Explorer 16

Question

Question 3.
We want to send a character B (ASCII CODE 42H) through UART2 with 1 stop bit and using odd parity.
Q1. Write the packet. (3 Marks)
Q2. If baud rate is 9600, then, to transmit this packet, UART2 will take how much time? (3 Marks)

Answer 1

Q1. The packet to send the character B (ASCII CODE 42H) through UART2 with 1 stop bit and odd parity would consist of the following:

1. Start bit: Logic low (0)
2. Data bits: The character B in binary is 01000010.
3. Parity bit: An odd parity bit is calculated based on the data bits. Since the number of 1s in the data bits is odd (2), the parity bit would be 0.
4. Stop bit: Logic high (1)

Therefore, the packet would be: 0 01000010 0

Q2. To calculate the time required to transmit the packet, we need to consider the baud rate. Baud rate represents the number of bits transmitted per second.

Given that the baud rate is 9600, it means 9600 bits will be transmitted per second.

The packet contains a total of 10 bits (1 start bit + 8 data bits + 1 stop bit).

Therefore, the time required to transmit the packet can be calculated as:

Time = Number of bits / Baud rate
Time = 10 bits / 9600 bits per second

Time = 0.00104 seconds (rounded to 5 decimal places)

So, to transmit this packet, UART2 will take approximately 0.00104 seconds.