测量频率在 1k-2k之间，要求精度0.01%，也就是2kHZ, 0.2HZ。在做的过程中，发现最大跳动在0.8HZ，不知道代码哪有问题，之后就用avr做了一个，能满足要求，
不甘心啊，为什么当时用STM8就没做出来呢，求大家指点
代码如下：
#include "TIM1_Capture.h"
#include "includes.h"
#include "delay.h"
//u16 tab[200];
INT_CHARS ICValue1,ICValue2;
u8 cnt;
u8 TIM1_OVF_Num;
bool flag;
void TIM1_Init(void)
{
GPIOC->DDR &= ~(1<<1); // 配置PC1为输入
GPIOC->CR1 |= 1<<1;
GPIOC->CR2 &= ~(1<<1);
// 开启TIM1时钟
CLK->PCKENR1 |= 0X80;
TIM1->CNTRH = 0X00;
TIM1->CNTRL = 0X00;
TIM1->ARRH = 0Xff;
TIM1->ARRL = 0Xff;
TIM1->CR1 = 0X80; // 自动重装
TIM1->CCMR1 = 0X00;
TIM1->CCER1 = 0X00;
TIM1->CCMR1 |= 0X01;
TIM1->CCER1 &= 0XFD; // 上升沿捕获
TIM1->IER |= 0X03; // 允许TIM1捕获中断，更新中断
TIM1->SR1 = 0x00;
TIM1->SR2 = 0x00;
TIM1->CR1 |= 0X01; // 使能TIM1
TIM1->CCER1 = 0X01; // 使能捕获
}
u16 ComputeFreq(void)
{
u16 temp;
u32 sum;
flag = FALSE;
cnt = 0;
TIM1_Init();
while(!flag); // 等待100次捕获结束
sum = TIM1_OVF_Num;
sum *= 65535;
sum += ICValue2.x;
sum -= ICValue1.x;
sum /= 10;
temp = 1600000000/sum;
return temp;
}
#pragma vector=0x0E
__interrupt void TIM1_CAP_COM_IRQHandler(void)
{
disableInterrupts();
if(cnt==5)
{
ICValue1.FJ[0] = TIM1->CCR1H;
ICValue1.FJ[1] = TIM1->CCR1L;
flag = FALSE;
TIM1_OVF_Num = 0;
}
else if(cnt==105) // 100次捕获结束
{
ICValue2.FJ[0] = TIM1->CCR1H;
ICValue2.FJ[1] = TIM1->CCR1L;
flag = TRUE;
TIM1->CR1 &= 0XFE; // 失能TIM1
TIM1->CCER1 = 0X00;
}
TIM1->SR1 &= 0xFD;
// TIM1->SR2 = 0x00;
cnt++;
enableInterrupts();
}
#pragma vector=0x0D
__interrupt void TIM1_OVF_IRQHandler(void)
{
disableInterrupts();
TIM1_OVF_Num++;
TIM1->SR1 &= 0xFE;
// TIM1->SR2 = 0x00;
enableInterrupts();
}
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