Question

Derive the circuits for a three-bit parity generator and four-bit parity checker using an odd parity bit.

Answer 1

Solution

3 bit parity generator using odd parity

A	B	C	P
0	0	0	1
0	0	1	0
0	1	0	0
0	1	1	1
1	0	0	0
1	0	1	1
1	1	0	1
1	1	1	0

We need to make the total number of bits odd.

If number of 1s is odd, parity bit value is 0. If number of 1s is even, parity bit value is 1

We can use karnaug map to simplify the values

Or simply see the 1 values in the truth table(P)

For e.g

A B с P о 0 0 1

This is A’B’C’

0 1 1 1

This is

A’BC

1 0 1 1

This is

AB’C

1 1 0 1

ABC’

Therefore

P=A’B’C’+A’BC+AB’C’+ABC’

=A(B’C’+BC)+A(B’C+BC’)

=A(B ⊙ C)+A(B ⊙ C)’

P=A⊕B⊙C

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4-bit parity checker using odd parity

A	B	C	P	C
0	0	0	0	1
0	0	0	1	0
0	0	1	0	0
0	0	1	1	1
0	1	0	0	0
0	1	0	1	1
0	1	1	0	1
0	1	1	1	0
1	0	0	0	0
1	0	0	1	1
1	0	1	0	1
1	0	1	1	0
1	1	0	0	1
1	1	0	1	0
1	1	1	0	0
1	1	1	1	1

Similarly if we solve

C=A’B’C’P’+A’B’CP+A’BC’P+A’BCP’+ABC’P’+ABCP+AB’C’P+AB’CP’

=A’B’(C’P’+CP)+A’B(C’P+CP’)+AB(C’P’+CP)+AB’(C’P+CP’)

=A’B’(C⊙P)+A’B(C⊕P)+AB(⊙)+AB’(C⊕P)

=(A’B’+AB)(C⊙P)+(A’B+AB’)(C⊕P)

=(A⊙B)(C⊙P)+(A⊕B)(C⊕P)

C=(A⊙B)⊙(C⊙P)

NOW we will draw the logic circuit

P=A⊕B⊙C

А Р B с

C=(A⊙B)⊙(C⊙P)

В - С P7

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ALL THE BEST