A system transmits ASCII characters using 7 data bits plus 1 even parity bit (8 bits total). The parity bit is the most significant (leftmost) bit. (a) The ...
Question 1 Report
A system transmits ASCII characters using 7 data bits plus 1 even parity bit (8 bits total). The parity bit is the most significant (leftmost) bit.
(a) The ASCII codes for the characters A, B, C are 1000001, 1000010, 1000011. Add the even parity bit to each and show the complete 8-bit code. [3]
| Character | ASCII (7 bits) | Number of 1s | Parity bit | Complete byte |
|---|---|---|---|---|
| A | 1000001 | |||
| B | 1000010 | |||
| C | 1000011 |
Answer Details
(a) Adding even parity bits to ASCII codes (parity bit is the leftmost bit): [3]
| Character | ASCII (7 bits) | Number of 1s | Parity bit | Complete byte |
|---|---|---|---|---|
| A | 1000001 | 2 (even) | 0 | 01000001 |
| B | 1000010 | 2 (even) | 0 | 01000010 |
| C | 1000011 | 3 (odd) | 1 | 11000011 |
[1] for A (parity 0, byte 01000001), [1] for B (parity 0, byte 01000010), [1] for C (parity 1, byte 11000011).
For even parity, the total number of 1s in the complete 8-bit byte (including the parity bit) must be even. Characters A and B each have two 1s in their ASCII codes (already even), so the parity bit is 0. Character C has three 1s (odd), so the parity bit must be 1 to bring the total to four (even).
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