[PPT] - Administrative Notes Note: New homework instructions starting with PowerPoint Presentation

SLIDE 1

Chapter 2 <9>

Note: New homework instructions

starting with HW03

Homework is due at the beginning of

class

Homework must be organized, legible

(messy is not), and stapled to be graded

Administrative Notes

SLIDE 2

Chapter 2 <10>

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

SLIDE 3

Chapter 2 <11>

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

SLIDE 4

Chapter 2 <12>

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

SLIDE 5

Chapter 2 <13>

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

SLIDE 6

Chapter 2 <14>

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

SLIDE 7

Chapter 2 <15>

Aside: Precedence

AND has precedence over OR
In other words:
AND is performed before OR
Example:
𝑍 = 𝐵 ⋅ 𝐶 + 𝐵 ⋅ 𝐶
Equivalent to:
𝑍 = 𝐵 𝐶 + (𝐵𝐶)

SLIDE 8

Chapter 2 <16>

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

SLIDE 9

Chapter 2 <17>

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 ⋅ (𝐵 + 𝐶)

SLIDE 10

Chapter 2 <18>

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

SLIDE 11

Chapter 2 <19>

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

SLIDE 12

Chapter 2 <20>

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

SLIDE 13

Chapter 2 <21>

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

SLIDE 14

Chapter 2 <22>

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

SLIDE 15

Chapter 2 <23>

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

SLIDE 16

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

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

SOP & POS Form

SLIDE 17

Chapter 2 <25>

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

SLIDE 18

Chapter 2 <26>

Boolean Axioms

SLIDE 19

Chapter 2 <27>

Duality in Boolean axioms and theorems:

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

Duality

SLIDE 20

Chapter 2 <28>

Boolean Axioms

SLIDE 21

Chapter 2 <29>

Boolean Axioms

Dual: Exchange:

and +

0 and 1

SLIDE 22

Chapter 2 <30>

Boolean Axioms

Dual: Exchange:

and +

0 and 1

SLIDE 23

Chapter 2 <31>

Basic Boolean Theorems

B = B

SLIDE 24

Chapter 2 <32>

Basic Boolean Theorems: Duals

Dual: Exchange:

and +

0 and 1

SLIDE 25

Chapter 2 <33>

B 1 = B
B + 0 = B

T1: Identity Theorem

SLIDE 26

Chapter 2 <34>

1

= =

B B B B

B 1 = B
B + 0 = B

T1: Identity Theorem

SLIDE 27

Chapter 2 <35>

Simplification of digital logic  connecting

wires with a on/off switch

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

Switching Algebra

SLIDE 28

Chapter 2 <36>

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

Series Switching Network: AND

SLIDE 29

Chapter 2 <37>

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

Parallel Switching Network: OR

SLIDE 30

Chapter 2 <38>

1

= =

B B B B

B 1 = B
B + 0 = B

T1: Identity Theorem

SLIDE 31

Chapter 2 <39>

B 0 = 0
B + 1 = 1

T2: Null Element Theorem

SLIDE 32

Chapter 2 <40>

= =

B 1 B 1

B 0 = 0
B + 1 = 1

T2: Null Element Theorem

SLIDE 33

Chapter 2 <41>

B B = B
B + B = B

T3: Idempotency Theorem

SLIDE 34

Chapter 2 <42>

B

= =

B B B B B

B B = B
B + B = B

T3: Idempotency Theorem

SLIDE 35

Chapter 2 <43>

B = B

T4: Involution Theorem

SLIDE 36

Chapter 2 <44>

= B

B

B = B

T4: Involution Theorem

1 1 1

SLIDE 37

Chapter 2 <45>

B B = 0
B + B = 1

T5: Complements Theorem

SLIDE 38

Chapter 2 <46>

B

= =

B B B 1

B B = 0
B + B = 1

T5: Complements Theorem

SLIDE 39

Chapter 2 <47>

Recap: Basic Boolean Theorems

SLIDE 40

Chapter 2 <48>

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

SLIDE 41

Chapter 2 <49>

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?

SLIDE 42

Chapter 2 <50>

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

SLIDE 43

Chapter 2 <51>

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

SLIDE 44

Chapter 2 <52>

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

SLIDE 45

Chapter 2 <53>

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 0 0 1 1

SLIDE 46

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

SLIDE 47

Chapter 2 <55>

T7: Associativity

Number Theorem Name

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

SLIDE 48

Chapter 2 <56>

T8: Distributivity

Number Theorem Name

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

SLIDE 49

Chapter 2 <57>

T9: Covering

Number Theorem Name

T9 B• (B+C) = B Covering

SLIDE 50

Chapter 2 <58>

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

SLIDE 51

Chapter 2 <59>

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

SLIDE 52

Chapter 2 <60>

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

SLIDE 53

Chapter 2 <61>

T9: Covering

Number Theorem Name

T9 B• (B+C) = B Covering

Method 2: Prove true using other axioms and theorems.

SLIDE 54

Chapter 2 <62>

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

SLIDE 55

Chapter 2 <63>

T10: Combining

Number Theorem Name

T10 (B•C) + (B•C) = B Combining

Prove true using other axioms and theorems:

SLIDE 56

Chapter 2 <64>

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

SLIDE 57

Chapter 2 <65>

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.

SLIDE 58

Chapter 2 <66>

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

SLIDE 59

Chapter 2 <67>

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

SLIDE 60

Chapter 2 <68>

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 (•)

SLIDE 61

Chapter 2 <69>

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.