Boolean Algebra The Fundamental postulates of Boolean Algebra
Boolean Algebra The Fundamental postulates of Boolean Algebra contains Ø 2 distinct values(0, 1) Ø 2 Binary operations—(AND, OR (or) *, +) Ø 1 Unary operator---(Complement ‘ )
There are 7 basic fundamental postulates given as 1. Identity 2. Commutative 3. Associative 4. Additive 5. Multiplicative 6. Complementation 7. Distributive
1. Identity Property A+A=A A. A=A 2. Commutative property A+B=B+A A. B=B. A
3. Associative Property (A+B)+C=A+(B+C) (A. B). C=A. (B. C)
6. Complementation property A+A’=1 A. A’=0 7. Distributive property A. (B+C)=(A. B)+(A. C) A+(B. C)=(A+B). (A+C) Note: All the above properties are restricted only for Boolean functions of digital circuits. These are incorrect for real one’s.
Complementation Laws The term complement simply means to invert i. e to change 0’s to 1’s and 1’s to 0’s. The five laws of complementation are Ø 0’=1 Ø 1’=0 ØIf A=0 then A’=1 ØIf A=1 then A’=0 ØA”=A Double complementation law
AND Laws ØA. 0=0 Null law ØA. 1=A Identity law Ø A. A=A ØA. A’=0
OR Laws ØA+0=A Null law ØA+1=1 Identity law Ø A+A=A ØA+A’=1
Basic Theorems Duality theorem or Duality Principal The principle of duality theorem says that, starting with a Boolean relation you can derive another Boolean relation by ØChanging each OR sign to an AND sign ØChanging each AND sign to an OR sign ØComplementing any 0 or 1 appearing in the expression.
Ex: A+A’=1 Apply the Duality Principal. A. A’=0 Ex: A. (B. C)=(A. B). C A+(B+C)=(A+B)+C A+A’=1 A. A’=0
De Morgan's Theorems De Morgan's suggested 2 theorems De Morgan's 1’st Theorem It states that the complement of a product is equal to sum of individual complements. (A. B)’=A’+B’
De Morgan's 1’st Theorem Verification A B A’ B’ A. B 0 0 1 1 1 0 0 1 (A. B)’ 1 1 1 0 A’+B’ 1 1 1 0
De Morgan's 2’nd Theorem It states that the complement of a sum is equal to product of individual complements. (A+B)’=A’. B’
De Morgan's 2’nd Theorem Verification A B A’ B’ A+B 0 0 1 1 1 0 0 1 1 1 (A+B)’ A’. B’ 1 0 0 0
Redundant Literal Rule (RLR) 1. A+A’B=A+B 2. A(A’+B)=AB Absorption Laws 1. A+A. B=A 2. A(A+B)=A
State and prove the associative law (A+B)+C=A+(B+C) (A. B). C=A. (B. C)
Steps to reducing the Boolean expression ØMultiply all variables necessary to remove parentheses. ØLook for identical terms. Only one of those terms be retained and all others dropped. Ex: AB+AB+AB=AB ØLook for variable and its negation in the same term, this term can be dropped. Ex: 1. A. BB’=A. 0=0 2. ABCC’=AB. 0=0
ØLook for pairs of terms that are identical except for one variable which may be missing in one of the terms, the large term can be dropped. Ex: ABC’D’+ABC’= ABC’(D’+1)= ABC’. 1= ABC’ ØLook for pairs of terms which are having the same variables, with one or more variables complemented. If variable in one term of such a pair is complemented while in the second term it is not, such terms can be combined in to a single term with that variable dropped. Ex: 1. ABC’D’+ABC’D= ABC’(D’+D)= ABC’. 1= ABC’ 2. AB(C+D)+AB(C+D)’= AB((C+D)+(C+D)’)= AB. 1= AB
Digital Logic Gates The gate is a digital circuit with one or more input voltages but only one out put voltage The types of logic gates available are 1. NOT 2. AND 3. OR 4. NAND 5. NOR 6. Ex-OR 7. Ex-NOR
NOTE: The operation of the logic gate can be easily understood with the help of Truth Table: A Truth table is a table that shows all the input-output possibilities of a logic circuit. ie the truth table indicates the outputs for different possibilities of the input.
Alternative Gate representation ØBubbled NOR gate is nothing but AND gate ØBubbled NAND gate is nothing but OR gate
ØBubbled OR gate is nothing but NAND gate ØBubbled AND gate is nothing but NOR gate
Draw the circuit shown in below using alternative symbols.
Universal gates. NAND and NOR gates are called universal gates because we can implement or realise any basic gate using these gates. NAND Realisation. NAND to NOT AND OR NOR
NOR Realisation. NOT NOR to AND OR NAND
Switching Functions/ Boolean Functions are used to describe Boolean expressions. Ex: If the Boolean expression (A+B’)C is used to describe the function f, then Boolean Function can be written as f(A, B, C)=(A+B’)C Or f=(A+B’)C
Boolean Functions are represented in two forms. ØSum Of Product forms (SOP) ØProduct Of Sum forms (POS) Sum Of Product forms (SOP): A sum of product is a group of product terms OR ed together. Ex: f(A, B, C)=ABC+AB’C’ Ex: f(P, Q, R, S)=P’Q+QR+RS
Product Of Sum forms (POS): A product of sum form is a group of sum terms AND ed together. Ex: f(A, B, C)=(A+B). (B’+C) Ex: f(P, Q, R, S)=(P+Q). (R+S’). (P+S)
Canonical forms(Standard SOP and POS forms): Standard SOP form or Minterm canonical form: If each term in SOP form contains all the literals or variables then the SOP form is known as Standard or Canonical SOP form. Each individual term in the Standard SOP form is called “minterm”. Ex: f(A, B, C)=AB’C+ABC+A’BC’
Standard POS form or Maxterm canonical form: If each term in POS form contains all the literals or variables then the POS form is known as Standard or Canonical POS form. Each individual term in the Standard POS form is called “Maxterm”. Ex: f(A, B, C)=(A+B+C). (A+B’+C). (A’+B’+C’)
Steps to convert SOP to Standard SOP form. ØFind the missing literal in each product term if any. ØAND each product term with (missing literal + its complement). ØExpand the terms by applying distributive law and record the literals in the product terms. ØReduce the expression by omitting repeated product terms if any. Because A+A=A.
Steps to convert POS to Standard POS form. ØFind the missing literal in each sum term if any. ØOR each product term with (missing literal * its complement). ØExpand the terms by applying distributive law and record the literals in the sum terms. ØReduce the expression by omitting repeated sum terms if any. Because A. A=A.
M-Notations: Minterms and Maxterms Variables ABC 000 001 010 011 100 101 110 111 Minterms (mi) A’B’C’=m 0 A’B’C=m 1 A’BC’=m 2 A’BC=m 3 AB’C’=m 4 AB’C=m 5 ABC’=m 6 ABC=m 7 Maxterms (Mi) A+B+C=M 0 A+B+C’=M 1 A+B’+C=M 2 A+B’+C’=M 3 A’+B+C=M 4 A’+B+C’=M 5 A’+B’+C=M 6 A’+B’+C’=M 7
Shorthand notation of SOP & POS forms: Ex: Write the shorthand notation for the following Boolean function. f(A, B, C)=A’B’C’+A’B’C+A’BC+ABC’ Sol: f(A, B, C)=m 0+m 1+m 3+m 6 =Σm(0, 1, 3, 6) Σ=Sum of product terms
Ex: Write the shorthand notation for the following Boolean function. f(A, B, C)=(A+B+C’)(A+B’+C’)(A’+B’+C) Sol: f(A, B, C)=M 1 M 3 M 6 =ΠM(1, 3, 6) Π =Product of sum terms
Conversion from min to Max & Max to min: Ex: Convert the following min terms to max terms f(A, B, C)=m 0+m 1+m 3+m 4+m 6+m 7. Sol: f(A, B, C)=m 0+m 1+m 3+m 4+m 6+m 7 =Σm(0, 1, 3, 4, 6, 7) =ΠM(2, 5)
Ex: Convert the following max terms to min terms f(A, B, C, D)=M 1 M 3 M 5 M 7 M 9 M 11 M 13 M 15. =ΠM(1, 3, 5, 7, 9, 11, 13, 15) =Σm(0, 2, 4, 6, 8, 10, 12, 14
Ex: Simply the following three variable expression using Boolean algebra. Y=Σm(1, 3, 5, 7) Sol: Y=A’B’C+A’BC+AB’C+ABC =A’C(B’+B)+AC(B’+B) =A’C(1)+AC(1) =C(A’+A) =C
Ex: Convert the expression given below in to min terms using complementary property and simply the expression using Boolean algebra. Y=ΠM(3, 5, 7) Sol: Y=Σm(0, 1, 2, 4, 6) =A’B’C’+A’B’C+A’BC’+AB’C’+ABC’ =A’B’C’+A’BC’+A’B’C =B’C’(A’+A)+BC’(A’+A)+A’B’C =B’C’+BC’+A’B’C =C’(B’+B)+A’B’C =C’+A’B’ A+A’B=A+B
- Slides: 51