A1 - Rev. 4.1 (2006-07) by Enrico Nardelli1 Logic Circuits and Computer Architecture Appendix A...

Rev. 4.1 (2006-07) by Enrico Nardelli 1A1 -

Logic Circuits and Computer Architecture

Appendix ADigital Logic Circuits

Part 1: Combinational Circuitsand Minimization


Structured organization

• Problem-oriented language level• Assembly language level• Operating system machine level• Instruction set architecture level• Microarchitecture level• Digital logic level A

bst

ract

ion

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el


Digital Logic level

• Digital circuits Only two logical levels present (i.e., binary) low/high voltage

• Basic gates AND, OR, NOT

• Basic circuits Combinational (without memory, stateless) Sequential (with memory, state dependent

behaviour)


Boolean Algebra

• Variables: A, B, …• Domain of variables: 2 values

1 or 0; Y or N; true or false; …

• Fundamental Operations AND, OR, NOT

• Intended meaning (for humans - Laws of Thought) AND: both inputs are true OR: at least one input is true NOT: negate the input

• Named from George Boole


George Boole (1815-1864)

An Investigation of the Laws of Thought, on Which are founded the Mathematical Theories of Logic and Probabilities (1854)


Claude Shannon(1916-2001)

A Symbolic Analysis of Relay and Switching Circuits (1938)

ENIAC (Electronic Numerical Integrator And Calculator) (1946)


Formal definition of functions (1)

• By means of “truth tables” Explicit representation of the output for all

possible inputs

A B AND

0 0 0

0 1 0

1 0 0

1 1 1

A B OR

0 0 0

0 1 1

1 0 1

1 1 1

A NOT

0 1

1 0


…truth table for a 3-variable function…

f(A,B,C)= 1 if and only if at least 2 variables are equal to 1

A B C f(A,B,C)

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1



• By means of “boolean equation”• boolean equation consists of:

variables constants 0 and 1 boolean operations (AND, OR, NOT) parentheses

M = OR( AND(NOT(A),NOT(B)), AND(A,B) )


Boolean functions

• Conventions NOT (negation): NOT(A) = A’ = A AND (conjunction): AND(A,B) = AB = A.B OR (disjunction): OR(A,B) = A+B

M = OR( AND(NOT(A),NOT(B)), AND(A,B) )

M = (((A)’(B)’) + (AB))


Boolean Operator Precedence

• The order of evaluation in boolean expression is:1. Parentheses2. NOT3. AND4. OR

• Consequence: parantheses appear around OR expressions

• Example: F=A(B+C)(C+D’)

M = (((A)’(B)’) + (AB))

M = A’B’ + AB


NOT gate - the simplest one

• NOT gate - inverts the signal If A is 0, X is 1 If A is 1, X is 0

• A NOT gate is also called an inverter


AND gate

• Output is 1 if all inputs are 1 In general, if the AND gate has N inputs, both

input 1 AND input 2 AND … AND input N must be 1 for the output to be 1

• 2-input AND gate


OR gate

• Output is 1 if at least one input is 1 In general, if the OR gate has N inputs, input 1

OR input 2 OR … OR input N must be 1 for the output to be 1

• 2-input OR gate


A more complex example

• 2-input “equivalence” circuit• The output is 1 if the inputs are the same

(i.e., both 0 or both 1)

Truth table• Boolean function:

M = A’B’ + ABA B M

0 0 1

0 1 0

1 0 0

1 1 1



• By means of logic circuits Combination of logic gates joined by wires


Conventions for logic circuits


Exercise (1)

• Write the truth table and the logic circuit for

F = X + Y’Z


Truth table

X Y Z F

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1


Logic Circuit


Exercise (2)

• Write the boolean function and its truth table for the following logic circuit


Function and Truth Table

• F = Y’ + X’YZ’ + XY

X Y Z F

0 0 0 1

0 0 1 1

0 1 0 1

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1


Exercise (3)

• Write the boolean function and its truth table for the following logic circuit


Function and Truth Table

• F = X’YZ + X’YZ’ + XZ


Conversion between represent.

• Circuit ->-> Boolean formula (left-to-right inspection)-> Truth table (explicit case-by-case computation)

• Boolean formula ->-> Circuit (bottom-up construction)-> Truth table (explicit case-by-case evaluation)

• Truth table -> -> Circuit (through boolean formula)-> Boolean formula (through canonical form – see

later)


Boolean Identities

1A = A 0+A = A Identity0A = 0 1+A = 1 NullAA = A A+A = A IdempotentAA’ = 0 A+A’ = 1 InverseAB = BA A+B = B+A Commutative(AB)C = A(BC) (A+B)+C = A+(B+C)

AssociativeA+BC = (A+B)(A+C) A(B+C) = AB+AC DistributiveA(A+B) = A A+AB = A Absorption(AB)’ = A’+B’ (A+B)’ = A’B’ De Morgan

duality principle: any algebraic equality remains true when the operators OR and AND, and the elements 0 and 1 are interchanged


Truth tables to verify De Morgan’s theorem


RemarkEach equality remains true if you sobstitute any variable withany expression

Examples

(A+B)(A+CD’) = A + BCD’ (distributive)

((A+BC)(D+A))’ = (A+BC)’ + (D+A)’ (De Morgan) = A’ (BC)’ + D’A’ (De Morgan) = A’(B’+C’) + D’A’


F = X’YZ + X’YZ’ + XZ (distributive) = X’Y(Z + Z’) + XZ (inverse) = X’Y 1 + XZ (identity) = X’Y + XZ

… algebraic manipulation…


Boolean Algebra Vs Switching Algebra (1)

A Boolean Algebra is a structure A = <A, +, · , ’, 0, 1> where

• A is a set• + and · are binary operations• ‘ is a unary operation• 0, 1 A

satisfying the following axioms

(i) + and · are commutative

(ii) 0 and 1 satisfy: a·1=a and a+0=a, a A

(iii) + and · distribute over each other

(iv) for each element a A, there exists an element a’ A such that a + a’= 1 and a·a’=0


Boolean Algebra Vs Switching algebra (2)

Switching Algebra is the following boolean algebra

A = <{0,1}, +, · , ’, 0, 1>


ObservationAxioms (i)-(iv) can be used to prove all the other identities

An example: Idempotent X + X = X

X + X = (X + X)·1 (ii) = (X + X)(X + X’) (iv) = X + (X·X’) (iii) = X + 0 (iv) = X (ii)


De Morgan circuit equivalents

• AND/OR can be interchanged if you invert the inputs and outputs

bubble means inversion


NAND gate - the negation of AND

• The opposite of the AND gate is the NAND gate (output is 0 if all inputs are 1)

Truth table• Logic diagram A B NAN

D

0 0 1

0 1 1

1 0 1

1 1 0


NOR gate - the negation of OR

• The opposite of the OR gate is the NOR gate (output is 0 if any input is 1) Truth table

• Logic diagram A B NOR

0 0 1

0 1 0

1 0 0

1 1 0


Exercise

• Write the truth table for: a 3 input NAND gate a 3 input NOR gate


XOR gate - the exclusive OR

• For a 2-input gate Output is 1 if exactly one of the inputs is 1

Truth table• Logic diagram

• For > 2 inputs: output is 1 if an odd number of inputs is 1

A B XOR

0 0 0

0 1 1

1 0 1

1 1 0


Universal Gates

• How many logical functions there are with n input?

• With n inputs there are 2(2n ) possible logical functionsA B 0 1 2 3 4 5 6 7 8 9 1

011

12

13

14

15

0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1


Universal Gates (2)

• AND, OR, NOT can generate all possible boolean functions (see later)

• Is it possible to use fewer basic operations?

• Universal gate: a gate type that can implement any Boolean function


Universal Gates (3)

• AND, NOT are enough !

• OR, NOT are enough !

• Even NAND alone or NOR alone are enough !


• Simulation of NOT ???

How NAND simulates AND, OR


Alternative NAND representations


• Simulation of NOT ???

How NOR simulates AND, OR


Alternative NOR representations


Gate equivalence

• Any AND, OR, NOT gate can be obtained using just NAND gates or just NOR gates

• Consequence: any circuit can be constructed using just NAND gates or just NOR gates (easier to build)


Equivalence modifications (1)


Equivalence modifications (2)

• Substitute equivalent gates


Transforming OR, AND to NAND

• Transform the following circuit


Solution


Exercise

• Write a NAND only logic circuit for

F = XY’ + X’Y + Z


Solution


Exercise

• Write a NAND only logic circuit for the exclusive OR function (XOR)

XOR(A,B) = A’B + AB’


Solution (1)

• Truth table initial circuit

A B XOR

0 0 0

0 1 1

1 0 1

1 1 0


Solution (2): equivalence transform.

A’BAB’

A’BAB’

A’BAB’


Exercise

• Write a NOR only logic circuit for

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


Solution

• Direct realization


SOP and POS rapresentations

• It is useful to specify boolean function in a particular form

• sum of products (SOP) rapresentation it is an OR of AND combinations of its inputs F1=A’B + BC it is a SOP rapresentation F2= A’B + (BC)’ it is not a SOP rapresentation

• product of sums (POS) rapresentation it is an AND of OR combiations of its inputs F1=A’(B + C’)(B+A) it is a POS rapresentation F2=A’B + BC it is not a POS rapresentation


Exercise

• Express Z=(A(B+C(A’+B’)))’ as sum of products


Solution

Z=(A(B+C(A’+B’)))’= A’ + (B+C(A’+B’))’= A’ + B’(C(A’+B’))’= A’ + B’(C’+(A’+B’)’)= A’ + B’(C’+AB)= A’ + B’C’ + ABB’= A’ + B’C’ + A0= A’ + B’C’ + 0 = A’ + B’C’


Canonical Form for boolean functions

• It is a “standard” way of expressing SOP or POS • It is:

a sum of minterms, for SOP a product of maxterms for POS

• A minterm is a product containing all variables, either in the positive form or in the negative form

• A maxterm is a sum containing all variables, either in the positive or in the negative form.

• Examples:F = (A’+B+C) (B’+C) is not in a POS canonical formM = AB + A’BC is not in a SOP canonical form


Minterms

• Given that each variable may appear normal (e.g. X) or complemented (e.g. X’), there are 2n minterms for n variables

• Example: Two variables (X and Y) produce 4 combinations: XY X’Y XY’ X’Y’

• Thus there are 4 minterms of 2 variables


Maxterms

• Given that each variable may appear normal (e.g. X) or complemented (e.g. X’), there are 2n maxterms for n variables

• Example: Two variables (X and Y) produce 4 combinations: X+Y X+Y’ X’+Y X’+Y’

• Thus there are 4 maxterms of 2 variables


Maxterms and Minterms

• Examples: 2 variable minterms and maxterms

• The index above is important for describing which variables in the terms are true and which are complemented

Index Minterm Maxterm

0 X’Y’ X+Y

1 X’Y X+Y’

2 XY’ X’+Y

3 XY X’+Y’


(Number)r = j = - m

jj

i

i = 0i rArA

(Integer Portion) + (Fraction Portion)

i = n - 1 j = - 1

• Positive radix, positional number systems• A number with radix r is represented by a

string of digits: An - 1An - 2 … A1A0 . A- 1 A- 2 … A- m 1 A- m

in which 0 Ai < r and . is the radix point.

• The string of digits represents the power series:

Number Systems – Representation


General Decimal Binary

Radix (Base)

r 10 2

Digits 0 => r - 1 0 => 9 0 => 1

0123

4 5

-1-2-3-4-5

r0

r1

r2

r3

r4

r5

r -1

r -2

r -3

r -4

r -5

110

1001000

10,000100,000

0.10.01

0.0010.0001

0.00001

1248

16320.5

0.250.1250.0625

0.03125

Number Systems – Examples

powers of

radix


Standard Order

• Minterms and Maxterms are designated with subscript• The subscript is a numer, corresponding to a binary pattern• The bits in the pattern represent the complemented or

normal state of each variable listed in a standard order• All the variables will be present in a minterm or maxterm

and will be listed in the same order (usually alphabetically)• Example: For variables A, B, C:

Maxterms: (A + B + C’), (A + B + C) Terms: (B + A + C), AC’B, and (C + B + A) are NOT in

standard order Minterms: AB’C, ABC, AB’C’ Terms: (A+C), B’C, and (A+B’) do not contain all

variables


Purpose of the Index

• The index for the minterm or maxterm, expressed as a binary number, is used to determine whether the variable is shown in the true form or complemented form.

• For Minterms: “1” means the variable is “Not Complemented” and “0” means the variable is “Complemented”

• For Maxterms: “0” means the variable is “Not Complemented” and “1” means the variable is “Complemented”


X Y Z Minterm

Symbol m0 m1

m2

m3

m4

m5

m6

m7

0 0 0 X’Y’Z’ m0 1 0 0 0 0 0 0 0

0 0 1 X’Y’Z m1 0 1 0 0 0 0 0 0

0 1 0 X’YZ’ m2 0 0 1 0 0 0 0 0

0 1 1 X’YZ m3 0 0 0 1 0 0 0 0

1 0 0 XY’Z’ m4 0 0 0 0 1 0 0 0

1 0 1 XY’Z m5 0 0 0 0 0 1 0 0

1 1 0 XYZ’ m6 0 0 0 0 0 0 1 0

1 1 1 XYZ m7 0 0 0 0 0 0 0 1


X Y Z Maxterm

Symbol M0 M1 M2 M3 M4 M5 M6 M7

0 0 0 X+Y+Z M0 0 1 1 1 1 1 1 1

0 0 1 X+Y+Z’

M1 1 0 1 1 1 1 1 1

0 1 0 X+Y’+Z

M2 1 1 0 1 1 1 1 1

0 1 1 X+Y’+Z’

M3 1 1 1 0 1 1 1 1

1 0 0 X’+Y+Z

M4 1 1 1 1 0 1 1 1

1 0 1 X’+Y+Z’

M5 1 1 1 1 1 0 1 1

1 1 0 X’+Y’+Z

M6 1 1 1 1 1 1 0 1

1 1 1 X’+Y’+Z’

M7 1 1 1 1 1 1 1 0


Minterm and Maxterm Relationship

• DeMorgan's Theorem:

• Two-variable example:

• Thus M2 is the complement of m2 and vice-versa.

• Since DeMorgan's Theorem holds for n variables, the above holds for terms of n variables

• giving:

• Thus Mi is the complement of mi.

yx y· x yxyx

y x M2 yx· m2

i mM i ii Mm


Boolean function implementation

• Any function can be implemented as the OR of the AND combinations of its inputs canonical SOP

• Start from the truth table For each 1 in the output Write its inputs in AND Write these in OR

• M=A’BC+AB’C+ABC’+ABC = m3+m5+m6+m7

A B C M

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1


Equivalent Logic Circuit


Boolean function implem. (2)

• Any function can be implemented as the AND of the OR combinations of its inputs canonical POS

• Start from the truth table For each 0 in the output Write its inputs in AND Write these negated in AND Obtain F = Z0’ . Z1’ . Z2’ ... Finally, apply De Morgan to F It is an AND combination of maxterms

A B C F

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 1

1 0 0 0

1 0 1 0

1 1 0 1

1 1 1 0



• De Morgan (general): (ABC)’=A’+B’+C’

• F=(A’B’C’)’.(A’B’C)’.(AB’C’)’.(AB’C)’.(ABC)’=(A’’+B’’+C’’).(A’’+B’’+C’).(A’+B’’+C’’).

.(A’+B’’+C’).(A’+B’+C’)=(A+B+C).(A+B+C’).(A’+B+C).

.(A’+B+C’).(A’+B’+C’) =M0·M1·M4·M5·M7



• Alternative procedure for POS form (use only if you know what you are doing!) Complement the table by substituting

everywhere a 0 with a 1 and a 1 with a 0 Write a SOP form for the complemented table Complement the formula by substituting

everywhere and AND with an OR and an OR with an AND

Why does it work ???


Example: canonical SOP

X Y Z F

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 0

1 1 0 1

1 1 1 1

m1

m3

m6

m7

0 0 0 0

1 0 0 0

0 0 0 0

0 1 0 0

0 0 0 0

0 0 0 0

0 0 1 0

0 0 0 1

F = m1 + m3+ m6 + m7

= X’Y’Z + X’YZ + XYZ’ +XYZ

canonical SOP


Example: canonical POS

X Y Z F

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 0

1 1 0 1

1 1 1 1

M0

M2

M4

M5

0 1 1 1

1 1 1 1

1 0 1 1

1 1 1 1

1 1 0 1

1 1 1 0

1 1 1 1

1 1 1 1

F = M0 · M2 · M4 · M5

= (X+Y+Z) · (X+Y’+Z) (X’+Y+Z) · (X’+Y+Z’)

canonical POS


F = X’YZ + X’YZ’ + XZ = X’Y(Z + Z’) + XZ = X’Y 1 + XZ = X’Y + XZ

… algebraic manipulation…

…a simpler SOP representation leads to a simpler circuit…


Circuit Optimization

• Goal: To obtain the simplest implementation for a given function

• Optimization is a more formal approach to simplification that is performed using a specific procedure or algorithm

• Optimization requires a cost criterion to measure the simplicity of a circuit

• Optimization for two-level (SOP and POS) circuits: minimal SOP

• minimum number of pruduct terms• minimum number of literals for each product term

similarly for POS


Minimization procedures

• Karnaugh’s maps (by hand)• Used to minimize boolean functions of up

to 4 input variables• For more variables use the method of

Quine-McKluskey (programmable)


Karnaugh’s Maps (KM)

• Grid-like representation for boolean functions• Each cell represents a minterm• The K-map can be viewed as a reorganized version

of the truth table• Minterms with just one variable different occupies

adjacent cells• Alternative algebraic expressions for the same

function are derived by recognizing patterns of squares

• Consider only 1s in the representation (focusing on a SOP representation)

• IDEA: if 2 adjacent cells have a 1 the function can be simplified


A KM for 2-variable functions

• The generic KM

• Function F = XY Function F = X + Y


example

X

Y

Z

0

1

00 01 11 10XYZ X Y Z F=

Z

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 1

1 1 0 0

1 1 1 1

1

1 1

1


X

Y

Z

0

1

00 01 11 10XYZ X Y Z F=

Z’

0 0 0 1

0 0 1 0

0 1 0 1

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0

1

11

1


X

Y

Z

0

1

00 01 11 10XYZ X Y Z F=Y

Z

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 0

1 1 0 0

1 1 1 1

1

1


An example• F = X’YZ + X’YZ’ + XY’Z + XY’Z’

Idea: • we want to cover all 1s by using rectangles of adjacent cells• each rectangle of 2k adjacent cells (for some k) represents a literal product term• bigger rectangles correspond to simpler product terms

X Y Z F

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 1

1 0 0 1

1 0 1 1

1 1 0 0

1 1 1 0

= X’Y(Z+Z’) + XY’(Z+Z’)= X’Y + XY’


Combining Squares

• By combining squares, we reduce number of literals in a product term

• On a 3-variable K-Map: One square represents a minterm with three

variables Two adjacent squares represent a product

term with two variables Four “adjacent” terms represent a product

term with one variable Eight “adjacent” terms is the function of all

ones (no variables) = 1.


Circular adjacencies for 3 variables

labels are useful to get the expressioncorresponding to a given rectangle


Three-Variable Maps

• Example Shapes of 2-cell Rectangles:y

0 1 3 2

5 64 7x

z

X’Y’

YZ

X’Z’


Three-Variable Maps

• Example Shapes of 4-cell Rectangles:y

0 1 3 2

5 64 7x

z


Four 1 adjacents

F = X’Y’Z’ + XY’Z’ + X’YZ’ + XYZ’ = Y’Z’(X’+X) + YZ’ (X’+X) = Y’Z’ + YZ’ = Z’(Y’+Y) = Z’


Exercise (1)

• Which is the minimal SOP expression for function F1=m3+m4+m6+m7?


Solution

• F1 = YZ + XZ’


Exercise (2)

• Which is the minimal SOP expression for function F2=m0+m2+m4+m5+m6?


Solution

• F2 = Z’ + XY’


k-cube of 1s

• Is a set of 2k adjacent cells• 0-cube, 1 cell, a minterm• 1-cube, 2 adjacent cells• 2-cube, 4 adjacent cells• 3-cube, 8 adjacent cells• ….


Prime implicants

• F = P1 + P2 + P3 + ...

• The term corresponding to a k-cube is called an implicant Infact, it is a term Pn which implies the function F,

i.e. if Pn is true then F is true

• An implicant is said to be a prime implicant for F if it does not imply any other implicant of F

• A prime implicant can be chosen by selecting a maximal k-cube, i.e. a k-cube in the KM which is not contained in any larger h-cube (h>k)


Minimal representation

F = P1 + P2 + P3 + ...

has a minimal SOP representation if:1. Each Pi is a prime implicant

2. There is a minimum number of them


Minimality procedure

1. Find maximal k-cubes (prime implicants)2. If a 1 is covered by only one maximal k-

cube this has to be chosen (essential prime implicants)

3. Select a minimum number of the remaining k-cube so to cover all 1s not covered by essential prime implicants


Exercise (3)

• Find the minimal SOP expression for F=m1+m3+m4+m5+m6

X

Y

Z

0

1

00 01 11 10XYZ

1 1

1 11

F=X’Z+XZ’+XY’


Exercise (3)

• Find the minimal SOP expression for F=m1+m2+m3+m5+m7

X

Y

Z

0

1

00 01 11 10XYZ

1 1

1

1

1

F=Z+X’Y


Circular adjacenciesfor 4 variables


Four Variable Terms

Four variable maps can have rectangles corresponding to:

• A single 1 = 4 variables, (i.e. Minterm)• Two 1s = 3 variables,• Four 1s = 2 variables• Eight 1s = 1 variable,• Sixteen 1s = zero variables (i.e. Constant "1")


Four-Variable Maps

• Example Shapes of Rectangles:

8 9 1011

12 13 1415

0 1 3 2

5 64 7

X

Y

Z

W

XZ

X’Z’YW’


Four-Variable Maps

• Example Shapes of Rectangles:

X

Y

Z

8 9 1011

12 13 1415

0 1 3 2

5 64 7

W


Simplify F(A, B, C, D) given on the K-map. Example:

1

1

1

1 1

1

1

B

D

A

C

1

1

1

1

1

1 1

1

1

B

D

A

C

1

1

Essential

Minterms covered by essential prime implicants

Selected

minimal SOP: A’B+A’CD+AC’D+B’C’D’


One more example

DB

CB

1 1

1 1

1 1

B

D

A

1 1

1 1

1

ESSENTIAL Prime Implicants

C

BD

CD

BD

Minterms covered by single prime implicant

DB

1 1

1 1

1 1

B

C

D

A

1 1

1 1

1

AD

BA

CD

BA

minimal SOP: BD+B’D’+CD+AB’


Five Variable or More K-Maps• For five variable problems, we use two adjacent K-maps. It becomes

harder to visualize adjacent minterms for selecting k-cubes.

• You can extend the problem to six variables by using four K-Maps.

X

Y

Z

W

V = 0

X

Z

W

V = 1Y


The KM method for POS

• Which is the POS expression of function F represented by this KM?

• Use the same method used for build POS canonical form from truth tables Find the minimal SOP for F’ apply DeMorgan


Example

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

B

D

A

C

0 0

0

00

0

0

0 0

1 11

1

1 1 1


Don't Cares in K-Maps

• Sometimes a function table or map contains entries for which it is known: the input values for the minterm will never occur, or the output value for the minterm is not used

• In these cases, the output value need not be defined• Instead, the output value is defined as a “don't care”• By placing “don't cares” ( an “x” entry) in the

function table or map, the cost of the logic circuit may be lowered.


Example 1

• A logic function having the binary codes for the BCD digits as its inputs.

• Only the codes for 0 through 9 are used.

• The six codes, 1010 through 1111 never occur, so the output values for these codes are “x” to represent “don’t cares.”

A B C D f(A,B,C,D)

0 0 0 0 f(0,0,0,0)

0 0 0 1 f(0,0,0,1)

0 0 1 0 f(0,0,1,0)

0 0 1 1 f(0,0,1,1)

0 1 0 0 f(0,1,0,0)

0 1 0 1 f(0,1,0,1)

0 1 1 0 f(0,1,1,0)

0 1 1 1 f(0,1,1,1)

1 0 0 0 f(1,0,0,0)

1 0 0 1 f(1,0,0,1)

1 0 1 0 x1 0 1 1 x1 1 0 0 x1 1 0 1 x1 1 1 0 x1 1 1 1 x


Example 2

• Consider the following function f(A,B) f(A,B)=1 if A=B=0 f(A,B)=0 if A B

A B f

0 0 1

0 1 0

1 0 0

1 1 x

A B f

0 0 1

0 1 0

1 0 0

1 1 0

A B f

0 0 1

0 1 0

1 0 0

1 1 1

Truth table on the left may be substitued by anyone on the right


Example 3

• Consider the following function f(A,B) f(A,B)=1 if A=B=0 f(A,B)=0 if A B f(A,B) is used just as input for another function

g(f,A,B)=(A’+B) f(A,B)

A B f

0 0 1

0 1 0

1 0 x

1 1 x

Notice that:for A=1 and B=0g(f,A,B) = (1’+0) f(A,B) = 0


Example

X

Y

Z

W

X

X

X1 1

1

1

1

f(W,X,Y,Z)= YZ + X’W’

Simplify the choice, since each X can be

considered

a 0 or a 1


X

Y

Z

W

X

X

X1 1

1

1

1

f(W,X,Y,Z)= YZ + ZW’

Example

a different choise

is possible

A1 - Rev. 4.1 (2006-07) by Enrico Nardelli1 Logic Circuits and Computer Architecture Appendix A...

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