Problem with receiving binary data

Projects Programming
1

Hi guys,

I have a problem with receiving data. In the code I am trying to send binary data to digital pin thorugh which I want to receive and store them. This setup is on Arduino UNO where the two pins are connected together. I want to store them somehow through the pin change interrupt. I know that there have to be some kind of delay to know if there are two or more "1s" or "0s". I measured the high pulse width before and I got two different times when I tried it with ISR (16µs)and without ISR(8µs) .

Thank you in advance.

#define TxPin 9
#define RxPin 3

uint8_t data = B10110010;
String str, string;

void sendSignal(String string){
  for(int i=0;i<8;i++){
    if(string[i]==0){
      digitalWrite(TxPin, LOW);
    }
    else{
      digitalWrite(TxPin, HIGH);
    }
  }
}

void isrhandler(){
  int pinState, state;
  pinState = digitalRead(RxPin);
  if(pinState == HIGH || pinState == LOW){
    delayMicroseconds(16);
    state = digitalRead(RxPin);
    if(state == HIGH){
      str = str + 1;
    }else{
      str = str + 0;
    }
  }
}

void setup() {
  Serial.begin(9600);
  pinMode(TxPin, OUTPUT); 
  pinMode(RxPin, INPUT_PULLUP);

  attachInterrupt(digitalPinToInterrupt(RxPin), isrhandler, CHANGE);

  string = String(data, BIN);
  sendSignal(string);

  Serial.print(str);
}

void loop() {
  
}
2

Why would you convert a byte to String just to get the bits? if(string[i]==0) should probably be if(string[i]=='0'). Try to use bitRead instead:

for (int i = 0; i < sizeof(data); i++)
  digitalWrite(TxPin, bitRead(data, i));
3

What other values could pinState be?

1 Like
4

Thank you, I changed it.

5

so if the data is all 1s or all 0s, you might only get one interrupt

it might help to understand conventional serial transmission

a UART sends a ~byte of data, prefixed by a start bit and suffixed with one or 2 stops bits.

the start bit is the opposite of the idle line value so that there is a well defined change to indicate when a transmission begins.

bit times are well defined (e.g. 9600). bit timing is determined using a a timer

each bit is sampled one or more times (possibly voted to determine its state. if a single sample is used, the first delay based on the beginning of the start bit is 1.5 bit periods, and 1 bit period after that

stop bits are opposite the start bit state and guarantee that there is a change when start occurs and guarantees some amount of time between transissions

the # of bits transmitted is limited to tolerate ~2% overspeed and underspeed by both TX and RX.

this approach only relies on recognizing the change in state when the start bit occurs and uses timing to detect each bit.

6

Ok guys thank you all for your advice. I did it a little differently but it is done.Here is the code

#define TxPin 9
#define RxPin 3
#define ClockPin 2

uint8_t data = B10001010;
String str, string;

void sendSignal(uint8_t data){
  for(int i=0;i<8;i++){
    digitalWrite(TxPin, bitRead(data,i));
    digitalWrite(ClockPin, HIGH);
    digitalWrite(ClockPin, LOW);
  }
}

void isrhandler1(){
  int pinState;
  pinState = digitalRead(RxPin);
  if(pinState==HIGH){
    str = str + "1";
  }else{
    str = str + "0";
  }
}

void setup() {
  Serial.begin(9600);
  pinMode(TxPin, OUTPUT); 
  pinMode(RxPin, INPUT_PULLUP);
  pinMode(ClockPin, OUTPUT);

  attachInterrupt(digitalPinToInterrupt(ClockPin), isrhandler1, FALLING);

  noInterrupts();
  digitalWrite(ClockPin, LOW);
  interrupts();

  sendSignal(data);

  Serial.println(str);
}

void loop() {
  
}
7

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