Logic Design I (17.341)

Fall 2011

Lecture Outline

lClass # 03

September 26, 2011

Dohn Bowden

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Today’s Lecture

• Administrative

• Main Logic Topic

• Homework

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CCoursed iAdmin

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Administrative

• Admin for tonight …

– Syllabus Review

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Syllabus

• Syllabus

– Exam #1 … next week

• Chapters 1 - 3

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Syllabus ReviewWeek Date Topics Chapter Lab Report Due

1 09/12/11 Introduction to digital systems and number systems 1

2 09/19/11 Binary Codes and Boolean Algebra 2

3 09/26/11 Boolean Algebra (continued) 3

4 10/03/11 Examination 1

X 10/10/11 No Class - Holiday

5 10/17/11 Application of Boolean Algebra. Lab lecture 4

6 10/24/11 K h M 56 10/24/11 Karnaugh Maps 5

7 10/31/11 Multi-Level Gate Circuits. NAND and NOR Gates 7 1

8 11/07/11 Examination 2

9 11/14/11 Combinational Circuit Design and Simulation Using 8 29 11/14/11 Combinational Circuit Design and Simulation Using Gates

8 2

10 11/23/11 Multiplexers, Decoders. Encoder, and PLD 9

11 11/28/11 Introduction to VHDL 10 3

12 12/05/11 Examination 3

13 12/12/11 Review 4

14 12/19/11 Final Exam

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Questions?

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Chapter 3 …Chapter 3 …

Boolean AlgebraBoolean Algebra(Continued)( )

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Objectives

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Objectives

• Understand and apply Chapter 2 Boolean algebra Laws and TheoremsTheorems

– Apply laws and theorems to the manipulation of algebraic expressions includingexpressions including …

• Simplifying an expression

• Finding the complement of an expression

• Multiplying out and factoring an expression

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Objectives

• Continued …

– Prove theorems using a truth table or algebraic proof

– Define the exclusive-OR and equivalence operations

– Use the consensus theorem to delete terms from and add terms to a switching expression

– Given an equation …Given an equation … • Prove algebraically that it is valid … or …

• Show that it is not valid• Show that it is not valid

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Multiplying Outandand

Factoring Expressions

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Multiplying OutMultiplying Out

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Multiplying Out

• Given an expression in product-of-sums form …

– The corresponding sum-of-products expression can be obtained by …

– Multiplying out … using … the following two distributive laws …

X(Y + Z) = XY + XZ (3-1)

(X + Y)(X + Z) = X + YZ (3-2)(X Y)(X Z) X YZ (3 2)

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Multiplying Out

• In addition … the following theorem is very useful for factoring and multiplying out …multiplying out …

(X + Y)(X′ + Z) = XZ + X′Y (3-3)

• Note that the variable that is paired with X on one side of the equation is …

» Paired with X on the other side … and … vice versa

• This theorem can be applied when we have …– Two terms … one which contains a variable … and … another

which contains its complementwhich contains its complement

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Multiplying Out

• Example … Multiply out … AB + A’C

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Multiplying Out

• Example … Multiply out … (Q + AB’)(C’D + Q’)

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Multiplying Out

• Example … Multiply out …

(A + B + C’)(A + B + D)(A + B + E)(A + D’ + E)(A’ +C)

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FactoringFactoring

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Factoring

• Any expression can be converted to a product-of-sums form by using …using …

– The same theorems that are useful for multiplying out expressions Byexpressions … By …

• Repeatedly applying … (3-1) … (3-2) … and (3-3)

( ) ( )X(Y + Z) = XY + XZ (3-1)

(X + Y)(X + Z) = X + YZ (3-2)

(X + Y)(X′ + Z) = XZ + X′Y (3-3)

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Factoring

• Example … Factor … AC + A’BD’ + A’BE + A’C’DE

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Exclusive-OR

andand

E i l O iEquivalence Operations

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Exclusive-ORExclusive OR

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Exclusive-OR

• The exclusive-OR operation ( ) is defined as follows …

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Exclusive-OR

• Exclusive OR can be expressed in terms of AND and OR

• Because X Y = 1 … iff …» X is 0 … and … Y is 1 … or …» X is 1 … and … Y is 0, we can write

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Exclusive-OR

• The following theorems apply to exclusive OR …

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Equivalence OperationsEquivalence Operations

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Equivalence Operation

• Equivalence is the complement of exclusive-OR …

• The equivalence operation ( ) is defined by …

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Equivalence Operation

• Because equivalence is the complement of exclusive-OR … an …

– Alternate symbol of the equivalence gate is an exclusive-OR gate with a … complemented output …

• The equivalence gate is also called an …The equivalence gate is also called an …

» Exclusive-NOR gate

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ExamplesExamples …

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Examples

• Example … Simplify …

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Examples

• Example … Simplify … A’ B C

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The Consensus TheoremThe Consensus Theorem

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Consensus Theorem

• The consensus theorem is very useful in simplifying Boolean expressionsexpressions

• Given an expression of the form …

• The term YZ is redundant … and …

XY + X'Z + YZ

– Can be eliminated to form the equivalent expression …

XY + X'Z

– The term that was eliminated is referred to as the consensus term

XY + X Z

consensus term

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Consensus Theorem

• Given a pair of terms for which a …

– Variable appears in one term … and …

– The complement of that variable in another

– The consensus term is formed by multiplying the two original g gterms together …

• Leaving out the selected variable and its complementLeaving out the selected variable and its complement

XY + X'Z + YZ = XY + X'Z

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Dual Form - Consensus Theorem

• The Dual Form of the consensus theorem is …

(X + Y)(X’ + Z)(Y + Z) = (X + Y)(X’ + Z)

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Basic Methods

forfor

Si lif i F iSimplifying Functions

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Combining TermsCombining Terms

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Combining Terms

1. Combining terms …

• Use the theorem …

XY + XY′ = X

• To combine two terms

• For example,

abc′d′ + abcd′ = abd′ Let x = abd′ and Y = c

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Eliminating TermsEliminating Terms

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Eliminating Terms

2. Eliminating terms …

– Use the theorem …

X + XY = X

• To eliminate redundant terms if possible … then …

• Try to apply the consensus theorem …

XY + X′Z + YZ = XY + X′Z

T li i• To eliminate any consensus terms41

Eliminating Terms

• For example … use X + XY = X … for …

a′b + a′bc Let x = a′b

a′b + a′bc = a′b

• Example … use the consensus theorem (XY + X′Z + YZ = XY + X′Z)

a b + a bc = a b

We get

a′bc′ + bcd + a′bd … Let x = c … Y = bd … Z = a′b

We get …

a′bc′ + bcd + a′bd = a′bc′ + bcd

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Eliminating LiteralsEliminating Literals

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Eliminating Literals

3. Eliminating literals …

– Use the theorem …

X + X’Y = X + Y

– … To eliminate redundant literals

– Simple factoring may be necessary before the theorem is applied

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Eliminating Literals

• Example … Simplify … A′B + A′B′C′D′ + ABCD′

• Using … X + X’Y = X + Y

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Adding Redundant TermsAdding Redundant Terms

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Adding Redundant Terms

4. Adding redundant terms … Redundant terms can be introduced in several ways such as …several ways such as …

– Adding xx′

– Multiplying by (x + x′)

– Adding yz to xy + x′z

– Or adding xy to xOr adding xy to x

• When possible … the added terms should be chosen so that they will combine with or eliminate other termswill combine with or eliminate other terms

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Adding Redundant Terms

• Example … WX + XY + X′Z′ + WY′Z′

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Adding Redundant Terms

• Example …

A’B’C’D’ + A’BC’D’ + A’BD + A’BC’D +ABCD + ACD’ + B’CD’

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Proving Validity

of anof an

E iEquation

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Proving Validity of an Equation

• There are Several methods that can be used to determine if an equation is valid for all combinations of values of the variables …equation is valid for all combinations of values of the variables …

• Method 1 …

– Construct a truth table and evaluate both sides of the equation for all combinations of values of the variables

• This method is rather tedious if the number of variables is large

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Proving Validity of an Equation

• Method 2 …

– Manipulate one side of the equation by applying various theorems until it is identical with the other side

• Method 3 …

– Reduce both sides of the equation independently to the same expression

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Proving Validity of an Equation

• Method 4 …

– Perform the same operation on both sides of the equation provided that the operation is reversible

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Proving Validity of an Equation

• Method 4 … Examples …

– Complement both sides of the equation

– CANNOT multiply both sides of the equation by the same expression

• Multiplication is not reversible because division is not defined for Boolean algebra

– CANNOT add the same term to both sides of the equation

• Because subtraction is not defined for Boolean algebra• Because subtraction is not defined for Boolean algebra

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Proving That An Equation Is

NOTNOT

V lidValid

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Proving That An Equation Is NOT Valid

• To prove that an equation is not valid …

– It is sufficient to show …

• One combination of values of the variables for which the …

– Two sides of the equation have different values

– When using method 2 or 3 to prove that an equation is valid, a useful strategy is touseful strategy is to

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Proving That An Equation Is NOT Valid

• When using method 2 or 3 to prove that an equation is valid …

– A useful strategy is to …

1. First reduce both sides to a sum of products (or a product of sums)

2. Compare the two sides of the equation to see how they d ffdiffer

3. Then try to add terms to one side of the equation that are present on the other side

4. Finally try to eliminate terms from one side that are not present on the other

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Proving That An Equation Is NOT Valid

• Whatever method is used …

– Frequently compare both sides of the equation and let the different between them serve as a guide for what steps to take nextnext

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Example

• Show that … A'BD' + BCD + ABC' + AB'D = BC'D' + AD + A'BC

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Differences Between

Boolean algebraBoolean algebra and

O di Al bOrdinary Algebra

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Differences between Boolean algebra and ordinary algebra

• Some of the theorems of Boolean algebra are not true for ordinary algebraalgebra

• Similarly … some of the theorems of ordinary algebra are not true for Boolean algebrafor Boolean algebra

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Differences between Boolean algebra and ordinary algebra

• For example … the cancellation law for ordinary algebra …

If x + y = x + z then y = z

• The cancellation law is not true for Boolean algebra

• For example … x + y = x + z but y ≠ z.

– Let x = 1 y = 0 z = 1 ThenLet x = 1 y = 0 z = 1 … Then …

1 + 0 = 1 + 1 both side = 1 … but (y ≠ z … 0 ≠ 1)

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Differences between Boolean algebra and ordinary algebra

• In ordinary algebra … the cancellation law for multiplication is

If xy = xz, then y = z (3-32)

• This law is valid provided x ≠ 0

• In Boolean algebra … the cancellation law for multiplication is also gnot valid when x = 0

Let x = 0, y = 0, z = 1 … then 0 • 0 = 0 • 1, but 0 ≠ 1Let x 0, y 0, z 1 … then 0 0 0 1, but 0 ≠ 1

• Because x = 0 about half the time in switching algebra … the cancellation law for multiplication cannot be usedcancellation law for multiplication cannot be used

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Similarities Between

Boolean AlgebraBoolean Algebra and

O di Al bOrdinary Algebra

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Similarities Between Boolean Algebra and Ordinary Algebra

• Even though the statements in the previous slides are generally false for Boolean algebra … the converses are true …false for Boolean algebra … the converses are true …

If y = z, then x + y = x + z

If y = z, then xy = xz

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LabLab

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Lab

• No topics this week

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Next Week …

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Next Week Topics

• Exam #1 …

– Chapters 1 - 3 …

• I will provide a list of Laws and Theorems (page 55)

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Home Work

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Homework

1. Study for Exam #1 …

• Chapters 1 – 3

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References

1. None

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