Skip to main content

Parity bits and 3-bit odd parity generator

Short answer

Parity bits enable single-bit error detection. A 3-bit odd parity generator outputs P = (A⊕B⊕C)' to make total number of 1s odd. For even number of 1s in ABC, P=1; for odd, P=0.

Describe the importance of parity bits in communication system. Explain 3 bits odd parity generator circuit clearly.

Importance of parity bits

A parity bit is one extra bit attached to a data word so that the total count of 1s is either even (even parity) or odd (odd parity). In communication systems it matters because:

  • Error detection: noise or interference may flip a bit. The receiver checks the parity; a mismatch shows an error, and the data can be re-requested.
  • Low cost: only one extra bit per word and a few XOR gates.
  • Single-bit errors are always detected (in fact any odd number of errors).
  • Limits: an even number of bit errors goes unnoticed, and the faulty bit cannot be located or corrected. Codes like Hamming code extend the parity idea to correct errors.
  • Odd parity is often preferred because an all-zero word (e.g. a dead line) still carries a 1, so a stuck-at-0 line is detected.

3-bit odd parity generator

Inputs A,B,CA, B, C; output PP is chosen so that A,B,C,PA, B, C, P together contain an odd number of 1s.

ABCP
0001
0010
0100
0111
1000
1011
1101
1110
P=Σm(0,3,5,6)=A′B′C′+A′BC+AB′C+ABC′P = \Sigma m(0,3,5,6) = A'B'C' + A'BC + AB'C + ABC'

The K-map has no adjacent 1s, so it is simplified with XOR/XNOR:

P=A′(B⊙C)+A(B⊕C)=A⊕B⊕C‾\begin{aligned} P &= A'(B \odot C) + A(B \oplus C) \\ &= \overline{A \oplus B \oplus C} \end{aligned}
 A --+
     +--[XOR]--+
 B --+         +--[XNOR]--- P
 C ------------+

Working: if the data already has an odd number of 1s, P=0P = 0; if even, P=1P = 1. Example: data 110110 (two 1s) gives P=1⊕1⊕0‾=1P = \overline{1 \oplus 1 \oplus 0} = 1, so the word 11011101 has three 1s (odd). The receiver's odd parity checker computes A⊕B⊕C⊕P‾\overline{A \oplus B \oplus C \oplus P}, which is 0 when no error is present.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Digital Logic chapter 4 questions →