CODEWITHSUNDEEP
arduino-unobaud-rate
Anne
Anne

Reputation: 21

Baud rate on Arduino

I am implementing a simple visible light communication module with two Arduinos, as a transmitter and a receiver, with a short text message consisting of 120 characters. I have used Manchester encoding with on-off -keying modulation.

Altogether, in my message frame, with Manchester encoding and with preambles and end-of-frame byte, there are 2480 bits. I have set one bit period to be 500 microseconds. On the receiver side, I am sampling this bit four times, at (500/4) 125 microseconds. From my knowledge, since each bit is 500 μs, there are 2000 bits/s that are being transmitted from the transmitter so a baud rate of 9600 bits/s should work. However, 9600 is not working and any baud rate above 38400 up to 115200 is working, and I can properly decode this short message in my receiver.

What is the explanation for this behavior? Why is baud rate of 9600 not working though I am only transmitting 2000 bits per second?

Further information: I have set the prescalar to 128, so the ADC sampling frequency is (16 MHz/128)/13 = 9.6 kHz.

Upvotes: 2

Views: 2799

Answers (1)

John Burger
John Burger

Reputation: 3672

When you suddenly started talking about "baud rates", it implied that you're using the hardware serial port on the Arduino. If so, then realise that feeding 2,000 bit/s into a device expecting 9,600 bit/s has problems.

The way that an asynchronous UART works is that it waits for a start signal (bit), then decodes the next (typically) 9 signals at the current bit rate. It then checks that the 9th bit is a stop bit; if it isn't, it discards the byte.

Since you're only changing every 9600/2000 = 4.8 bits, then odds on the 9th "stop" bit will be of the wrong sense, and the data will be lost.

Below is an ASCII diagram for the timing that I'm talking about.

  • I'll use the bitstream 00101101 for the signal produced by the circuit, with a . as a 0 ms separator between bits;
  • I'll use a ^ to point out where the UART is sampling the bits;
  • I'll use a * to indicate a "correct" byte (insofar as the byte ends in a correct stop bit);
  • I'll use a ! to indicate an "incorrect" byte (insofar as the byte ends in an incorrect stop bit);

Of course, I'll assume a baud rate of 10,000 bit/s (5 rather than 4.8...)

00000.00000.11111.00000.11111.11111.00000.11111
^^^^^.^^^^!.......^^^^^.^^^^*.......^^^^^.^^^^*

This sequence would result in the UART recording the following 3 bytes:

  1. Error
  2. 0xF0 (LSB first is defined...)
  3. 0xF0 (LSB first is defined...)

Upvotes: 1

Related Questions