Binary logic. Dr.AbuArqoub


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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. minmax 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 (Minterms) Any Boolean function can be expressed as a sum of minterms ( 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 minterm (m j ) or Standard of Product. 36
37 Boolean Functions (Minterms) 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 (Minterms) 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 (Minterms) 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 (Maxterms) A Boolean function can be expressed as the product of maxterms ( 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 maxterm (M j ) or Standard Sum. 4
41 Boolean Functions (Maxterms) 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 (Maxterms) 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 (Maxterms) 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 (Maxterms) 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 minterms or maxterms 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 minterms (SOM) 4. Express F (X, Y) in product of maxterms (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 Maxterms (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 minterms (SOM). 2. Product of maxterms (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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