Textbook: Digital Design, 3rdrd. Edition . Edition M. …soc.eecs.yuntech.edu.tw/Course/Digital Logic Design/Logic_Design... · Gate-Level Minimization Chapter 3. P-3/70 Outline of

P-1/702005/5/11

Textbook: Textbook: Digital Design, 3Digital Design, 3rdrd. Edition. EditionM. Morris Mano

Prentice-Hall, Inc.

教師 : 蘇慶龍INSTRUCTOR : CHING-LUNG SU

課程名稱課程名稱: : 數位邏輯設計數位邏輯設計

E-mail: [email protected]

P-2/702005/5/11

Chapter 3Chapter 3GateGate--Level MinimizationLevel Minimization

Chapter 3Chapter 3

P-3/702005/5/11Outline of Chapter 3Outline of Chapter 3

3.1 The Map Method3.2 Four-Variable Map3.3 Five-Variable Map3.4 Product of Sums Simplification3.5 Don’t-Care Conditions3.6 NAND and NOR Implementation3.7 Other Two-Level Implementations3.8 Exclusive-OR Function3.9 Hardware Description Language (HDL)

P-4/702005/5/113.1 3.1 The Map MethodThe Map Method


P-5/702005/5/11

nn Map Simplification Methods of the Boolean Map Simplification Methods of the Boolean FunctionFunction

3.1 3.1 The Map MethodThe Map Method

Karnaugh Map (K-Map): 1. is a pictorial form of the truth table2. express any Boolean function as a sum of minterms3. presents all possible Boolean function in standard

form4. Simplify a Boolean function in two standard form:

sum of products or product of sums5. the simplest K-map will be implemented the

minimum number of gates

P-6/702005/5/11

nn TwoTwo--Variable KVariable K--MapMap


Karnaugh Map (K-Map) presents the four minterms of a two-variable function as follows

m0 m1

m2 m3

xy xy

xy xy

yx 0

0

x=1

y=1

P-7/702005/5/11

nn Presentation of Functions in a KPresentation of Functions in a K--MapMap


Example: F=xy

1

yx 0

0

x=1

y=1

Example: F = x + y= x’y + xy’+ xy= m1 + m2 + m3

1

1 1

yx 0

0

x=1

y=1

P-8/702005/5/11

nn ThreeThree--Variable KVariable K--MapMap


Karnaugh Map (K-Map) presents the eight mintermsof a three-variable function as follows

m0 m1

m4 m5

yzx 00

0

x=1

y=1

m3 m2

m7 m6

01 11 10

1

Gray-Code(Only one variable changes

between two adjacent position)

z=1

y=0

x=0

z=0z=0

P-9/702005/5/11

nn ThreeThree--Variable KVariable K--Map with Map with MintermsMinterms


m0 m1

m4 m5

m3 m2

m7 m6

xyz xyz

xyz xyz

yzx 00

0

x=1

y=1

xyz xyz

xyz xyz

01 11 10

1

z=1

P-10/702005/5/11

nn ThreeThree--Variable KVariable K--Map with Map with MintermsMinterms


Example for a 3-variable function with its K-map:F(x,y,z) = m2 + m3 + m4 + m5

= Σ(2,3,4,5)= x’y + xy’

1 1

yzx 00

0

x=1

y=1

1 1

01 11 10

1

z=1

y=0

m2+m3

m4+m5

P-11/702005/5/11

nn Simplify a ThreeSimplify a Three--Variable Boolean Variable Boolean Function Using KFunction Using K--MapMap


F(x,y,z) = Σ(3,4,6,7)= yz + z’x

1

yzx 00

0

x=1

y=1

1

1 1

01 11 10

1

z=1

m3+m7

m4+m6

P-12/702005/5/11



F(x,y,z) = Σ(0,2,4,6)= z’

1

1

yzx 00

0

x=1

y=1

1

1

01 11 10

1

z=1

(m0+m2)

(m4+m6)+

P-13/702005/5/11

nn KK--Map Simplify for a ThreeMap Simplify for a Three--Variable FunctionVariable Function


1 minterm represents a term of 3 literals2 adjacent squares represents a term of 2 literals4 adjacent squares represents a term of 1 literal8 adjacent squares represents the function = 1

P-14/702005/5/11



F(x,y,z) = Σ(0,2,4,5,6)= z’ + xy’

1

1 1

yzx 00

0

x=1

y=1

1

1

01 11 10

1

z=1

m4+m5

m0+m2+m4+m6

P-15/702005/5/11

nn Simplify Boolean Function Using KSimplify Boolean Function Using K--MapMap


F(A,B,C) = A’C + A’B + AB’C + BC= Σ (1,2,3,5,7)= C + A’B

1

1

BCA 00

0

A=1

B=1

1 1

1

01 11 10

1

C=1

A B

C

P-16/702005/5/113.2 3.2 FourFour--Variable MapVariable Map



nn FourFour--Variable KVariable K--MapMap

Karnaugh Map (K-Map) presents the 16 minterms of a four-variable function as follows

m0 m1

m4 m5

yzwx 00

00

w=1

y=1

m3 m2

m7 m6

01 11 10

01

Gray-Code

z=1

m12 m13

m8 m9

m15 m14

m11 m10

11

10

x=1

Gray-Code

P-18/702005/5/11

nn FourFour--Variable KVariable K--Map with Map with MintermsMinterms

3.2 3.2 FourFour--Variable MapVariable Map

m0 m1

m4 m5

yzwx 00

00

w=1

y=1

m3 m2

m7 m6

01 11 10

01

z=1

m12 m13

m8 m9

m15 m14

m11 m10

11

10

x=1

wxyz

yzwx 00

00

01 11 10

01

wxyz11

10

P-19/702005/5/11

nn KK--Map Simplify for a FourMap Simplify for a Four--Variable FunctionVariable Function

1 minterm represents a term of 4 literals2 adjacent squares represents a term of 3 literals4 adjacent squares represents a term of 2 literals8 adjacent squares represents a term of 1 literal16 adjacent squares represents the function = 1

3.2 3.2 FourFour--Variable MapVariable Map


nn Simplify a FourSimplify a Four--Variable Boolean Variable Boolean Function Using KFunction Using K--MapMap

F(w,x,y,z) = Σ(0,1,2,4,5,6,8,9,12,13,14)= y’ + w’z’ + xz’

1 1

1 1

yzwx 00

00

w=1

y=1

1

1

01 11 10

01

z=1

1 1

1 1

111

10

x=1

y'

z w

xz


nn Simplify a FourSimplify a Four--Variable Boolean Function Using KVariable Boolean Function Using K--MapMap

F = A’B’C’ + B’CD’ + A’BCD’ + AB’C’= B’D’ + B’C’ + A’CD’

1 1

CDAB 00

00

A=1

C=1

1

1

01 11 10

01

D=1

1 1 1

11

10

B=1

B C

A CD

B D


nn Prime Prime Implicants Implicants and Essential Prime and Essential Prime ImplicansImplicans

1. Prime Implicant : A product term containing the maximum possible number of adjacent squares in the K-map

2. Essential Prime Implicant : A prime implicantcontains a minterm that is covered by only one prime implicant


nn Function Simplification Using Prime Function Simplification Using Prime ImplicantImplicantEssential Prime Implicants:BD and B’D’

Prime Implicants:CD, B’C, AD, and AB’

1

1

CDAB 00

00

A

C

1

1

01 11 10

01

D

1

1

1

1

11

10

B

1

1

CDAB 00

00

A

C

1 1

1

01 11 10

01

D

1

1 1

1

1 1

11

10

B


nn Simplify Boolean Function Using KSimplify Boolean Function Using K--MapMap

F = Σ(0,2,3,5,7,8,9,10,11,13,15)= BD+B’D’+CD+AD= BD+B’D’+CD+AB’= BD+B’D’+B’C+AD= BD+B’D’+B’C+AB’

Essential Prime Implicants

P-25/702005/5/113.3 3.3 FiveFive--Variable MapVariable Map



nn FiveFive--Variable KVariable K--Map: F(A,B,C,D,E)Map: F(A,B,C,D,E)

m0 m1

m4 m5

DEBC 00

00

B

D

m3 m2

m7 m6

01 11 10

01

E

m12 m13

m8 m9

m15 m14

m11 m10

11

10

C

m16 m17

m20 m21

DEBC 00

00

B

D

m19 m18

m23 m22

01 11 10

01

E

m28 m29

m24 m25

m31 m30

m27 m26

11

10

C

A=0 A=1


nn The Relationship between the Number of Adjacent The Relationship between the Number of Adjacent Squares and the Number of Literal in the TermSquares and the Number of Literal in the Term

K 2k n=2 n=3 n=4 n=5

0 1 2 3 4 51 2 1 2 3 42 4 0 1 2 33 8 0 1 24 16 0 15 32 0

Number of Literals in aTerm in an n-variable Map

Number ofAdjacent Square


nn Simplify the 5Simplify the 5--variable Function Using Kvariable Function Using K--MapMapF = Σ(0,2,4,6,9,13,21,23,25,29,31)

= A’B’E’+BD’E+ACE

1

1

DEBC 00

00

B

D

1

1

01 11 10

01

E

1

1

11

10

C1

DEBC 00

00

B

D

1

01 11 10

01

E

1

1

111

10

C

A=0 A=1

BD E

ACE

P-29/702005/5/113.4 3.4 Product of Sums SimplificationProduct of Sums Simplification



nn SOP vs. POSSOP vs. POS

1. Review DeMorgen’s Theorem:A Function can be express as SOP or POS

2. Example:

F(x,y,z) = Σ(1,3,6,7)= Π(0,2,4,5)x y z F

0 0 0 0 Π0 0 1 1 Σ0 1 0 0 Π0 1 1 1 Σ1 0 0 0 Π1 0 1 0 Π1 1 0 1 Σ1 1 1 1 Σ


nn Simplify Function in SOP and POSSimplify Function in SOP and POS

F(A,B,C,D) = Σ(0,1,2,5,8,9,10)= B’D’ + B’C’ + A’C’D

F=(F’)’=(AB+CD+BD’)’= (A’+B’)(C’+D’)(B’+D)

1 1

0 1

CDAB 00

00

A

C

0 1

0 0

01 11 10

01

D

0 0

1 1

0 0

0 1

11

10

B


nn Gate Implementation in SOP and POSGate Implementation in SOP and POS

F(A,B,C,D) = B’D’ + B’C’ + A’C’D = (A’+B’)(C’+D’)(B’+D)

AB

D

D

FC

BD

CAD

F

P-33/702005/5/113.5 3.5 DonDon’’tt--Care ConditionsCare Conditions


P-34/702005/5/113.5 3.5 DonDon’’tt--Care ConditionsCare Conditions

nn DonDon’’tt--Care Terms (x): Care Terms (x): A don’t-care term is a combination of variables whose logical value is not

specified. Eg. BCD has 6 don’t-care terms.Decimal BCD0 00001 00012 00103 00114 01005 01016 01107 01118 10009 100110 x11 x12 x13 x14 x15 x

P-35/702005/5/113.5 3.5 DonDon’’t Care Conditionst Care Conditions

nn DonDon’’tt--Care Terms (x): Care Terms (x): A don’t-care term can be assigned 1 or 0 for simplification functions.

F(w,x,y,z) = Σ(1,3,7,11,15)= yz + w’x’ = yz + w’zDon’t-care terms: d = Σ(0,2,5)

x 1

0 x

yzwx 00

00

w

y

1 x

1 0

01 11 10

01

z

0 0

0 0

1 0

1 0

11

10

x

x 1

0 x

yzwx 00

00

w

y

1 x

1 0

01 11 10

01

z

0 0

0 0

1 0

1 0

11

10

x

P-36/702005/5/113.6 3.6 NAND and NOR ImplementationNAND and NOR Implementation



nn NAND and NOR GatesNAND and NOR Gates

1. NAND and NOR gates are easier to fabricate with electronic components.

2. NAND and NOR gates are the basic gates in all IC digital family.

3. The Function implemented in NAND and NOR gates is important.


nn NAND GatesNAND Gates1. NAND gate is a universal gate2. Any digital system can be implemented with NAND

gates3. Logical Operation using NAND Gates

Inverter

AND

OR

x

xy

x

y

x

xy

(x y ) = x+y


nn Two Implementations for NAND GatesTwo Implementations for NAND Gates

DeMorgans’ Theorem: (xyz)’ = x’+y’+z’

xyz

(xyz)

xyz

x+y+z


nn TwoTwo--Level ImplementationLevel Implementation

F= AB + CD = [(AB)’(CD)’]’

AB

CD

F

AB

CD

F

AB

CD

F


nn Example: TwoExample: Two--Level Implementation with Level Implementation with NAND GatesNAND Gates

F(x,y,z) = Σ(1,2,3,4,5,7)= xy’ + x’y + z

1

1 1

yzx 00

0

x=1

y=1

1 1

1

01 11 10

1

z=1


nn Example: TwoExample: Two--Level Implementation with Level Implementation with NAND GatesNAND Gates

F(x,y,z) = xy’ + x’y + z

xyxy

z

F

xyxy

z

F


nn The Procedure for Implementation The Procedure for Implementation Function with 2Function with 2--level NAND Gateslevel NAND Gates

1. Simplify the function and express it in sum of products.2. Draw a NAND gate for each product term of the

expression that has least two literals.The inputs to each NAND gate are the literals of the term. This constitutes a group of first-level gates.

3. Draw a single gate using the AND-invert or invert-OR graphic symbol in the second level, with inputs coming from outputs of first level gates.

4. A term with a single literal requires an inverter in the first level. However, if the single literal is complemented, it can be connected directly to an input of the second level NAND gates.


nn Multilevel NAND CircuitsMultilevel NAND Circuits

F(A,B,C,D) = A(CD+B) + BC’CDBABC

F

CDBABC

F


nn Multilevel NAND CircuitsMultilevel NAND CircuitsF(A,B,C,D) = (AB’+ A’B)( C+D’ )

ABABCD

F

ABABCD

F


nn The Procedure for Converting a Function into The Procedure for Converting a Function into an Allan All--NAND Gate RepresentationNAND Gate Representation

1. Convert all AND gates to NAND gates with AND-invert graphic symbols.

2. Convert all OR gates to NAND gates with invert-OR graphic symbols.

3. Check all the bubbles in the diagram. For every bubble that is not compensated by another small circle along the same line, insert an invert (one-input NAND gate) or complement the input literal.


nn NOR ImplementationNOR Implementation

1. NOR gate is a universal gate2. Any digital system can be implemented with NOR

gates

Inverter

OR

AND

x

xy

x

y

x

x+y

(x +y ) = xy

P-48/702005/5/11

nn Two Implementations for NOR GatesTwo Implementations for NOR Gates

DeMorgans’ Theorem: (x+y+z)’ = x’y’z’

3.6 3.6 NAND and NOR ImplementationNAND and NOR Implementation

xyz

(x+y+z)

xyz

xyz=(x+y+z)


nn Example: TwoExample: Two--Level Implementation with NOR GatesLevel Implementation with NOR Gates

F = (A+B)(C+D)E

ABCD

E

F

ABCD

E

F


nn Multilevel NOR CircuitsMultilevel NOR CircuitsF(A,B,C,D) = (AB’+ A’B)( C+D’ )

ABABCD

F

ABABCD

F

P-51/702005/5/113.7 3.7 Other TwoOther Two--Level ImplementationsLevel Implementations



nn Wired LogicWired Logic

F = (AB)’(CD)’ = (AB+CD)’ Wired-AND= (A+B)’+ (C+D)’ = [(A+B)(C+D)]’ Wired-OR

AB

CD

F=(AB+CD) : Wired-AND

AB

CD

F=[(A+B)(C+D)] : Wired-OR


nn NonNon--degenerate Formsdegenerate Forms

Two-level Function Implementation with1. AND 2. OR 3. NAND 4. NORTherefore, 16 possible combinations.

The 8 non-degenerate combinations are1. AND-OR (3.4) 5. OR-AND (3.4)2. NAND-NAND (3.6) 6. NOR-NOR (3.6)3. NOR-OR 7. NAND-AND4. OR-NAND 8. AND-NOR


nn ANDAND--OROR--INVERT ImplementationINVERT Implementation

ABCD

E

F

AND-NOR(AND-OR-INVERT)

ABCD

E

F

AND-NOR

ABCD

E

F

NAND-AND

F = (AB+CD+E)’


nn OROR--ANDAND--INVERT ImplementationINVERT Implementation

F = [(A+B)(C+D)E]’

ABCD

E

F

OR-NAND(OR-AND-INVERT)

ABCD

E

F

OR-NAND

ABCD

E

F

NOR-OR


nn Implementation with Other TwoImplementation with Other Two--Level FormsLevel Forms

Equivalent Implements Simplify To GetNon-degenerate the F and OutputForm Function in of

(a) (b)

AND-NOR NAND-AND AND-OR-INVERT F

OR-NAND NOR-OR OR-AND-INVERT F

Sum of productsby combining 0's

in the K-map

Product of sum bycombining 1's in the

K-map and thencomplementing

Table 3-3


nn ExampleExampleImplementation F with the 4 two-level forms in Table 3-3

1 0

0 0

yzx 00

0

x=1

y=1

0 0

0 1

01 11 10

1

z=1

F = x y z + xyzF = x y + xy + z

xyxy

z

F

AND-NOR

xyxy

z

F

NAND-AND

F= (x’y+xy’+z)’


nn ExampleExampleImplementation F with the 4 two-level forms in Table 3-3

1 0

0 0

yzx 00

0

x=1

y=1

0 0

0 1

01 11 10

1

z=1

F = x y z + xyzF = x y + xy + z

xyz

F

OR-NAND NOR-OR

xyz

xyz

Fxyz

F= [(x+y+z)(x’+y’+z)]’

P-59/702005/5/113.8 3.8 ExclusiveExclusive--OR FunctionOR Function



nn ExclusiveExclusive--OR (XOR)OR (XOR)

x y+

(x y)+

xy + x y

xy + x y

Exclusive-OR

Exclusive-NOR

x y+Commutative

Associative

y x+=

(x y) z = x (y z) = x y z+ + + + + +


nn ExclusiveExclusive--OR ImplementationsOR Implementations

x

y

x y+

x

y

x y+ : AND-OR-NOT

: NAND Gates


nn Primitive ExclusivePrimitive Exclusive--OR OperationsOR Operations

1. x 0 = x2. x 1 = x3. x x = 04. x x = 15. x y = x y = (x y)

+++++ + +


nn Odd FunctionOdd Function

Function with odd number of input variables is equal to 1

A B C = (AB + A B)C + (AB + A B )C = AB C + A BC + ABC + A B C = Σ (1,2,4,7)

+ +

1

1

yzx 00

0

x=1

y=1

1

1

01 11 10

1

z=1

1

1

yzx 00

0

x=1

y=1

1

1

01 11 10

1

z=1

Odd Function : F = A B C Even Function : F = (A B C)+ + + +


nn Logic ImplementationLogic Implementation

AB

C3 Input Odd

FunctionF:

AB

C3 Input Even

FunctionF:


nn FourFour--Variable ExclusiveVariable Exclusive--OR FunctionOR FunctionA B C D = Σ (1,2,4,7,8,11,13,14)+ + +

1

1

yzwx 00

00

w=1

y=1

1

1

01 11 10

01

z=1

1

1

1

1

11

10

x=1

1

1

yzwx 00

00

w=1

y=1

1

1

01 11 10

01

z=1

1

1

1

1

11

10

x=1

Even FunctionOdd Function


nn Parity Generation and CheckingParity Generation and Checking

1. XOR is used in error-detection and error-correction.2. Parity bits are extra bits with the data to make its

number of 1’s either odd or even.3. The odd/even parity data is transmitted for error

detection.4. The circuit that generates the parity bit in the

transmitter is called a parity generator.5. The circuit that checks the parity in the receiver is

called a parity checker.


nn Parity Data Transmission SystemParity Data Transmission System

Source Data

Parity Generator

n bits

n+1 bitsEven/Odd Data

Data Transmission

Parity Checker

Source Data

n+1 bitsEven/Odd Data

n bits

Parity bits:1 bit


nn EvenEven--ParityParity--Generator Truth TableGenerator Truth Table

Three-Bit Data Parity Bit

x y z P

0 0 0 00 0 1 10 1 0 10 1 1 01 0 0 11 0 1 01 1 0 01 1 1 1


nn Logic Implementation of Parity Generator and CheckerLogic Implementation of Parity Generator and Checker

P = x y zC = x y z P

+ +

xy

z

P

xy

zP

C

Data Transmission+ + +

P-70/702005/5/113.9 3.9 Hardware Description Language (HDL)Hardware Description Language (HDL)


Documents

Textbook: Digital Design, 3rdrd. Edition . Edition M. …soc.eecs.yuntech.edu.tw/Course/Digital Logic Design/Logic_Design... · Gate-Level Minimization Chapter 3. P-3/70 Outline of