# Binary logic. Dr.Abu-Arqoub

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1 Binary logic Binary logic deals with variables like (a, b, c,, x, y) that take on two discrete values (, ) and with operations that assume logic meaning ( AND, OR, NOT) Truth table is a table of all possible combinations of the variables showing the relation between the values that the variables may take and the result of operation. Ex. AND (. ) z= x. y z= x and y z=xy z=x and y

2 AND GATE x y Z=x.y off on X Y ~` 2

3 OR GATE OR (+) X Y Z=X+Y Z=X+Y Z= x or y 3

4 OR GATE X ~` Y 4

5 NOT GATE NOT (Bar), (Prime) Z = X, Z = X NOT X, X Bar, X Prime X Z=X 5

6 NOT GATE ` ~ X 6

7 Boolean algebra and logic gates Boolean algebra is an algebraic structure defined on a set of elements B together with two binary operators +(OR) and.(and) providing the following postulates:. closed with (مغلقة) respect to operator +(OR) a + b=c a, b, c Є B and closer with respect to operator.(and) a. b=c a, b, c Є B 2. An identity (محايد) element with respect to +(OR) by and an identity element with respect to.(and) by x+=+x=x x. =. x=x 3. (تبديلي) Commutative with respect to +: x +y=y + x Commutative with respect to. : x. y=y. x (توزيعي) 4. Distributive. (AND) is distributive over +(OR): x (y + z)=x. y + x. z +(OR) is distributive over.(and): x +(y. z)=(x + y). (x + z) 7

8 Boolean algebra and logic gates 5. For every element x Є B, there exist an element x Є B called the complement of x such that : x+x = and x.x = 6. There exist at least two elements x, y Є B such that x y 8

9 Boolean algebra and logic gates Example : a two valued Boolean algebra is defined on a set of two elements B={,}, with rules for the two binary operators (+) & (.) such. Closed results or and (,) Є B 2. It has the two identity elements for (+) and for (.) 3 commutative since x+y=y+x & x.y=y.x next slide 5. For each x there is x : x+x = += += 6. Two elements (,) with x y x.y x+y x x+x x.x 9

10 Boolean algebra and logic gates 4. Distributive since x.(y+z)=x.y+x.z xy+xz x.(y+z) Y+z x.z x.y z y x

11 Basic theorems and properties of Boolean algebra Duality :(ازدواجية) in duality leave the element (variables) of the set of B the same, then change the (+) by (.) operators, the (.) by (+) & replace the s by zeros, and zeros by ones. Ex x+=x by duality x.=x

12 Postulates & Theorems of Boolean Algebra. X + = X Dual X. = X 2. X + X = X. X = 3. X + X = X X. X = X 4. X + = X. = 5. (X ) = X 6. X + Y = Y + X X. Y = Y. X 7. X + ( Y + Z ) = ( X + Y ) + Z X. ( Y. Z ) = ( X. Y ). Z 2

13 Postulates & Theorems of Boolean Algebra 8. X. ( Y + Z ) = X. Y + X. Z X + ( Y. Z ) = X + Y. X + Z 9. ( X + Y ) = X. Y (De Morgan Theorem) ( X. Y ) = X + Y. X + XY = X X. ( X + Y ) = X X. ( X + Y ) = X. X + X. Y = X + XY = X ( + Y ) = X. = X 3

14 Postulates & Theorems of Boolean Algebra 4

15 Boolean Functions A Boolean function is an expression formed with binary variables, the two binary operators OR & AND, the unary operator NOT, parentheses, and equal sign. A Boolean function may be represented in a truth table. 5

16 Boolean Functions Ex: Suppose we have three Boolean functions: F, F 2, and F 3. F = F 2 & F F 3 X Y F (X, Y) F 2 (X, Y) F 3 (X, Y) 6

17 Boolean Functions Two functions of n variables are said to be equal if they have the same value for all possible 2 n combinations of n variables. The complement of a function F is F and it s obtained from an interchange of s for s and s for s in the value of F. 7

18 Boolean Functions Ex: Note: (F ) (F 2 ),(F dual F2) but (F) =(F3) F 3 (X, Y, Z) = X + Y + Z F 2 (X, Y, Z) = X + Y + Z F (X, Y, Z) = X. Y. Z Z Y X 8

19 Examples on functions Draw a logic circuit that represents the following function: F=x+y z 9

20 Examples on functions Draw a logic circuit that represents the following function: F=xy +x z 2

21 Examples on functions Draw a logic circuit that represents the following functions: 2

22 Simplification of Boolean Functions Simplify the following Boolean functions:. x(x +y) 2. x+x y 3. (x+y).(x+y ) 4. xy+x z+yz 22

23 Simplification of Boolean Functions Simplify the following Boolean functions: F=x y z+x yz+xy =x z(y +y)+xy =x z.+xy = x z+xy 23

24 Complement of a Boolean function To find the complement of any Boolean function we can use:. Direct method (using Demorgan law) 2. Dual of the function and complement of each literal. 3. min-max terms 24

25 Complement a function using Demorgan law In this method. the relation between elements changed from AND to OR and vice versa 2. each element individually inversed Examples: find the complement of the following functions:. F=x yz +x y z F 25

26 Examples 2. F 3. F2=x.(y z+yz ) F2 =x +(y z+yz ) =x +(y z).(yz ) =x +(y+z ).(y +z) 26

27 Complement a function using Duality In this method :. take the dual of the function 2. take the complement of each literal Example: F=x.(y z+yz ) find F. dual x+(y +z).(y+z ) 2. complement of each litiral x +(y+z ).(y +z) 27

28 Example Given F (X, Y, Z) = XY Z + XYZ, Find F using:.de Morgan s theorem 2.Dual & then complement the variables. 28 Continue

29 Example. Using De Morgan s theorem F (X, Y, Z) = (XY Z + XYZ ) = (X + Y + Z ). (X + Y + Z) 2. Using dual & then complement the variables Duality of F (X + Y + Z). (X + Y + Z ) Then complement the variables F (X, Y, Z) = (X + Y + Z ). (X + Y + Z) 29

30 Digital logic gates 3

31 Digital logic gates 3

32 NAND (NOT AND) Gate F (X, Y, Z) F (X, Y, Z) We can describe it using the NAND gate as follows: 32 Continue

33 NAND Gate 33 Continue

34 NOR (NOT OR) Gate F 2 (X, Y, Z) F 2 (X, Y, Z) We can describe it using the NOR gate as follows: 34 Continue

35 NOR Gate Note: A simple way for deriving the complement of a function is to take the dual & complement each literal. 35

36 Boolean Functions (Min-terms) Any Boolean function can be expressed as a sum of min-terms ( sum : ORing of terms) and the function will be in sum of Product (SOP). n variables forming an AND term with each variable being primed (if equal to ) or unprimed (if equal to ) providing 2 n possible combinations called min-term (m j ) or Standard of Product. 36

37 Boolean Functions (Min-terms) Ex: X Y m j F(X, Y) = m + m2 m m m 2 m 3 m = X. Y m 2 = X. Y m = X. Y m 3 = X. Y F(X, Y) = m + m 2 = X Y + XY = (, 2) (SOP) (Canonical form) 37

38 Boolean Functions (Min-terms) Ex: X Y Z m j F(X, Y, Z) m m m 2 m 3 m 4 m 5 m 6 m 7 38

39 Boolean Functions (Min-terms) m = X. Y. Z m = X. Y. Z m 2 = X. Y. Z m 3 = X. Y. Z m 4 = X. Y. Z m 5 = X. Y. Z m 6 = X. Y. Z m 7 = X. Y. Z F(X, Y, Z) = m + m 3 + m 5 + m 6 + m 7 = (, 3, 5, 6, 7) 39

40 Boolean Functions (Max-terms) A Boolean function can be expressed as the product of max-terms ( Product = ANDing of the terms) and the function would be then in the product of Sum (POS). n variables forming an OR term with each variable being primed (if equal to ) or unprimed (if equal to ) providing 2 n possible combinations called max-term (M j ) or Standard Sum. 4

41 Boolean Functions (Max-terms) Ex: X Y M j F ( X, Y ) M M M 2 M 3 M = X + Y M 2 = X + Y M = X + Y M 3 = X + Y F(X, Y) = M. M 3 = (X + Y). (X + Y ) = (,3) (POS) 4

42 Boolean Functions (Max-terms) Note that (,2) is a complement of (,2) (m j ) = M j ( M j) = m j (m ) = M (X + Y) = X. Y If given F(x,Y)= m + m2 = X Y + XY = (, 2) = M. M3 = (X + Y). (X + Y )= (,3) then F (x,y)=m+m3=x y +xy= (, 3) =M.M2=(x+y )(X +y)= (,2) 42

43 Conversion between canonical forms m M F 3,5,6,7,,2,4 F,,2,4 3,5,6,7 43

44 Boolean Functions (Max-terms) Ex: X Y Z M j F (X, Y, Z) M M M 2 M 3 M 4 M 5 M 6 M 7 44

45 Boolean Functions (Max-terms) M = X + Y + Z M = X + Y + Z M 2 = X + Y + Z M 3 = X + Y + Z M 4 = X + Y + Z M 5 = X + Y + Z M 6 = X + Y + Z M 7 = X + Y + Z F (X, Y, Z) = M. M 2. M 4 F (X, Y, Z) = (,2,4) 45

46 Boolean Functions To convert from one Canonical form to another, interchange the symbols with and list those numbers missing from the original form. The total number of min-terms or max-terms is 2 n where n is the number of binary variables in the function. Ex: Given F(X, Y) = (, 3), Find F(X, Y) = (?) F(X, Y) = (, 2) (POS) 46

47 Boolean Functions Ex: Given F(X, Y, Z) = (, 3, 7), Find F(X, Y, Z) = (?) F(X, Y, Z) = (, 2, 4, 5, 6) (SOP) 47

48 Boolean Functions If the Boolean function is not in SOP, it can be made so by first expanding the expression into a sum of AND terms, then if any term missing one or more variables it s ANDed with an expression such as (X + X ) ( X is one of the missing variables). Ex: F(A, B, C) = B + A C, Find the SOP form of F(A, B, C) 48

49 Boolean Functions F(A, B, C) = B (A + A ) (C + C ) + A C (B + B ) = (BA + BA ) (C + C ) + A BC + A B C = ABC + ABC + A BC + A BC + A BC + A B C = m 7 + m 6 + m 3 + m 2 + m 3 + m = (, 2, 3, 6, 7) (SOP) F(A, B, C) = (, 4, 5) (POS) = (A + B + C ). (A + B + C ). (A + B + C ) 49

50 Boolean Functions If the Boolean function is not in POS, it can be made so by first expanding the expression into a product of OR terms (using the distributive rule), then if any term missing one or more variables it s ORed with an expression such as (X X ) ( X is one of the missing variables). Ex: F(A, B, C) = B + A C, Find the POS form of F(A, B, C) 5

51 F(A,B,C)=B+A C =(B+A )(B+C) =(B+A +CC )(B+C+AA ) =(B+A +C)(B+A +C )(B+C+A) (B+C+A ) =(A +B+C)(A +B+C )(A+B+C) =M4.M5.M =M.M4.M5 = (,4,5) 5

52 Boolean Functions Ex: X Y F (X, Y). Find F 2 (X, Y) = F (X, Y) 2. Find F 3 (X, Y) = (F ) (X, Y) 3. Express F (X, Y) in sum of min-terms (SOM) 4. Express F (X, Y) in product of max-terms (POM) 5. Express F 3 in SOM 6. Implement F (X, Y) using logic gates (Logic Diagram). 52

53 Boolean Functions X Y F (X, Y) F 2 (X, Y) F 3 (X, Y) Note: & 2 answers are shown in the table. 3. F (X, Y) = m + m 3 = X Y + XY = (, 3) 4. F (X, Y) = M. M 2 = (X + Y ). (X + Y) = (, 2) 53

54 Boolean Functions 5. F 3 (X, Y) = m + m 2 = X Y + XY = (, 2) 6. F (X, Y) = X Y + XY X Y 54

55 Boolean Functions Ex: Given F (X, Y, Z) = X YZ + XY Z + XYZ.. Find Truth Table for F (X, Y, Z). 2. Express F (X, Y, Z) in Product of Max-terms (POM). 3. Find F 2 (X, Y, Z) = (F ) (X, Y, Z). 4. Express F 2 (X, Y, Z) in SOM. 5. Implement F (X, Y, Z) using logic gates. 55

56 Boolean Functions. F (X, Y, Z) = m 3 + m 5 + m 7 = (3, 5, 7) F 2 (X, Y, Z) F (X, Y, Z) Z Y X 56

57 Boolean Functions 2. F (X, Y, Z) = (,, 2, 4, 6) = M. M. M 2. M 4. M 6 = (X + Y + Z). (X + Y + Z ). (X + Y + Z). (X + Y + Z). (X + Y + Z) 3. F 2 (X, Y, Z) = (F ) (X, Y, Z) =[ (X YZ) + (XY Z) + (XYZ) ] = (X + Y + Z ). (X + Y + Z ). (X + Y + Z ) = M 3. M 5. M 7 (Look at Truth Table) 57

58 Boolean Functions 4. F 2 (X, Y, Z) = m + m + m 2 + m 4 + m 6 = (X.Y. Z ) + (X.Y. Z) + 5. Logic Diagram (X.Y.Z ) + (X.Y.Z ). (X.Y.Z ) 58

59 Boolean Functions Logical Function can be expressed in Canonical forms:. Sum of min-terms (SOM). 2. Product of max-terms (POM). Or the logical Function can be expressed in the standard forms:. Sum of Product (SOP). 2. Product of Sum (POS). 59

60 Boolean Functions Ex: F(X, Y, Z) = X + XZ + XYZ Sum of Products Ex: F(X, Y, Z) = Y (X + Y) (X + Y + Z) Product of Sums Ex: Given F(X, Y, Z) = XY + XZ + YZ Find. The truth table of F(X, Y, Z). 2. Express F(X, Y, Z) in SOM. SOP 3. Implement F(X, Y, Z) using logic gates. 6

61 Boolean Functions. The answer is shown in the table. F(X, Y, Z) Z Y X 6

62 Boolean Functions 2. F(X, Y, Z) = XY (Z + Z ) + XZ (Y + Y ) + YZ (X + X ) = XYZ + XYZ + XYZ + XY Z + XYZ + X YZ = m 7 + m 6 + m 5 + m 3 3. Logic Diagram 62

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