arduino声调命令的问题

Question

我正试着用模拟传感器做一架阿迪诺钢琴作为钥匙。但是，所有的键都产生相同的音调，生成的音调在程序中不会出现在任何地方。当同时按下多个键时，所述音调在两个预期音调之间交替，而未定义的音调不产生。我检查了在触摸键时是否激活了其他命令，只有当按下相应的键时，它们才会激活。

#define Note_C 65.41   //Hz   
#define Note_D 73.42   //Hz   
#define Note_E 82.41   //Hz
#define Note_F 87.31   //Hz
#define Note_G 98      //Hz
#define Note_A 110     //Hz  
#define threshold 1000  


const int speaker=3;  
const int B_1=A0;    // pins A0-A5 have sensors attatched to them 
const int B_2=A1;    // pins 13-8 are being used to power each sensor
const int B_3=A2;     
const int B_4=A3;     
const int B_5=A4;
const int B_6=A5;
const int P_1=13;
const int P_2=12;
const int P_3=11;
const int P_4=10;
const int P_5=9;
const int P_6=8;


int val_1=0;
int val_2=0;
int val_3=0;
int val_4=0;
int val_5=0;
int val_6=0; 


void setup()
{   
Serial.begin(9600);
pinMode(P_1, OUTPUT);      
digitalWrite(P_1, HIGH);
pinMode(P_2, OUTPUT);
digitalWrite(P_2, HIGH);
pinMode(P_3, OUTPUT);
digitalWrite(P_3, HIGH);
pinMode(P_4, OUTPUT);
digitalWrite(P_4, HIGH);
pinMode(P_5, OUTPUT);
digitalWrite(P_5, HIGH);
pinMode(P_6, OUTPUT);
digitalWrite(P_6, HIGH);
}


void loop()
{
  analogRead(B_1); // checks each sensor value and stores it
  delay(1);              // each value must be checked twice with a 
  val_1=analogRead(B_1); // delay inbetween to provide consistant values
  analogRead(B_2);
  delay(1);
  val_2=analogRead(B_2);
  analogRead(B_3);
  delay(1);
  val_3=analogRead(B_3);
  analogRead(B_4);
  delay(1);
  val_4=analogRead(B_4);
  analogRead(B_5);
  delay(1);
  val_5=analogRead(B_5);
  analogRead(B_6);
  delay(1);
 val_6=analogRead(B_6);

 if (val_1 < threshold)     
  {
  tone (speaker, Note_C);
  Serial.println("1");
}
    else
   {
     noTone (speaker);   
   }     
 if (val_2 < threshold)     
  {
    Serial.println("2");
  tone (speaker, Note_D);
} 
    else
{ 
  noTone(speaker);
}
 if (val_3 < threshold)     
  {
    Serial.println("3");
  tone (speaker, Note_E);
} 
    else
   { 
     noTone (speaker);
   }
 if (val_4 < threshold) 
  {
  Serial.println("4");
  tone (speaker, Note_F);
}
    else 
    {
      noTone (speaker);
    }
  if (val_5 < threshold)
  {
  Serial.println("5");
  tone (speaker, Note_G);
}
    else
   {
     noTone (speaker);
   }
  if (val_6 < threshold) 
   {
   Serial.println("6");
   tone (speaker, Note_A);
   }
    else
   {
     noTone (speaker);
   }
    noTone(speaker);
  }

我知道数组的效率会更高，但我想先让它正常工作。我也是一个初级程序员，所以任何其他建议都将不胜感激。

Answer 1

我怀疑问题是，每次在循环中，你都在激活和消除音调。我认为更好的方法是在特定输入被激活时启动音调，并一直播放到输入被停用为止。(不能同时使用tone()播放多个音符。)

您需要在变量中存储当前播放音调的数量。像这样的东西可能会起作用：

int currentTone = 0;

void loop()
{
    // (take your analog readings here)


    if (val_1 < threshold)
    {
        if (currentTone == 0)
        {
            tone(speaker, Note_C);
            currentTone = 1;
        }
    }
    else if (currentTone == 1)
    {
        noTone(speaker);
        currentTone = 0;
    }

    if (val_2 < threshold)
    {
        if (currentTone == 0)
        {
            tone(speaker, Note_D);
            currentTone = 2;
        }
    }
    else if (currentTone == 2)
    {
        noTone(speaker);
        currentTone = 0;
    }

    // etc.
}

这通过每个输入来检查它是否已被激活。如果是这样，如果目前没有其他音调，则开始播放相应的音符。如果输入不是活动的，但其相应的音符正在播放，则停止输入。