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1 CK Cheng CSE Dept. UC San Diego CS 140, Lecture 2 Combinational Logic

2 Outlines Review of Boolean Algebra Specification Synthesis

3 Combinational Logic vs Boolean Algebra Schematic Diagram: 5 primary inputs 4 components 9 signal nets 12 pins ab + cd abab cdcd e cd ab e (ab+cd) Boolean Algebra: 5 literals 4 operators

Copyright © 2007 Elsevier 2- Some Definitions 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)

5Copyright © 2007 Elsevier 1- Digital Discipline: Binary Values Typically consider only two discrete values: –1’s and 0’s –1, TRUE, HIGH –0, FALSE, LOW 1 and 0 can be represented by specific voltage levels, rotating gears, fluid levels, etc. Digital circuits usually depend on specific voltage levels to represent 1 and 0 Bit: Binary digit

1- Copyright © 2007 Elsevier1- Born to working class parents Taught himself mathematics and joined the faculty of Queen’s College in Ireland. Wrote An Investigation of the Laws of Thought (1854) Introduced binary variables Introduced the three fundamental logic operations: AND, OR, and NOT. George Boole, 1815 - 1864

1- Boolean Algebra Set of axioms and theorems to simplify Boolean equations Like regular algebra, but in some cases simpler because variables can have only two values (1 or 0) Axioms and theorems obey the principles of duality: –ANDs and ORs interchanged, 0’s and 1’s interchanged

1- Boolean Axioms

1- T1: Identity Theorem B 1 = B + 0 =

1- T1: Identity Theorem B 1 = B B + 0 = B

1- T2: Null Element Theorem B 0 = B + 1 =

1- T2: Null Element Theorem B 0 = 0 B + 1 = 1

1- T3: Idempotency Theorem B B = B + B =

1- T3: Idempotency Theorem B B = B B + B = B

1- T4: Identity Theorem B =

1- T4: Identity Theorem B = B

1- T5: Complement Theorem B B = B + B =

1- T5: Complement Theorem B B = 0 B + B = 1

1- Boolean Theorems: Summary

20 AND, OR, NOT A B C 0 0 0 0 1 0 1 0 0 1 1 1 AND A1A1 A A B C 0 0 0 0 1 1 1 0 1 1 1 1 OR A1A1 1 A0A0 0 A0A0 A A C 0 1 1 0 NOT Review of Boolean algebra and switching functions 0 dominates in AND 1 dominates in OR

1- Boolean Theorems of Several Variables

22 1. Identity A * 1 = A A + 1 = 1 A * 0 = 0 A + 0 = A 2. Complement A + A’ = 1 A * A’ = 0 T8. Distributivity Law A(B+C) = AB + AC A+BC = (A+B)(A+C) A BCBC ACAC ABAB A BCBC ACAC ABAB

23 T7. Associativity (A+B) + C = A + (B+C) (AB)C = A(BC) C ABAB A BCBC C ABAB A BCBC T12. DeMorgan’s Law(A+B)’ = A’B’ (AB)’ = A’ + B’ T11. AC + AB + BC’ = AC + BC’

1- Simplifying Boolean Expressions: Example 1 Y = AB + AB

1- Simplifying Boolean Expressions: Example 1 Y = AB + AB = B(A + A)T8 = B(1)T5’ = BT1

1- Simplifying Boolean Expressions: Example 2 Y = A(AB + ABC)

1- Simplifying Boolean Expressions: Example 2 Y = A(AB + ABC) = A(AB(1 + C))T8 = A(AB(1))T2’ = A(AB)T1 = (AA)BT7 = ABT3

1- DeMorgan’s Theorem Y = AB = A + B Y = A + B = A B

1- Bubble Pushing Pushing bubbles backward (from the output) or forward (from the inputs) changes the body of the gate from AND to OR or vice versa. Pushing a bubble from the output back to the inputs puts bubbles on all gate inputs. Pushing bubbles on all gate inputs forward toward the output puts a bubble on the output and changes the gate body.

30 Part I. Combinational Logic I) Specification –a. Language –b. Truth Table –c. Boolean Algebra –d. Incompletely Specified Function

31 Binary Addition 5 + 7 1 2 Carry Sum 1 1 1 1 0 1 + 1 1 1 1 1 0 0 Carryout Sums Carry bits 5 7 12

32 Half Adder a b carry sum 0 0 0 0 0 1 0 1 1 0 0 1 1 1 1 0 Truth Table a b Sum Carry

33 Switching Function Sum (a,b) = a’b + ab’ Ex: Sum (0,0) = 0’0 + 0*0’ = 0 + 0 = 0 Sum (0,1) = 0’1 + 0*1’ = 1 + 0 = 1 Sum (1,1) = 1’1 + 1*1’ = 0 + 0 = 0 Carry (a, b) = a*b a b sum a b carry

34 Full Adder Id a b c in carry sum 0 0 0 0 0 0 1 0 0 1 0 1 2 0 1 0 0 1 3 0 1 1 1 0 4 1 0 0 0 1 5 1 0 1 1 0 6 1 1 0 1 0 7 1 1 1 1 1 Truth Table a b Sum Carry C in

35 Minterm and Maxterm Id a b c in carryout 0 0 0 0 0 a+b+c 1 0 0 1 0 a+b+c’ 2 0 1 0 0 a+b’+c 3 0 1 1 1 a’ b c 4 1 0 0 0 a’+b+c 5 1 0 1 1 a b’c 6 1 1 0 1 a b c’ 7 1 1 1 1 a b c minterm maxterm

36 f 1 (a,b,c) = a’bc + ab’c + abc’ + abc a’bc = 1 iff (a,b,c,) = (0,1,1) ab’c = 1 iff (a,b,c,) = (1,0,1) abc’ = 1 iff (a,b,c,) = (1,1,0) abc = 1 iff (a,b,c,) = (1,1,1) f 1 (a,b,c) = 1 iff (a,b,c) = (0,1,1), (1,0,1), (1,1,0), or (1,1,1) Minterms Ex: f 1 (1,0,1) = 1’01 + 10’1 + 101’ + 101 = 1 f 1 (1,0,0) = 1’00 + 10’0 + 100’ + 100 = 0

37 f 2 (a,b,c) = (a+b+c)(a+b+c’)(a+b’+c)(a’+b’+c) a + b + c = 0 iff (a,b,c,) = (0,0,0) a + b + c’ = 0 iff (a,b,c,) = (0,0,1) a + b’ + c = 0 iff (a,b,c,) = (0,1,0) a’ + b + c = 0 iff (a,b,c,) = (1,0,0) f 2 (a,b,c) = 0 iff (a,b,c) = (0,0,0), (0,0,1), (0,1,0), (1,0,0) Maxterms Ex: f 2 (1,0,1) = (1+0+1)(1+0+1’)(1+0’+1)(1’+0+1) = 1 f 2 (0,1,0) = (0+1+0)(0+1+0’)(0+1’+0)(0’+1+0) = 0

38 f 1 (a,b,c) = a’bc + ab’c + abc’ + abc f 2 (a,b,c) = (a+b+c)(a+b+c’)(a+b’+c)(a’+b’+c) f 1 (a, b, c) = m 3 + m 5 + m 6 + m 7 =  m(3,5,6,7) f 2 (a, b, c) = M 0 M 1 M 2 M 4 =  M(0, 1, 2, 4) Question: Does f 1 = f 2 ? Yes! Prove using Boolean algebra.

39 Id a b c in carry minterm 4 = ab’c’ 0 0 0 0 0 0 1 0 0 1 0 0 2 0 1 0 0 0 3 0 1 1 1 0 4 1 0 0 0 1 5 1 0 1 1 0 6 1 1 0 1 0 7 1 1 1 1 0 Minterm 4 = ab’c’ Only one row has a 1.

40 Id a b c in carry maxterm 4 = a+b+c 0 0 0 0 0 1 1 0 0 1 0 1 2 0 1 0 0 1 3 0 1 1 1 1 4 1 0 0 0 0 5 1 0 1 1 1 6 1 1 0 1 1 7 1 1 1 1 1 Maxterm 4 = a’+b+c Only one row has a 0.

41 Id a b f (a, b) 0 0 0 1 1 0 2 1 0 1 3 1 1 - 1)The input does not happen. 2)The input happens, but the output is ignored. Examples: -Decimal number 0… 9 uses 4 bits. (1,1,1,1) does not happen. -Final carry out bit (output is ignored). Don’t care set is important because it allows us to minimize the function Incompletely Specified Function

42 Incompletely Specified Function Id a b c g(a,b,c) 0 0 0 0 0 1 0 0 1 1 2 0 1 0 - 3 0 1 1 1 4 1 0 0 1 5 1 0 1 - 6 1 1 0 0 7 1 1 1 1 g 1 (a,b,c) canonical form w/ minterms g 2 (a,b,c) canonical form w/ maxterms Does g 1 (a,b,c) = g 2 (a,b,c)? No! Because there is a group of Don’t Care set. (g 1 only covers the onset, g 2 only covers the offset).

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