
The diagram and the text are not explicit about the bit order. By that I mean, is the chracter in the above example 0x51 = ‘Q’, or reading in the opposite direction 0x8a = Š (Latin capital letter S with caron). The latter seemed an unlikely choice for this example, which implies MSB-first is more likely.
The serial port has many pins. We will discuss the transmit and receive pin first. Electrically speaking, whenever the serial port sends a logical one (1) a negative voltage is effected on the transmit pin. Whenever the serial port sends a logical zero (0) a positive voltage is effected. When no data is being sent, the serial port’s transmit pin’s voltage is negative (1) and is said to be in aMARK state. Note that the serial port can also be forced to keep the transmit pin at a positive voltage (0) and is said to be the SPACE or BREAK state. (The terms MARK and SPACE are also used to simply denote a negative voltage (1) or a positive voltage(0) at the transmit pin respectively).When transmitting a byte, the UART (serial port) first sends a START BIT which is a positive voltage (0), followed by the data (general 8 bits, but could be 5, 6, 7, or 8 bits) followed by one or two STOP BITs which is a negative(1) voltage. The sequence is repeated for each byte sent.

Assume we want to send the letter ‘A’ over the serial port. The binary representation of the letter ‘A’ is 01000001. Remembering that bits are transmitted from least significant bit (LSB) to most significant bit (MSB), the bit stream transmitted would be as follows for the line characteristics 8 bits, no parity, 1 stop bit, 9600 baud.
LSB (0 1 0 0 0 0 0 1 0 1) MSB
Here for comparison is my scope image of ‘A’ allegedly with even parity.
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Character ‘A’ |
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Character ‘C’ |
