ECE- 1551 DIGITAL LOGIC LECTURE 8: BOOLEAN FUNCTIONS Assistant Prof. Fareena Saqib Florida Institute of Technology Spring 2016, 02/11/2016.

Slides:



Advertisements
Similar presentations
Boolean Algebra and Logic Gates
Advertisements

ECE 331 – Digital System Design Boolean Algebra (Lecture #3) The slides included herein were taken from the materials accompanying Fundamentals of Logic.
ECE 301 – Digital Electronics Minterm and Maxterm Expansions and Incompletely Specified Functions (Lecture #6) The slides included herein were taken from.
ECE 331 – Digital System Design
Logic Gate Level Part 2. Constructing Boolean expression from truth table First method: write nonparenthesized OR of ANDs Each AND is a 1 in the result.
Chapter 2 – Combinational Logic Circuits Part 1 – Gate Circuits and Boolean Equations Logic and Computer Design Fundamentals.
CS 151 Digital Systems Design Lecture 6 More Boolean Algebra A B.
Boolean Algebra and Logic Gates
ECE 301 – Digital Electronics Boolean Algebra and Standard Forms of Boolean Expressions (Lecture #4) The slides included herein were taken from the materials.
28/06/041 CSE-221 Digital Logic Design (DLD) Lecture-5: Canonical and Standard forms and Integrated Circuites.
Chapter Two Boolean Algebra and Logic Gate
Chapter 2: Boolean Algebra and Logic Functions
Chapter 2 Boolean Algebra and Logic Gates
Logic Design CS221 1 st Term Boolean Algebra Cairo University Faculty of Computers and Information.
1 Logic Gates Digital Computer Logic Kashif Bashir WWW:
BOOLEAN ALGEBRA Saras M. Srivastava PGT (Computer Science)
Boolean Algebra and Digital Circuits
Switching functions The postulates and sets of Boolean logic are presented in generic terms without the elements of K being specified In EE we need to.
ECE 331 – Digital System Design
1 Representation of Logic Circuits EE 208 – Logic Design Chapter 2 Sohaib Majzoub.
F = ∑m(1,4,5,6,7) F = A’B’C+ (AB’C’+AB’C) + (ABC’+ABC) Use X’ + X = 1.
Combinational Logic 1.
Chapter 2: Boolean Algebra and Logic Gates. F 1 = XY’ + X’Z XYZX’Y’XY’X’ZF1F
CHAPTER 3: PRINCIPLES OF COMBINATIONAL LOGIC
LOGIC GATES & BOOLEAN ALGEBRA
Venn Diagram – the visual aid in verifying theorems and properties 1 E.
Boolean Algebra and Logic Gates
Based on slides by:Charles Kime & Thomas Kaminski © 2004 Pearson Education, Inc. ECE/CS 352: Digital System Fundamentals Lecture 6 – Canonical Forms.
Chap 2. Combinational Logic Circuits
ENGIN112 L6: More Boolean Algebra September 15, 2003 ENGIN 112 Intro to Electrical and Computer Engineering Lecture 6 More Boolean Algebra A B.
ece Parity Used to check for errors Can be either ODD or EVEN Left most bit used as the indicator For EVEN, insert a 0 or a 1 so as to make the.
LOGIC CIRCUITLOGIC CIRCUIT. Goal To understand how digital a computer can work, at the lowest level. To understand what is possible and the limitations.
A. Abhari CPS2131 Chapter 2: Boolean Algebra and Logic Gates Topics in this Chapter: Boolean Algebra Boolean Functions Boolean Function Simplification.
1 Lect # 2 Boolean Algebra and Logic Gates Boolean algebra defines rules for manipulating symbolic binary logic expressions. –a symbolic binary logic expression.
Module –I Switching Function
Binary Logic and Gates Boolean Algebra Canonical and Standard Forms Chapter 2: Boolean Algebra and Logic Gates.
Boolean Algebra & Logic Circuits Dr. Ahmed El-Bialy Dr. Sahar Fawzy.
Module 5.  In Module 3, you have learned the concept of Boolean Algebra which consists of binary variables and binary operator.  A binary variable x,
Chapter 2: Basic Definitions BB inary Operators ●A●AND z = x y = x yz=1 if x=1 AND y=1 ●O●OR z = x + yz=1 if x=1 OR y=1 ●N●NOT z = x = x’ z=1 if x=0.
ECEN 248: INTRODUCTION TO DIGITAL SYSTEMS DESIGN Lecture 4 Dr. Shi Dept. of Electrical and Computer Engineering.
Review. Boolean Algebra.
ECE DIGITAL LOGIC LECTURE 6: BOOLEAN ALGEBRA Assistant Prof. Fareena Saqib Florida Institute of Technology Fall 2016, 02/01/2016.
ECE 301 – Digital Electronics Minimizing Boolean Expressions using K-maps, The Minimal Cover, and Incompletely Specified Boolean Functions (Lecture #6)
School of Computer and Communication Engineering, UniMAP DKT 122/3 - DIGITAL SYSTEM I Chapter 4A:Boolean Algebra and Logic Simplification) Mohd ridzuan.
Boolean Algebra. BOOLEAN ALGEBRA Formal logic: In formal logic, a statement (proposition) is a declarative sentence that is either true(1) or false (0).
Lecture 5 More Boolean Algebra A B. Overview °Expressing Boolean functions °Relationships between algebraic equations, symbols, and truth tables °Simplification.
CSE 260 BRAC University.
Fuw-Yi Yang1 數位系統 Digital Systems Department of Computer Science and Information Engineering, Chaoyang University of Technology 朝陽科技大學資工系 Speaker: Fuw-Yi.
Digital Logic & Design Dr. Waseem Ikram Lecture 09.
Mu.com.lec 9. Overview Gates, latches, memories and other logic components are used to design computer systems and their subsystems Good understanding.
CHAPTER 2 Boolean algebra and Logic gates
Speaker: Fuw-Yi Yang 楊伏夷 伏夷非征番, 道德經 察政章(Chapter 58) 伏者潛藏也
CHAPTER 3 Simplification of Boolean Functions
ECE 20B, Winter 2003 Introduction to Electrical Engineering, II LECTURE NOTES #2 Instructor: Andrew B. Kahng (lecture)
CS 105 Digital Logic Design
Complement of a Function
Princess Sumaya University
ECE 331 – Digital System Design
Speaker: Fuw-Yi Yang 楊伏夷 伏夷非征番, 道德經 察政章(Chapter 58) 伏者潛藏也
ECE Digital logic Lecture 10: Karnaugh MAps
17-Nov-18 Logic Algebra 1 Combinational logic.
Boolean Algebra.
Boolean Algebra.
Chapter 2 Boolean Algebra and Logic Gate
2. Boolean Algebra and Logic Gates
MINTERMS and MAXTERMS Week 3
Digital Logic Chapter-2
Digital Logic Chapter-2
Analysis of Logic Circuits Example 1
Analysis of Logic Circuits Example 1
Presentation transcript:

ECE DIGITAL LOGIC LECTURE 8: BOOLEAN FUNCTIONS Assistant Prof. Fareena Saqib Florida Institute of Technology Spring 2016, 02/11/2016

Recap  Chapter 2: Boolean Functions  Operator Precedence  Demorgan’s Theorem  Simplification Theorems  Complement of Function using Demorgan’s Theorem  Canonical and Standard Forms

Agenda  Complement of a Function with more than 2 variables  Representation of Boolean Expression in Canonical form  Minterms and Maxterms  Representation of Boolean Expression in Standard Forms  Sum of Products  Product of Sums  Representation of Boolean Expression in Non-standard form  Conversion between different forms.

Complement of a Boolean Function  The complement of a function F is F  Is obtained from an interchange of 0’s for 1’s and 1’s for 0’s in the value of F in truth table.  The complement of a function may be derived algebraically through DeMorgan’s theorems as described for two variables and can be extended for three or more variables.  The three ‐ variable form of the first  DeMorgan’s theorem is derived as follows, from postulates and theorems listed in Table 2.1 :  (A + B + C) = (A + x) let B + C = x = Ax by theorem 5(a) (DeMorgan) = A(B + C) substitute B + C = x = A(BC) by theorem 5(a) (DeMorgan) = ABC by theorem 4(b) (associative)  The generalized form of DeMorgan’s theorems states that the complement of a function is obtained by interchanging AND and OR operators and complementing each literal. (NOTE)

Complement of Function: Example  F1 = x’yz’ + x’y’z and F2 = x(y’z’ + yz). F1’ =? F1’ =?  F1’ = (x’yz’ + x’y’z )’ = (x’yz’ )’(x’y’z)’ = (x + y’ + z)(x + y + z’ )  F2’ = [x(y’z’ + yz)]’ = x’ + (y’z’ + yz)’ = x’ + (y’z’ ) (yz)’ = x’ + (y + z)(y’ + z’ ) = x’ + yz’ + y’z

Complement of Function: Example  Note: A simpler procedure for deriving the complement of a function is to take the dual of the function and complement each literal.  Dual of a function is obtained from the interchange of AND and OR operators and interchanging identity elements1’s and 0’s.  F1 = x’yz’ + x’y’z and F2 = x(y’z’ + yz). F1’ =? F1’ =? F1’ = (x’yz’ + x’y’z)’ = (x’yz’)’ (x’y’z)’ = ((x’)’ + y’ + (z’)’ )(x + y + z’) = (x+y’+z)(x+y+z’) F2’ = x(y’z’ + yz) = x’ + (y’z’+yz)’ = x’ + [(y’z’) (yz) ] = x’ + (y+z)(y’+z’) = x’ + yy’ + zy’ +yz’+zz’ = x’ + zy’ + yz’

Complement of Function: Example  A simpler procedure for deriving the complement of a function is to take the dual of the function and complement each literal. Where dual of a function is obtained from the interchange of AND and OR operators and 1’s and 0’s.  F1 = x’yz’ + x’y’z and F2 = x(y’z’ + yz).  F1’ =? F1’ =? F1’ = (x’yz’ + x’y’z) = (x+y’+z) (x+y+z’) F2’ = x(y’z’ + yz) = x’+ (y+z) (y’+z’)

Canonical and Standard Forms  In canonical form, boolean function is expressed using min terms and max terms.  In Standard form, boolean function is expressed in the form of sum of products and products of sums.

Canonical and Standard Forms  For a given operation “and” with two variables x and y, there are four possible combinations: xy, x’y, xy,’ and x’y.’  Each of these four AND terms is called a minterm, or a standard product.  In a similar fashion, n variables forming an OR term  Each variable being primed or unprimed, provide 2 n possible combinations, called maxterms, or standard sums  Boolean function in terms of minterms from truth table  Any Boolean function can be expressed as a sum of minterms (with “sum” meaning the ORing of terms).  Boolean function in terms of maxterm from truth table:  Any Boolean function can be expressed as a product of maxterms (with “product” meaning the Anding of terms).  Conversion between canonical forms  Boolean function in terms of POS from a given SOP

Canonical Form using minterm and maxterms A Boolean function can be expressed algebraically from a given truth table by forming a minterm for each combination of the variables that produces a 1 in the function and then taking the OR of all those terms. Example: f1 = x’y’z + xy’z’ + xyz It may be read from the truth table by forming a minterm for each combination that produces a 1 in the function and then ORing those terms Example: f1 = x’y’z + xy’z’ + xyz = m1 + m4 + m7 = M0. M2. M3. M5. M6

Canonical Form using minterm and maxterms  A Boolean function can be expressed algebraically from a given truth table by forming a maxterm for each combination of the variables that produces a 0 in the function.  Each maxterm is obtained from an OR term of the n variables, with each variable being unprimed if the corresponding bit is a 0 and primed if a 1 Example: f1 = x’y’z + xy’z’ + xyz = m1 + m4 + m7 = M0. M2. M3. M5. M6

Conversion Between Canonical Forms: From SOP to POS  Example: F1 = x’y’z + xy’z’ + xyz = m1 + m4 + m7  F1’ = m0 + m2+m3+m5+m6  F1’ = x’y’z’ + x’yz’+ x’yz +xy’z+xyz’  Take complement of f1’ = ?  F1 = x’y’z’ + x’yz’+ x’yz +xy’z+xyz’ F1 = x’y’z’ + x’yz’+ x’yz +xy’z+xyz’ F1 = (x+y+z)(x+y’+z)(x+y’+z’)(x’+y+z’)(x’+y’+z)

Conversion Between Canonical Forms: … Contd  The complement of a function expressed as the sum of minterms equals the sum of minterms missing from the original function.  Consider an example F(A, B, C) = ∑ (1, 4, 5, 6, 7)  This function has a complement that can be expressed as  F’ (A, B, C) = ∑ (0, 2, 3) = m0 + m2 + m3  Complement of F’ by demorgan’s theorem, we obtain F in form of maxterms  F = (m0 + m2 + m3) = (m’0 m’2 m’3) = M0M2M3 = ∏ (0, 2, 3) where following relation holds: m i ’ =M i  Example F= xy+x’z??

Practice Example  Examples  F1 = x’y’z + xy’z’ + xyz  F2 = x’yz + xy’z + xyz’ + xyz  Part 1) Express Boolean function in terms of minterms  Part 2) Express Boolean function in terms of maxterms

Standard Form  Another way to express Boolean functions is in standard form.  The two canonical forms of Boolean algebra are basic forms that one obtains from reading a given function from the truth table.  These forms are very seldom the ones with the least number of literals, because each minterm or maxterm must contain, by definition, all the variables, either complemented or uncomplemented  The terms that form the function may contain one, two, or any number of literals. There are two types of standard forms: the sum of products and products of sums.  The sum of products is a Boolean expression containing AND terms, called product terms, with one or more literals each.  The sum denotes the ORing of these terms.

Standard Form : Sum of Products - Example  An example of a function expressed as a sum of products is  F1 = y’ + xy + x’yz’  The expression has three product terms, with one, two, and three literals.  Their sum is, in effect, an OR operation.  Logic diagram of a sum-of-products?  The logic sum is formed with an OR gate whose inputs are the outputs of the AND gates and the single literal.  This circuit configuration is referred to as a two ‐ level implementation. (assumed that the input variables are directly available in their complements)

Standard Form : Products of Sum - Example  An example of a function expressed as a sum of products is  F2 = x(y’ + z)(x’ + y + z’)  The expression has three sum terms, with one, two, and three literals.  Their product is, in effect, an AND operation.  Logic diagram of a sum-of-products?  The logic sum is formed with an OR gate whose inputs are the outputs of the AND gates and the single literal.  This circuit configuration is referred to as a two ‐ level implementation as well. (assumed that the input variables are directly available in their complements)

Example 2.4 is conversion from standard form to canonical form  Express the Boolean function F = A + B’C as a sum of minterms.  The function has three variables: A, B, and C.  The first term A is missing two variables; therefore  A = A(B + B‘ ) = AB + AB‘ AB = AB(C+C‘) AB‘ = AB‘(C+C‘) AB = ABC+ABC‘ AB‘ = AB‘C + AB‘C‘ A = ABC+ABC‘ + AB‘C + AB‘C‘  The second term B‘C is missing one variable; therefore  B‘C = B‘C(A+A‘) B‘C = B‘CA + B‘CA‘  F = A + B‘C = ABC+ABC‘ + AB‘C + AB‘C‘ + B‘CA + B‘CA‘ = ABC + ABC‘ + AB‘C + AB‘C‘ + B‘CA‘  F= ∑(1,4,5,6,7) = m1+m4+m5+m6+m7

Example 2.5 is conversion from standard form to canonical form  Express the Boolean function F = xy + x’z as a product of maxterms.  First, convert the function into OR terms by using the distributive law:  F = xy + x’ z = (xy + x’ )(xy + z) = (x + x’ )(y + x’ )(x + z)(y + z) = (x’ + y)(x + z)(y + z)  The function has three variables: x, y, and z. Each OR term is missing one variable  X’ + y = x’ + y + zz’ = (x’ + y + z)(x’ + y + z’ )  x + z = x + z + yy’ = (x + y + z)(x + y’ + z)  y + z = y + z + xx’ = (x + y + z)(x’ + y + z)  Combining all the terms and removing those which appear more than once, we obtain  F = (x + y + z)(x + y’ + z)(x’ + y + z)(x’ + y + z’ ) = M 0 M 2 M 4 M 5

Next Class  Karnaugh Maps or K-map  2,3 and 4 input K-Map  Don’t Care conditions