2

#include <Wire.h>
#include <LiquidCrystal_I2C.h>
#include "digitalWriteFast.h"

#define BUTTONPIN 2
#define FDCSPIN 6
#define CIPIN 5
#define SPEAKERPIN 4

float RPM=1000.0;
int PWus=1.5*1000.0;
int FDCSoffset=(1000.0/RPM/60.0/4.0) - (PWus/1000.0);
int CiCnt;
int totalCnt;
int CIState=B01100000;
bool Buttonstate=0;
LiquidCrystal_I2C lcd(0x27, 20, 4);
 
void setup() 
{
  Serial.begin(9600);
  lcd.begin();
  lcd.backlight();
  lcd.clear();
  lcd.print("READY TO TEST"); // Waiting for button press
  pinMode(BUTTONPIN, INPUT);// Setting button as Input
  pinMode(FDCSPIN, OUTPUT); // Fuel Delivery Control Signal
  pinMode(CIPIN, OUTPUT); // Cylinder Identification
  digitalWriteFast(FDCSPIN, LOW);
  digitalWriteFast(CIPIN, LOW);
}

void loop() 
{
 Buttonstate=digitalReadFast(BUTTONPIN);
 if(Buttonstate==1)
 {
  delay(5);
  if(Buttonstate==1)
  {
    runInjectorTest();
  }
 }
}
void runInjectorTest()
{
  lcd.print("Test Run");
  playtesttone();
  
    
    for(totalCnt=0;totalCnt<8000;totalCnt++)
    {
    
      PORTD=CIState;
      delayMicroseconds(FDCSoffset);
      PORTD=PORTD^B01000000;
      delayMicroseconds(PWus);
      if(totalCnt%4==0)
      {
        CIState=CIState^B00100000;
      }
    }
  totalCnt=0;
  playfinishtone();
}
void playtesttone()
{
  tone(SPEAKERPIN, 500, 500);
  delay(700);
  tone(SPEAKERPIN, 500, 500);
  delay(700);
  tone(SPEAKERPIN, 500, 500);
  delay(700);
  tone(SPEAKERPIN, 2000, 500);
}
void playfinishtone()
{

  tone(SPEAKERPIN, 2000, 500);
  delay(700);
  tone(SPEAKERPIN, 2000, 500);
  delay(700);
  tone(SPEAKERPIN, 2000, 500);
  delay(700);
  tone(SPEAKERPIN, 500, 500);

}

I am using this code to generate 2 signals, called CI and FDCS. CI stays on until the 5th FDCS pulse. I checked the output which is connected to a driver using the scope and I got this.The signal The rising edge and falling edge of the longer signal (CI, dark green) are supposed to sync with the rising edge of the FDCS signal (Lime green),but they are inline with the falling edge. Note that the output is inverted (when compared to the code) hence the signal generated by the Arduino pins is according to the code.

Improve this question
7
  • so, what is your question? – jsotola Dec 1 '20 at 15:59
  • @jsotola Why is the signal not turning on at the exact same time? according to the code, I am writing the PORTD register at the same time for both pin 6 and 5, so they should sync up right? – Redlion11 Dec 1 '20 at 16:01
  • which code line writes both bits to PORTD? – jsotola Dec 1 '20 at 16:07
  • @jsotola PORTD=CIState, under runInjectortest(), in the for loop. CIState=B01100000 – Redlion11 Dec 1 '20 at 16:08
  • 1
    If the output is inverted, and you turn the higher frequent signal on it is displayed on the scope as a falling edge. (because it is inverted: rising edge becomes a falling edge.) You got what you programmed. ;-) – Peter Paul Kiefer Dec 1 '20 at 16:39
2

I did not tried it out. But something like this might help to synchronize the inverted signals to the rising edge of the high frequent signal.

I removed the FDCSoffset delay, because it is negative according to the variables at the top.

// initialize CIState with 00 
int CIState=B00000000;

// lines not shown .....

for( totalCnt=0; totalCnt<8000; totalCnt++ )
{
  CIState=CIState^B01000000;
  if(totalCnt%4==0)
  {
    CIState=CIState^B00100000;
  }
  // both signals are set at the same time
  PORTD=CIState;
  delayMicroseconds(PWus);
}
Improve this answer

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.