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標(biāo)題: [求助]6位數(shù)顯頻率計數(shù)器顯示不正常 [打印本頁]

作者: liangzhenyun    時間: 2009-3-29 10:33
標(biāo)題: [求助]6位數(shù)顯頻率計數(shù)器顯示不正常

 

 


    
 6位數(shù)顯頻率計數(shù)器 (圖片在附件)顯示不正常,有該程序的好人希望能發(fā)你的程序過來,liangzhenyuang@163.com
  1.實驗任務(wù)
              利用AT89S51單片機的T0、T1的定時計數(shù)器功能,來完成對輸入的信號進行頻率計數(shù),計數(shù)的頻率結(jié)果通過8位動態(tài)數(shù)碼管顯示出來。要求能夠?qū)?-250KHZ的信號頻率進行準(zhǔn)確計數(shù),計數(shù)誤差不超過±1HZ。
  2.電路原理圖
3.系統(tǒng)板上硬件連線
            (1).把“單片機系統(tǒng)”區(qū)域中的P0.0-P0.7與“動態(tài)數(shù)碼顯示”區(qū)域中的ABCDEFGH端口用8芯排線連接。
            (2).把“單片機系統(tǒng)”區(qū)域中的P2.0-P2.7與“動態(tài)數(shù)碼顯示”區(qū)域中的S1S2S3S4S5S6S7S8端口用8芯排線連接。
            (3).把“單片機系統(tǒng)”區(qū)域中的P3.4(T0)端子用導(dǎo)線連接到“頻率產(chǎn)生器”區(qū)域中的WAVE端子上。
            4.程序設(shè)計內(nèi)容
            (1).定時/計數(shù)器T0和T1的工作方式設(shè)置,由圖可知,T0是工作在計數(shù)狀態(tài)下,對輸入的頻率信號進行計數(shù),但對工作在計數(shù)狀態(tài)下的T0,最大計數(shù)值為fOSC/24,由于fOSC=12MHz,因此:T0的最大計數(shù)頻率為250KHz。對于頻率的概念就是在一秒只數(shù)脈沖的個數(shù),即為頻率值。所以T1工作在定時狀態(tài)下,每定時1秒中到,就停止T0的計數(shù),而從T0的計數(shù)單元中讀取計數(shù)的數(shù)值,然后進行數(shù)據(jù)處理。送到數(shù)碼管顯示出來。
            (2).T1工作在定時狀態(tài)下,最大定時時間為65ms,達不到1秒的定時,所以采用定時50ms,共定時20次,即可完成1秒的定時功能。
            5.C語言源程序
            #include<reg52.h>
            unsigned char code
            dispbit[]={0xfe,0xfd,0xfb,0xf7,0xef,0xdf,0xbf,0x7f};
            unsigned char code dispcode[]={0x3f,0x06,0x5b,0x4f,0x66,
                                           0x6d,0x7d,0x07,0x7f,0x6f,0x00,0x40};
            unsigned char dispbuf[8]={0,0,0,0,0,0,10,10};
            unsigned char temp[8];
            unsigned char dispcount;
            unsigned char T0count;
            unsigned char timecount;
            bit flag;
            unsigned long x;
            void main(void)
            {
              unsigned char i;
              TMOD=0x15;
              TH0=0;
              TL0=0;
              TH1=(65536-4000)/256;
              TL1=(65536-4000)%6;
              TR1=1;
              TR0=1;
              ET0=1;
              ET1=1;
              EA=1;
              while(1)
               {
                 if(flag==1)
                   {
                     flag=0;
                     x=T0count*65536+TH0*256+TL0;
                     for(i=0;i<8;i++)
                       {
                         temp=0;
                       }
                     i=0;
                     while(x/10)
                       {
                         temp=x++;
                         x=x/10;
                         i++;
                       }
                     temp=x;
                     for(i=0;i<6;i++)
                       {
                         dispbuf=temp;
                       }
                     timecount=0;
                     T0count=0;
                     TH0=0;
                     TL0=0;
                     TR0=1;
                   }
               }
            }
            void t0(void) interrupt 1 using 0
            {
              T0count++;
            }
            void t1(void) interrupt 3 using 0
            {
              TH1=(65536-4000)/256;
              TL1=(65536-4000)%6;
              timecount++;
              if(timecount==250)
                {
                  TR0=0;
                  timecount=0;
                  flag=1;
                }
              P0=dispcode[dispbuf[dispcount]];
              P2=dispbit[dispcount];
              dispcount++;
              if(dispcount==8)
                {
                  dispcount=0;
                }
            }

         







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