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I am trying to average input values for a number of inputs and have it then store those averages into a string to write to a txt doc on a sd card.

I am using:

// read 6 sensors and append to the string:
for (int analogPin = 0; analogPin < 6; analogPin++) {
  int sensor = analogRead(analogPin);
  dataString += String(sensor);
  if (analogPin < 6) {
    dataString += ",";
  }
}

to read the pins at the moment, but I want to change the analogRead(analogPin) to an average value and then use the averages to create the string.

but when I try to average the data I run into errors, or I get output that is 0 for something that should be higher (3.3V).

I am doing this to stabilize my values a bit better and remove jitter.

Any suggestions?

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This function will average readings on an analog pin, include it and use it in place of analogRead(...) and it should do what you want.

int aveAnalogRead(int pin, int ave=10, int delayms = 0)
{
    int tot = 0;
    for(int i=0; i<ave; i++)
    {
        tot += analogRead(pin);
        if(delayms) delay(delayms);
    }
    return tot/ave;
}

in your loop():

for (int analogPin = 0; analogPin < 6; analogPin++) 
{
   int sensor = aveAnalogRead(analogPin);
     <<do your work here>>
}
|improve this answer|||||
  • 1
    There is almost never a good reason to add a delay. – Edgar Bonet Jun 21 '18 at 9:10
  • With this though I have to run it multiple times, one for each input. Is it possible to have it also cycle though the inputs such as a loop within a loop? – Michael H. Jun 21 '18 at 13:42
  • yes use it inplace of your analogInput call above loop within a loop – esoterik Jun 21 '18 at 14:04
  • so this:int pin, for (int n=0, n<10, n++){ pin = n; int aveAnalogRead( int ave=10, int delayms = 50) { int tot = 0; for(int i=0; i<ave; i++) { tot += analogRead(pin); delay(delayms); } return tot/ave; }} – Michael H. Jun 21 '18 at 14:39
  • that's not how C++ works; you put this function somewhere outside your loop() on your sketch, then call then function. you don't need to copy the definition of the function to call it. – esoterik Jun 21 '18 at 14:47

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