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Chapter 2 Professor Brendan Morris, SEB 3216, - - PowerPoint PPT Presentation

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Chapter 2 Professor Brendan Morris, SEB 3216, brendan.morris@unlv.edu http://www.ee.unlv.edu/~b1morris/cpe100/ CPE100: Digital Logic Design I Combinational Logic Design Chapter 2 <1> Chapter 2 :: Topics Introduction Boolean

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SLIDE 1

Chapter 2 <1>

Professor Brendan Morris, SEB 3216, brendan.morris@unlv.edu http://www.ee.unlv.edu/~b1morris/cpe100/

Chapter 2

CPE100: Digital Logic Design I

Combinational Logic Design

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SLIDE 2

Chapter 2 <2>

  • Introduction
  • Boolean Equations
  • Boolean Algebra
  • From Logic to Gates
  • Multilevel Combinational Logic
  • X’s and Z’s, Oh My
  • Karnaugh Maps
  • Combinational Building Blocks
  • Timing

Chapter 2 :: Topics

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SLIDE 3

Chapter 2.1

Introduction

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SLIDE 4

Chapter 2 <4>

A logic circuit is composed of:

  • Inputs
  • Outputs
  • Functional specification
  • Timing specification

inputs

  • utputs

functional spec timing spec

Introduction

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SLIDE 5

Chapter 2 <5>

  • Nodes
  • Inputs: A, B, C
  • Outputs: Y, Z
  • Internal: n1
  • Circuit elements
  • E1, E2, E3
  • Each a circuit

A E1 E2 E3 B C n1 Y Z

Circuits

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SLIDE 6

Chapter 2 <6>

  • Combinational Logic (Ch 2)
  • Memoryless
  • Outputs determined by current values of inputs
  • Sequential Logic (Ch 3)
  • Has memory
  • Outputs determined by previous and current values
  • f inputs

inputs

  • utputs

functional spec timing spec

Types of Logic Circuits

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SLIDE 7

Chapter 2 <7>

  • Every element is combinational
  • Every node is either an input or connects

to exactly one output

  • The circuit contains no cyclic paths

– E.g. no connection from output to internal node

  • Example:

Rules of Combinational Composition

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SLIDE 8

Chapter 2.2

Boolean Equations

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SLIDE 9

Chapter 2 <9>

  • Functional specification of outputs in terms
  • f inputs
  • Example: S

= F(A, B, Cin) Cout = F(A, B, Cin)

A S S = A  B  Cin Cout = AB + ACin + BCin B Cin C L Cout

Boolean Equations

A B Cin S Cout 1 1 1 1 1 1 1 1 1 1 1 1

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SLIDE 10

Chapter 2 <10>

Goals:

  • Systematically express logical functions using

Boolean equations

  • To simplify Boolean equations

Functional specification

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SLIDE 11

Chapter 2 <11>

  • Complement: variable with a bar over it

A, B, C

  • Literal: variable or its complement

A, A, B, B, C, C

  • Implicant: product of literals

ABC, AC, BC

  • Minterm: product that includes all input

variables ABC, ABC, ABC

  • Maxterm: sum that includes all input variables

(A+B+C), (A+B+C), (A+B+C)

Some Definitions

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SLIDE 12

Chapter 2 <12>

  • All equations can be written in SOP form
  • Each row has a minterm
  • A minterm is a product (AND) of literals
  • Each minterm is TRUE for that row (and only that row)

A B Y 1 1 1 1 1 1 minterm A B A B A B A B minterm name m0 m1 m2 m3

Canonical Sum-of-Products (SOP) Form

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SLIDE 13

Chapter 2 <13>

Y = F(A, B) =

  • All equations can be written in SOP form
  • Each row has a minterm
  • A minterm is a product (AND) of literals
  • Each minterm is TRUE for that row (and only that row)
  • Form function by ORing minterms where the output is TRUE

A B Y 1 1 1 1 1 1 minterm A B A B A B A B minterm name m0 m1 m2 m3

Canonical Sum-of-Products (SOP) Form

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SLIDE 14

Chapter 2 <14>

Y = F(A, B) = AB + AB = Σ(m1, m3)

Canonical Sum-of-Products (SOP) Form

  • All equations can be written in SOP form
  • Each row has a minterm
  • A minterm is a product (AND) of literals
  • Each minterm is TRUE for that row (and only that row)
  • Form function by ORing minterms where the output is TRUE
  • Thus, a sum (OR) of products (AND terms)

A B Y 1 1 1 1 1 1 minterm A B A B A B A B minterm name m0 m1 m2 m3

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SLIDE 15

Chapter 2 <15>

Y = F(A, B) =

SOP Example

  • Steps:
  • Find minterms that result in Y=1
  • Sum “TRUE” minterms

A B Y 1 1 1 1 1 1

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SLIDE 16

Chapter 2 <16>

Y = F(A, B) =

SOP Example

  • Steps:
  • Find minterms that result in Y=1
  • Sum “TRUE” minterms

A B Y 1 1 1 1 1 1 minterm name minterm 𝑛0 ҧ 𝐵 ത 𝐶 𝑛1 ҧ 𝐵𝐶 𝑛2 𝑛3

𝑛0 + 𝑛1 = ҧ 𝐵 ത 𝐶 + ҧ 𝐵𝐶

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SLIDE 17

Chapter 2 <17>

Aside: Precedence

  • AND has precedence over OR
  • In other words:
  • AND is performed before OR
  • Example:
  • 𝑍 =

ҧ 𝐵 ⋅ 𝐶 + 𝐵 ⋅ 𝐶

  • Equivalent to:
  • 𝑍 =

ҧ 𝐵𝐶 + (𝐵𝐶)

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SLIDE 18

Chapter 2 <18>

  • All Boolean equations can be written in POS form
  • Each row has a maxterm
  • A maxterm is a sum (OR) of literals
  • Each maxterm is FALSE for that row (and only that row)

Canonical Product-of-Sums (POS) Form

A + B A B Y 1 1 1 1 1 1 maxterm A + B A + B A + B maxterm name M0 M1 M2 M3

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SLIDE 19

Chapter 2 <19>

  • All Boolean equations can be written in POS form
  • Each row has a maxterm
  • A maxterm is a sum (OR) of literals
  • Each maxterm is FALSE for that row (and only that row)
  • Form function by ANDing the maxterms for which the
  • utput is FALSE
  • Thus, a product (AND) of sums (OR terms)

Canonical Product-of-Sums (POS) Form

A + B A B Y 1 1 1 1 1 1 maxterm A + B A + B A + B maxterm name M0 M1 M2 M3

𝑍 = 𝑁0 ⋅ 𝑁2 = 𝐵 + 𝐶 ⋅ ( ҧ 𝐵 + 𝐶)

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SLIDE 20

Chapter 2 <20>

  • Sum of Products (SOP)
  • Implement the “ones” of the output
  • Sum all “one” terms  OR results in “one”
  • Product of Sums (POS)
  • Implement the “zeros” of the output
  • Multiply “zero” terms  AND results in “zero”

SOP and POS Comparison

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SLIDE 21

Chapter 2 <21>

  • You are going to the cafeteria for lunch

– You will eat lunch (E=1) – If it’s open (O=1) and – If they’re not serving corndogs (C=0)

  • Write a truth table for determining if you

will eat lunch (E).

O C E 1 1 1 1

Boolean Equations Example

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SLIDE 22

Chapter 2 <22>

  • You are going to the cafeteria for lunch

– You will eat lunch (E=1) – If it’s open (O=1) and – If they’re not serving corndogs (C=0)

  • Write a truth table for determining if you

will eat lunch (E).

O C E 1 1 1 1 1

Boolean Equations Example

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SLIDE 23

Chapter 2 <23>

  • SOP – sum-of-products
  • POS – product-of-sums

O C E 1 1 1 1 minterm O C O C O C O C O + C O C E 1 1 1 1 maxterm O + C O + C O + C

SOP & POS Form

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SLIDE 24

Chapter 2 <24>

  • SOP – sum-of-products
  • POS – product-of-sums

O + C O C E 1 1 1 1 1 maxterm O + C O + C O + C

O C E 1 1 1 1 1 minterm O C O C O C O C

SOP & POS Form

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SLIDE 25

Chapter 2 <25>

  • SOP – sum-of-products
  • POS – product-of-sums

O + C O C E 1 1 1 1 1 maxterm O + C O + C O + C

O C E 1 1 1 1 1 minterm O C O C O C O C

E = OC = Σ(m2)

SOP & POS Form

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SLIDE 26

Chapter 2 <26>

  • SOP – sum-of-products
  • POS – product-of-sums

O + C O C E 1 1 1 1 1 maxterm O + C O + C O + C

O C E 1 1 1 1 1 minterm O C O C O C O C

E = (O + C)(O + C)(O + C) = Π(M0, M1, M3) E = OC = Σ(m2)

SOP & POS Form

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SLIDE 27

Chapter 2.3

Boolean Algebra

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SLIDE 28

Chapter 2 <28>

  • Axioms and theorems to simplify Boolean

equations

  • Like regular algebra, but simpler: variables

have only two values (1 or 0)

  • Duality in axioms and theorems:

– ANDs and ORs, 0’s and 1’s interchanged

Boolean Algebra

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SLIDE 29

Chapter 2 <29>

Boolean Axioms

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SLIDE 30

Chapter 2 <30>

Duality in Boolean axioms and theorems:

– ANDs and ORs, 0’s and 1’s interchanged

Duality

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SLIDE 31

Chapter 2 <31>

Boolean Axioms

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SLIDE 32

Chapter 2 <32>

Boolean Axioms

Dual: Exchange:• and + 0 and 1

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SLIDE 33

Chapter 2 <33>

Boolean Axioms

Dual: Exchange:• and + 0 and 1

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SLIDE 34

Chapter 2 <34>

Basic Boolean Theorems

B = B

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SLIDE 35

Chapter 2 <35>

Basic Boolean Theorems: Duals

Dual: Exchange:• and + 0 and 1

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SLIDE 36

Chapter 2 <36>

  • B 1 = B
  • B + 0 = B

T1: Identity Theorem

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SLIDE 37

Chapter 2 <37>

1

= =

B B B B

  • B 1 = B
  • B + 0 = B

T1: Identity Theorem

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SLIDE 38

Chapter 2 <38>

1

= =

B B B B

  • B 1 = B
  • B + 0 = B

T1: Identity Theorem

1 1 1 1

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SLIDE 39

Chapter 2 <39>

  • Simplification of digital logic  connecting

wires with a on/off switch

  • X = 0 (switch open)
  • X = 1 (switch closed)

Switching Algebra

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SLIDE 40

Chapter 2 <40>

  • Switching circuit in series performs AND
  • 1 is connected to 2 iff A AND B are 1

Series Switching Network: AND

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SLIDE 41

Chapter 2 <41>

  • Switching circuit in parallel performs OR
  • 1 is connected to 2 if A OR B is 1

Parallel Switching Network: OR

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SLIDE 42

Chapter 2 <42>

1

= =

B B B B

  • B 1 = B
  • B + 0 = B

T1: Identity Theorem

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SLIDE 43

Chapter 2 <43>

  • B 0 = 0
  • B + 1 = 1

T2: Null Element Theorem

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SLIDE 44

Chapter 2 <44>

= =

B 1 B 1

  • B 0 = 0
  • B + 1 = 1

T2: Null Element Theorem

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SLIDE 45

Chapter 2 <45>

  • B B = B
  • B + B = B

T3: Idempotency Theorem

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SLIDE 46

Chapter 2 <46>

B

= =

B B B B B

  • B B = B
  • B + B = B

T3: Idempotency Theorem

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SLIDE 47

Chapter 2 <47>

  • B = B

T4: Involution Theorem

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SLIDE 48

Chapter 2 <48>

= B

B

  • B = B

T4: Involution Theorem

1 1 1

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SLIDE 49

Chapter 2 <49>

  • B B = 0
  • B + B = 1

T5: Complements Theorem

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SLIDE 50

Chapter 2 <50>

B

= =

B B B 1

  • B B = 0
  • B + B = 1

T5: Complements Theorem

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SLIDE 51

Chapter 2 <51>

Recap: Basic Boolean Theorems

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SLIDE 52

Chapter 2.3.3

Theorems of Several Variables

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SLIDE 53

Chapter 2 <53>

Boolean Theorems of Several Vars

Number Theorem Name

T6 B•C = C•B Commutativity T7 (B•C) • D = B • (C • D) Associativity T8 B • (C + D) = (B•C) + (B•D) Distributivity T9 B• (B+C) = B Covering T10 (B•C) + (B•C) = B Combining T11 B•C + (B•D) + (C•D) = B•C + B•D Consensus

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SLIDE 54

Chapter 2 <54>

Boolean Theorems of Several Vars

Number Theorem Name

T6 B•C = C•B Commutativity T7 (B•C) • D = B • (C • D) Associativity T8 B • (C + D) = (B•C) + (B•D) Distributivity T9 B• (B+C) = B Covering T10 (B•C) + (B•C) = B Combining T11 B•C + (B•D) + (C•D) = B•C + B•D Consensus

How do we prove these are true?

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SLIDE 55

Chapter 2 <55>

How to Prove Boolean Relation

  • Method 1: Perfect induction
  • Method 2: Use other theorems and axioms

to simplify the equation

  • Make one side of the equation look like

the other

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SLIDE 56

Chapter 2 <56>

Proof by Perfect Induction

  • Also called: proof by exhaustion
  • Check every possible input value
  • If two expressions produce the same value

for every possible input combination, the expressions are equal

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SLIDE 57

Chapter 2 <57>

Example: Proof by Perfect Induction

Number Theorem Name

T6 B•C = C•B Commutativity

0 0 0 1 1 0 1 1 B C BC CB

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SLIDE 58

Chapter 2 <58>

Example: Proof by Perfect Induction

Number Theorem Name

T6 B•C = C•B Commutativity

0 0 0 1 1 0 1 1 B C BC CB 0 0 0 0 1 1

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SLIDE 59

Chapter 2 <59>

Boolean Theorems of Several Vars

Number Theorem Name

T6 B•C = C•B Commutativity T7 (B•C) • D = B • (C • D) Associativity T8 B • (C + D) = (B•C) + (B•D) Distributivity T9 B• (B+C) = B Covering T10 (B•C) + (B•C) = B Combining T11 B•C + (B•D) + (C•D) = B•C + B•D Consensus

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SLIDE 60

Chapter 2 <60>

T7: Associativity

Number Theorem Name

T7 (B•C) • D = B • (C • D) Associativity

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SLIDE 61

Chapter 2 <61>

T8: Distributivity

Number Theorem Name

T8 B • (C + D) = (B•C) + (B•D) Distributivity

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SLIDE 62

Chapter 2 <62>

T9: Covering

Number Theorem Name

T9 B• (B+C) = B Covering

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SLIDE 63

Chapter 2 <63>

T9: Covering

Number Theorem Name

T9 B• (B+C) = B Covering

Prove true by:

  • Method 1: Perfect induction
  • Method 2: Using other theorems and axioms
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SLIDE 64

Chapter 2 <64>

T9: Covering

Number Theorem Name

T9 B• (B+C) = B Covering

0 0 0 1 1 0 1 1 B C (B+C) B(B+C)

Method 1: Perfect Induction

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SLIDE 65

Chapter 2 <65>

T9: Covering

Number Theorem Name

T9 B• (B+C) = B Covering

Method 1: Perfect Induction

0 0 0 1 1 0 1 1 B C (B+C) B(B+C) 1 1 1 1 1

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SLIDE 66

Chapter 2 <66>

T9: Covering

Number Theorem Name

T9 B• (B+C) = B Covering

Method 2: Prove true using other axioms and theorems.

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SLIDE 67

Chapter 2 <67>

T9: Covering

Number Theorem Name

T9 B• (B+C) = B Covering

Method 2: Prove true using other axioms and theorems. B•(B+C) = B•B + B•C T8: Distributivity = B + B•C T3: Idempotency = B•(1 + C) T8: Distributivity = B•(1) T2: Null element = B T1: Identity

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SLIDE 68

Chapter 2 <68>

T10: Combining

Number Theorem Name

T10 (B•C) + (B•C) = B Combining

Prove true using other axioms and theorems:

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SLIDE 69

Chapter 2 <69>

T10: Combining

Number Theorem Name

T10 (B•C) + (B•C) = B Combining

Prove true using other axioms and theorems: B•C + B•C = B•(C+C) T8: Distributivity = B•(1) T5’: Complements = B T1: Identity

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SLIDE 70

Chapter 2 <70>

T11: Consensus

Number Theorem Name

T11 (B•C) + (B•D) + (C•D) = (B•C) + B•D Consensus

Prove true using (1) perfect induction or (2) other axioms and theorems.

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SLIDE 71

Chapter 2 <71>

Recap: Boolean Thms of Several Vars

Number Theorem Name

T6 B•C = C•B Commutativity T7 (B•C) • D = B • (C • D) Associativity T8 B • (C + D) = (B•C) + (B•D) Distributivity T9 B• (B+C) = B Covering T10 (B•C) + (B•C) = B Combining T11 B•C + (B•D) + (C•D) = B•C + B•D Consensus

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SLIDE 72

Chapter 2 <72>

Boolean Thms of Several Vars: Duals

# Theorem Dual Name T6 B•C = C•B B+C = C+B Commutativity T7 (B•C) • D = B • (C•D) (B + C) + D = B + (C + D) Associativity T8 B • (C + D) = (B•C) + (B•D) B + (C•D) = (B+C) (B+D) Distributivity T9 B • (B+C) = B B + (B•C) = B Covering T10 (B•C) + (B•C) = B (B+C) • (B+C) = B Combining T11 (B•C) + (B•D) + (C•D) = (B•C) + (B•D) (B+C) • (B+D) • (C+D) = (B+C) • (B+D) Consensus

Dual: Replace: • with + 0 with 1

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SLIDE 73

Chapter 2 <73>

Boolean Thms of Several Vars: Duals

# Theorem Dual Name T6 B•C = C•B B+C = C+B Commutativity T7 (B•C) • D = B • (C•D) (B + C) + D = B + (C + D) Associativity T8 B • (C + D) = (B•C) + (B•D) B + (C•D) = (B+C) (B+D) Distributivity T9 B • (B+C) = B B + (B•C) = B Covering T10 (B•C) + (B•C) = B (B+C) • (B+C) = B Combining T11 (B•C) + (B•D) + (C•D) = (B•C) + (B•D) (B+C) • (B+D) • (C+D) = (B+C) • (B+D) Consensus

Dual: Replace: • with + 0 with 1

Warning: T8’ differs from traditional algebra: OR (+) distributes over AND (•)

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SLIDE 74

Chapter 2 <74>

Boolean Thms of Several Vars: Duals

# Theorem Dual Name T6 B•C = C•B B+C = C+B Commutativity T7 (B•C) • D = B • (C•D) (B + C) + D = B + (C + D) Associativity T8 B • (C + D) = (B•C) + (B•D) B + (C•D) = (B+C) (B+D) Distributivity T9 B • (B+C) = B B + (B•C) = B Covering T10 (B•C) + (B•C) = B (B+C) • (B+C) = B Combining T11 (B•C) + (B•D) + (C•D) = (B•C) + (B•D) (B+C) • (B+D) • (C+D) = (B+C) • (B+D) Consensus Axioms and theorems are useful for simplifying equations.

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SLIDE 75

Chapter 2.3.5

Simplifying Equations

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SLIDE 76

Chapter 2 <76>

Simplifying an Equation

  • Reducing an equation to the fewest

number of implicants, where each implicant has the fewest literals

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SLIDE 77

Chapter 2 <77>

Simplifying an Equation

  • Reducing an equation to the fewest

number of implicants, where each implicant has the fewest literals

Recall:

  • Implicant: product of literals

ABC, AC, BC

  • Literal: variable or its complement

A, A, B, B, C, C

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SLIDE 78

Chapter 2 <78>

Simplifying an Equation

  • Reducing an equation to the fewest

number of implicants, where each implicant has the fewest literals

Recall:

  • Implicant: product of literals

ABC, AC, BC

  • Literal: variable or its complement

A, A, B, B, C, C

  • Also called: minimizing the equation
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SLIDE 79

Chapter 2 <79>

Simplification methods

  • Distributivity (T8, T8’)

B (C+D) = BC + BD B + CD = (B+ C)(B+D)

  • Covering (T9’)

A + AP = A

  • Combining (T10)

PA + PA = P

𝑄 𝐵 Minterm ത 𝑄 ҧ 𝐵 1 ത 𝑄𝐵 1 𝑄 ҧ 𝐵 1 1 𝑄𝐵

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SLIDE 80

Chapter 2 <80>

Simplification methods

  • Distributivity (T8, T8’)

B (C+D) = BC + BD B + CD = (B+ C)(B+D)

  • Covering (T9’)

A + AP = A

  • Combining (T10)

PA + PA = P

  • Expansion

P = PA + PA A = A + AP

  • Duplication

A = A + A

𝑄 𝐵 Minterm ത 𝑄 ҧ 𝐵 1 ത 𝑄𝐵 1 𝑄 ҧ 𝐵 1 1 𝑄𝐵

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SLIDE 81

Chapter 2 <81>

Simplification methods

  • Distributivity (T8, T8’)

B (C+D) = BC + BD B + CD = (B+ C)(B+D)

  • Covering (T9’)

A + AP = A

  • Combining (T10)

PA + PA = P

  • Expansion

P = PA + PA A = A + AP

  • Duplication

A = A + A

  • A combination of Combining/Covering

PA + A = P + A

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SLIDE 82

Chapter 2 <82>

Simplification methods

  • A combination of Combining/Covering

PA + A = P + A

Proof: PA + A = PA + (A + AP) T9’ Covering = PA + PA + A T6 Commutativity = P(A + A) + A T8 Distributivity = P(1) + A T5’ Complements = P + A T1 Identity

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SLIDE 83

Chapter 2 <83>

Recap: Boolean Thms of Several Vars

# Theorem Dual Name T6 B•C = C•B B+C = C+B Commutativity T7 (B•C) • D = B • (C•D) (B + C) + D = B + (C + D) Associativity T8 B • (C + D) = (B•C) + (B•D) B + (C•D) = (B+C) (B+D) Distributivity T9 B • (B+C) = B B + (B•C) = B Covering T10 (B•C) + (B•C) = B (B+C) • (B+C) = B Combining T11 (B•C) + (B•D) + (C•D) = (B•C) + (B•D) (B+C) • (B+D) • (C+D) = (B+C) • (B+D) Consensus

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SLIDE 84

Chapter 2 <84>

T11: Consensus

Number Theorem Name

T11 (B•C) + (B•D) + (C•D) = (B•C) + B•D Consensus

Prove using other theorems and axioms:

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SLIDE 85

Chapter 2 <85>

T11: Consensus

Number Theorem Name

T11 (B•C) + (B•D) + (C•D) = (B•C) + B•D Consensus

Prove using other theorems and axioms: B•C + B•D + C•D = BC + BD + (CDB+CDB) T10: Combining = BC + BD + BCD+BCD T6: Commutativity = BC + BCD + BD + BCD T6: Commutativity = (BC + BCD) + (BD + BCD) T7: Associativity = BC + BD T9’: Covering

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SLIDE 86

Chapter 2 <86>

Simplification methods

  • Distributivity (T8, T8’)

B (C+D) = BC + BD B + CD = (B+ C)(B+D)

  • Covering (T9’)

A + AP = A

  • Combining (T10)

PA + PA = P

  • Expansion

P = PA + PA A = A + AP

  • Duplication

A = A + A

  • A combination of Combining/Covering

PA + A = P + A

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SLIDE 87

Chapter 2 <87>

Y = AB + AB

Simplifying Boolean Equations

Example 1:

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SLIDE 88

Chapter 2 <88>

Y = AB + AB Y = A T10: Combining

  • r

= A(B + B) T8: Distributivity = A(1) T5’: Complements = A T1: Identity

Simplifying Boolean Equations

Example 1:

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SLIDE 89

Chapter 2 <89>

Simplification methods

  • Distributivity (T8, T8’)

B (C+D) = BC + BD B + CD = (B+ C)(B+D)

  • Covering (T9’)

A + AP = A

  • Combining (T10)

PA + PA = P

  • Expansion

P = PA + PA A = A + AP

  • Duplication

A = A + A

  • A combination of Combining/Covering

PA + A = P + A

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SLIDE 90

Chapter 2 <90>

Y = A(AB + ABC)

Example 2:

Simplifying Boolean Equations

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SLIDE 91

Chapter 2 <91>

Y = A(AB + ABC) = A(AB(1 + C)) T8: Distributivity = A(AB(1)) T2’: Null Element = A(AB) T1: Identity = (AA)B T7: Associativity = AB T3: Idempotency

Example 2:

Simplifying Boolean Equations

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SLIDE 92

Chapter 2 <92>

Simplification methods

  • Distributivity (T8, T8’)

B (C+D) = BC + BD B + CD = (B+ C)(B+D)

  • Covering (T9’)

A + AP = A

  • Combining (T10)

PA + PA = P

  • Expansion

P = PA + PA A = A + AP

  • Duplication

A = A + A

  • A combination of Combining/Covering

PA + A = P + A

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SLIDE 93

Chapter 2 <93>

Y = A’BC + A’ Recall: A’ = A

Example 3:

Simplifying Boolean Equations

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SLIDE 94

Chapter 2 <94>

Y = A’BC + A’ Recall: A’ = A = A’ T9’ Covering: X + XY = X

  • r

= A’(BC + 1) T8: Distributivity = A’(1) T2’: Null Element = A’ T1: Identity

Example 3:

Simplifying Boolean Equations

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SLIDE 95

Chapter 2 <95>

Simplification methods

  • Distributivity (T8, T8’)

B (C+D) = BC + BD B + CD = (B+ C)(B+D)

  • Covering (T9’)

A + AP = A

  • Combining (T10)

PA + PA = P

  • Expansion

P = PA + PA A = A + AP

  • Duplication

A = A + A

  • A combination of Combining/Covering

PA + A = P + A

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SLIDE 96

Chapter 2 <96>

Y = AB’C + ABC + A’BC

Example 4:

Simplifying Boolean Equations

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SLIDE 97

Chapter 2 <97>

Y = AB’C + ABC + A’BC

= AB’C + ABC + ABC + A’BC T3’: Idempotency = (AB’C+ABC) + (ABC+A’BC) T7’: Associativity = AC + BC T10: Combining

Example 4:

Simplifying Boolean Equations

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SLIDE 98

Chapter 2 <98>

Simplification methods

  • Distributivity (T8, T8’)

B (C+D) = BC + BD B + CD = (B+ C)(B+D)

  • Covering (T9’)

A + AP = A

  • Combining (T10)

PA + PA = P

  • Expansion

P = PA + PA A = A + AP

  • Duplication

A = A + A

  • A combination of Combining/Covering

PA + A = P + A

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SLIDE 99

Chapter 2 <99>

Y = AB + BC +B’D’ + AC’D’

Example 5:

Simplifying Boolean Equations

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SLIDE 100

Chapter 2 <100>

Y = AB + BC +B’D’ + AC’D’

Method 1: Y = AB + BC + B’D’ + (ABC’D’ + AB’C’D’) T10: Combining = (AB + ABC’D’) + BC + (B’D’ + AB’C’D’) T6: Commutativity T7: Associativity = AB + BC + B’D’ T9: Covering

Example 5:

Simplifying Boolean Equations

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SLIDE 101

Chapter 2 <101>

Y = AB + BC +B’D’ + AC’D’

Method 1: Y = AB + BC + B’D’ + (ABC’D’ + AB’C’D’) T10: Combining = (AB + ABC’D’) + BC + (B’D’ + AB’C’D’) T6: Commutativity T7: Associativity = AB + BC + B’D’ T9: Covering Method 2: Y = AB + BC + B’D’ + AC’D’ + AD’ T11: Consensus = AB + BC + B’D’ + AD’ T9: Covering = AB + BC + B’D’ T11: Consensus

Example 5:

Simplifying Boolean Equations

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SLIDE 102

Chapter 2 <102>

Simplification methods

  • Distributivity (T8, T8’)

B (C+D) = BC + BD B + CD = (B+ C)(B+D)

  • Covering (T9’)

A + AP = A

  • Combining (T10)

PA + PA = P

  • Expansion

P = PA + PA A = A + AP

  • Duplication

A = A + A

  • A combination of Combining/Covering

PA + A = P + A

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SLIDE 103

Chapter 2 <103>

Example 6:

Simplifying Boolean Equations

Y = (A + BC)(A + DE)

Apply T8’ first when possible: W+XZ = (W+X)(W+Z)

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SLIDE 104

Chapter 2 <104>

Y = (A + BC)(A + DE)

Apply T8’ first when possible: W+XZ = (W+X)(W+Z) Make: X = BC, Z = DE and rewrite equation Y = (A+X)(A+Z) substitution (X=BC, Z=DE) = A + XZ T8’: Distributivity = A + BCDE substitution

  • r

Y = AA + ADE + ABC + BCDE T8: Distributivity = A + ADE + ABC + BCDE T3: Idempotency = A + ADE + ABC + BCDE = A + ABC + BCDE T9’: Covering = A + BCDE T9’: Covering

Example 6:

Simplifying Boolean Equations

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SLIDE 105

Chapter 2 <105>

Y = (A + BC)(A + DE)

Apply T8’ first when possible: W+XZ = (W+X)(W+Z) Make: X = BC, Z = DE and rewrite equation Y = (A+X)(A+Z) substitution (X=BC, Z=DE) = A + XZ T8’: Distributivity = A + BCDE substitution

  • r

Y = AA + ADE + ABC + BCDE T8: Distributivity = A + ADE + ABC + BCDE T3: Idempotency = A + ADE + ABC + BCDE = A + ABC + BCDE T9’: Covering = A + BCDE T9’: Covering

Example 6:

Simplifying Boolean Equations

This is called multiplying out an expression to get sum-of-products (SOP) form.

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SLIDE 106

Chapter 2 <106>

An expression is in simplified sum-of- products (SOP) form when all products contain literals only.

  • SOP form: Y = AB + BC’ + DE
  • NOT SOP form: Y = DF + E(A’+B)
  • SOP form: Z = A + BC + DE’F

Multiplying Out: SOP Form

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SLIDE 107

Chapter 2 <107>

Y = (A + C + D + E)(A + B)

Apply T8’ first when possible: W+XZ = (W+X)(W+Z)

Example:

Multiplying Out: SOP Form

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SLIDE 108

Chapter 2 <108>

Y = (A + C + D + E)(A + B)

Apply T8’ first when possible: W+XZ = (W+X)(W+Z) Make: X = (C+D+E), Z = B and rewrite equation Y = (A+X)(A+Z) substitution (X=(C+D+E), Z=B) = A + XZ T8’: Distributivity = A + (C+D+E)B substitution = A + BC + BD + BE T8: Distributivity

  • r

Y = AA+AB+AC+BC+AD+BD+AE+BE T8: Distributivity = A+AB+AC+AD+AE+BC+BD+BE T3: Idempotency = A + BC + BD + BE T9’: Covering

Example:

Multiplying Out: SOP Form

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SLIDE 109

Chapter 2 <109>

  • SOP – sum-of-products
  • POS – product-of-sums

O + C O C E 1 1 1 1 1 maxterm O + C O + C O + C

O C E 1 1 1 1 1 minterm O C O C O C O C

E = (O + C)(O + C)(O + C) = Π(M0, M1, M3) E = OC = Σ(m2)

Canonical SOP & POS Form

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SLIDE 110

Chapter 2 <110>

An expression is in simplified product-

  • f-sums (POS) form when all sums

contain literals only.

  • POS form: Y = (A+B)(C+D)(E’+F)
  • NOT POS form: Y = (D+E)(F’+GH)
  • POS form: Z = A(B+C)(D+E’)

Factoring: POS Form

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SLIDE 111

Chapter 2 <111>

Y = (A + B’CDE)

Apply T8’ first when possible: W+XZ = (W+X)(W+Z)

Example 1:

Factoring: POS Form

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SLIDE 112

Chapter 2 <112>

Y = (A + B’CDE)

Apply T8’ first when possible: W+XZ = (W+X)(W+Z) Make: X = B’C, Z = DE and rewrite equation Y = (A+XZ) substitution (X=B’C, Z=DE) = (A+B’C)(A+DE) T8’: Distributivity = (A+B’)(A+C)(A+D)(A+E) T8’: Distributivity

Example 1:

Factoring: POS Form

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SLIDE 113

Chapter 2 <113>

Y = AB + C’DE + F

Apply T8’ first when possible: W+XZ = (W+X)(W+Z)

Example 2:

Factoring: POS Form

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SLIDE 114

Chapter 2 <114>

Y = AB + C’DE + F

Apply T8’ first when possible: W+XZ = (W+X)(W+Z) Make: W = AB, X = C’, Z = DE and rewrite equation Y = (W+XZ) + F substitution W = AB, X = C’, Z = DE = (W+X)(W+Z) + F T8’: Distributivity = (AB+C’)(AB+DE)+F substitution = (A+C’)(B+C’)(AB+D)(AB+E)+F T8’: Distributivity = (A+C’)(B+C’)(A+D)(B+D)(A+E)(B+E)+F T8’: Distributivity = (A+C’+F)(B+C’+F)(A+D+F)(B+D+F)(A+E+F)(B+E+F) T8’: Distributivity

Example 2:

Factoring: POS Form

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SLIDE 115

Chapter 2 <115>

Boolean Thms of Several Vars: Duals

# Theorem Dual Name T6 B•C = C•B B+C = C+B Commutativity T7 (B•C) • D = B • (C•D) (B + C) + D = B + (C + D) Associativity T8 B • (C + D) = (B•C) + (B•D) B + (C•D) = (B+C) (B+D) Distributivity T9 B • (B+C) = B B + (B•C) = B Covering T10 (B•C) + (B•C) = B (B+C) • (B+C) = B Combining T11 (B•C) + (B•D) + (C•D) = (B•C) + (B•D) (B+C) • (B+D) • (C+D) = (B+C) • (B+D) Consensus Axioms and theorems are useful for simplifying equations.

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SLIDE 116

Chapter 2 <116>

Simplification methods

  • Distributivity (T8, T8’)

B (C+D) = BC + BD B + CD = (B+ C)(B+D)

  • Covering (T9’)

A + AP = A

  • Combining (T10)

PA + PA = P

  • Expansion

P = PA + PA A = A + AP

  • Duplication

A = A + A

  • A combination of Combining/Covering

PA + A = P + A

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SLIDE 117

Chapter 2 <117>

DeMorgan’s Theorem

Number Theorem Name

T12 B0•B1•B2… = B0+B1+B2… DeMorgan’s Theorem

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SLIDE 118

Chapter 2 <118>

DeMorgan’s Theorem

Number Theorem Name

T12 B0•B1•B2… = B0+B1+B2… DeMorgan’s Theorem

The complement of the product is the sum of the complements

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SLIDE 119

Chapter 2 <119>

# Theorem Dual Name

T12 B0•B1•B2… = B0+B1+B2… B0+B1+B2… = B0•B1•B2… DeMorgan’s Theorem

DeMorgan’s Theorem: Dual

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SLIDE 120

Chapter 2 <120>

# Theorem Dual Name

T12 B0•B1•B2… = B0+B1+B2… B0+B1+B2… = B0•B1•B2… DeMorgan’s Theorem

DeMorgan’s Theorem: Dual

The complement of the product is the sum of the complements.

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SLIDE 121

Chapter 2 <121>

DeMorgan’s Theorem: Dual

# Theorem Dual Name

T12 B0•B1•B2… = B0+B1+B2… B0+B1+B2… = B0•B1•B2… DeMorgan’s Theorem

The complement of the product is the sum of the complements. Dual: The complement of the sum is the product of the complements.

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SLIDE 122

Chapter 2 <122>

  • Y = AB = A + B
  • Y = A + B = A B

A B Y A B Y A B Y A B Y

DeMorgan’s Theorem

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SLIDE 123

Chapter 2 <123>

Y = (A+BD)C

DeMorgan’s Theorem Example 1

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SLIDE 124

Chapter 2 <124>

Y = (A+BD)C = (A+BD) + C = (A•(BD)) + C = (A•(BD)) + C = ABD + C

DeMorgan’s Theorem Example 1

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SLIDE 125

Chapter 2 <125>

Y = (ACE+D) + B

DeMorgan’s Theorem Example 2

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SLIDE 126

Chapter 2 <126>

Y = (ACE+D) + B = (ACE+D) • B = (ACE•D) • B = ((AC+E)•D) • B = ((AC+E)•D) • B = (ACD + DE) • B = ABCD + BDE

DeMorgan’s Theorem Example 2

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SLIDE 127

Chapter 2 <127>

  • SOP – sum-of-products
  • POS – product-of-sums

O + C O C E 1 1 1 1 1 maxterm O + C O + C O + C

O C E 1 1 1 1 1 minterm O C O C O C O C

E = (O + C)(O + C)(O + C) = Π(M0, M1, M3) E = OC = Σ(m2)

Canonical SOP & POS Form Revisited

How do we implement this logic function with gates?

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SLIDE 128

Chapter 2 <128>

  • “Add” literal/complement terms to reverse

simplification (expand literal)

  • Example
  • 𝑍 = 𝐷
  • 𝑍 = 𝐷 + 𝐵 ҧ

𝐵

  • 𝑍 = 𝐷 + 𝐵 ⋅ (𝐷 +

ҧ 𝐵)

  • 𝑍 = [ 𝐷 + 𝐵 + 𝐶 ത

𝐶](𝐷 + ҧ 𝐵)

  • 𝑍 =

𝐷 + 𝐵 + 𝐶 𝐷 + 𝐵 + ത 𝐶 𝐷 + ҧ 𝐵

  • …

Canonical POS Expansion

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SLIDE 129

Chapter 2 <129>

Midterm 1: Wednesday, Sep. 30

  • 1 hour and 15 minutes + 15 mins for scan and

upload on Canvas

  • Chap 1 – 2.6
  • Closed book, closed notes
  • No calculator
  • Boolean Theorems & Axioms document will

be attached to the exam for your convenience

Reminder

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SLIDE 130

Chapter 2.4

From Logic to Gates

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SLIDE 131

Chapter 2 <131>

  • Two-level logic: ANDs followed by ORs
  • Example: Y = ABC + ABC + ABC

B A C Y minterm: ABC minterm: ABC minterm: ABC A B C

From Logic to Gates

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SLIDE 132

Chapter 2 <132>

  • Inputs on the left (or top)
  • Outputs on right (or bottom)
  • Gates flow from left to right
  • Straight wires are best

Circuit Schematics Rules

𝑍 = ത 𝐶 ҧ 𝐷 + 𝐵 ത 𝐶

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SLIDE 133

Chapter 2 <133>

  • Wires always connect at a T junction
  • A dot where wires cross indicates a

connection between the wires

  • Wires crossing without a dot make no

connection

wires connect at a T junction wires connect at a dot wires crossing without a dot do not connect

Circuit Schematic Rules (cont.)

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SLIDE 134

Chapter 2 <134>

A1 A 1 1 1 1 Y3 Y2 Y1 Y0 A

3

A2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 A0 A1 PRIORITY CiIRCUIT A2 A3 Y0 Y1 Y2 Y3

  • Example: Priority Circuit

Output asserted corresponding to most significant TRUE input

Multiple-Output Circuits

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SLIDE 135

Chapter 2 <135>

A1 A 1 1 1 1 Y3 Y2 Y1 Y0 1 1 1 A

3

A2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 A0 A1 PRIORITY CiIRCUIT A2 A3 Y0 Y1 Y2 Y3

  • Example: Priority Circuit

Output asserted corresponding to most significant TRUE input

Multiple-Output Circuits

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SLIDE 136

Chapter 2 <136>

A1 A 1 1 1 1 Y3 Y2 Y1 Y0 1 1 1 A

3

A2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

A3A2A1A0 Y3 Y2 Y1 Y0

Priority Circuit Hardware

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SLIDE 137

Chapter 2 <137>

A1 A 1 1 1 1 Y3 Y2 Y1 Y0 1 1 1 A

3

A2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

A1 A0 1 1 X X X Y3 Y2 Y1 Y0 1 1 1 A3 A2 1 X X 1 1 X

Don’t Cares

A3A2A1A0 Y3 Y2 Y1 Y0

  • Simplify truth table by ignoring entries

Much easier to read off Boolean equations = 𝐵3 = 𝐵3𝐵2 = 𝐵3 𝐵2 𝐵1 = 𝐵3 𝐵2 𝐵1 𝐵0

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SLIDE 138

Chapter 2.5

Multilevel Combinational Logic

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SLIDE 139

Chapter 2 <139>

  • Two-level logic: One level-ANDs, one-level

OR

  • Can have high hardware requirements
  • Multiple inputs to AND and OR gates
  • Multilevel design can simplify schematic

Multilevel Hardware Simplification

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SLIDE 140

Chapter 2 <140>

  • 3-input XOR example (odd parity)
  • 𝑍 =

ҧ 𝐵 ത 𝐶𝐷 + ҧ 𝐵𝐶 ҧ 𝐷 + 𝐵 ത 𝐶 ҧ 𝐷 + 𝐵𝐶𝐷

Multilevel Hardware Simplification

4x 3-input AND 1x 4-input OR 2x 2-input XOR

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SLIDE 141

Chapter 2 <141>

  • Backward:

– Body changes – Adds bubbles to inputs

  • Forward:

– Body changes – Adds bubble to output

A B Y A B Y A B Y A B Y

Bubble Pushing

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SLIDE 142

Chapter 2 <142>

A B Y C D

  • What is the Boolean expression for this

circuit?

Bubble Pushing

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SLIDE 143

Chapter 2 <143>

A B Y C D

  • What is the Boolean expression for this

circuit? Y = AB + CD

Bubble Pushing

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SLIDE 144

Chapter 2 <144>

A B C D Y

  • Begin at output, then work toward inputs
  • Push bubbles on final output back
  • Draw gates in a form so bubbles cancel

Bubble Pushing Rules

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SLIDE 145

Chapter 2 <145>

A B C Y D

Bubble Pushing Example

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SLIDE 146

Chapter 2 <146>

A B C Y D no output bubble

Bubble Pushing Example

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SLIDE 147

Chapter 2 <147>

bubble on input and output A B C D Y A B C Y D no output bubble

Bubble Pushing Example

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SLIDE 148

Chapter 2 <148>

A B C D Y bubble on input and output A B C D Y A B C Y D Y = ABC + D no output bubble no bubble on input and output

Bubble Pushing Example

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SLIDE 149

Chapter 2.6

X’s and Z’s, Oh My

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SLIDE 150

Chapter 2 <150>

  • Contention: circuit tries to drive output to 1 and 0

– Actual value somewhere in between – Could be 0, 1, or in forbidden zone – Might change with voltage, temperature, time, noise – Often causes excessive power dissipation

  • Warnings:

– Contention usually indicates a bug. – X is used for “don’t care” and contention - look at the context to tell them apart

A = 1 Y = X B = 0

Contention: X

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SLIDE 151

Chapter 2 <151>

  • Floating, high impedance, open, high Z
  • Floating output might be 0, 1, or

somewhere in between

– A voltmeter won’t indicate whether a node is floating Tristate Buffer

E A Y Z 1 Z 1 1 1 1 A E Y

Floating: Z

Note: tristate buffer has an enable bit (𝐹) to turn

  • n the gate
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SLIDE 152

Chapter 2 <152>

  • Floating nodes are used in tristate

busses

– Many different drivers – Exactly one is active at

  • nce

en1 to bus from bus en2 to bus from bus en3 to bus from bus en4 to bus from bus

shared bus processor video Ethernet memory

Tristate Busses

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SLIDE 153

Chapter 2.7

Karnaugh Maps

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SLIDE 154

Chapter 2 <154>

  • Boolean expressions can be minimized by

combining terms

  • PA + PA = P
  • K-maps minimize equations graphically
  • Put terms to combine close to one another

C 00 01 1 Y 11 10 AB 1 1 C 00 01 1 Y 11 10 AB ABC ABC ABC ABC ABC ABC ABC ABC B C 1 1 1 1 A 1 1 1 1 1 1 1 1 1 1 Y

Karnaugh Maps (K-Maps)

𝑍 = ҧ 𝐵 ത 𝐶 ҧ 𝐷 + ҧ 𝐵 ത 𝐶𝐷 = ҧ 𝐵 ത 𝐶(𝐷 + ҧ 𝐷)

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SLIDE 155

Chapter 2 <155>

C 00 01 1 Y 11 10 AB 1 1

B C 1 1 1 1 A 1 1 1 1 1 1 1 1 1 1 Y

  • Circle 1’s in adjacent squares
  • In Boolean expression, include only

literals whose true and complement form are not in the circle

𝑍 = ҧ 𝐵 ത 𝐶

K-Map

𝐷 not included because both 𝐷 and ҧ 𝐷 included in circle

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SLIDE 156

Chapter 2 <156>

C 00 01 1 Y 11 10 AB ABC ABC ABC ABC ABC ABC ABC ABC

1 B C Y 1 1 1 1 1 Truth Table C 00 01 1 Y 11 10 AB A 1 1 1 1 1 1 1 1 1 K-Map

3-Input K-Map

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SLIDE 157

Chapter 2 <157>

  • Complement: variable with a bar over it

A, B, C

  • Literal: variable or its complement

A, A, B, B, C, C

  • Implicant: product of literals

ABC, AC, BC

  • Prime implicant: implicant corresponding to

the largest circle in a K-map

K-Map Definitions

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SLIDE 158

Chapter 2 <158>

  • Every 1 must be circled at least once
  • Each circle must span a power of 2 (i.e. 1, 2,

4) squares in each direction

  • Each circle must be as large as possible
  • A circle may wrap around the edges
  • A “don't care” (X) is circled only if it helps

minimize the equation

K-Map Rules

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SLIDE 159

Chapter 2 <159>

01 11 01 11 10 00 00 10 AB CD Y C D 1 1 1 1 B 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Y A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

4-Input K-Map

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SLIDE 160

Chapter 2 <160>

01 11 1 1 1 1 01 1 1 1 1 1 11 10 00 00 10 AB CD Y C D 1 1 1 1 B 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Y A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

4-Input K-Map

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SLIDE 161

Chapter 2 <161>

01 11 1 1 1 1 01 1 1 1 1 1 11 10 00 00 10 AB CD Y Y = AC + ABD + ABC + BD C D 1 1 1 1 B 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Y A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

4-Input K-Map

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SLIDE 162

Chapter 2 <162>

C D 1 1 1 1 B 1 1 1 1 1 1 1 1 1 1 1 X 1 1 Y A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 X X X X X X 01 11 01 11 10 00 00 10 AB CD Y

K-Maps with Don’t Cares

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SLIDE 163

Chapter 2 <163>

C D 1 1 1 1 B 1 1 1 1 1 1 1 1 1 1 1 X 1 1 Y A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 X X X X X X 01 11 1 X X X 1 1 01 1 1 1 1 X X X X 11 10 00 00 10 AB CD Y

K-Maps with Don’t Cares

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SLIDE 164

Chapter 2 <164>

C D 1 1 1 1 B 1 1 1 1 1 1 1 1 1 1 1 X 1 1 Y A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 X X X X X X 01 11 1 X X X 1 1 01 1 1 1 1 X X X X 11 10 00 00 10 AB CD Y Y = A + BD + C

K-Maps with Don’t Cares

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SLIDE 165

Chapter 2 <165>

01 11 1 1 1 1 01 1 1 1 1 1 11 10 00 00 10 AB CD Y C D 1 1 1 1 B 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Y A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

4-Input K-Map: POS & SOP Form

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SLIDE 166

Chapter 2 <166>

01 11 1 1 1 1 01 1 1 1 1 1 11 10 00 00 10 AB CD Y Y = AC + ABD + ABC + BD C D 1 1 1 1 B 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Y A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

4-Input K-Map: POS Form

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SLIDE 167

Chapter 2 <167>

01 11 1 1 1 1 01 1 1 1 1 1 11 10 00 00 10 AB CD Y C D 1 1 1 1 B 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Y A 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

4-Input K-Map: POS Form

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SLIDE 168

Chapter 2.8

Combinational Building Blocks

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SLIDE 169

Chapter 2 <169>

  • Multiplexers
  • Decoders

Combinational Building Blocks

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SLIDE 170

Chapter 2 <170>

  • Selects between one of N inputs to connect

to output

  • log2N-bit required to select input – control

input S

  • Example:

2:1 Mux (2 inputs to 1 output)

  • 𝑂 = 2
  • log2 2 = 1 control bit required

Multiplexer (Mux)

Y 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 S D0 Y D1 D1 D0 S Y 1 D1 D0 S

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SLIDE 171

Chapter 2 <171>

  • Logic gates
  • Sum-of-products form

Y D0 S D1

D1 Y D0 S S 00 01 1 Y 11 10 D0 D1 1 1 1 1 Y = D0S + D1S

  • Tristates
  • For an N-input mux, use N

tristates

  • Turn on exactly one to

select the appropriate input

Multiplexer Implementations

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SLIDE 172

Chapter 2 <172>

A B Y 1 1 1 1 1 Y = AB

00

Y

01 10 11

A B

  • Using the mux as a lookup

table

  • Zero outputs tied to GND
  • One output tied to VDD

Logic using Multiplexers

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SLIDE 173

Chapter 2 <173>

A B Y 1 1 1 1 1 Y = AB A Y 1 1 A B Y B

  • Reducing the size of the mux

Logic using Multiplexers

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SLIDE 174

Chapter 2 <174>

2:4 Decoder A1 A0 Y3 Y2 Y1 Y0 00 01 10 11 1 1 1 1 1 Y3 Y2 Y1 Y0 A0 A1 1 1 1

  • N inputs, 2N outputs
  • One-hot outputs: only
  • ne output HIGH at once
  • Example

2:4 Decoder (2 inputs to 4 outputs)

  • 𝐵𝑗 decimal value selects the

corresponding output

Decoders

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SLIDE 175

Chapter 2 <175>

Y3 Y2 Y1 Y0 A0 A1

Decoder Implementation

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SLIDE 176

Chapter 2 <176>

2:4 Decoder A B 00 01 10 11 Y = AB + AB Y AB AB AB AB Minterm = A  B

  • OR minterms

Logic Using Decoders

XNOR function

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SLIDE 177

Chapter 2.9

Timing

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SLIDE 178

Chapter 2 <178>

  • Delay between input change and output

changing

  • How to build fast circuits?

A Y Time delay A Y

Timing

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SLIDE 179

Chapter 2 <179>

A Y Time A Y tpd tcd

  • Propagation delay: tpd = max delay from input to final output
  • Contamination delay: tcd = min delay from input to initial output

change

Propagation & Contamination Delay

Note: Timing diagram shows a signal with a high and low and transition time as an ‘X’. Cross hatch indicates unknown/changing values

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SLIDE 180

Chapter 2 <180>

  • Delay is caused by
  • Capacitance and resistance in a circuit
  • Speed of light limitation
  • Reasons why tpd and tcd may be different:
  • Different rising and falling delays
  • Multiple inputs and outputs, some of which are

faster than others

  • Circuits slow down when hot and speed up when

cold

Propagation & Contamination Delay

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SLIDE 181

Chapter 2 <181>

A B C D Y Critical Path Short Path n1 n2

Critical (Long) Path: tpd = 2tpd_AND + tpd_OR Short Path: tcd = tcd_AND

Critical (Long) & Short Paths

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SLIDE 182

Chapter 2 <182>

  • When a single input change causes an output

to change multiple times

Glitches

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SLIDE 183

Chapter 2 <183>

A B C Y 00 01 1 Y 11 10 AB 1 1 1 1 C Y = AB + BC

  • What happens when A = 0, C = 1, B falls?

Glitch Example

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SLIDE 184

Chapter 2 <184>

A = 0 B = 1 0 C = 1 Y = 1 0 1 Short Path Critical Path B Y Time 1 0 0 1 glitch

n1 n2

n2 n1

Glitch Example (cont.)

Note: n1 is slower than n2 because of the extra inverter for B to go through

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SLIDE 185

Chapter 2 <185>

00 01 1 Y 11 10 AB 1 1 1 1 C Y = AB + BC + AC AC

B = 1 0 Y = 1 A = 0 C = 1

Fixing the Glitch

Consensus term

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SLIDE 186

Chapter 2 <186>

  • Glitches shouldn’t cause problems because
  • f synchronous design conventions (see

Chapter 3)

  • It’s important to recognize a glitch: in

simulations or on oscilloscope

  • Can’t get rid of all glitches – simultaneous

transitions on multiple inputs can also cause glitches

Why Understand Glitches?