成标识

unsigned char T0RH = 0xff; //T0重载值高字节
unsigned char T0RL = 0x00; //T0重载值低字节
void PlayTwoTiger(); void main()
{   
	unsigned int i;
    EA = 1;
	TMOD =0x01;  //模式1	
	TH0 = T0RH;
	TL0 = T0RL;  
	ET0 = 1;  //使能T0中断
    TR0 = 1;  //启动
    while(1)
	{  
		PlayTwoTiger();
		for(i=0;i<40000;i++); 
    }
}
/**音乐函数**/
void PlayTwoTiger()
{
	unsigned char beat;    //节拍索引
	unsigned char note;    //节拍对应音符
	unsigned int time=0;   //节拍计时
	unsigned int beattime=0;  //总时间计时
	unsigned int soundtime=0; //没拍发声计时
	unsigned char code PlayTwoTigerNote[]={ //音符表
		3,3,5,6,8,8,6,5,5,6,5,
		3,3,5,6,8,8,6,5,5,6,5,
		5,5,5,3,5,6,6,5,
		3,2,3,5,3,2,1,1,2,1,
		3,2,1,3,2,3,5,6,8,5,
		2,3,5,2,3,1,20,19,
		20,1,2,3,
		1,2,1,20,19
		};  
	unsigned char code PlayTwoBeat[]={  //节拍表,4表示一拍，1表示1/4拍,8表示两拍
		4,2,2,2,2,2,2,4,2,2,8,
		4,2,2,2,2,2,2,4,2,2,8,
		4,4,4,2,2,4,4,8,
		4,2,2,4,2,2,4,2,2,8,
		2,2,2,2,6,2,4,2,2,8,
		4,2,2,2,2,2,2,8,
		4,4,6,2,
		2,2,2,2,16
		};   
	for(beat=0; beat<sizeof(PlayTwoTigerNote);) //节拍索引循环变量
	{		   
		while(!tmrflay);   //每次定时器中断完成 节拍处理 
		tmrflay = 0;       
		if(time == 0)      //节拍播放完成重启
		{
			note = PlayTwoTigerNote[beat]-1;
			T0RH = NoteReload[note]>>8;
			T0RL = NoteReload[note];     //计算总时间,右移2位等于除4,移位代替除法加快速度
			beattime = (PlayTwoBeat[beat]*NoteFrequ[note])>>2;   //计算发声时间,为总时间的0.75s
			soundtime =beattime - (beattime>>2);  
			enable = 1;  //开始发声  
			time++;   
		}
		else    //节拍播放未结束，则继续处理
		{
			if(time >= beattime) //当前时间清零  
			{
				time = 0;     //准备重新启动
				beat++;   
			}
			else      //累加时间
			{
				time++;   
				if(time == soundtime)    //发声时间到达；关闭蜂鸣器
				{
					enable =0;      //用以区分连续两个节拍  
				}
			}  
		}
	}
}

void InterRupt() interrupt 1   //中断服务
{
	TH0 =T0RH;
	TL0 =T0RL;
	tmrflay = 1;
	if(enable)
	{  
		Buzz=~Buzz;
	}
	else
	{   
		Buzz=1;
	}
}

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