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# 4. Boolean Function and Their Representation.pptx

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SOP POS, Minterm and Maxterm
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# 4. Boolean Function and Their Representation.pptx

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### 4. Boolean Function and Their Representation.pptx

1. 1. Boolean Functions and their Representation
2. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 19. Contd. 7. Expand A’+B+CA to minterms and maxterms 8. Expand A(A’+B) (A’+B+C’) to maxterms and minterms
20. 20. Thank You!!