Karnaugh-Maps September 14, 2006 Typeset by Foil T EX What are - PowerPoint PPT Presentation

Karnaugh-Maps September 14, 2006 – Typeset by Foil T EX –

What are Karnaugh Maps? A simpler way to handle most (but not all) jobs of manipulating logic functions. – Typeset by Foil T EX – 1

What are Karnaugh Maps? A simpler way to handle most (but not all) jobs of manipulating logic functions. Hooray!! – Typeset by Foil T EX – 2

Karnaugh Map Advantages • Can be completed more systematically • Much simpler to find minimum solutions • Easier to see what is happening (graphical) • Almost always use instead of boolean minimization... – Typeset by Foil T EX – 3

Truth Table Adjacencies A B F These are adjacent in a gray code 0 0 1 sense – they differ by only one bit. ✛ ✘ ✘ ✘ F = A’ 0 1 1 ✘ We can apply XY + XY’ = X ✘ ✘ ✘ ✘ ✾ 1 0 0 A’B’ + A’B = A’(B’+B) = A’(1) 1 1 0 = A’ A B F 0 0 0 Same idea: F = B 0 1 1 ② ❳ ❳ ❳ ❳ A’B + AB = B ❳ ❳ ❳ ❳ 1 0 0 ✘ ✘ ✘ ✘ ✘ ✘ ✘ ✾ ✘ 1 1 1 – Typeset by Foil T EX – 4

Truth Table Adjacencies Key Idea: Gray code adjacency allows use of simplification theorem (e.g., XY + XY’ = X). Problem: Physical adjacency in truth table is not equal to gray code adjacency. – Typeset by Foil T EX – 5

Two variable Karnaugh Map A 0 1 B 0 A=1,B=0 A=0,B=0 1 A=0,B=1 A=1,B=1 A different way to draw a truth table by folding it. – Typeset by Foil T EX – 6

K-Map Physical adjacency does imply Gray code adjacency. A A 0 1 0 1 B B 0 0 1 0 0 0 1 1 1 0 1 1 F=A’B’+A’B=A’ F = A’B + AB = B – Typeset by Foil T EX – 7

Two Variable Karnaugh Map A B F 0 0 0 0 1 0 1 0 1 1 1 1 – Typeset by Foil T EX – 8

Two Variable Karnaugh Map A B F A 0 0 1 0 1 B 0 1 1 0 1 1 0 0 1 1 0 1 1 – Typeset by Foil T EX – 9

Two Variable Karnaugh Map A B F A 0 0 1 0 1 B 0 1 1 0 1 0 1 0 0 1 1 0 1 1 0 – Typeset by Foil T EX – 10

Two Variable Karnaugh Map A B 0 1 A B F 0 1 0 0 0 1 0 1 1 1 1 0 1 0 0 1 1 0 F = A’B’+A’B = A’ – Typeset by Foil T EX – 11

Two Variable Karnaugh Map A B 0 1 A B F 0 1 0 0 0 1 A=0 0 1 1 1 1 0 B=0 or 1 1 0 0 1 1 0 F = A’B’+A’B = A’ – Typeset by Foil T EX – 12

Another Two Variable Karnaugh Map A B 0 1 A B F 0 0 1 0 0 0 0 1 1 1 1 1 1 0 1 1 1 1 F = A’B + AB’+ AB = A+B Each ’1’ must be covered at least once. – Typeset by Foil T EX – 13

Yet Another Two Variable Karnaugh Map A B 0 1 A B F 0 1 1 0 0 1 0 1 1 1 1 1 1 0 1 1 1 1 F = A’B’ + A’B + AB’+ AB = 1 Groups of more than two 1’s can be combined as well. – Typeset by Foil T EX – 14

Three Variable Karnaugh Map Showing Minterm Locations A 0 1 BC Note the order of the B C Variables: 00 m0 m4 B C 01 m1 m5 ABC=101 0 0 11 m3 m7 0 1 m2 m6 10 1 1 1 0 ABC=010 – Typeset by Foil T EX – 15

Adjacencies Adjacent squares differ by exactly one variable. A 0 1 BC 00 AB’C’ There is wrap-around: 01 A’B’C AB’C top and bottom rows are adjacent. 11 ABC 10 – Typeset by Foil T EX – 16

Truth Table to Karnaugh Map A B C F A 0 0 0 0 0 BC 0 1 0 0 1 0 1 00 0 0 0 4 0 1 0 1 2 01 0 1 0 1 1 1 3 1 5 1 0 0 0 4 ✲ 11 1 1 1 0 1 1 5 3 7 1 1 0 0 6 10 1 0 2 6 1 1 1 1 7 – Typeset by Foil T EX – 17

Solution Example A BC 0 1 00 0 0 01 0 1 A’BC+A’BC’ = A’B AB’C+ABC = AC → 11 1 1 ← 10 1 0 F = A’B + AC – Typeset by Foil T EX – 18

Another Example A BC 0 1 00 0 1 01 1 0 A’B’C+A’BC=A’C AB’C’+ABC’=AC’ 11 1 0 10 0 1 F = A’C + AC’ = A ⊕ C – Typeset by Foil T EX – 19

Minterm Expansion to K-Map F = � m (1 , 3 , 4 , 6) A A BC 0 1 BC 0 1 00 m0 m4 00 0 1 01 m1 m5 01 1 0 11 m3 m7 11 1 0 m2 m6 10 10 0 1 Minterms are the 1’s, everything else is 0. – Typeset by Foil T EX – 20

Maxterm Expansion to K-Map F = � M (0 , 2 , 5 , 7) A A BC 0 1 BC 0 1 00 M0 M4 00 0 1 01 M1 M5 01 1 0 11 M3 M7 11 1 0 10 10 M2 M6 0 1 Maxterms are the 0’s, everything else is 1. – Typeset by Foil T EX – 21

Yet Another Example 2 n 1’s can be circled at a time. This includes 1,2,4,8 ... . 3,... not OK. A BC 0 1 A’B’C’+AB’C’+ 00 1 1 AB’C + ABC = A’B’C+AB’C = AC B’ 01 1 1 11 0 1 10 0 0 F=B’+AC The larger the group of 1’s – the simpler the resulting product term. – Typeset by Foil T EX – 22

Boolean Algebra to Karnaugh Map Plot: A ab’c’ + bc + a’ 0 1 BC 00 01 11 10 – Typeset by Foil T EX – 23

Boolean Algebra to Karnaugh Map Plot: A ab’c’ + bc + a’ 0 1 BC 00 1 01 11 10 – Typeset by Foil T EX – 24

Boolean Algebra to Karnaugh Map Plot: A ab’c’ + bc + a’ 0 1 BC 00 1 01 11 1 1 10 – Typeset by Foil T EX – 25

Boolean Algebra to Karnaugh Map Plot: A ab’c’ + bc + a’ 0 1 BC 00 1 1 01 1 11 1 1 10 1 – Typeset by Foil T EX – 26

Boolean Algebra to Karnaugh Map Plot: A ab’c’ + bc + a’ 0 1 BC 00 1 1 Remaining spaces 01 1 0 are ’0’. 11 1 1 10 1 0 – Typeset by Foil T EX – 27

Boolean Algebra to Karnaugh Map F = B’C’ + BC + A’ A ab’c’ + bc + a’ BC 0 1 00 1 1 01 1 0 11 1 1 10 1 0 This is a simpler equation than we started with. How did we get here? – Typeset by Foil T EX – 28

Mapping Sum of Product Terms The three variable map has 12 possible groups of 2 spaces. These become terms with 2 literals. A A A 0 1 0 1 0 1 BC BC BC 00 00 00 01 01 01 11 11 11 10 10 10 – Typeset by Foil T EX – 29

Mapping Sum of Product Terms The three variable map has 6 possible groups of 4 spaces. These become terms with 1 literal. A A A 0 1 0 1 0 1 BC BC BC 00 00 00 01 01 01 11 11 11 10 10 10 – Typeset by Foil T EX – 30

Four Variable Karnaugh Map AB 00 01 11 10 CD AB 00 01 11 10 CD 00 0 0 0 1 00 m0 m4 m12 m8 01 1 1 1 1 D 01 m1 m5 m13 m9 AB’C’ 11 1 1 1 1 11 m3 m7 m15 m11 10 0 1 0 0 10 m2 m6 m14 m10 A’BC F = A’BC + AB’C + D Note the row and column numbering. This is required for adjacency. – Typeset by Foil T EX – 31

Find a POS Solution BC’ AB CD 00 01 11 10 00 1 0 0 1 01 0 0 0 0 C’D 11 1 1 1 1 10 1 1 1 0 AB’CD’ F’ = C’D + BC’ + AB’CD’ F = (C+D’)(B’+C)(A’+B+C’+D) Find solutions to groups of 0’s to find F’. Invert to get F using DeMorgan’s. – Typeset by Foil T EX – 32

Dealing With Don’t Cares F = � m (1 , 3 , 7) + � d (0 , 5) A BC 0 1 00 X 0 A’BC+AB’C+ 01 1 X A’BC+ABC= C 11 1 1 10 0 0 F = C Circle the x’s that help get bigger groups of 1’s (or 0’s if POS). Don’t circle the x’s that don’t help. – Typeset by Foil T EX – 33

Minimal K-Map Solutions Some Terminology and An Algorithm to Find Them – Typeset by Foil T EX – 34

Prime Implicants • A group of 1’s which are adjacent and can be combined on a Karnaugh Map is called an implicant. • The biggest group of 1’s which can be circled to cover a 1 is called a prime implicant . – They are the only implicants we care about. – Typeset by Foil T EX – 35

Prime Implicants AB Prime Implicants Non−prime Implicants CD 00 01 11 10 00 0 0 0 1 01 0 0 1 1 11 0 1 1 1 10 0 1 1 1 Are there any additional prime implicants in the map that are not shown above? – Typeset by Foil T EX – 36

All the Prime Implicants AB Prime Implicants CD 00 01 11 10 00 0 0 0 1 01 0 0 1 1 11 0 1 1 1 10 0 1 1 1 When looking for a minimal solution – only circle prime implicants... A minimal solution will never contain non-prime implicants – Typeset by Foil T EX – 37

Essential Prime Implicants AB CD 00 01 11 10 Not all prime implicants are required ... 00 0 0 0 1 A prime implicant which is the only cover 01 0 0 1 1 of some 1’s is essential – a minimal 11 0 1 1 1 solution requires it. 10 0 1 1 1 Essential Prime Implicants Non−essential Prime Implicants – Typeset by Foil T EX – 38

A Minimal Solution Example AB CD 00 01 11 10 00 0 0 0 1 01 0 0 1 1 Not required. 11 0 1 1 1 10 0 1 1 1 F = AB’+BC+AD Minimum – Typeset by Foil T EX – 39

Another Example AB CD 00 01 11 10 Not required. 00 1 0 0 1 01 1 1 0 0 A’B’ 11 1 1 1 0 10 1 0 0 1 F = A’D+BCD+B’D’ Minimum Every one one of F’s locations is covered by multiple implicants. After choosing essentials, everything is covered... – Typeset by Foil T EX – 40

Finding the Minimum Sum of Products 1. Find each essential prime implicant and include it in the solution. 2. Determine if any minterms are not yet covered. 3. Find the minimal number of remaining prime implicants which finish the cover. – Typeset by Foil T EX – 41