Boolean Function SOP & POS
- 1. Boolean Function Dr. (Mrs.) Gargi Khanna Associate Professor Electronics & Communication Engineering Department National Institute of Technology, Hamirpur
- 2. Boolean function • A Boolean function has: • At least one Boolean variable, • At least one Boolean operator, and • At least one input from the set {0,1}. • It produces an output that is also a member of the set {0,1}.
- 3. Truth Table • Addition of 2 bits: A B Sum Carry 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 Sum= 𝐴′ 𝐵 + 𝐴𝐵′ 𝐶𝑎𝑟𝑟𝑦 = 𝐴𝐵 Sum= 𝐴 + 𝐵 . (𝐴′ + 𝐵′) 𝐶𝑎𝑟𝑟𝑦 = 𝐴 + 𝐵 . 𝐴 + 𝐵′ . (𝐴′ + 𝐵)
- 4. Truth Table • Subtraction of 2 bits: A B Difference Borrow 0 0 0 0 0 1 1 1 1 0 1 0 1 1 0 0 Difference= 𝐴′ 𝐵 + 𝐴𝐵′ 𝐵𝑜𝑟𝑟𝑜𝑤 = 𝐴′𝐵 Difference = 𝐴 + 𝐵 . (𝐴′ + 𝐵′) 𝐵𝑜𝑟𝑟𝑜𝑤 = 𝐴 + 𝐵 . 𝐴′ + 𝐵 . (A’ +B’)
- 5. Truth Table • Addition of 3 bits: A B C Sum Carry 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Sum= 𝐴′ 𝐵′ 𝐶 + 𝐴′ 𝐵𝐶′ + 𝐴𝐵′ 𝐶′ + 𝐴𝐵𝐶 𝐶𝑎𝑟𝑟𝑦 = 𝐴′ 𝐵𝐶 + 𝐴𝐵′ 𝐶 + 𝐴𝐵𝐶′ +ABC Sum= (𝐴 + 𝐵 + 𝐶). (𝐴 + 𝐵′+𝐶′)(𝐴′ + 𝐵 + 𝐶′)(𝐴′ + 𝐵′ + 𝐶)
- 6. Two-Level Implementations • Boolean functions in either SOP or POS forms can be implemented using 2-Level implementations. • For SOP forms AND gates will be in the first level and a single OR gate will be in the second level. • For POS forms OR gates will be in the first level and a single AND gate will be in the second level. • SOP forms can be implemented using only NAND gates, while POS forms can be implemented using only NOR gates.
- 8. F = XZ + Y’Z + X’YZ XZ Y’Z X’YZ XZ . Y’Z . X’YZ = XZ + Y’Z + X’YZ F = XZ + Y’Z + X’YZ
- 9. Implement the following POS function F = (X+Z) (Y’+Z) (X’+Y+Z)
- 10. Standard SOP • Write minterms of BC + A. Solution : BC + A Write terms • XBC + AXX Add Xs where letters are missing • A’BC, ABC Vary Xs in XBC • AB’C’, AB’C, ABC’, ABC Vary Xs in AXX • Thus BC + A = A’ BC +ABC+ AB’C’ + AB’C + A B C’ + A B C • or BC + A = A’BC + ABC + AB’C’ + AB’C + ABC’
- 11. Standard SOP F(A.B,C) = Σm(3, 7, 4, 5, 6)
- 12. Complementary Nature of Min & Max terms
- 13. F=Σm( 15, 11, 14, 13, 12) F=П M (0,1,2,3,4,5,6,7,8,9,10)
- 14. Standard POS Write full form of the expression Y(A,B,C) = П M (0, 1, 3, 4)
- 15. Standard SOP & POS • F= B’+A’C • X0X, 0X1 • (000, 001,100,101), (001,011) • F= Ʃm(0,1,4,5,3,8) F(A,B,C)=(A+B)(A’+C) • (0+0+X) (1+X+0)=(0+0+0),(0+0+1),(1+0+0),(1+1+0) • F=ПM(1,0,4,6)
- 16. Happy Learning