1. Chapter 3 Boolean Algebra and Logic Gate (Part 2)

2. 1. Identity Elements 2. Inverse Elements 1. A = A A. A = 0 0 + A = A A + A = 1 3. Idempotent Laws 4. Boundess Laws A + A = A A + 1 = 1 A. A = A A. 0 = 0 5. Distributive Laws 6. Order Exchange Laws A. (B + C) = A.B + A.C A. B = B. A A + (B. C) = (A+B). (A+C) A + B = B + A 7. Absorption Laws 8. Associative Laws A + (A. B) = AA + (B + C) = (A + B) + C A. (A + B) = AA. (B. C) = (A. B). C 9. Elimination Laws 10. De Morgan Theorem     A + (A. B) = A + B (A + B) = A. B     A. (A + B) = A. B (A. B) = A + B Basic Theorems of Boolean Algebra

3. Truth Table

4. ABCQ 0010 1010 0110 1101 A = 0, B = 0, C = 1, Q = 0 A = 1, B = 0, C = 1, Q = 0 A = 0, B = 1, C = 1, Q = 0 A = 1, B = 1, C = 0, Q = 1 Truth Table

5. ABQ 000 101 010 110

6. pqr p. q q q + r q + r 000 01 1 001 01 1 010 00 0 011 00 1 100 0 1 1 101 0 11 110 10 0 111 1 0 1

7. pqp+qp.(p+q)p.qp.p(p.p)+(p.q) 0000000 0110000 1011011 1111111 Logically equivalent p.(p+q) = (p.p)+(p.q)

8. Relationship Between Boolean Function and Logic Circuit A B F A.B = AB C D C + D = AB + C + D

9. G = A. (B + C + D) A B C D C + D B + C + D

10. A B Q A AB B = AB + B Produce a truth table from the logic circuit ABAABQ 00100 01111 10000 11001

11. Elimination Laws:A.(A + B) = A.B Proof using truth table. ABAA + B A.BA.(A + B) 001100 011100 100000 110111 Draw the logic circuit for output = A.(A + B)

12. A A A.(A+B) A A + B Draw the logic circuit for output = A.(A + B) B

13. De Morgan Theorem : (A + B) = A. B Proof using truth table. ABA+BAB A.B(A + B) 0001111 0111000 1010100 1110000 Draw the logic circuit for output = (A + B)

14. Karnaugh Map A graphical way of depicting the content of a truth table where the adjacent expressions differ by only one variable For the purposes simplification, the Karnaugh map is a convenient way of representing a Boolean function of a small number (up to four) of variables The map is an array of 2 n squares, representing all possible combination of values of n binary variables Example: 2 variables, A and B A B 0001 1011 B A B A B A B A 10 1 0

15. The number of squares in Karnaugh map depends on the number of variables e.g., if 2 variables, A, and B, there are 2 2 = 4 squares in the Karnaugh map e.g., if 3 variables, A, B, and C, there are 2 3 =8 squares 000001 010011 110111 100101 AB C C C or 000 001 AB C C C

16. 00000001 0100 1100 1000 AB C D A B C D A B CD A B C D 4 variables, A, B, C, D  2 4 = 16 squares

17. 000010110100 001011111101 AB C C C 000001 010011 110111 100101 ABC C 00011110 0101 00 01 11 10 01 The adjacent differ by only one variable List combinations in the order 00, 01, 11, 10 C

18. ABCF 0001 0010 0100 0111 1001 1011 1100 1110 Truth Table Karnaugh Map 11 11 0 00 11 11 0 BC A 0 1 A A How to create Karnaugh Map 1.Place 1 in the corresponding square

19. 11 0 00 11 11 0 AB F = AB + AB A B Karnaugh Maps to Represent Boolean Functions

20. 11 1 0 00 11 11 0 BC A A A ABC F=ABC + ABC + ABC

21. 1 1 1 0 00 11 11 0 CD AB ABCD A B 00 01 11 10 ABCD F = + ABCD +

22. ABCF 0001 0010 0100 0111 1001 1011 1100 1110 Truth Table Karnaugh Map 11 11 0 00 11 11 0 BC A 0 1 A A Create Karnaugh Map 1.Place 1 in the corresponding square

23. 2.Group the adjacent squares: Begin grouping square with 2 n-1 for n variables e.g. 3 variables, A, B, and C 2 3-1 = 2 2 = 4 = 2 1 = 2 = 2 0 = 1 11 11 0 00 11 11 0 BC A 0 1 A A AB BC ABC F = BCABABC + + Represent Boolean Functions

24. 11 111 0 00 11 11 0 BC A 0 1 A A 3 variables: 2 3-1 = 2 2 = 4 2 2-1 = 2 1 = 2 2 1-1 = 2 0 = 1 C AB F = C + AB

25. 1 1111 0 00 11 11 0 BC A 0 1 A A 3 variables: 2 3-1 = 2 2 = 4 2 2-1 = 2 1 = 2 2 1-1 = 2 0 = 1 A BC F = A + BC

26. 11 1 1 111 AB 01 00 CD 11 10 11 4 variables, A, B, C, D  2 4-1 = 2 3 = 8 ( maximum ); 2 2 = 4; 2 1 = 2; 2 0 = 1 ( minimum ); CD +BDABC+F =

27. 11 AB 01 00 CD 11 10 11 ABD F = 1 1 AB 01 00 CD 11 10 11 F = BCD

28. 11 11 AB 01 00 CD 11 10 11 BC F = 1111 1111 AB 01 00 CD 11 10 11 F = A

29. 11 11 11 11 AB 01 00 CD 11 10 11 F = D