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

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 convenience Chapter 2 <99>

Multiplying Out: SOP Form 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 Chapter 2 <100>

Multiplying Out: SOP Form Example: Y = (A + C + D + E)(A + B) Apply T8 ’ first when possible: W+XZ = (W+X)(W+Z) Chapter 2 <101>

Multiplying Out: SOP Form Example: 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 or 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 Chapter 2 <102>

Canonical SOP & POS Form • SOP – sum-of-products O C E minterm 0 0 0 O C 0 1 0 O C E = OC 1 0 1 O C = Σ (m 2 ) 1 1 0 O C • POS – product-of-sums maxterm O C E 0 0 0 O + C E = ( O + C )( O + C )( O + C ) 0 1 0 O + C = Π (M0, M1, M3) 1 0 1 O + C 1 1 0 O + C Chapter 2 <103>

Factoring: POS Form An expression is in simplified product- of-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’) Chapter 2 <104>

Factoring: POS Form Example 1: Y = (A + B’CDE) Apply T8 ’ first when possible: W+XZ = (W+X)(W+Z) Chapter 2 <105>

Factoring: POS Form Example 1: 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 Chapter 2 <106>

Factoring: POS Form Example 2: Y = AB + C’DE + F Apply T8 ’ first when possible: W+XZ = (W+X)(W+Z) Chapter 2 <107>

Factoring: POS Form Example 2: 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) T 8 ’: Distributivity Chapter 2 <108>

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) • (C+D) = Consensus (B • C) + (B • D) (B+C) • (B+D) Axioms and theorems are useful for simplifying equations. Chapter 2 <109>

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 Chapter 2 <110>

DeMorgan’s Theorem Number Theorem Name T12 B 0 •B 1 • B 2 … = B 0 +B 1 +B 2 … DeMorgan’s Theorem Chapter 2 <111>

DeMorgan’s Theorem Number Theorem Name T12 B 0 •B 1 • B 2 … = B 0 +B 1 +B 2 … DeMorgan’s Theorem The complement of the product is the sum of the complements Chapter 2 <112>

DeMorgan’s Theorem: Dual # Theorem Dual Name T12 B 0 •B 1 • B 2 … = B 0 +B 1 +B 2 … = DeMorgan’s B 0 +B 1 +B 2 … B 0 • B 1 • B 2 … Theorem Chapter 2 <113>

DeMorgan’s Theorem: Dual # Theorem Dual Name T12 B 0 •B 1 • B 2 … = B 0 +B 1 +B 2 … = DeMorgan’s B 0 +B 1 +B 2 … B 0 • B 1 • B 2 … Theorem The complement of the product is the sum of the complements Chapter 2 <114>

DeMorgan’s Theorem: Dual # Theorem Dual Name T12 B 0 •B 1 • B 2 … = B 0 +B 1 +B 2 … = DeMorgan’s B 0 +B 1 +B 2 … B 0 • B 1 • B 2 … Theorem The complement of the product is the sum of the complements. Dual: The complement of the sum is the product of the complements . Chapter 2 <115>

DeMorgan’s Theorem • Y = AB = A + B A Y B A Y B • Y = A + B = A B A Y B A Y B Chapter 2 <116>

DeMorgan’s Theorem Example 1 Y = (A+BD)C Chapter 2 <117>

DeMorgan’s Theorem Example 1 Y = (A+BD)C = (A+BD) + C = (A•(BD)) + C = ( A•(BD)) + C = ABD + C Chapter 2 <118>

DeMorgan’s Theorem Example 2 Y = (ACE+D) + B Chapter 2 <119>

DeMorgan’s Theorem Example 2 Y = (ACE+D) + B = (ACE+D) • B = (ACE•D) • B = ((AC+E)•D) • B = ((AC+E)•D) • B = (ACD + DE) • B = ABCD + BDE Chapter 2 <120>

Bubble Pushing • Backward: – Body changes – Adds bubbles to inputs A A Y Y B B • Forward: – Body changes – Adds bubble to output A A Y Y B B Chapter 2 <121>

Bubble Pushing • What is the Boolean expression for this circuit? A B Y C D Chapter 2 <122>

Bubble Pushing • What is the Boolean expression for this circuit? A B Y C D Y = AB + CD Chapter 2 <123>

Bubble Pushing Rules • Begin at output, then work toward inputs • Push bubbles on final output back • Draw gates in a form so bubbles cancel A B C Y D Chapter 2 <124>

Bubble Pushing Example A B C Y D Chapter 2 <125>

Bubble Pushing Example no output A bubble B C Y D Chapter 2 <126>

Bubble Pushing Example no output A bubble B C Y D bubble on A input and output B C Y D Chapter 2 <127>

Bubble Pushing Example no output A bubble B C Y D bubble on A input and output B C Y D no bubble on input and output A B C Y D Y = ABC + D Chapter 2 <128>

Canonical SOP & POS Form Revisited • SOP – sum-of-products How do we implement this logic function with gates? O C E minterm 0 0 0 O C 0 1 0 O C E = OC 1 0 1 O C = Σ (m 2 ) 1 1 0 O C • POS – product-of-sums maxterm O C E 0 0 0 O + C E = ( O + C )( O + C )( O + C ) 0 1 0 O + C = Π (M0, M1, M3) 1 0 1 O + C 1 1 0 O + C Chapter 2 <129>

From Logic to Gates • Two-level logic: ANDs followed by ORs • Example: Y = ABC + ABC + ABC A B C A B C minterm: ABC minterm: ABC minterm: ABC Y Chapter 2 <130>

Circuit Schematics Rules • Inputs on the left (or top) • Outputs on right (or bottom) • Gates flow from left to right • Straight wires are best 𝐷 + 𝐵𝐶 𝑍 = 𝐶 Chapter 2 <131>

Circuit Schematic Rules (cont.) • 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 crossing wires connect wires connect without a dot do at a T junction at a dot not connect Chapter 2 <132>

Multiple-Output Circuits • Example: Priority Circuit Y 3 Y 2 Y 1 Y 0 A A 2 A 1 A 3 0 Output asserted 0 0 0 0 0 0 0 1 corresponding to 0 0 1 0 0 0 1 1 most significant 0 1 0 0 0 1 0 1 TRUE input 0 1 1 0 0 1 1 1 A 3 1 0 0 0 Y 3 1 0 0 1 1 0 1 0 A 2 Y 2 1 0 1 1 1 1 0 0 Y 1 A 1 1 1 0 1 1 1 1 0 Y 0 A 0 1 1 1 1 PRIORITY CiIRCUIT Chapter 2 <133>

Multiple-Output Circuits • Example: Priority Circuit Y 3 Y 2 Y 1 Y 0 A A 2 A 1 A 3 0 Output asserted 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 corresponding to 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 0 most significant 0 1 0 0 0 1 0 0 0 1 0 1 0 1 0 0 TRUE input 0 1 1 0 0 1 0 0 0 1 1 1 0 1 0 0 A 3 1 0 0 0 1 0 0 0 Y 3 1 0 0 1 1 0 0 0 1 0 1 0 1 0 0 0 A 2 Y 2 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 0 Y 1 A 1 1 1 0 1 1 0 0 0 Y 0 1 1 1 0 1 0 0 0 A 0 1 1 1 1 1 0 0 0 PRIORITY CiIRCUIT Chapter 2 <134>

Priority Circuit Hardware A A 2 A 1 A Y 3 Y 2 Y 1 Y 0 3 0 0 0 0 0 0 0 0 0 A 3 A 2 A 1 A 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 Y 3 0 0 1 1 0 0 1 0 0 1 0 0 0 1 0 0 Y 2 0 1 0 1 0 1 0 0 0 1 1 0 0 1 0 0 0 1 1 1 0 1 0 0 1 0 0 0 1 0 0 0 Y 1 1 0 0 1 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 1 1 0 0 0 Y 0 1 1 0 0 1 0 0 0 1 1 0 1 1 0 0 0 1 1 1 0 1 0 0 0 1 1 1 1 1 0 0 0 Chapter 2 <135>