Conversion from one number base to another Binary arithmetic Equation simplification DeMorgan’s Laws Conversion to/from SOP/POS Reading equations from Truth Tables Boolean expression to Karnaugh Map Minimization using Karnaugh Maps Minterm and Maxterm Equations Minimization using don’t cares Logic to Boolean Expression conversion Word problems Determining how many gates and inputs a boolean expression has Determining Prime Implicants and Essential Prime Implicants Logical completeness Review for Exam 1

Conversion from one number base to another to Hexadecimal (2 digits)

Conversion from one number base to another

Binary arithmetic 23 6 |

Equation simplification Simplify and convert to SOP (A’ + B + C’)(A’ + C’ + D)(B’ + D’) Y = (AB’ + (AB + B)) B + A

Equation simplification (X + Y)(X + Z) = (X + YZ) X + XY = X X + X’Y = X + Y X + XY = X

DeMorgan’s Laws G = {[(R + S + T)’ PT(R + S)’]’T}’

DeMorgan’s Laws G = {[(R + S + T)’ PT(R + S)’]’T}’ = [(R + S + T)’ PT(R + S)’] + T’ = [ R’S’T’ PT(R’S’)] + T’ = R’S’T’PTR’S’ + T’ = R’S’P(T’T) + T’ = T’

Conversion to/from SOP/POS (X + YZ) = (X + Y)(X + Z)

Reading equations from Truth Tables ABCDF

ABCDF 00001A’B’C’D’ A’BC’D’ A’BCD’ 01111A’BCD ABC’D’ ABCD’ 11110

Boolean expression to Karnaugh Map AB CD AB + C’D + A’B’C + ABCD + AB’C

Boolean expression to Karnaugh Map AB CD AB + C’D + A’B’C + ABCD + AB’C

Minimization using Karnaugh Maps AB CD AB + C’D + A’B’C + ABCD + AB’C AB + C’D + B’C

Minterm and Maxterm Equations F(ABCD) =  m (0,2,4,7,9,12,14,15) AB CD BC’D’ + BCD + ABC + A’B’D’ + AB’C’D

Minterm and Maxterm Equations F(ABCD) =  m (0,2,4,7,9,12,14,15) AB CD BC’D’ + BCD + ABC + A’B’D’ + AB’C’D

Minimization using don’t cares AB CD A’B’ + AD F(ABCD) =  m (0,1,2,11,13) +  d (3,9,12,15)

Minimization using don’t cares AB CD x 0111x 11xx1 101 A’B’ + AD F(ABCD) =  m (0,1,2,11,13) +  d (3,9,12,15)

Logic to Boolean Expression conversion

F = (XY + W)Z + V F = (B+C)A + BC

Word problems

Determining how many gates and inputs a boolean expression has levels gates inputs transistors inputs/gate max levels gates inputs transistors inputs/gate max F = (XY + W)Z + V Z = A’B’C’ + ABC + BCD +B’C’D’

Determining how many gates and inputs a boolean expression has 4 levels 4 gates 8 inputs 16 transistors 2 inputs/gate max 2 levels 5 gates 16 inputs 32 transistors 4 inputs/gate max F = (XY + W)Z + V Z = A’B’C’ + ABC + BCD +B’C’D’

Determining Prime Implicants and Essential Prime Implicants AB CD x 11xx1 101

Determining Prime Implicants and Essential Prime Implicants AB CD x 11xx prime implicants 3 essential prime implicants

Logical completeness Inverter AND gate OR gate

Logical completeness Inverter AND gate NAND NAND gate Inverter OR gate