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單片機波形發生器(6種)源程序+仿真電路
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作者:
ziyin
時間:
2019-12-20 14:22
標題:
單片機波形發生器(6種)源程序+仿真電路
波形發生器(6種)源程序+仿真電路
51hei.png
(18.11 KB, 下載次數: 61)
下載附件
2019-12-20 14:24 上傳
單片機源程序如下:
#include<reg51.h>
#include<absacc.h>
#define uchar unsigned char
#define uint unsigned int
#define DAC0832 XBYTE[0xfffe]
sbit k0=P1^0; //產生方波
sbit k1=P1^1; //產生鋸齒波
sbit k2=P1^2; //產生三角波
sbit k3=P1^3; //產生階梯波
sbit k4=P1^4; //產生正弦波
sbit k5=P1^5;//產生梯形波
uchar mm=0,x=0,y=0;
uchar flag=0;//不同值對應不同波形
uchar code SETTAB[256]={0x80,0x82,0x85,0x88,0x8b,0x8e,0x91,0x94,0x97,0x9a,0x9d,
0xa0,0xa3,0xa6,0xa9,0xac,0xaf,0xb2,0xb6,0xb9,0xbc,0xbf,0xc2,0xc5,0xc7,0xca,0xcc,0xcf,
0xd1,0xd4,0xd6,0xd8,0xda,0xdd,0xdf,0xe1,0xe3,0xe5,0xe7,0xe9,0xea,0xec,
0xee,0xef,0xf1,0xf2,0xf4,0xf5,0xf6,0xf7,0xf8,0xf9,0xfa,0xfb,0xfc,0xfd,
0xfd,0xfe,0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,0xff,
0xfe,0xfd,0xfd,0xfc,0xfb,0xfa,0xf9,0xf8,0xf7,0xf6,0xf5,0xf4,0xf2,0xf1,
0xef,0xee,0xec,0xea,0xe9,0xe7,0xe5,0xe3,0xe1,0xde,0xdd,0xda,0xd8,0xd6,
0xd4,0xd1,0xcf,0xcc,0xca,0xc7,0xc5,0xc2,0xbf,0xbc,0xba,0xb7,0xb4,0xb1,
0xae,0xab,0xa8,0xa5,0xa2,0x9f,0x9c,0x99,0x96,0x93,0x90,0x8d,0x89,0x86,
0x83,0x80,0x80,0x7c,0x79,0x76,0x72,0x6f,0x6c,0x69,0x66,0x63,0x60,0x5d,
0x5a,0x57,0x55,0x51,0x4e,0x4c,0x48,0x45,0x43,0x40,0x3d,0x3a,0x38,0x35,
0x33,0x30,0x2e,0x2b,0x29,0x27,0x25,0x22,0x20,0x1e,0x1c,0x1a,0x18,0x16,
0x15,0x13,0x11,0x10,0x0e,0x0d,0x0b,0x0a,0x09,0x08,0x07,0x06,0x05,0x04,
0x03,0x02,0x02,0x01,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x01,0x02,0x02,0x03,0x04,0x05,0x06,0x07,0x08,0x09,0x0a,0x0b,
0x0d,0x0e,0x10,0x11,0x13,0x15,0x16,0x18,0x1a,0x1c,0x1e,0x20,0x22,0x25,
0x27,0x29,0x2b,0x2e,0x30,0x33,0x35,0x38,0x3a,0x3d,0x40,0x43,0x45,0x48,
0x4c,0x4e,0x51,0x55,0x57,0x5a,0x5d,0x60,0x63,0x66,0x69,0x6c,0x6f,0x72,
0x76,0x79,0x7c,0x80};
void delay(uint ms)
{uchar i;
while(ms--)for(i=0;i<120;i++);}
void sin() //正弦波發生函數
{DAC0832=SETTAB[y];
x++;y++;}
void fangbo( ) //方波發生函數
{ DAC0832=y;
x++;
if(flag==0)
{y=0xff;
if(x==128){flag=1;y=0x00;}}//方波+
else {y=0x00;
if(x==0){flag=0;y=0xff;}}}//方波-
void sanjiaobo( ) //三角波發生函數
{ DAC0832=y;
x++;
if(flag==0)
{y+=2;
if(x==128){flag=1;y-=2;}}//三角波+
else {y-=2;
if(x==0){flag=0;y=0;}}}//三角波-
void juchibo() //鋸齒波發生函數
{DAC0832=y; y++;}
void jietibo()//階梯波發生函數
{DAC0832=y;
y+=25;}
void tixing( ) //梯形波發生函數
{uint i;
DAC0832=y;
if(flag==0){y+=2;x++;if(x==128){flag=1;y-=2;}}
else {y-=2;x--;if(x==0){for(i=0;i<5000;i++)y=0x00;flag=0;}}}
void read_key()
{if(k0==0){delay(10);if(k0==0){mm=1;while(k0==0);}}
else if(k1==0){delay(10);if(k1==0){mm=2;while(k1==0);}}
else if(k2==0){delay(10);if(k2==0){mm=3;while(k2==0);}}
else if(k3==0){delay(10);if(k3==0){mm=4;while(k3==0);}}
else if(k4==0){delay(10);if(k4==0){mm=5;while(k4==0);}}
else if(k5==0){delay(10);if(k5==0){mm=6;while(k5==0);}}}
void main()
{TMOD=0x02;
TH0=TL0=156;
TR0=1;ET0=1;EA=1;
mm=0;
for(;;)read_key();}
void timer0() interrupt 1 using 2
{{switch(mm)
{case 1:fangbo();break;
case 2:juchibo();break;
case 3:sanjiaobo();break;
case 4:jietibo();break;
case 5:sin();break;
case 6:tixing();break;}} }
復制代碼
所有資料51hei提供下載:
波形發生器(6種).zip
(52.42 KB, 下載次數: 116)
2019-12-20 14:21 上傳
點擊文件名下載附件
下載積分: 黑幣 -5
作者:
保持沉默
時間:
2019-12-25 11:00
不好用,只顯示一種錯誤波形啊,
作者:
113265
時間:
2019-12-25 13:49
剛好需要謝謝啦
作者:
HWL0541
時間:
2019-12-27 12:05
謝謝樓主分享
作者:
pangshe
時間:
2019-12-27 16:48
仿真只有一個波形,有錯誤。
作者:
埃斯頓林米
時間:
2020-2-25 18:55
感謝樓主分享
作者:
p1364090187
時間:
2020-4-25 04:12
雖然可以參考,但是仿真的只有一個波形,應該是只編寫一個波形的么?
歡迎光臨 (http://www.raoushi.com/bbs/)
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