Lecture 4

III - Working with Combinational Logic © Copyright 2004, Gaetano Borriello and Randy H. Katz 1

ECE2072/TRC2300/TEC2172 Digital Systems:Working with combinational logic

based on textbook

Contemporary Logic Design, 2nd Editionby R. H. Katz and G. Borriello

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Working with combinational logic Summary: (Ch 3 Katz 2e)

Simplificationtwo-level simplificationexploiting don’t caresalgorithm for simplification

Logic realizationtwo-level logic and canonical forms realized with NANDs and NORsmulti-level logic, converting between ANDs and ORs

Time behavior


we'll need a 4-variable Karnaugh map for each of the 3 output functions

Design example: two-bit comparator (Ex3.1 Katz 2e)

block diagram

LTEQGT

A B < C DA B = C DA B > C D

ABCD

N1

N2

A B C D LT EQ GT0 0 0 0 0 1 0

0 1 1 0 01 0 1 0 01 1 1 0 0

0 1 0 0 0 0 10 1 0 1 01 0 1 0 01 1 1 0 0

1 0 0 0 0 0 10 1 0 0 11 0 0 1 01 1 1 0 0

1 1 0 0 0 0 10 1 0 0 11 0 0 0 11 1 0 1 0

andtruth table


A' B' D + A' C + B' C D

B C' D' + A C' + A B D'

LT =

EQ =

GT =

K-map for EQK-map for LT K-map for GT

0 0

1 0

0 0

0 0

Design example: two-bit comparator (cont’d)

D

A

1 1

1 1

0 1

0 0

B

C

1 0

0 1

0 0

0 0D

A

0 0

0 0

1 0

0 1

B

C

0 1

0 0

1 1

1 1D

A

0 0

0 0

0 0

1 0

B

C

= (A xnor C) • (B xnor D)

LT and GT are similar (flip A/C and B/D)

A' B' C' D' + A' B C' D + A B C D + A B' C D’


two alternativeimplementations of EQwith and without XOR

XNOR is implemented with at least 3 simple gates

A B C D

EQ

EQ

Design example: two-bit comparator (cont’d)


block diagramand

truth table

4-variable K-mapfor each of the 4output functions

A2 A1 B2 B1 P8 P4 P2 P10 0 0 0 0 0 0 0

0 1 0 0 0 01 0 0 0 0 01 1 0 0 0 0

0 1 0 0 0 0 0 00 1 0 0 0 11 0 0 0 1 01 1 0 0 1 1

1 0 0 0 0 0 0 00 1 0 0 1 01 0 0 1 0 01 1 0 1 1 0

1 1 0 0 0 0 0 00 1 0 0 1 11 0 0 1 1 01 1 1 0 0 1

Design example: 2x2-bit multiplier

P1P2P4P8

A1A2B1B2


K-map for P8 K-map for P4

K-map for P2 K-map for P1

Design example: 2x2-bit multiplier (cont’d)

0 0

0 0

0 0

0 0B1

A2

0 0

0 0

0 1

1 1

A1

B2

0 0

0 1

0 0

1 0B1

A2

0 1

0 0

1 0

0 0

A1

B2

0 0

0 0

0 0

1 1B1

A2

0 1

0 1

0 1

1 0

A1

B2

0 0

0 0

0 0

0 0B1

A2

0 0

0 0

1 0

0 0

A1

B2 P8 = A2A1B2B1

P4 = A2B2B1'+ A2A1'B2

P2 = A2'A1B2+ A1B2B1'+ A2B2'B1+ A2A1'B1

P1 = A1B1


I8 I4 I2 I1 O8 O4 O2 O10 0 0 0 0 0 0 10 0 0 1 0 0 1 00 0 1 0 0 0 1 10 0 1 1 0 1 0 00 1 0 0 0 1 0 10 1 0 1 0 1 1 00 1 1 0 0 1 1 10 1 1 1 1 0 0 01 0 0 0 1 0 0 11 0 0 1 0 0 0 01 0 1 0 X X X X1 0 1 1 X X X X1 1 0 0 X X X X1 1 0 1 X X X X1 1 1 0 X X X X1 1 1 1 X X X Xblock diagram

andtruth table

4-variable K-map for each of the 4 output functions

O1O2O4O8

I1I2I4I8

Design example: BCD increment by 1(Ex3.3 Katz Ed2)


O8 = I4 I2 I1 + I8 I1'

O4 = I4 I2' + I4 I1' + I4’ I2 I1

O2 = I8’ I2’ I1 + I2 I1'

O1 = I1'

O8 O4

O2 O1

Design example: BCD increment by 1 (cont’d)

0 0

0 0

X 1

X 0I1

I8

0 1

0 0

X X

X X

I4

I2

0 0

1 1

X 0

X 0I1

I8

0 0

1 1

X X

X X

I4

I2

0 1

0 1

X 0

X 0I1

I8

1 0

0 1

X X

X X

I4

I2

1 1

0 0

X 1

X 0I1

I8

0 0

1 1

X X

X X

I4

I2


Definition of terms for two-level simplification

Implicantsingle element of ON-set or DC-set or any group of these elements that can be combined to form a grouping

Prime implicantimplicant that can't be combined with another to form a larger grouping

Essential prime implicantprime implicant is essential if it alone covers an element of ON-setwill participate in ALL possible covers of the ON-setDC-set used to form prime implicants but not to make implicant essential

Objective:grow implicant into prime implicants(minimize literals per term)cover the ON-set with as few prime implicants as possible(minimize number of product terms)


0 X

1 1

1 0

1 0D

A

1 0

0 0

1 1

1 1

B

C

5 prime implicants:BD, ABC', ACD, A'BC, A'C'D

Examples to illustrate terms

0 0

1 1

1 0

1 0D

A

0 1

0 1

1 1

0 0

B

C

6 prime implicants:A'B'D, BC', AC, A'C'D, AB, B'CD

minimum cover: AC + BC' + A'B'D

essential

minimum cover: 4 essential implicants

essential


Algorithm for two-level simplification

Algorithm: minimum sum-of-products expression from a Karnaugh map

Step 1: choose an element of the ON-setStep 2: find "maximal" groupings of 1s and Xs adjacent to that element

consider top/bottom row, left/right column, and corner adjacenciesthis forms prime implicants (number of elements always a power of 2)

Repeat Steps 1 and 2 to find all prime implicants

Step 3: revisit the 1s in the K-mapif covered by single prime implicant, it is essential, and participates in final cover1s covered by essential prime implicant do not need to be revisited

Step 4: if there remain 1s not covered by essential prime implicantsselect the smallest number of prime implicants that cover the remaining 1s


X 1

0 1

0 1

1 1D

A

0 X

0 1

X 0

0 1

B

C

3 primes around AB'C'D'

X 1

0 1

0 1

1 1

Algorithm for two-level simplification (example) (Ex3.4 Katz Ed2)

D

A

0 X

0 1

X 0

0 1

B

C

2 primes around A'BC'D'

X 1

0 1

0 1

1 1D

A

0 X

0 1

X 0

0 1

B

C

2 primes around ABC'D

X 1

0 1

0 1

1 1D

A

0 X

0 1

X 0

0 1

B

C

minimum cover (3 primes)

X 1

0 1

0 1

1 1D

A

0 X

0 1

X 0

0 1

B

C

X 1

0 1

0 1

1 1D

A

0 X

0 1

X 0

0 1

B

C

2 essential primes

X 1

0 1

0 1

1 1D

A

0 X

0 1

X 0

0 1

B

C


Activity

X 0

0 1

X 0

X 1D

A

0 X

X 1

X 0

1 1

B

C

BC BD AB AC’DCD’

BDCD’ AC’D

BDCD’ AC’D

List all prime implicants for the following K-map:

Which are essential prime implicants?

What is the minimum cover?

X 0

0 1

X 0

X 1D

A

0 X

X 1

X 0

1 1

B

C


Gate Logic: Two-Level Simplification5-Variable K-maps

ƒ(A,B,C,D,E) = Σm(2,5,7,8,10,13,15,17,19,21,23,24,29 31)

=

BC DE

BC DE

A =0

A =1

00 01 11 10 00

01

11

10

00 01 11 10 00

01

11

10

0 4 12 8

1 5 13 9

3 7 15 11

2 6 14 10

16 20 28 24

17 21 29 25

19 23 31 27

18 22 30 26

1

00

1

10

1

10 BC

DE 00 01 11 00

01

11

10

A=0

BC DE 00 01 11

01

11

10

A=1

1 1

1

1 1

1

1 1

1 1 1


Gate Logic: Two-Level Simplification5-Variable K-maps

ƒ(A,B,C,D,E) = Σm(2,5,7,8,10,13,15,17,19,21,23,24,29 31)

= C E + A B' E + B C' D' E' + A' C' D E'

BC DE 00 01 11 10

00

01

11

10

A=0

BC DE 00 01 11 10

00

01

11

10

A=1

1 1

1

1

1 1

1

1

1 1 1

1 1 1

BC DE

BC DE

A =0

A =1

00 01 11 10 00

01

11

10

00 01 11 10 00

01

11

10

0 4 12 8

1 5 13 9

3 7 15 11

2 6 14 10

16 20 28 24

17 21 29 25

19 23 31 27

18 22 30 26


Gate Logic: Two-Level Simplification6- Variable K-Maps

ƒ(A,B,C,D,E,F) =Σm(2,8,10,18,24,

26,34,37,42,45,50,53,58,61)

=

CD EF

CD EF

AB =00

AB =01

00 01 11 10 00

01

11

10

00 01 11 10 00

01

11

10

0 4 12 8

1 5 13 9

3 7 15 11

2 6 14 10

16 20 28 24

17 21 29 25

19 23 31 27

18 22 30 26

CD EF

AB =11

00 01 11 10 00

01

11

10

48 52 60 56

49 53 61 57

51 55 63 59

50 54 62 58

CD EF

AB =10

00 01 11 10 00

01

11

10

32 36 44 40

33 37 45 41

35 39 47 43

34 38 46 42

CD EF

CD EF

AB =00

AB =01

00 01 11 10 00

01

11

10

00 01 11 10 00

01

11

10

CD EF

AB =11

00 01 11 10 00

01

11

10

CD EF

AB =10

00 01 11 10 00

01

11

10

1

1 1

1

1 1

1 1

1 1

1 1

1 1


Gate Logic: Two-Level Simplification6- Variable K-Maps

ƒ(A,B,C,D,E,F) =Σm(2,8,10,18,24,

26,34,37,42,45,50,53,58,61)

= D' E F' + A D E' F+ A' C D' F'

CD EF

CD EF

AB =00

AB =01

00 01 11 10 00

01

11

10

00 01 11 10 00

01

11

10

0 4 12 8

1 5 13 9

3 7 15 11

2 6 14 10

16 20 28 24

17 21 29 25

19 23 31 27

18 22 30 26

CD EF

AB =11

00 01 11 10 00

01

11

10

48 52 60 56

49 53 61 57

51 55 63 59

50 54 62 58

CD EF

AB =10

00 01 11 10 00

01

11

10

32 36 44 40

33 37 45 41

35 39 47 43

34 38 46 42

CD EF

CD EF

AB =00

AB =01

00 01 11 10 00

01

11

10

00 01 11 10 00

01

11

10

CD EF

AB =11

00 01 11 10 00

01

11

10

CD EF

AB =10

00 01 11 10 00

01

11

10

1

1 1

1

1 1

1 1

1 1

1 1

1 1


Gate Logic: CAD Tools for SimplificationQuine-McCluskey Method

Tabular method to systematically find all prime implicants

Implication Table

Column I 0000

0100 1000

0101 0110 1001 1010

0111 1101

1111

ƒ(A,B,C,D) = Σ m(4,5,6,8,9,10,13) + Σ d(0,7,15)

Step 1: Fill Column 1 with ON-set andDC-set minterm indices. Groupby number of 1's.

Stage 1: Find all prime implicants




Column II 0-00 -000

010-01-0 100-10-0

01-1 -101 011-1-01

-111 11-1

ƒ(A,B,C,D) = Σm(4,5,6,8,9,10,13) + Σd(0,7,15)

Step 2: Apply Uniting Theorem—Compare elements of group w/N 1's against those with N+1 1's.Differ by one bit implies adjacent.Eliminate variable and place innext column.

E.g., 0000 vs. 0100 yields 0-000000 vs. 1000 yields -000

When used in a combination,mark with a check. If cannot becombined, mark with a star. These are the prime implicants.

Repeat until no further combinations can be made.

Stage 1: Find all prime implicantsStep 1: Fill Column 1 with ON-set and

DC-set minterm indices. Groupby number of 1's.

Column 0000

0100 1000

0101 0110 1001 1010

0111 1101

1111

Implication Table




Implication Table

Column I Column II Column III0000 0-00 * 01-- *

-000 *0100 -1-1 *1000 010-

01-0 0101 100- *0110 10-0 *1001 1010 01-1

-101 0111 011-1101 1-01 *

1111 -111 11-1

ƒ(A,B,C,D) = Σm(4,5,6,8,9,10,13) + Σd(0,7,15)

Step 1: Fill Column 1 with ON-set andDC-set minterm indices. Groupby number of 1's.

Step 2: Apply Uniting Theorem—Compare elements of group w/N 1's against those with N+1 1's.Differ by one bit implies adjacent.Eliminate variable and place innext column.

E.g., 0000 vs. 0100 yields 0-000000 vs. 1000 yields -000

When used in a combination,mark with a check. If cannot becombined, mark with a star. Theseare the prime implicants.

Repeat until no further combinations can be made.

Stage 1: Find all prime implicants


Gate Logic: CAD Tools for SimplificationQuine-McCluskey Method Continued

Prime Implicants:

0-00 = A' C' D'

100- = A B' C'

1-01 = A C' D

-1-1 = B D

-000 = B' C' D'

10-0 = A B' D'

01-- = A' B

AB CD 00 01 11 10

00

01

11

10

D

B

C

A

X 1 0 1

0 1 1 1

0 X X 0

0 1 0 1


Gate Logic: CAD Tools for SimplificationQuine-McCluskey Method Continued

Prime Implicants:

0-00 = A' C' D'

100- = A B' C'

1-01 = A C' D

-1-1 = B D

-000 = B' C' D'

10-0 = A B' D'

01-- = A' B

Stage 2: find smallest set of prime implicants that cover the ON-set

recall that essential prime implicants must be in all covers

another tabular method– the prime implicant chart

AB CD 00 01 11 10

00

01

11

10

D

B

C

A

X 1 0 1

0 1 1 1

0 X X 0

0 1 0 1


Gate Logic: CAD Tools for Simplification

rows = prime implicantscolumns = ON-set elementsplace an "X" if ON-set element is

covered by the prime implicant

Prime Implicant Chart



rows = prime implicantscolumns = ON-set elementsplace an "X" if ON-set element is

covered by the prime implicant

Prime Implicant Chart

If column has a single X, than theimplicant associated with the rowis essential. It must appear inminimum cover


Gate Logic: CAD Tools for SimplificationPrime Implicant Chart (Continued)

Eliminate all columns covered byessential primes


Gate Logic: CAD Tools for SimplificationPrime Implicant Chart (Continued)

Eliminate all columns covered byessential primes

Find minimum set of rows thatcover the remaining columns

ƒ = A B' D' + A C' D + A' Bƒ = A B' D' + A C' D + A' B


Gate Logic: CAD Tools for SimplificationESPRESSO Method

Problem with Quine-McCluskey: the number of prime implicantsgrows rapidly with the number of inputs

upper bound: 3 /n, where n is the number of inputsn

finding a minimum cover is NP-complete, i.e., a computationalexpensive process not likely to yield to any efficientalgorithm

Espresso: trades solution speed for minimality of answer

don't generate all prime implicants (Quine-McCluskey Stage 1)

judiciously select a subset of primes that still covers the ON-set

operates in a fashion not unlike a human finding primes in a K-map


Gate Logic: CAD Tools for SimplificationEspresso Method: Overview

1. Expands implicants to their maximum sizeImplicants covered by an expanded implicant are removed from

further considerationQuality of result depends on order of implicant expansionHeuristic methods used to determine orderStep is called EXPAND

Irredundant cover (i.e., no proper subset is also a cover) is extractedfrom the expanded primes

Just like the Quine-McCluskey Prime Implicant ChartStep is called IRREDUNDANT COVER

Solution usually pretty good, but sometimes can be improvedMight exist another cover with fewer terms or fewer literalsShrink prime implicants to smallest size that still covers ON-setStep is called REDUCE

Repeat sequence REDUCE/EXPAND/IRREDUNDANT COVER to findalternative prime implicants

Keep doing this as long as new covers improve on last solution

A number of optimizations are tried, e.g., identify and remove essential primes early in the process

2.

3.

4.

5.


Gate Logic: CAD Tools for SimplificationEspresso: Why Iterate on Reduce, Irredundant Cover, Expand?

Initial Set of Primes found bySteps1 and 2 of the Espresso

Method

4 primes, irredundant cover,but not a minimal cover!

Result of REDUCE:Shrink primes while still

covering the ON-set

Choice of order in which to perform shrink is important

AB CD 00 01 11 10

00

01

11

10

D

B

C

A

1 1 0 0

1 1 1 1

0 0 1 1

1 1 1 1

AB CD 00 01 11 10

00

01

11

10

D

B

C

A

1 1 0 0

1 1 1 1

0 0 1 1

1 1 1 1



Espresso Iteration (Continued)

Second EXPAND generates adifferent set of prime implicants

IRREDUNDANT COVER found byfinal step of espresso

Only three prime implicants!

AB CD 00 01 11 10

00

01

11

10

D

B

C

A

1 1 0 0

1 1 1 1

0 0 1 1

1 1 1 1

AB CD 00 01 11 10

00

01

11

10

D

B

C

A

1 1 0 0

1 1 1 1

0 0 1 1

1 1 1 1


Gate Logic: CAD Tools for SimplificationEspresso Inputs and Outputs

.i 4

.o 1

.ilb a b c d

.ob f

.p 100100 10101 10110 11000 11001 11010 11101 10000 -0111 -1111 -.e

-- # inputs-- # outputs-- input names-- output name-- number of product terms-- A'BC'D'-- A'BC'D-- A'BCD'-- AB'C'D'-- AB'C'D-- AB'CD'-- ABC'D-- A'B'C'D' don't care-- A'BCD don't care-- ABCD don't care-- end of list

ƒ(A,B,C,D) = Σm(4,5,6,8,9,10,13) + Σd(0,7,15)

Espresso Input Espresso Output.i 4.o 1.ilb a b c d.ob f.p 31-01 110-0 101-- 1.e

ƒ = A C' D + A B' D' + A' B

A 1 in an output columnmeans ONLY that this row’s

product term is includedin the sum of products

for that output


Interpreting Espresso Output Files

Example: Construct a truth table for f and g

.i 3

.o 2

.ilb a b c

.ob f g

.p 301- 1 00-1 0 1-11 1 1.e

a b c f g0 0 0 0 0 10 1 00 1 11 0 01 0 11 1 01 1 1

?




.i 3

.o 2

.ilb a b c

.ob f g

.p 301- 1 00-1 0 1-11 1 1.e

a b c f g0 0 0 0 0 10 1 0 10 1 1 11 0 01 0 11 1 01 1 1

?




.i 3

.o 2

.ilb a b c

.ob f g

.p 301- 1 00-1 0 1-11 1 1.e

a b c f g0 0 0 0 0 1 10 1 0 10 1 1 11 111 0 01 0 11 1 01 1 1 1 1

?




.i 3

.o 2

.ilb a b c

.ob f g

.p 301- 1 00-1 0 1-11 1 1.e

a b c f g0 0 0 0 00 0 1 0 10 1 0 1 00 1 1 1 11 0 0 0 01 0 1 0 01 1 0 0 01 1 1 1 1

?Remaining terms 0

ƒ = A’B + BC = A’BC’+A’BC+ABC

g = A’ C + BC = A’B’C+A’BC+ABC


Two-Level Logic: SummaryPrimitive logic building blocks

INVERTER, AND, OR, NAND, NOR, XOR, XNOR

Canonical FormsSum of Products, Products of Sums

Incompletely specified functions/don't cares

Logic MinimizationGoal: two-level logic realizations with fewest gates and fewest

number of gate inputs

Obtained via Laws and Theorems of Boolean Algebra

or the Uniting Theorem

or K-map Methods up to 6 variables

or Quine-McCluskey Algorithm (optimal but slow for large problems)

or Espresso CAD Tool

Lecture 4

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