logic design i (17.341) fall 2011 lecture...

Logic Design I (17.341)

Fall 2011

Lecture Outline

lClass # 02

September 19, 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

– 17 341 Web Site17.341 Web Site

– Email List creation

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Syllabus

• Syllabus

– One update … remove some material for class #1

• Too much for the first class!!!

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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 09/28/11 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


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


13 12/12/11 Review 4

14 12/19/11 Final Exam

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Access to Computer Lab and/or Engineering Lab

• To gain access to the Computer Lab (BL-420) … and/OR … the Engineering Lab (EB-321) …Engineering Lab (EB 321) …

– You will need to have an access cards

– Access cards are given out at Access Services

• Arrangements are done through the Continuing Education Office

– Then I will need your UMS# (will be on your Access Card)

– I will then send your name and UMS# to the Room POC

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I will then send your name and UMS# to the Room POC

Email Distribution List

• Did you receive my test email????

– No … then see me and resend another email …

• Email me at:

[email protected]

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Questions?

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Ch t 2Chapter 2 …

Boolean Algebra

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Objectives

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Objectives

• Familiar with first 12 of 15 laws and theorems of Boolean algebra

• You should be able to …

1. Understand the basic operations and laws of Boolean algebra

2. Relate these operations and laws to … • Circuits composed of AND gates, OR gates, and INVERTERS• Circuits composed of switches

3. Prove any of these laws using a truth table

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Objectives

4. Apply these laws to the manipulation of algebraic expressions including …including …

a) Multiplying out an expression to obtain a sum of products (SOP)(SOP)

b) Factoring an expression to obtain a product of sums (POS)

c) Simplifying an expression by applying one of the laws

d) Finding the complement of an expression

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Introduction

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Introduction

• Boolean algebra …

– The basic mathematics needed for the study of the logic design of digital systems

• Switching circuits are essentially …

– Two-state devices …such as …

• A transistor with high or low output voltageA transistor with high or low output voltage

• We will study the special case of Boolean algebra in which all of the variables assume only one of two values Referred to as switchingvariables assume only one of two values … Referred to as switching algebra

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Introduction

• Boolean variable … such as X or Y …

– Represents the input or output of a switching circuit

• Assume these variables can take on only two different values

– Symbols “ 0” and “ 1”

• If X is a Boolean (switching) variable …then …

– Either X = 0 … or … X = 1

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Introduction

• Symbols “ 0” and “ 1” … do not …

– Have a numeric value

• They represent two different states

– In a logic gate circuit … “0” represents a range of low g g gvoltages … and …

– “1” represents a range of high voltages1 represents a range of high voltages

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Introduction

• For Switch circuits …

– “0” … represents an open switch … and … – “1” represents a closed circuit

• In general “0” and “1” can be used to represent …

– The two states in any binary-valued system

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Basic Operations

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Basic Operations

• Basic operations …

– AND

– OR

– Complement .. or … Inverse

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Complement

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Complement

• The complement of …

– 0 … is … 1

– 1 … is … 0

• Symbolically … where the prime (') denotes complementation

0' = 1

and

1' 01' = 0 22

Complement

• X is a switching variable … therefore …

X' = 1 … if … X = 0

and

X' = 0 … if … X = 1

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Complement

• Complementation is inversion

• Electronic circuit which forms the inverse of X is …

– An inverter

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AND operation

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AND operation

• The AND operation … where “•” denotes AND …

0 • 0 = 0 0 • 1 = 0 1 • 0 = 0 1 • 1 = 1

• This is not binary multiplication …

– Because 0 and 1 are Boolean constants rather than binaryBecause 0 and 1 are Boolean constants rather than binary numbers

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AND operation

• Boolean expression …

C = A • B

• Given the values of A and B …

– C can be determine from the following table …g

• C = 1 … if and only if … A and B are both 1 27

AND operation

• Logic gate that performs the AND operation …

• C = 1 … if and only if …

» A and B are both 1» A and B are both 1

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AND operation

• The dot symbol (•) is frequently omitted

• Write …

AB … instead of … A•B

• The AND operation is also referred to as logical (or Boolean) gmultiplication

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OR operation

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OR operation

• The OR operation can be defined as follows …

0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 1

• “+” denotes OR

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OR operation

• C = A + B … C can be determine from the following table …

• C = 1 if and only if A or B (or both) are 1

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OR operation

• Logic gate that performs the OR operation …

• C = 1 if and only if …

» A or B (or both) are 1» A or B (or both) are 1

• The OR operation is also referred to as logical (or Boolean) addition

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Switches

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Switches

• Applying switching algebra to circuits containing switches

• Each switch is labeled with a variable

– If switch X is open … the value of X is 0

– If switch X is closed … the value of X is 1

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Switches

• Two switches in series …

• Closed circuit between terminals 1 and 2 (T = 1) iff (if and only if) switch A is closed and switch B is closedswitch A is closed and switch B is closed

• Algebraically …

» T AB» T = AB

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Switches

• Two switches in parallel …

• Closed circuit … between terminals 1 and 2 … iff switch A is closed … or … switch B is closed

• Algebraically …

» T A + B» T = A + B

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Switches

• Therefore …

– Switches in a series perform the AND operation …

– Switches in parallel perform the OR operation

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Boolean ExpressionsBoolean Expressions and

Truth Tables

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Boolean Expressions

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Boolean Expressions

• Boolean expressions are formed by …

– Application of the basic operations to one or more variables or constants

• The simplest expressions consist of a …

– Single constant or variable …

• 00• X• Y’

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Boolean Expressions

• More complicated expressions are formed by …

– Combining two or more other expressions using …

» AND» OR» Or by complementing another expressiong

• Examples …

AB' + C

[A(C D)]' BE[A(C + D)]' + BE 42

Boolean Expressions

• Order in which the operations are performed …

– Parentheses are added as needed to specify the order

– When parentheses are omitted … the order is …

» Complementation

» AND

» OR

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Boolean Expressions

• Example … the order in which the operations are performed for …

AB' + C

– B' … is formed first … then …

– AB' … and finally …

– AB' + C is performed

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Boolean Expressions

• Circuit of logic gates for the expression … AB' + C

F = AB' + C

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Boolean Expressions

• Circuit of logic gates for the expression … [A(C + D)]' + BE

F = [A(C + D)]' + BE

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Literal

• Literal …

– Each appearance of a variable or its complement in an expression

• Example … ab'c + a’b + a’bc' +b’c'

» three variables

» 10 literals10 literals

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Truth Table

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Truth Table

• Truth table …

– Specifies the values of a Boolean expression for every possible combination of values of the variables in the expression

– Specify the output values for a circuit of logic gates in terms of the values of the input variables

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Truth Table

• Example … F = A’ + B

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Truth Table

• Example … F = AB’ + C

A B C B’ AB’ AB’+C A+C B’+C (A+C)(B’+C)0 0 0 1 0 0 0 1 00 0 0 1 0 0 0 1 00 0 1 1 0 1 1 1 1 0 1 0 0 0 0 0 0 0 0 1 1 0 0 1 1 1 1 1 0 0 1 1 1 1 1 11 0 0 1 1 1 1 1 11 0 1 1 1 1 1 1 1 1 1 0 0 0 0 1 0 0 1 1 1 0 0 1 1 1 1

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Truth Table

• Two expressions are equal … if …

– They have the same value for every possible combination of the variables

• For example … lets prove … using the truth table …

AB’ + C = (A + C)(B’ + C)

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Truth Table

• Prove … AB’ + C = (A + C)(B’ + C) … using the truth table …


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Truth Table

• Prove … AB’ + C = (A + C)(B’ + C) … using the truth table …

Values for all 8 combinations are equal


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Truth Table

• If an expression has …

– n variables … and … each variable can have the value 0 or 1

– Then …

• The number of different combinations of values of the variables is …

2 x 2 x 2 x . . . = 2n

i

• Therefore … a truth table for an n-variable expression will have 2n

rows

n times

rows

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Truth Table

• For … AB’ + C

– We have 3 variables … A, B’, and C … therefore …

3n = 3 … # of rows in truth table is 2n = 23 = 8–


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Basic Theorems

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Operations with 0 and 1Operations with 0 and 1

Theorem 2-4

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Single Variable --- Operations with 0 and 1

• Single Variable --- Operations with 0 and 1

X + 0 = X (Theorem 2-4)

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• Lets think about (2-4) …

X + 0 = X

0 + 0 = 00 + 1 = 1 1 + 0 = 11 + 1 = 1

X + 0 = XX + 0 = X

• As you see … whatever the value of X is … is also the equivalent valuevalue

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Dual

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Dual

• Dual …

– Is formed by replacing …

» AND with OR» OR with AND» 0 with 1» 1 with 0

– Variables and complements are left unchangedVariables and complements are left unchanged

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Duality

• Most theorems have duals

• Theorem duals are formed by replacing …


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Theorem 2-4D

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X + 0 = X (2-4)

X • 1 = X (2-4D)

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• Lets think about (2-4D) …

X • 1 = X

0 • 0 = 0 0 • 1 = 01 • 0 = 0 1 • 1 = 1

X • 1 = XX • 1 = X

• As you see … whatever the value of X is … is also the equivalent valuevalue

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Theorem 2-5

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X + 1 = 1 (2-5)

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Theorem 2-5D

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X + 1 = 1 (2-5)

X • 0 = 0 (2-5D)

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Idempotent LawsIdempotent Laws

Theorem 2-6

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Idempotent Law

• Idempotent Law

X + X = X (2-6)

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Idempotent LawsIdempotent Laws

Theorem 2-6D

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Idempotent Law

• Idempotent Law

X + X = X (2-6)

X • X = X (2-6D)

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Involution LawInvolution Law

Theorem 2-7

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Involution Law

• Involution Law

(X’)’ = X (2-7)

• There is NO DUAL for 2-7

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Laws of ComplementarityLaws of Complementarity

Theorem 2-8

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Laws of Complementarity

• Laws of Complementarity

X + X’ = 1 (2-8)

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Laws of ComplementarityLaws of Complementarity

Theorem 2-8D

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Laws of Complementarity

• Laws of Complementarity

X + X’ = 1 (2-8)

X • X’ = 0 (2-8D)

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General Theorem

InformationInformation

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Proving Theorems

• Theorems can easily be proved by …

– Showing that it is valid for both of the possible values of X

• Example … To prove X + X′ = 1 … we observe that if …

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Use of Theorems

• Any expression can be substituted for …

– The variable X in the theorems

• Example … (AB' + D)E + 1 … would equal 1 … by …

• Using Theorem (2- 5) … X + 1 = 1g

(AB' + D)E + 1

Would equal … 1

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Use of Theorems

• Example … (AB' + D)(AB' + D)’ … would equal 0 … by …

• Using Theorem (2- 8D) … X + X’ = 1

(AB' + D)(AB' + D)’

Would equal … 0

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Switches

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Switches

• If two switches are both labeled with the variable A …

– This means that both switches are open when A = 0 … and …– Both are closed when A = 1

• Then the following circuits are equivalent … (Idempotent law)

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Switches

• Idempotent law

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Switches

• Operations with 0

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Switches

• Operations with 1

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Switches

• A in parallel with A’ can be replaced with a closed circuit because …

– One or the other of the two switches is always closed … Law of Complementarity

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Switches

• Switch A in series with A’ can be replaced with an open circuit because …because …

– One or the other of the two switches is always open … Law of ComplementarityComplementarity

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CommutativeCommutative,Associative, and

Distributive Laws

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Commutative, Associative, and Distributive Laws

• Many of the laws of ordinary algebra …such as …

– The commutative and associative laws …

• Also apply to Boolean algebra

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Commutative LawsCommutative Laws

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Commutative Law

• The commutative laws for AND and OR …

– Which follow directly from the definitions of the AND and OR operations … are …

XY = YX (2-9)

X + Y = Y + X (2-9D)

• This means that … the order in which the variables are written …This means that … the order in which the variables are written …

– Will not affect the result of applying the AND and OR operations

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Associative LawsAssociative Laws

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Associative Laws

• The associate laws also apply to AND and OR …

(XY)Z = X(YZ) = XYZ (2-10)

(X + Y) + Z + X + (Y + Z) = X + Y + Z (2-10D)

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Associative Laws

• Prove the associative law (2-10) using a truth table …

(XY)Z = X(YZ) = XYZ

– List all combinations of values of the variables X, Y, and Z

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Associative Laws


(XY)Z = X(YZ) = XYZ

– Compute XY and YZ for each combination of values of X, Y, Z

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Associative Laws


(XY)Z = X(YZ) = XYZ

– Compute (XY)Z and X(YZ)

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Associative Laws


(XY)Z = X(YZ) = XYZ

– Because (XY)Z and X(YZ) are equal for all combinations … Equation (2-10) is valid

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Associative Laws

• Two … 2-input AND gates … can be replaced with ….

– A single … 3-input AND gate

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Associative Laws

• Two … 2-input OR gates … can be replaced with …

– A single … 3-input OR gate

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Associative Laws

• When two or more variables are ANDed together …

– The value of the result will be 1 …

• iff … all of the variables have the value 1

– If any of the variables have the value 0 … then

• The result of the AND operation will be 0

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Associative Laws

• When two or more variables are ORed together …

– The value of the result will be 1 if …

• Any of the variables have the value 1

– The result of the OR operation will be 0 …

• iff all of the variables have the value 0

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Distributive LawsDistributive Laws

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Distributive Law

• Ordinary (algebra) distributive law …

X(Y + Z) = XY + XZ

– The ordinary distributive law states that …

• The AND operation distributes over OR

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Distributive Law

• Second distributive law valid for Boolean algebra ONLY …

X + YZ = (X + Y)(X + Z)

– The second distributive law states that OR distributes over AND

• This law is very useful in manipulating Boolean expressionsg

• Example … A + BC … cannot be factored in ordinary algebra

– Is easily factored using the second distributive law …

A BC (A B)(A C)A + BC = (A +B)(A + C)108

Simplification Theorems

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Simplification Theorems

• The following theorems are useful in simplifying Boolean expressions …expressions …

XY + XY’ = X (2-12) (X + Y)(X + Y’) = X (2-12D)

X + XY = X (2-13) X(X + Y) = X (2-13D)

(X + Y’)Y = XY (2-14) XY’ + Y = X + Y (2-14D)(X Y )Y XY (2 14) XY Y X Y (2 14D)

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Illustration of Theorem (2-14D) … XY’ + Y = X + Y

• T = Y + XY' because there is a closed circuit between the terminals if switch Y is closed or switch X is closed and switch Y' is closedif switch Y is closed or switch X is closed and switch Y is closed

• The second circuit is equivalent because … if Y is closed (Y =1) both circuits have a transmission of 1; if Y is open (Y' = 1) both circuitscircuits have a transmission of 1; if Y is open (Y = 1) both circuits have a transmission of X

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Simplification of Logic Gates

• Simplify the logic gates for … F = A(A’ + B)

• By Theorem (2-14) … (X + Y’) = XY …

F = A(AB) = A • A • B = AB (using 2-3D)

=

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Simplification ExamplesSimplification Examples …

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Example 1

• Simplify Z = A′BC + A′ … rearranging … Z = A’ + A’BC

Using 2-13 … X + XY = X

Let X = A′ and Y = BC …

The expression simplifies to …

Z = A’ + A’BC = X + XY = X = A′

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Example 2

• Simplify Z = [A + B’C + D + EF][A + B’C + (D + EF)’] …

Using 2-12D … (X + Y)(X + Y’) = X

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Example 3

• Simplify Z = (AB + C)(B’D + C’E’) + (AB + C)’ …

Using 2-14D … XY’ + Y = X + Y

Let Y = (AB + C)′ rather than (AB + C) … To match form of (2-14D)

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Multiplying Out

and

FactoringFactoring

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Sum-Of-Products (SOP)Sum Of Products (SOP) …

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Sum-Of-Products (SOP)

• The two distributive laws are used to multiply out an expression to obtain a sum-of-products (SOP) formobtain a sum of products (SOP) form

• An expression is said to be in sum-of-products (SOP) form when …

– All products are the products of single variables

• This form is the end result when an expression is fully multiplied out

• It is usually easy to recognize a sum-of-products expression because it consists of a sum … of product terms

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• Sum-of-products examples …

AB’ + CD’E + AC’E’

’ ’A + B’ + C + D’E “degenerated” example

• Examples that are NOT sum-of-products

(A + B)CD + EF (A + B) is not a single variable

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• A sum-of-products expression can always be realized directly by …

– One or more AND gates feeding a single OR gate at the circuit output

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• The circuits below are referred to as two-level circuits …

– They have a maximum of two gates in series between an input and the circuit output

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Product-Of-Sums (POS)Product Of Sums (POS) …

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Product-Of-Sums (POS)

• Both distributive laws can be used to factor an expression to obtain a product-of-sums forma product of sums form

• An expression is in product-of-sums (POS) form when …

– All sums are the sums of single variables

• It is usually easy to recognize a product-of-sums expression since it consists of a product … of sum terms

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• Product-of-sums examples …

(A + B’)(C + D’ + E)(A + C’ + E’)

• Some “degenerated” examples …

(A + B)(C + D + E)F F is a single variable

AB’C(D’ + E)

• NOT a Product-of-sums

(A B)(C D) EF(A + B)(C + D) + EF125

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• An expression is fully factored … iff …

– It is in product-of-sums form

• Any expression not in this form can be factored further!

• When multiplying out an expression …g

– Apply the second distributive law first … when possible

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• Example … Multiplying out (A + BC)(A + D + E) …

Let … X = A Y = BC Z = D + E

Then …

(A + BC)(A + D + E) = (X + Y)(X + Z) = X + YZ =

= A + BC(D + E) = A + BCD + BCE

• The same result could be obtained the hard way by multiplying out the original expression completely … then eliminating redundant termsterms

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• EXAMPLE … Factor … A + B'CD

A + B'CD … is of the form X + YZ (second distributive law)

where X = A Y = B‘ Z = CD … therefore …

A + B'CD = X + YZ = (X + Y)(X + Z) = (A + B')(A + CD)

A + CD can be factored … use … second distributive law

A + CD = (A + C)(A + D) … therefore …

A B'CD (A B')(A C)(A D)A + B'CD = (A + B')(A + C)(A + D)128


• EXAMPLE … Factor … AB' + C'D

X = AB′ Y = C’ Z = D second distributive law

AB′ + C’D = X + YZ = (X + Y)(X + Z) =

= (AB′ + C′)(AB′ + D) factor each term using second distributive law

(AB′ + C′) ... Let X = C′ Y = A Z = B′(AB′ + C′) = X + YZ = (X + Y)(X + Z) = (A + C′)(B′ + C′)(AB + C ) = X + YZ = (X + Y)(X + Z) = (A + C )(B + C )

Same for (AB′ + D) = (A + D)(B′ + D) ... Which gives us ...

AB' C'D (A C′)(B′ C′)(A D)(B′ D)AB' + C'D = (A + C′)(B′ + C′)(A + D)(B′ + D)129


• Example … Factor … C'D + C'E' + G'H

First … apply ordinary distribution law … X(Y + Z) = XY + XZ

C'D + C'E' + G'H = C′(D + E′) + G′H

Next … apply the second distribution law … X + YZ = (X + Y)(X + Z)

Let X = G′H Y = C′ Z = (D + E′)

(X + Y)(X + Z) = (C′ + G′H)(D + E′ + G′H)Y X Z X

C i dContinued ....130


• Example … continued … Factor … C'D + C'E' + G'H

(C′ + G′H)(D + E′ + G′H) ... Factor each separately------------(C′ + G′H) ... Let X = C′ Y = G’ Z = H

X + YZ = (X + Y)(X + Z) = (C′ + G′)(C′ + H)------------(D + E′ + G′H) ... Let X = D + E′ Y = G’ Z = H

second distributive law

X + YZ = (X + Y)(X + Z) = (D + E′ + G′)(D + E′ + H)------------S C'D C'E' G'H (C′ G′)(C′ H)(D E′ G′)(D E′ H)So ... C'D + C'E' + G'H = (C′ + G′)(C′ + H)(D + E′ + G′)(D + E′ + H)

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• A product-of-sums expression can always be realized directly by …

– One or more OR gates feeding a single AND gate at the circuit output

• The circuits below are referred to as two- level circuits because they have a maximum of two gates in series between an input and the circuit outputcircuit output

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• Again … The circuits below are referred to as two-level circuits …

– They have a maximum of two gates in series between an input and the circuit output

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DeMorgan’s LawsDeMorgan s Laws

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DeMorgan’s Laws

• DeMorgan’s laws …

– The inverse or complement of any Boolean expression can easily be found by successively applying the following theorems …

(X + Y)' = X' Y' (2-21)

(XY)' = X' + Y' (2-22)

• DeMorgan’s laws are easily generalized to n variables …DeMorgan s laws are easily generalized to n variables …

(2-23)(2 24)(2-24)

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DeMorgan’s Laws

• Referring to the …

– OR operation as the logical sum … and …

– The AND operation as logical product …

• DeMorgan’s laws can be stated as …g

– The complement of the product is the sum of the complements– The complement of the sum is the product of the complementsThe complement of the sum is the product of the complements

• To form the complement of an expression containing both OR and AND operations DeMorgan’s laws are applied alternatelyAND operations … DeMorgan s laws are applied alternately

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DeMorgan’s Laws

• Example … Find the complement of … (A’ + B)C’

First … apply (2-22) … (XY)' = X' + Y'

[(A’ + B)C’]’ = (A’ + B)’ + (C’)’

Now apply (2-21) … (X + Y)' = X' Y'

(A’ + B)’ + (C’)’ = (A’)’B’ + (C’)’ = AB’ + C

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DeMorgan’s Laws

• Example … Find the complement of … (AB’ + C)D’ + E

First … apply (2-21) … (X + Y)' = X' Y'

[(AB’ + C)D’ + E]’ = [(AB’ + C)D’]’E’

Now apply (2-22) … (XY)' = X' + Y'

[(AB’ + C)D’]’E’ = [(AB’ + C)’ + (D’)’]E’ = [(AB’ + C)’ + D]E’

Now apply (2-21) … (X + Y)' = X' Y'

[(AB’ C)’ D]E’ [(AB’)’C’ D]E’ i d[(AB’ + C)’ + D]E’ = [(AB’)’C’ + D]E’ … continued …138

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DeMorgan’s Laws

• Example … continued … Find the complement of … (AB’ + C)D’ + E

Now apply (2-22) … (XY)' = X' + Y'

[(AB’)’C’ + D]E’ = [(A’ + B)C’ + D]E’

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DualityDuality

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Duality

• Duals are formed by replacing …


• The dual of an expression may be found by complementing the entire expression and then complementing each individual variables

• For example …

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LAWS AND THEOREMS

Page 55

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LAWS AND THEOREMS

• Compilation of all the Laws and Theorems are listed on page 55 of the textthe text

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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

• Boolean Algebra continued …

– Chapter 3 … Boolean Algebra (Continued)

• Pages 56 - 78

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Home Work

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Homework

1. Read …

• Chapter 3 … pages 56 -78

2 Solve the following Chapter 2 problems2. Solve the following Chapter 2 problems …

• See next slide

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Homework

2. (continued) … Solve the following Chapter 2 problems …

• 2.1 … (d)• 2.2 … (b)• 2.3 … (a), (b), (c), and (f)• 2.4 … (a)• 2.5 … (a)• 2.6 … (a), (c), and (e)• 2.7 … (a)• 2 8 (a) and (b)• 2.8 … (a) and (b)• 2.9 … (a)• 2.15 … (a)

2 16 ( )151

• 2.16 … (a)

References

1. None

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logic design i (17.341) fall 2011 lecture...

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