ECE 171 Lecture Notes 10

9/5/12 ECE 171 Lecture Notes 10

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ECE 171: Introduction to Digital Circuits Fall 1999 Rev: 11.2.99

Lecture Notes 10

Last Time

Project 1 Assigned

4 More BA Rules

TT POS Simplify w/BA NOR LD

Karnaugh Maps

Introduction

2,3, & 4 Variables

Review of Exam 1

This Time

Boolean Algebra - 4 More Rules

Factoring

Karnaugh Maps

Definitions

Many Examples

5 & 6 Variables

Boolean Algebra

29. A XOR A = 030. A XOR A' = 131. (A + B)(B' + C)(A + C) = (A + B)(B' + C)32. AB + B'C + AC = AB + B'C

Derive rule 31 by a truth table proof.

Derive rule 32 by a karnaugh map.

Factoring

Use rule 14 to help reduce the gate count.

14. AB + AC = A(B + C)

Example

Y = A' D' + A' C'

A straightforward implementation requires:

3 2-Input NAND's

3 Inverters

After factoring, we have

Y = A' (C'+D')

This can be implemented more efficiently as shown below



This requires

2 2-Input NAND's

2 Inverters

Example

Y = A' C' + A' D' + AEF + A B'

A straightforward implementation requires:

3 2-Input NAND's

1 3-Input NAND

1 4-Input NAND (to combine all four products)

4 Inverters (for A, B, C, and D)

This would take 3-4 IC's to implement depending on whether you used one of the spare NAND's to

implement the inverter.

After factoring, we have

Y = A' (C' + D') + A (EF + B')

This can be implemented as follows

This requires

6 2-Input NAND's

1 Inverter

Assuming you used standard IC's that have 4 2-Input NAND's per an IC, this would take 2 IC's to

implement if you used one of the spare NAND's to implement the inverter.

Karnaugh Maps



Today we will discuss Karnaugh maps more formally than last time and discuss a more orderly method foroptaining the minimal sum.

LiteralA variable or complement of a variable.Examples: A, B', C', D

Normal Product TermA product term in which no variable appears more than once.Examples: ABC, AB'C', A'B'C'Counter Examples: AA'BC, AB'B'C', ABCC

n-variable Minterm

A normal product term with n literals. There are 2n such terms.Example:

AB

0 0 A'B'

0 1 A'B

1 0 AB'

1 1 ABEach of the product terms corresponding to a row in the truth table is a minterm.

Minimal SumA sum of products (SOP) expression such that no SOP expression for Y has fewer product terms and anySOP expression with the same number of product terms has at least as many literals.This is what we are trying to produce through the use of Karnaugh maps.

ImplicantA normal product term that implies Y.Example: For the function Y = AB + ABC + BC, the implicants are AB, ABC, and BC because if any oneof those terms are true, then Y is true.

Prime ImplicantAn implicant of Y such that if any variable is removed from the implicant, the resulting term does not implyY.Example: Y = AB + ABC + BCPrime Implicants: AB, BCNot a prime implicant: ABCABC is not a prime implicant because the literal A can be removed to give BC and BC still implies Y.Conversely AB is not a prime implicant because you can't remove either A or B and have the remainingterm still imply Y.In truth tables the prime implicants are represented by the largest rectangular groups of ones that can becircled. If a smaller subgroup is circled, the smaller group is an implicant, but not a prime implicant.

PI TheoremA minimal sum is a sum of prime implicants.

Distinguished 1-CellAn input combination that is covered by 1 prime implicant. In terms of Karnaugh maps, distinguished 1-cells are 1's that are circled by only 1 prime implicant.

Essential Prime ImplicantA prime implicant that that includes one or more distinguished one cells. Essential prime implicants areimportant because a minimal sum contains all essential prime implicants.

Karnaugh Map Examples

In the following examples the distinguished 1-cells are marked in the upper left corner of the cell with an asterisk(*). The essential prime implicants are circled in blue, the prime implicants are circled in black, and the non-essential prime implicants included in the minimal sum are shown in red.

Example 1

Prime Implicants: 5Distinguished 1-Cells: 2



Essential Prime Implicants: 2Minimal Sums: 1

Y = A'CD' + AC'D + BCD

Example 2

Prime Implicants: 7Distinguished 1-Cells: 2Essential Prime Implicants: 2Minimal Sums: 1

Y = B'D' + AD' + A'C'D + BCD

Example 3




Y = AB'C' + A'CD' + AC'D + BCD

Y = AB'C' + A'CD' + ABD + A'BC

Y = AB'C' + A'CD' + ABD + BCD

Example 4

Prime Implicants: 5Distinguished 1-Cells: 3Essential Prime Implicants: 3



Minimal Sums: 1

Y = A'B' + A'C' + ABC + A'D

Example 5


Y = A'C + A'B + BD + CD

Example 6




Y = B'D + BC' + AB

Example 7


Y = A'B'C + A'BD + ABC' + AB'D'



Y = B'CD' + A'CD + BC'D + AC'D'

Example 8


Y = B'C + D + BC'

5-Variable Karnaugh Maps

For these you must circle the prime implicants on each map individually and then the prime implicants on the jointmap. The joint essential prime implicants are shown in green.


Y = A'B'C' + BE + ABC' + ACE + A'DE



Y = A'B'C' + BE + ABC' + ACE + CDE

Note that the joint map can help you identify the joint prime implicants.

6-Variable Karnaugh Maps



The prime implicants unique to each map are shown in black.The prime implicants shared between maps 0 and 1 (A=0) are shown in aqua.The prime implicants shared between maps 0 and 2 (B=0) are shown in violet.The prime implicants shared between maps 1 and 3 (B=1) are shown in olive.The prime implicants shared between maps 2 and 3 (A=1) are shown in brown.The prime implicants shared between all 4 maps are shown in orange.

To find the prime implicants shared among maps it may help to draw out each of the 5 joint maps.

Distinguished 1-Cells: 10Essential Prime Implicants: 5Minimal Sums: 2

Essential Prime ImplicantsA'EF (Maps 0 & 1)BCD' (Maps 1 & 3)B'D'F' (Maps 0 & 2)ACE' (Maps 2 & 3)ABDE' (Map 3)



Y = A'EF + BCD' + B'D'F' + ACE' + ABDE' + B'DE'F + A'B'C'F



Y = A'EF + BCD' + B'D'F' + ACE' + ABDE' + B'DE'F + A'B'C'D'

ECE 171 Lecture Notes 10

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