1. Boolean Algebra and Logic Gates 1 Computer Engineering (Logic Circuits) Lec. # 4 Dr. Tamer Samy Gaafar Dept. of Computer & Systems Engineering Faculty of Engineering Zagazig University

2. Boolean Algebra and Logic Gates 2 http://www.tsgaafar.faculty.zu.edu.eg Email: tsgaafar@yahoo.com Course Web Page

3. Announcements  Only 2 Students > 10 in Quizz.  1- Abdallah Salah Eldeen (13)  2- Mohamed Nagi Mahmoud (12)  Only 5 <2 Students < 10 in Quizz.  1- Mohamed Samir Abd-Ellatif (9)  2-Heba Allah Abdo Ali (8)  Every Lec. There will be a Quizz Boolean Algebra and Logic Gates 3

4. Lec. # 4 Binary Logic & Gates (Contd.) Boolean Algebra and Logic Gates 4

5. Boolean Algebra and Logic Gates 5 1. 3. 5. 7. 9. 11. 13. 15. 17. Commutative Associative Distributive DeMorgan’s 2. 4. 6. 8. X. 1 X = X. 00 = X. XX = 0 = Boolean Algebra 10. 12. 14. 16. X + YY + X = (X + Y)Z + X + (YZ)Z) += X(Y + Z)XYXZ += X + YX. Y = XYYX = (XY)ZX(YX(YZ)Z) = X+ YZ(X + Y)(X + Z)= X. YX + Y = X + 0 X = + X 11 = X + XX = 1 = X = X  An algebraic structure defined by a set B = {0, 1}, together with two binary operators (+ and ·) and a unary operator ( ) Idempotence Complement Involution Identity element

6. Boolean Algebra and Logic Gates 6 Some Properties of Boolean Algebra  Boolean Algebra is defined in general by a set B that can have more than two values  A two-valued Boolean algebra is also know as Switching Algebra. The Boolean set B is restricted to 0 and 1. Switching circuits can be represented by this algebra.  Sometimes, the dot symbol ‘  ’ (AND operator) is not written when the meaning is clear.

7. Boolean Algebra and Logic Gates 7 Boolean Operator Precedence  The order of evaluation is: 1.Parentheses 2.NOT 3.AND 4.OR  Consequence: Parentheses appear around OR expressions  Example: F = A(B + C)(C + D)

8. Boolean Algebra and Logic Gates 8 Boolean Algebraic Proof – Example 1  A + A · B = A (Absorption Theorem) Proof StepsJustification A + A · B = A · 1 + A · B Identity element: A · 1 = A = A · ( 1 + B) Distributive = A · 1 1 + B = 1 = A Identity element  Our primary reason for doing proofs is to learn: Careful and efficient use of the identities and theorems of Boolean algebra, and How to choose the appropriate identity or theorem to apply to make forward progress, irrespective of the application.

9. Boolean Algebra and Logic Gates 9  AB + AC + BC = AB + AC (Consensus Theorem) Proof StepsJustification = AB + AC + BC = AB + AC + 1 · BCIdentity element = AB + AC + (A + A) · BCComplement = AB + AC + ABC + ABCDistributive = AB + ABC + AC + ACBCommutative = AB · 1 + ABC + AC · 1 + ACBIdentity element = AB (1+C) + AC (1 + B)Distributive = AB. 1 + AC. 11+X = 1 = AB + ACIdentity element Boolean Algebraic Proof – Example 2

10. Boolean Algebra and Logic Gates 10 Useful Theorems  Minimization X Y + X Y = Y  Absorption X + X Y = X  Simplification X + X Y = X + Y  DeMorgan’s  X + Y = X · Y  Minimization (dual) (X+Y)(X+Y) = Y  Absorption (dual) X · (X + Y) = X  Simplification (dual) X · (X + Y) = X · Y  DeMorgan’s (dual)  X · Y = X + Y

11. Boolean Algebra and Logic Gates 11 Truth Table to Verify DeMorgan’s XYX·YX·YX+YXY X · YX·YX·YX+Y 0000111111 0101100011 1001010011 1111000000 X + Y = X · Y X · Y = X + Y  Generalized DeMorgan’s Theorem: X 1 + X 2 + … + X n = X 1 · X 2 · … · X n X 1 · X 2 · … · X n = X 1 + X 2 + … + X n

12. Boolean Algebra and Logic Gates 12 Complementing Functions  Use DeMorgan's Theorem: 1.Interchange AND and OR operators 2.Complement each constant and literal  Example: Complement F = F = (x + y + z)(x + y + z)  Example: Complement G = (a + bc)d + e G = (a (b + c) + d) e x  zyzyx

13. Boolean Algebra and Logic Gates 13 Expression Simplification  An application of Boolean algebra  Simplify to contain the smallest number of literals (variables that may or may not be complemented) = AB + ABCD + A C D + A C D + A B D = AB + AB(CD) + A C (D + D) + A B D = AB + A C + A B D = B(A + AD) +AC = B (A + D) + A C (has only 5 literals)  DCBADCADBADCABA

14. Boolean Algebra and Logic Gates 14 Next … Canonical Forms  Minterms and Maxterms  Sum-of-Minterm (SOM) = Sum-of-Products (SOP)  Canonical Sum  Product-of-Maxterm(POM)= Product of Sum (POS)  Canonical Form

15.  Representation of Complements of Functions  Conversions between Representations Boolean Algebra and Logic Gates 15

16. Boolean Algebra and Logic Gates 16 Minterms  Minterms are AND terms with every variable present in either true or complemented form.  Given that each binary variable may appear normal (e.g., x) or complemented (e.g., ), there are 2 n minterms for n variables.  Example: Two variables (X and Y) produce 2 x 2 = 4 combinations: (both normal) (X normal, Y complemented) (X complemented, Y normal) (both complemented)  Thus there are four minterms of two variables. YX XY YX YX x

17. Boolean Algebra and Logic Gates 17 Maxterms  Maxterms are OR terms with every variable in true or complemented form.  Given that each binary variable may appear normal (e.g., x) or complemented (e.g., x), there are 2 n maxterms for n variables.  Example: Two variables (X and Y) produce 2 x 2 = 4 combinations: (both normal) (x normal, y complemented) (x complemented, y normal) (both complemented) YX  YX  YX  YX 

18. Boolean Algebra and Logic Gates 18  Two variable minterms and maxterms.  The minterm m i should evaluate to 1 for each combination of x and y.  The maxterm is the complement of the minterm Minterms & Maxterms for 2 variables xyIndexMintermMaxterm 000m 0 = x yM 0 = x + y 011m 1 = x yM 1 = x + y 102m 2 = x yM 2 = x + y 113m 3 = x yM 3 = x + y

19. Boolean Algebra and Logic Gates 19 Minterms & Maxterms for 3 variables M 3 = x + y + zm 3 = x y z3110 M 4 = x + y + zm 4 = x y z4001 M 5 = x + y + zm 5 = x y z5101 M 6 = x + y + zm 6 = x y z6011 1 1 0 0 y 1 0 0 0 x 1 0 1 0 z M 7 = x + y + zm 7 = x y z7 M 2 = x + y + zm 2 = x y z2 M 1 = x + y + zm 1 = x y z1 M 0 = x + y + zm 0 = x y z0 MaxtermMintermIndex Maxterm M i is the complement of minterm m i M i = m i and m i = M i

20. Boolean Algebra and Logic Gates 20 Purpose of the Index  Minterms and Maxterms are designated with an index  The index number corresponds to a binary pattern  The index for the minterm or maxterm, expressed as a binary number, is used to determine whether the variable is shown in the true or complemented form  For Minterms: ‘1’ means the variable is “Not Complemented” and ‘0’ means the variable is “Complemented”.  For Maxterms: ‘0’ means the variable is “Not Complemented” and ‘1’ means the variable is “Complemented”.

21. Boolean Algebra and Logic Gates 21 Standard Order  All variables should be present in a minterm or maxterm and should be listed in the same order (usually alphabetically)  Example: For variables a, b, c: Maxterms (a + b + c), (a + b + c) are in standard order However, (b + a + c) is NOT in standard order (a + c) does NOT contain all variables Minterms (a b c) and (a b c) are in standard order However, (b a c) is not in standard order (a c) does not contain all variables

22. Boolean Algebra and Logic Gates 22 Sum-Of-Product (SOP)  Sum-Of-Product (SOP) = Canonical Sum: Sum of minterms of entries that evaluate to ‘1’ xyzFMinterm 0000 0011m 1 = x y z 0100 0110 1000 1010 1101m 6 = x y z 1111m 7 = x y z F = m 1 + m 6 + m 7 = ∑ (1, 6, 7) = x y z + x y z + x y z Focus on the ‘1’ entries

23. Sum-Of-Product (SOP) F = m1 + m6 + m7 = x y z + x y z + x y z Boolean Algebra and Logic Gates 23 Canonical Sum

24. Sum-Of-Product (SOP) F =∑ (1, 6, 7) Boolean Algebra and Logic Gates 24 On-Set

25. Boolean Algebra and Logic Gates 25 + a b c d Sum-Of-Products Examples  F(a, b, c, d) = ∑(2, 3, 6, 10, 11)  F(a, b, c, d) = m 2 + m 3 + m 6 + m 10 + m 11  G(a, b, c, d) = ∑(0, 1, 12, 15)  G(a, b, c, d) = m 0 + m 1 + m 12 + m 15 + a b c da b c d+ a b c d a b c d+ a b c d

26. Boolean Algebra and Logic Gates 26 Product-Of-Sum (POS)  Product-Of-Sum (POS) = Canonical Product: Product of maxterms of entries that evaluate to ‘0’ xyzFMaxterm 0001 0011 0100M 2 = (x + y + z) 0111 1000M 4 = (x + y + z) 1011 1100M 6 = (x + y + z) 1111 Focus on the ‘0’ entries F = M 2 ·M 4 ·M 6 = ∏ (2, 4, 6) = (x+y+z) (x+y+z) (x+y+z)

27. Boolean Algebra and Logic Gates 27 Product-Of-Sum (POS)  Canonical Product: xyzFMaxterm 0001 0011 0100M 2 = (x + y + z) 0111 1000M 4 = (x + y + z) 1011 1100M 6 = (x + y + z) 1111 F = M 2 ·M 4 ·M 6 = (x+y’+z) (x’+y+z) (x’+y’+z)

28. Boolean Algebra and Logic Gates 28 Product-Of-Sum (POS)  Off-Set xyzFMaxterm 0001 0011 0100M 2 = (x + y + z) 0111 1000M 4 = (x + y + z) 1011 1100M 6 = (x + y + z) 1111 F = M 2 ·M 4 ·M 6 = ∏ (2, 4, 6)

29. Boolean Algebra and Logic Gates 29  F(a, b, c, d) = ∏(1, 3, 6, 11)  F(a, b, c, d) = M 1 · M 3 · M 6 · M 11  G(a, b, c, d) = ∏(0, 4, 12, 15)  G(a, b, c, d) = M 0 · M 4 · M 12 · M 15 Product-Of-Sum Examples (a+b+c+d)

30. Boolean Algebra and Logic Gates 30 Observations  We can implement any function by "ORing" the minterms corresponding to the ‘1’ entries in the function table. A minterm evaluates to ‘1’ for its corresponding entry.  We can implement any function by "ANDing" the maxterms corresponding to ‘0’ entries in the function table. A maxterm evaluates to ‘0’ for its corresponding entry.  The same Boolean function can be expressed in two canonical ways: Sum-of-Products (SOP) and Product-of-Sum (POS).  If a Boolean function has fewer ‘1’ entries then the SOP canonical form will contain fewer literals than POS. However, if it has fewer ‘0’ entries then the POS form will have fewer literals than SOP.

31. Boolean Algebra and Logic Gates 31 Converting to Sum-of-Products Form  A function that is not in the Sum-of-Minterms form can be converted to that form by means of a truth table  Consider F = y + x z xyzFMinterm 0001m 0 = x y z 0011m 1 = x y z 0101m 2 = x y z 0110 1001m 4 = x y z 1011m 5 = x y z 1100 1110 F = ∑(0, 1, 2, 4, 5) = m 0 + m 1 + m 2 + m 4 + m 5 = x y z + x y z + x y z + x y z + x y z

32. Boolean Algebra and Logic Gates 32 Converting to Product-of-Sum Form  A function that is not in the Product-of-Minterms form can be converted to that form by means of a truth table  Consider again: F = y + x z xyzFMinterm 0001 0011 0101 0110M 3 = (x+y+z) 1001 1011 1100M 6 = (x+y+z) 1110M 7 = (x+y+z) F = ∏(3, 6, 7) = M 3 · M 6 · M 7 = (x+y+z) (x+y+z) (x+y+z)

33. Boolean Algebra and Logic Gates 33 Conversions Between Canonical Forms F = m 1 +m 2 +m 3 +m 5 +m 7 = ∑(1, 2, 3, 5, 7) = x y z + x y z + x y z + x y z + x y z F = M 0 · M 4 · M 6 = ∏(0, 4, 6) = (x+y+z)(x+y+z)(x+y+z) xyzFMintermMaxterm 0000M 0 = (x + y + z) 0011m 1 = x y z 0101m 2 = x y z 0111m 3 = x y z 1000M 4 = (x + y + z) 1011m 5 = x y z 1100M 6 = (x + y + z) 1111m 7 = x y z

34. Boolean Algebra and Logic Gates 34 Algebraic Conversion to Sum-of-Products  Expand all terms first to explicitly list all minterms  AND any term missing a variable v with (v + v)  Example 1: f = x + x y(2 variables) f = x (y + y) + x y f = x y + x y + x y f = m 3 + m 2 + m 0 = ∑(0, 2, 3)  Example 2: g = a + b c(3 variables) g = a (b + b)(c + c) + (a + a) b c g = a b c + a b c + a b c + a b c + a b c + a b c g = a b c + a b c + a b c + a b c + a b c g = m 1 + m 4 + m 5 + m 6 + m 7 = ∑ (1, 4, 5, 6, 7)

35. Boolean Algebra and Logic Gates 35 Algebraic Conversion to Product-of-Sum  Expand all terms first to explicitly list all maxterms  OR any term missing a variable v with v · v  Example 1: f = x + x y(2 variables) Apply 2 nd distributive law: f = (x + x) (x + y) = 1 · (x + y) = (x + y) = M 1  Example 2: g = a c + b c + a b(3 variables) g = (a c + b c + a) (a c + b c + b)(distributive) g = (c + b c + a) (a c + c + b)(x + x y = x + y) g = (c + b + a) (a + c + b)(x + x y = x + y) g = (a + b + c) (a + b + c) = M 5. M 2 = ∏ (2, 5)

36. Boolean Algebra and Logic Gates 36 Function Complements  The complement of a function expressed as a sum of products is constructed by selecting the minterms missing in the sum-of-products canonical form  Alternatively, the complement of a function expressed by a Sum of Products form is simply the Product of Sum with the same indices  Example: Given F(x, y, z) = ∑ (1, 3, 5, 7) F(x, y, z) = ∑ (0, 2, 4, 6) F(x, y, z) = ∏ (1, 3, 5, 7)

37. Boolean Algebra and Logic Gates 37 Summary of Minterms and Maxterms  There are 2 n minterms and maxterms for Boolean functions with n variables.  Minterms and maxterms are indexed from 0 to 2 n – 1  Any Boolean function can be expressed as a logical sum of minterms and as a logical product of maxterms  The complement of a function contains those minterms not included in the original function  The complement of a sum-of-products is a product-of- sum with the same indices

38. Boolean Algebra and Logic Gates 38  Standard Sum-of-Products (SOP) form: equations are written as an OR of AND terms  Standard Product-of-Sums (POS) form: equations are written as an AND of OR terms  Examples: SOP: POS:  These “mixed” forms are neither SOP nor POS Standard Forms B C B A C B A   C · ) C B (A · B) (A  C) (A C) B (A  B) (A C A C B A 

39. Boolean Algebra and Logic Gates 39 Standard Sum-of-Products (SOP)  A sum of minterms form for n variables can be written down directly from a truth table. Implementation of this form is a two-level network of gates such that: The first level consists of n-input AND gates The second level is a single OR gate  This form often can be simplified so that the corresponding circuit is simpler.

40. Boolean Algebra and Logic Gates 40  A Simplification Example:  Writing the minterm expression: F = A B C + A B C + A B C + ABC + ABC  Simplifying: F = A B C + A (B C + B C + B C + B C) F = A B C + A (B (C + C) + B (C + C)) F = A B C + A (B + B) F = A B C + A F = B C + A  Simplified F contains 3 literals compared to 15 Standard Sum-of-Products (SOP) )7,6,5,4,1()C,B,A(F  

41. Boolean Algebra and Logic Gates 41 AND/OR Two-Level Implementation  The two implementations for F are shown below It is quite apparent which is simpler!

42. Boolean Algebra and Logic Gates 42 SOP and POS Observations  The previous examples show that: Canonical Forms (Sum-of-minterms, Product-of-Maxterms), or other standard forms (SOP, POS) differ in complexity Boolean algebra can be used to manipulate equations into simpler forms Simpler equations lead to simpler implementations  Questions: How can we attain a “simplest” expression? Is there only one minimum cost circuit? The next part will deal with these issues