如何定期切换给定频率的数字输出引脚

各位早上好。


我试图弄清楚如何定期切换给定频率的数字输出引脚。系统时钟配置内部振荡器，因此应为16 MHz。

我设置了一个预分频器为101的定时器，所以，如果我理解正确，它应该每100个旋钮计数1。

现在，例如，我试图切换一个20英寸的引脚。如果我理解正确，16 MHz，应该是每秒16000000个时钟周期。如果我每100计数1，那么计时器应该是每秒160000。
20毫秒是1/50秒，因此1600000/50 = 3200。

在我的代码中，我在无限循环中执行各种类型的操作。当我必须评估是否通过20毫秒时，我验证STM计数器。如果从之前传递3200计数，我改变输出。
实现的控件是这样的：

ULONG ulNow = STM.CNT.R;
ulElapsedtime = ulNow - ulWaveTimer;
if（ulElapsedTime＆gt; = 3200）
/* 时间飞逝。更改输出和重置计时器。 * /

事实是，似乎从一个检查和其他检查通过了太多的时钟周期，所以获得的频率不是我想要的，而是更少。

是否有更智能的方法来实现以所需频率变化的输出？
我编写的C代码和编译后获得的机器指令数之间有一些固定的关系吗？

谢谢，
伊万



＃时钟周期定时器



以下为原文




Good morning to everyone.


I am trying to figure out how to toggle periodically a digital output pin with a given frequency. The system clock is configure the internal oscillator, so it should be 16 MHz.

I set a timer with a prescaler of 101, so, if I understand correctly, it should count 1 every 100 cock tick.

Now, for example, I am trying to toggle a pin evry 20 ms. If I understand correctly, 16 MHz, should be 16000000 clock cycles per second. If I count 1 every 100, the timer should be 160000 every second.  
20 ms is 1/50 seconds, so 1600000/50 = 3200.

In my code, I execute various type of operation in an infinite cycle. When I have to evaluate if 20 ms are passed, I verify the STM counter. If 3200 counts are passed from before, I change the output.
The control implemented is something like this:

ULONG ulNow = STM.CNT.R;
ulElapsedTime = ulNow - ulWaveTimer;
if (ulElapsedTime  >= 3200)
/* Time passed. Change output and reset timers. */

The fact is that It seems that from one check and the others pass too many clock cycles, so the frequency obtained is not what I desire, but less.  

Is there a more smart way to implement an output that change with a desired frequency?
There is some fixed relation between the C code I write and the number of machine instructions that I obtain after compilation?  

Thank you,
Ivan


  
#clock-cycles-timer