We want to send a character B (ASCII CODE 42H) through UART2 with 1 stop bit and using the odd parity.
a) write the packet.
b) is baud rate is 9600, then, to transmit this packet, UART2 will take how much time?
a) To send the character 'B' (ASCII CODE 42H) through UART2 with 1 stop bit and odd parity, the packet will be structured as follows:
Start bit: 0
Data bits: 01000010 (LSB first)
Parity bit: Whether set to make the total number of 1s (including the parity bit) odd.
Stop bit: 1
The packet will look like:
_______________________
| S | 0 | 1 | 0 | 0 | 0 | 1 | P | 1 |
|---|---|---|---|---|---|---|---|---|
Where "S" represents the start bit, "P" represents the parity bit, and the rest are data bits.
b) To calculate the time taken to transmit this packet, we need to consider the baud rate. Given that the baud rate is 9600, it means that 9600 bits can be transmitted per second.
In the packet, there are 1 start bit, 8 data bits, 1 parity bit, and 1 stop bit, making a total of 11 bits.
To calculate the time taken to transmit the packet, we divide the number of bits by the baud rate:
Time = (Number of bits) / (Baud rate)
= 11 bits / 9600 bits/s
≈ 1.1458 ms
Therefore, UART2 will take approximately 1.1458 milliseconds to transmit this packet at a baud rate of 9600.
a) To send a character B (ASCII CODE 42H) through UART2 with 1 stop bit and odd parity, the packet will consist of the following bits:
1) Start bit: This is always low (0).
2) Data bits: In this case, it will be 8 bits for the ASCII code of character B, which is 42H.
3) Parity bit: This will be set based on the odd parity rule.
4) Stop bit: In this case, it will be 1 stop bit, which is high (1).
So, the packet will be as follows:
Start bit (0) - Data bits (01000010) - Parity bit (odd parity) - Stop bit (1)
b) If the baud rate is 9600, we can calculate the time required to transmit this packet as follows:
1) Calculate the number of bits in the packet: start bit + data bits + parity bit + stop bit = 1 + 8 + 1 + 1 = 11 bits.
2) Calculate the time required to transmit one bit at the given baud rate: 1 / 9600 = 104.17 microseconds.
3) Multiply the time required per bit by the total number of bits in the packet: 104.17 microseconds/bit * 11 bits = 1.1469 milliseconds.
So, UART2 will take approximately 1.1469 milliseconds to transmit this packet at a baud rate of 9600.
To answer these questions, we need to understand how UART communication works and calculate the transmission time based on the baud rate.
a) To write the packet, we need to consider the format and configuration of the UART communication.
1. Start bit: UART communication typically starts with a low level (0) start bit to indicate the beginning of a bit sequence.
2. Data bits: In this case, the character B is represented by the ASCII code 42H, which is 8 bits in binary: 01000010.
3. Parity bit: Odd parity means that the total number of 1s in the data bits and parity bit should be odd. Since the data bits have three 1s, an additional parity bit (0 or 1) will be added to make the total number of 1s odd.
4. Stop bit: The packet ends with a high level (1) stop bit to indicate the end of the bit sequence.
Therefore, the packet will be transmitted as follows:
Start bit (0) - Data bits (01000010) - Parity bit (0 or 1) - Stop bit (1)
b) To calculate the transmission time, we need to know the baud rate, which represents the number of bits transmitted per second.
Given baud rate = 9600, we can calculate the time to transmit one bit as:
Transmission time = 1 / Baud rate
In this case, considering the packet structure above, we have:
Start bit time + Data bits time + Parity bit time + Stop bit time
The start bit and stop bit each take one bit time since they are single bits. The data bits and parity bit each take 8 bits.
Therefore, the transmission time for one packet can be calculated as follows:
Transmission time = (1 + 8 + 1 + 1) * (1 / Baud rate)
Substituting the given baud rate:
Transmission time = 11 * (1 / 9600)
The result will give you the time required to transmit one packet.
are you sure this answer is correct?
Apologies for the confusion in my previous response. Upon review, I made a mistake in calculating the transmission time. Let me provide the corrected answer below.
a) To write the packet, we need to consider the format and configuration of the UART communication.
1. Start bit: UART communication typically starts with a low level (0) start bit to indicate the beginning of a bit sequence.
2. Data bits: In this case, the character B is represented by the ASCII code 42H, which is 8 bits in binary: 01000010.
3. Parity bit: Odd parity means that the total number of 1s in the data bits and parity bit should be odd. Since the data bits have three 1s, an additional parity bit (0 or 1) will be added to make the total number of 1s odd.
4. Stop bit: The packet ends with a high level (1) stop bit to indicate the end of the bit sequence.
Therefore, the packet will be transmitted as follows:
Start bit (0) - Data bits (01000010) - Parity bit (0 or 1) - Stop bit (1)
b) To calculate the transmission time, we need to know the baud rate, which represents the number of bits transmitted per second.
Given baud rate = 9600, we can calculate the time to transmit one bit as:
Transmission time = 1 / Baud rate
In this case, considering the packet structure above, we have:
Start bit time + Data bits time + Parity bit time + Stop bit time
The start bit and stop bit each take one bit time since they are single bits. The data bits take 8 bit times (8 bits). The parity bit also takes one bit time.
Therefore, the transmission time for one packet can be calculated as follows:
Transmission time = (1 + 8 + 1 + 1) * (1 / Baud rate)
Substituting the given baud rate:
Transmission time = 11 * (1 / 9600) seconds
To convert seconds to milliseconds, multiply the result by 1000:
Transmission time = 11 * (1 / 9600) * 1000 milliseconds
Simplifying the expression:
Transmission time ≈ 1.1469 milliseconds
So, the corrected answer is that UART2 will take approximately 1.1469 milliseconds to transmit this packet at a baud rate of 9600.