Mtech1
Junior Member level 1
I'd like to understand the concepts. How can we send the higher nibble in a byte via UART when it's configured for 4N1 with a baud rate of 9600?
For example: When we shift the byte 10100010 right by 4 bits (>> 4), we indeed get 00001010. However, this result is still one byte. In UART communication with 4N1 configuration, each data frame consists of one start bit, four data bits, no parity bit, and one stop bit. So, the 00001010 we obtained is still one complete byte within this configuration.
For example: When we shift the byte 10100010 right by 4 bits (>> 4), we indeed get 00001010. However, this result is still one byte. In UART communication with 4N1 configuration, each data frame consists of one start bit, four data bits, no parity bit, and one stop bit. So, the 00001010 we obtained is still one complete byte within this configuration.