Reminder Midterm 1: Thursday, Oct. 5 th In class: 1 hour and 15 - - PowerPoint PPT Presentation

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Reminder Midterm 1: Thursday, Oct. 5 th In class: 1 hour and 15 - - PowerPoint PPT Presentation

Nov 08, 2023 441 likes •868 views

Reminder Midterm 1: Thursday, Oct. 5 th In class: 1 hour and 15 minutes Chap 1 2.6 Closed book, closed notes No calculator Boolean Theorems & Axioms document will be attached as last page of the exam for your

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SLIDE 1

Chapter 2 <99>

Midterm 1: Thursday, Oct. 5th

  • In class: 1 hour and 15 minutes
  • Chap 1 – 2.6
  • Closed book, closed notes
  • No calculator
  • Boolean Theorems & Axioms document

will be attached as last page of the exam for your convenience

Reminder

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SLIDE 2

Chapter 2 <100>

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 3

Chapter 2 <101>

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 4

Chapter 2 <102>

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 5

Chapter 2 <103>

  • 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 6

Chapter 2 <104>

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 7

Chapter 2 <105>

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 8

Chapter 2 <106>

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 9

Chapter 2 <107>

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 10

Chapter 2 <108>

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 11

Chapter 2 <109>

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 12

Chapter 2 <110>

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 13

Chapter 2 <111>

DeMorgan’s Theorem

Number Theorem Name

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

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SLIDE 14

Chapter 2 <112>

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 15

Chapter 2 <113>

# 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 16

Chapter 2 <114>

# 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 17

Chapter 2 <115>

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 18

Chapter 2 <116>

  • 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 19

Chapter 2 <117>

Y = (A+BD)C

DeMorgan’s Theorem Example 1

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SLIDE 20

Chapter 2 <118>

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 21

Chapter 2 <119>

Y = (ACE+D) + B

DeMorgan’s Theorem Example 2

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SLIDE 22

Chapter 2 <120>

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 23

Chapter 2 <121>

  • 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 24

Chapter 2 <122>

A B Y C D

  • What is the Boolean expression for this

circuit?

Bubble Pushing

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SLIDE 25

Chapter 2 <123>

A B Y C D

  • What is the Boolean expression for this

circuit? Y = AB + CD

Bubble Pushing

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SLIDE 26

Chapter 2 <124>

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 27

Chapter 2 <125>

A B C Y D

Bubble Pushing Example

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SLIDE 28

Chapter 2 <126>

A B C Y D no output bubble

Bubble Pushing Example

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SLIDE 29

Chapter 2 <127>

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 30

Chapter 2 <128>

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 31

Chapter 2 <129>

  • 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 32

Chapter 2 <130>

  • 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 33

Chapter 2 <131>

  • 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 = 𝐶 𝐷 + 𝐵𝐶

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SLIDE 34

Chapter 2 <132>

  • 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 35

Chapter 2 <133>

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 36

Chapter 2 <134>

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 37

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

A3A2A1A0 Y3 Y2 Y1 Y0

Priority Circuit Hardware

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SLIDE 38

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

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 39

Chapter 2 <137>

  • 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 40

Chapter 2 <138>

  • 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 41

Chapter 2 <139>

  • 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