Question

我正在尝试将Arduino Uno用作较大的电路的一部分，该电路可以同步两个平方电压波形（这些波形被馈送到Uno的引脚中）。 Uno将由较大电路另一部分的数字信号激活，并且该激活信号将一次持续几秒钟。当激活信号为HIGH时，Uno的工作就是检测其测量的方波何时同步：它将监视它们并在彼此相隔约100微秒内达到零交叉点时产生自己的脉冲（即波本身就是〜50Hz。

因为100us是一个很短的窗口，所以我认为我将尝试直接位操作，而不是使用DigitalRead（）或DigitalWrite（）。但是，这是我第一次尝试位操作，而且我的C经验通常很有限。我在这里有了代码的初稿-如果你们中的任何人都能告诉我它的逻辑是否合理，我将不胜感激！

它看起来很长，但是只有几行内容有价值，其余是一堆复制/粘贴的块（毫无疑问，这意味着还有很多更优雅的编写方法）。

// Synchronism detector

// This code is built to run on an Arduino Uno.
// The hardware timer runs at 16MHz. Using a
// divide by 8 on the counter each count is 
// 500 ns and a 50Hz wave cycle is 40000 clock cycles.


#include <avr/io.h>
#include <avr/interrupt.h>

void setup(){

  // Set clock prescaler
  // This is set to 'divide by 8' but if I can get away
  // with using delay statements that doesn't really matter
  TCCR1B &= ~(1 << CS12);
  TCCR1B != ~(1 << CS11);
  TCCR1B &= ~(1 << CS10);

  // Set up pins
  // 3rd bit of port D will be the activation flag
  // 4th and 5th bits will be read pins for the incoming waves
  DDRD = DDRD & B11100011
  // Note that I'll also have the incoming waves going into the int0 and int1 pins

  // 6th bit will be a write pin for the outgoing signal
  DDRD = DDRD | B00100000
  PORTD = PORTD & B11011111

  // Set initial values for zero-cross detect flags
  wave_0 = ((PIND & B00001000) >> 3);
  wave_0_prev = ((PIND & B00001000) >> 3);
  wave_1 = ((PIND & B00010000) >> 4);
  wave_1_prev = ((PIND & B00010000) >> 4);
}

void loop(){

  // Poll activation flag
  if (((PIND & B00000100) >> 2) == 1)

    // Poll wave input pins
    wave_0 = ((PIND & B00001000) >> 3);
    wave_1 = ((PIND & B00010000) >> 4);

    // Check zero crossing detection, 
    // Start with wave 0, rising edge
    if (wave_0 == 1 && wave_0_prev == 0))
      attachInterrupt(int1,sync_interrupt, RISING);
      delay(.0001);
      detachInterrupt(int1);

      // Reset pins
      wave_0 = ((PIND & B00001000) >> 3);
      wave_0_prev = ((PIND & B00001000) >> 3);
      wave_1 = ((PIND & B00010000) >> 4);
      wave_1_prev = ((PIND & B00010000) >> 4);
      )      

// Wave 0, falling edge
if (wave_0 == 0 && wave_0_prev == 1))
  attachInterrupt(int1,sync_interrupt, FALLING);
  delay(.0001);
  detachInterrupt(int1);

  // Reset pins
  wave_0 = ((PIND & B00001000) >> 3);
  wave_0_prev = ((PIND & B00001000) >> 3);
  wave_1 = ((PIND & B00010000) >> 4);
  wave_1_prev = ((PIND & B00010000) >> 4);
  )

// Wave 1, rising edge
if (wave_1 == 1 && wave_1_prev == 0))
  attachInterrupt(int0,sync_interrupt, RISING);
  delay(.0001);
  detachInterrupt(int0);

  // Reset pins
  wave_0 = ((PIND & B00001000) >> 3);
  wave_0_prev = ((PIND & B00001000) >> 3);
  wave_1 = ((PIND & B00010000) >> 4);
  wave_1_prev = ((PIND & B00010000) >> 4);
  )

// Wave 1, falling edge
if (wave_1 == 0 && wave_1_prev == 1))
  attachInterrupt(int0,sync_interrupt, FALLING);
  delay(.0001);
  detachInterrupt(int0);

  // Reset pins
  wave_0 = ((PIND & B00001000) >> 3);
  wave_0_prev = ((PIND & B00001000) >> 3);
  wave_1 = ((PIND & B00010000) >> 4);
  wave_1_prev = ((PIND & B00010000) >> 4);
  ) 
}

sync_interrupt(){
  // Set output bit high
  PORTD = PORTD | B00100000
  delay(.001}
  // Set output bit low
  PORTD = PORTD & B00100000

  // Reset pins
  wave_0 = ((PIND & B00001000) >> 3);
  wave_0_prev = ((PIND & B00001000) >> 3);
  wave_1 = ((PIND & B00010000) >> 4);
  wave_1_prev = ((PIND & B00010000) >> 4);
}

再次感谢您的帮助。

Answer 1

好，所以我不太确定此特定设备的中断标志到底是什么，但是我个人认为，最简单的方法是将其设置为为每个中断设置2个中断输入并设置每个中断代码块，以通过从输出引脚跳转来激活定时器中断。这样，如果一个在某个时间段内未激活而另一个被激活，则可以重置中断标志；如果第二个被激活，则可以在同步波形时运行希望运行的代码

Arduino：检测两个信号是否在彼此之间相差约100微秒内变化

1 个答案: