用于STC单片机内置10位AD转化的代码+【软件滤波】

上面这段GetADCResult()是本来的AD收集，没什么改变，看看就行了

GetResult()这个是软件滤波的，10位AD的之前弄滤波总犯错，今日刚弄好。
关键在于(unsigned long)(),不加这个就总犯错！核算过程中溢出，感觉好像是核算时主动降位了

ulong GetADCResult(uchar ch){ulong AD_result;ADC_CONTR = ADC_POWER | ADC_SPEEDLL | ch | ADC_START;_nop_();                        //Must wait before inquiry_nop_();_nop_();_nop_();while (!(ADC_CONTR & ADC_FLAG));//Wait complete flagADC_CONTR &= ~ADC_FLAG;         //Close ADCAD_result=ADC_RES*4+ADC_RESL; return   AD_result;             //Return ADC result 10bit}ulong GetResult(uchar ch)  //滤波段{ulong AD_result=0,AD_result1=0;ulong AD_max,AD_min,AD_temp;char AD_i,AD_o;for(AD_i=0;AD_i<4;AD_i++){AD_min=AD_max=GetADCResult(ch);AD_result=0;for(AD_o=0;AD_o<8;AD_o++){AD_temp=GetADCResult(ch);if(AD_tempAD_max){AD_result =(unsigned long)(AD_result + (unsigned long)AD_max); AD_max=AD_temp;}else AD_result =(unsigned long)(AD_result + (unsigned long)AD_temp);}AD_result=(unsigned long)(AD_result/8);AD_result1=AD_result1+ (unsigned long)(AD_result); }AD_result1=(unsigned long)(AD_result1/4);return   AD_result;             //Return ADC result 10bit}

办法是根据：去除最大最小值后均匀。共采样了33次去掉9个后的成果均匀值，我自己做电压表，速度满足！假如嫌慢的话能够把AD_i<4中的这个4改小点或许把这个循环去掉。