嗨piovan，
我不明白你的观点，但这是我对这个例子的理解：
使用TIM2_ICInit（）函数可以准确地配置捕获。
因此，如果使用TIM2_ICPSC_DIV，则在4个上升沿之后设置catpture事件（标志）（取决于TIM2_ICPOLARITY_FALLING / TIM2_ICPOLARITYRISING的降低边缘）
因此，为了计算信号的频率（在本例中它是LSI），你应该在下面的公式中乘以4（instaed为8），这样就变成了
LSIClockFreq =（4 * TIM2ClockFreq）/（ICValue2 - ICValue1）;
让我们处理时钟频率，计算频率值的一般规则是：
捕获分频器*定时器时钟/（第二次捕获 - 第一次捕获）
在ST提供的示例中，
1-捕获为8（TIM2_ICPSC_DIV8）
2-定时器时钟为2MHz（使用#define TIM2ClockFreq（（u32）2000000定义）
提醒：定时器的时钟频率与CPU相同，默认为STM8
似乎运行在2MHz，这就是它定义为2000000的原因。
我希望这清楚，它对你有用。
Herzlich
MCULüfter
以上来自于谷歌翻译
以下为原文
Hi piovan,
I didn't understand exactly your point but this is my understanding of the example:
using TIM2_ICInit() function you configure the capture when exactly it occurs.
So if you use TIM2_ICPSC_DIV, the catpture event (flag) is set after 4 rising edges (orfalling edges that depends on the TIM2_ICPOLARITY_FALLING/TIM2_ICPOLARITYRISING)
consequently to compute the frequency of a signal (in the example it is the LSI) you should multiply by 4 (instaed of 8)in the formula below so it becomes
LSIClockFreq = (4*TIM2ClockFreq) / (ICValue2 - ICValue1);
Let's process the clock frequency, the general rule to compute the frequency value is:
capture divider * timer clock / (second capture - first capture)
In the example provided by ST,
1- the capture is 8 (TIM2_ICPSC_DIV8)
2- the timer clock is 2MHz (It is defined using #define TIM2ClockFreq ((u32)2000000)
Reminder: the timer is clocked at the same clock than the CPU and by default STM8
seems to run at 2MHz that is why it is defined at 2000000.
I hope this clear and it is useful for you.
Herzlich
MCU Lüfter
嗨piovan，
我不明白你的观点，但这是我对这个例子的理解：
使用TIM2_ICInit（）函数可以准确地配置捕获。
因此，如果使用TIM2_ICPSC_DIV，则在4个上升沿之后设置catpture事件（标志）（取决于TIM2_ICPOLARITY_FALLING / TIM2_ICPOLARITYRISING的降低边缘）
因此，为了计算信号的频率（在本例中它是LSI），你应该在下面的公式中乘以4（instaed为8），这样就变成了
LSIClockFreq =（4 * TIM2ClockFreq）/（ICValue2 - ICValue1）;
让我们处理时钟频率，计算频率值的一般规则是：
捕获分频器*定时器时钟/（第二次捕获 - 第一次捕获）
在ST提供的示例中，
1-捕获为8（TIM2_ICPSC_DIV8）
2-定时器时钟为2MHz（使用#define TIM2ClockFreq（（u32）2000000定义）
提醒：定时器的时钟频率与CPU相同，默认为STM8
似乎运行在2MHz，这就是它定义为2000000的原因。
我希望这清楚，它对你有用。
Herzlich
MCULüfter
以上来自于谷歌翻译
以下为原文
Hi piovan,
I didn't understand exactly your point but this is my understanding of the example:
using TIM2_ICInit() function you configure the capture when exactly it occurs.
So if you use TIM2_ICPSC_DIV, the catpture event (flag) is set after 4 rising edges (orfalling edges that depends on the TIM2_ICPOLARITY_FALLING/TIM2_ICPOLARITYRISING)
consequently to compute the frequency of a signal (in the example it is the LSI) you should multiply by 4 (instaed of 8)in the formula below so it becomes
LSIClockFreq = (4*TIM2ClockFreq) / (ICValue2 - ICValue1);
Let's process the clock frequency, the general rule to compute the frequency value is:
capture divider * timer clock / (second capture - first capture)
In the example provided by ST,
1- the capture is 8 (TIM2_ICPSC_DIV8)
2- the timer clock is 2MHz (It is defined using #define TIM2ClockFreq ((u32)2000000)
Reminder: the timer is clocked at the same clock than the CPU and by default STM8
seems to run at 2MHz that is why it is defined at 2000000.
I hope this clear and it is useful for you.
Herzlich
MCU Lüfter
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