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- 1. Boolean Functions and their Representation
- 2. Normal Form • A Boolean variable in true form or in complemented form is called a literal. Thus a,b,c, (a’,b’,c’) and so on are literals. • The Boolean product of two or more literals is called a product term. • The Boolean sum of two or more literals is called a sum term. • When a Boolean function appears as a sum of several product terms, it is said to be expressed as a sum of products (SOP). The SOP form is also called the disjunctive normal form (DNF). • When a Boolean function appears as a product of several sum terms, it is said to be expressed as a product of sums (POS). The POS form is also called the conjunctive normal form (CNF).
- 3. Examples Example 1 f(a,b,c,d)=a’+b.c’+c.d is an expression in SOP form or DNF. Example 2 f(a,b,c,d)=(a+b).(b+c+d) is an expression in POS form or CNF.
- 4. Standard and Non-Standard Form • The standard form of the Boolean function is when it is expressed in sum of the products or product of the sums fashion. For example: Y =AB + BC + AC or Y = (A + B + C)(A + B′ + C)(A + B + C′) are the standard forms. • However, Boolean functions are also sometimes expressed in nonstandard forms like F = (AB + CD)(A′B′ + C′D′), which is neither a sum of products form nor a product of sums form. However, the same expression can be converted to a standard form with help of various Boolean properties, as: F = (AB + CD)(A′B′ + C′D′) = A′B′CD + ABC′D′
- 5. Minterm and Maxterm • Minterm: a product term in which all the variables appear exactly once, either complemented or uncomplemented. • Maxterm: a sum term in which all the variables appear exactly once, either complemented or uncomplemented
- 6. Minterm • Represents exactly one combination in the truth table. • Denoted by mj, where j is the decimal equivalent of the minterm’s corresponding binary combination (bj). • A variable in mj is complemented if its value in bj is 0, otherwise is uncomplemented. • Example: Assume 3 variables (A,B,C), and j=3. Then, bj = 011 and its corresponding minterm is denoted by mj = A’BC
- 7. Maxterm • Denoted Represents exactly one combination in the truth table. • by Mj, where j is the decimal equivalent of the maxterm’s corresponding binary combination (bj). • A variable in Mj is complemented if its value in bj is 1, otherwise is uncomplemented. • Example: Assume 3 variables (A,B,C), and j=3. Then, bj = 011 and its corresponding maxterm is denoted by Mj = A+B’+C’
- 8. Truth Table Notation for Minterms and Maxterms x y z Minterm Maxterm 0 0 0 x’y’z’ = m0 x+y+z = M0 0 0 1 x’y’z = m1 x+y+z’ = M1 0 1 0 x’yz’ = m2 x+y’+z = M2 0 1 1 x’yz = m3 x+y’+z’= M3 1 0 0 xy’z’ = m4 x’+y+z = M4 1 0 1 xy’z = m5 x’+y+z’ = M5 1 1 0 xyz’ = m6 x’+y’+z = M6 1 1 1 xyz = m7 x’+y’+z’ = M7
- 9. Canonical Form • When each of the terms of a Boolean functions is expressed either in SOP or POS form has all the variables in it it is said to be expressed in canonical form. • The canonical SOP form is called the disjunctive canonical form (DCF) and the canonical POS form is called the conjunctive canonical form (CCF).
- 10. Conversion of SOP from Standard to Canonical Form • Expand non-canonical terms by inserting equivalent of 1 in each missing variable x: (x + x’) = 1 • Remove duplicate minterms • f1(a,b,c) = a’b’c + bc’ + ac’ = a’b’c + (a+a’)bc’ + a(b+b’)c’ = a’b’c + abc’ + a’bc’ + abc’ + ab’c’ = a’b’c + abc’ + a’bc + ab’c’
- 11. Conversion of POS from Standard to Canonical Form • Expand noncanonical terms by adding 0 in terms of missing variables (e.g., xx’ = 0) and using the distributive law. • Remove duplicate maxterms • f1(a,b,c) = (a+b+c)•(b’+c’)•(a’+c’) = (a+b+c)•(aa’+b’+c’)•(a’+bb’+c’) = (a+b+c)•(a+b’+c’)•(a’+b’+c’)• (a’+b+c’)•(a’+b’+c’) = (a+b+c)•(a+b’+c’)•(a’+b’+c’)•(a’+b+c’)
- 12. Examples Express the following functions in canonical forms a) f1=a.b’.c+b.c’+a.c b) f2=(a+b).(b+c’) Solutions: a) f1 =a.b’.c+b.c’+a.c =a.b’.c+(a+a’).b.c’+a.(b+b’).c =a.b’.c+a. b.c’ +a’.b.c’+a.b.c+a.b’.c b) f2 = (a+b)(b+c’) = (a+b+c.c’)(a.a’+b+c’) = (a+b+c).(a+b+c’).(a+b+c’).(a’+b+c’) = (a+b+c).(a+b+c’).(a’+b+c’) f1 is expressed in DCF, f2 is expressed in CCF.
- 13. Shorthand: ∑ and ∏ • f1(a,b,c) = ∑ m(1,2,4,6), where ∑ indicates that this is a sum-of-products form, and m(1,2,4,6) indicates that the minterms to be included are m1, m2, m4, and m6. • f1(a,b,c) = ∏ M(0,3,5,7), where ∏ indicates that this is a product-of-sums form, and M(0,3,5,7) indicates that the maxterms to be included are M0, M3, M5, and M7. • Since mj = Mj’ for any j, f1(a,b,c) = ∑ m(1,2,4,6) = ∏ M(0,3,5,7)
- 14. Expansion of Boolean Expression to SOP Form Expand A’ + B’ to minterms and maxterms: A’+B’= A’(B+B’) + B’ (A+A’) = A’B+A’B’ + AB’+A’B’ = A’B+A’B’+AB’ = 01 + 00 + 10 = ∑ m(0,1,2) ∏ M(3)
- 15. Contd. Expand A+BC’+ABD’ to minterms and maxterms: A+BC’+ABD’ = A(B+B’) (C+C’) (D+D’) + BC’(A+A’) (D+D’) + ABD’ (C+C’) =ABCD+ABCD’+ABC’D+ABC’D’+AB’CD+AB’CD’+AB’C’D+ AB’C’D’ + ABC’D+ABC’D’+A’BC’D+AB’CD’ + ABCD’+ABC’D’ = ∑ m(4,5,8,9,10,11,12,13,14,15) ∏ M(0,1,2,3,6,7)
- 16. Expansion of Boolean Expression to POS Form Expand A(B’+A)B to maxterms and minterms: A(B’+A)B = (A+B)(A+B’) (A+B’) (A+B) (A’+B) = (A+B) (A+B’) (A’+B) = (00) (01) (10) = ∏ M(0,1,2) ∑ m(3)
- 17. Activity 1 1. The terms in SOP are called ___________ a) max terms b) min terms c) mid terms d) sum terms 1. Which of the following is an incorrect SOP expression? a) x+x.y b) (x+y)(x+z) c) x d) x+y
- 18. Contd. 3. The corresponding min term when x=0, y=0 and z=1. a) x.y.z’ b) X’+Y’+Z c) X+Y+Z’ d) x’.y’.z 4. Convert to canonical SOP function F = x y + x z + y z 5. Convert to canonical POS form F = (A’ + B + C) * (B’ + C + D’) * (A + B’ + C’ + D)
- 19. Contd. 7. Expand A’+B+CA to minterms and maxterms 8. Expand A(A’+B) (A’+B+C’) to maxterms and minterms
- 20. Thank You!!