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Sistem Digital

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Materi Kuliah ke-5

BOOLEAN ALGEBRA AND LOGICSIMPLICATION


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Boolean AlgebraBoolean Algebra

VERY nice machinery used to manipulate

(simplify) Boolean functions

George Boole (1815-1864): An

investigation of the laws of thought Terminology:

Literal:A variable or its complement

Product term:literals connected by Sum term:literals connected by +


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Boolean Algebra PropertiesBoolean Algebra PropertiesLet X: boolean variable, 0,1: constants

1. X + 0 = X -- Zero Axiom

2. X 1 = X -- Unit Axiom3. X + 1 = 1 -- Unit Property

4. X 0 = 0 -- Zero Property


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Boolean Algebra Properties (cont.)Boolean Algebra Properties (cont.)

Let X: boolean variable, 0,1: constants

5. X + X = X -- Idepotence

6. X X = X -- Idepotence7. X + X = 1 -- Complement

8. X X = 0 -- Complement

9. (X) = X -- Involution

Unchanged invalue following

multiplication byitself


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The Duality PrincipleThe Duality Principle The dual of an expression is obtained by

exchanging ( and +), and (1 and 0) in it,provided that the precedence of operations isnot changed.

Cannot exchange x with x

Example: Find H(x,y,z), the dual of F(x,y,z) = xyz + xyz H = (x+y+z) (x+y+ z)

Dual does not always equal the originalexpression

If a Boolean equation/equality is valid, its dual is alsovalid


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The Duality Principle (cont.)The Duality Principle (cont.)With respect to duality, Identities 1 8 have the

following relationship:

1. X + 0 = X 2. X 1 = X (dual of 1)3. X + 1 = 1 4. X 0 = 0 (dual of 3)5. X + X = X 6. X X = X (dual of 5)7. X + X = 1 8. X X = 0 (dual of 8)


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More Boolean Algebra PropertiesMore Boolean Algebra PropertiesLet X,Y, and Z: boolean variables

10. X + Y = Y + X 11. X Y = Y X -- Commutative

12. X + (Y+Z) = (X+Y) + Z 13. X(YZ) = (XY)Z -- Associative

14. X(Y+Z) = XY + XZ 15. X+(YZ) = (X+Y) (X+Z)

-- Distributive16. (X + Y) = X Y 17. (X Y) = X + Y -- DeMorgans

In general,( X1 + X2 + + Xn ) = X1X2 Xn, and

( X1X2 Xn ) = X1 + X2 + + Xn


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Absorption Property (Covering)Absorption Property (Covering)

1. x + xy = x

2. x(x+y) = x (dual)

Proof:

x + xy = x1 + xy= x(1+y)= x1

= xQED (2 true by duality)


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Consensus TheoremConsensus Theorem

1. xy + xz + yz = xy + xz

2. (x+y)(x+z)(y+z) = (x+y)(x+z) -- (dual)

Proof:

xy + xz + yz = xy + xz + (x+x)yz= xy + xz + xyz + xyz= (xy + xyz) + (xz + xzy)

= xy + xzQED (2 true by duality).


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Truth Tables (revisited)Truth Tables (revisited)

Enumerates all possible

combinations of variablevalues and the correspondingfunction value

Truth tables for some arbitraryfunctionsF1(x,y,z), F2(x,y,z), and

F3(x,y,z) are shown to the right.

101111

000011010101

010001

110110

100010

100100110000

F3F2F1zyx


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Truth Tables (cont.)Truth Tables (cont.) Truth table: a unique representation of a

Boolean function If two functions have identical truth tables, the

functions are equivalent (and vice-versa).

Truth tables can be used to prove equalitytheorems. However, the size of a truth table grows

exponentially with the number of variablesinvolved, hence unwieldy. This motivates theuse of Boolean Algebra.


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Boolean expressionsBoolean expressions--NOT uniqueNOT unique

Unlike truth tables, expressionsrepresenting a Boolean function areNOT unique.

Example:

F(x,y,z) = xyz + xyz + xyz G(x,y,z) = xyz + yz

The corresponding truth tables for F()and G() are to the right. They are

identical!

Thus, F() = G()

00111

1101100101

00001

00110

11010

00100

11000

GFzyx


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Algebraic ManipulationAlgebraic Manipulation Boolean algebra is a useful tool for

simplifying digital circuits. Why do it? Simpler can mean cheaper,

smaller, faster.

Example: Simplify F = xyz + xyz + xz.F = xyz + xyz + xz

= xy(z+z) + xz= xy1 + xz= xy + xz


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Algebraic Manipulation (cont.)Algebraic Manipulation (cont.) Example: Prove

xyz + xyz + xyz = xz + yz Proof:

xyz+ xyz+ xyz

= xyz + xyz + xyz + xyz= xz(y+y) + yz(x+x)= xz1 + yz1

= xz + yzQED.


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Complement of a FunctionComplement of a Function

The complement of a function is derivedby interchanging ( and +), and (1 and 0),and complementing each variable.

Otherwise, interchange 1s to 0s in thetruth table column showing F.

The complementof a function IS NOT

THE SAME as the dualof a function.


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Complementation: ExampleComplementation: Example

Find G(x,y,z), the complement ofF(x,y,z) = xyz + xyz

G = F = (xyz + xyz)

= (xyz) (xyz) DeMorgan= (x+y+z) (x+y+z) DeMorganagain

Note: The complement of a function can also bederived by finding the functions dual, and thencomplementing all of the literals


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Canonical and Standard FormsCanonical and Standard Forms

We need to consider formal techniques for

the simplification of Boolean functions. Minterms and Maxterms

Sum-of-Minterms and Product-of-Maxterms

Product and Sum terms

Sum-of-Products (SOP) and Product-of-Sums

(POS)


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DefinitionsDefinitions Literal:A variable or its complement

Product term:literals connected by Sum term:literals connected by +

Minterm:a product term in which all thevariables appear exactly once, eithercomplemented or uncomplemented

Maxterm:a sum term in which all the variablesappear exactly once, either complemented oruncomplemented


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MintermMinterm Represents exactly one combination in the truth

table. Denoted by mj, wherejis the decimal equivalentof the minterms corresponding binarycombination (bj).

A variable in mjis complemented if its value in bjis 0, otherwise is uncomplemented.

Example: Assume 3 variables (A,B,C), andj=3.Then, bj= 011 and its corresponding minterm isdenoted by mj= ABC


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MaxtermMaxterm Represents exactly one combination in the truth

table. Denoted by Mj, wherejis the decimal equivalentof the maxterms corresponding binarycombination (bj).

A variable in Mjis complemented if its value in bjis 1, otherwise is uncomplemented.

Example: Assume 3 variables (A,B,C), andj=3.Then, bj= 011 and its corresponding maxterm isdenoted by Mj= A+B+C


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Truth Table notation forTruth Table notation for

Minterms andMinterms and MaxtermsMaxterms

Minterms andMaxterms areeasy to denoteusing a truth

table. Example:

Assume 3variables x,y,z(order is fixed)

x+y+z = M7xyz = m7111

x+y+z = M6xyz = m6011

x+y+z = M5

xyz = m5

101

x+y+z = M4xyz = m4001

x+y+z= M3xyz = m3110

x+y+z = M2xyz = m2010

x+y+z = M1xyz = m1100

x+y+z = M0xyz = m0000

MaxtermMintermzyx


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Canonical Forms (Unique)Canonical Forms (Unique) Any Boolean function F( ) can be

expressed as a uniquesum of mintermsand a unique product of maxterms (undera fixed variable ordering).

In other words, every function F() has twocanonical forms: Canonical Sum-Of-Products (sum of

minterms) Canonical Product-Of-Sums (product ofmaxterms)


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Canonical Forms (cont.)Canonical Forms (cont.)

Canonical Sum-Of-Products:

The minterms included are those mjsuchthat F( ) = 1 in rowjof the truth table for F().

Canonical Product-Of-Sums:The maxterms included are those Mjsuch

that F( ) = 0 in rowjof the truth table for F().


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ExampleExample Truth table for f1(a,b,c) at right

The canonical sum-of-products form forf1 isf1(a,b,c) = m1 + m2 + m4 + m6

= abc + abc + abc + abc

The canonical product-of-sums form forf1 isf1(a,b,c) = M0 M3 M5 M7

= (a+b+c)(a+b+c)(a+b+c)(a+b+c).

Observe that: mj = Mj

0111

10110101

1001

01101010

1100

0000

f1cba


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Shorthand:Shorthand: andand f1(a,b,c) = m(1,2,4,6), where indicates that

this is a sum-of-products form, and m(1,2,4,6)indicates that the minterms to be included arem1, m2, m4, and m6.

f1(a,b,c) = M(0,3,5,7), where indicates that

this is a product-of-sums form, and M(0,3,5,7)indicates that the maxterms to be included areM0, M3, M5, and M7.

Since mj = Mj for anyj, m(1,2,4,6) = M(0,3,5,7) = f1(a,b,c)


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Conversion Between Canonical FormsConversion Between Canonical Forms

Replace with (or vice versa) and replace thosejs

that appeared in the original form with those that do not. Example:

f1(a,b,c) = abc + abc + abc + abc= m1 + m2 + m4 + m6

=(1,2,4,6)

=(0,3,5,7)= (a+b+c)(a+b+c)(a+b+c)(a+b+c)


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Standard Forms (NOT Unique)Standard Forms (NOT Unique) Standard forms are likecanonical forms,

except that not all variables need appearin the individual product (SOP) or sum(POS) terms.

Example:f1(a,b,c) = abc + bc + acis a standardsum-of-products form

f1(a,b,c) = (a+b+c)(b+c)(a+c)is a standardproduct-of-sums form.


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Conversion of SOP fromConversion of SOP from

standard to canonical formstandard to canonical form

Expand non-canonicalterms by inserting

equivalent of 1 in each missing variable x:(x + x) = 1

Remove duplicate minterms f1(a,b,c) = abc + bc + ac= abc + (a+a)bc + a(b+b)c

= abc + abc + abc + abc + abc= abc + abc + abc + abc


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Conversion of POS fromConversion of POS from

standard to canonical formstandard to canonical form Expand noncanonical terms by adding 0 in terms

of missing variables (e.g., xx = 0) and using thedistributive law

Remove duplicate maxterms

f1(a,b,c) = (a+b+c)(b+c)(a+c)= (a+b+c)(aa+b+c)(a+bb+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)


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KarnaughKarnaugh MapsMaps Karnaugh maps (K-maps) are graphical

representations of boolean functions. One map cellcorresponds to a row in the

truth table.

Also, one map cell corresponds to aminterm or a maxterm in the booleanexpression

Multiple-cell areas of the map correspondto standard terms.


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TwoTwo--Variable Map (cont.)Variable Map (cont.) Any two adjacent cells in the map differ by

ONLY one variable, which appearscomplemented in one cell anduncomplemented in the other.

Example:m0 (=x1x2) is adjacent to m1 (=x1x2) and

m2 (=x1x2) but NOT m3 (=x1x2)


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22--Variable MapVariable Map ---- ExampleExample f(x1,x2) = x1x2+ x1x2 + x1x2

= m0 + m1 + m2= x1 + x2

1s placed in K-map forspecified minterms m0, m1, m2

Grouping (ORing) of 1s allows

simplification What (simpler) function is

represented by each dashedrectangle?

a1 = m0 + m1 a2 = m0 + m2 Note m0 covered twice

011

110

10x1

x2

0 1

2 3


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Minimization as SOP using KMinimization as SOP using K--mapmap

Enter 1s in the K-map for each product

term in the function Group adjacentK-map cells containing 1s

to obtain a product with fewer variables.

Groups must be in power of 2 (2, 4, 8, ) Handle boundary wrap for K-maps of 3

or more variables.

Realize that answer may not be unique


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ThreeThree--Variable MapVariable Map

m6m7m5m41

m2m3m1m00

10110100yz

x0 1 3 2

4 5 7 6

-Note: variable ordering is (x,y,z); yz specifies

column, x specifies row.-Each cell is adjacent to threeother cells (left orright or top or bottom or edge wrap)


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ThreeThree--Variable Map (cont.)Variable Map (cont.)

The types of structuresThe types of structures

that are either minterms orthat are either minterms orare generated by repeatedare generated by repeatedapplication of theapplication of theminimization theorem on aminimization theorem on a

three variable map arethree variable map areshown at right.shown at right.Groups of 1, 2, 4, 8 areGroups of 1, 2, 4, 8 arepossible.possible.

minterm

group of 2 terms

group of 4 terms


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SimplificationSimplification Enter minterms of the Boolean function

into the map, then group terms Example: f(a,b,c) = ac + abc + bc

Result: f(a,b,c) = ac+ b

11

111

abc

11111

00 01 10 11

0

1

00 01 10 11

0

1


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More ExamplesMore Examples

f1(x, y, z) = m(2,3,5,7)

f2(x, y, z) = m (0,1,2,3,6)

ff11(x, y, z) =(x, y, z) = xxyy ++ xzxz

ff22(x, y, z) =(x, y, z) = xx+yz+yz

10110100yz

X

1

0

1111

1

1111


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FourFour--Variable MapsVariable Maps

Top cells are adjacent to bottom cells. Left-edgecells are adjacent to right-edge cells.

Note variable ordering (WXYZ).

m11m10m9m810

m14m15m13m1211

m6m7m5m401

m2m3m1m000

10110100WX

YZ


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FourFour--variable Map Simplificationvariable Map Simplification One square represents a minterm of 4 literals. A rectangle of 2 adjacent squares represents a

product term of 3 literals. A rectangle of 4 squares represents a product

term of 2 literals.

A rectangle of 8 squares represents a productterm of 1 literal.

A rectangle of 16 squares produces a function

that is equal to logic 1.


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ExampleExample Simplify the following Boolean function

(A,B,C,D) = m(0,1,2,4,5,7,8,9,10,12,13).

First put the function g( ) into the map, and thengroup as many 1s as possible.

cdab

111

11

111

111

g(A,B,C,D) =g(A,B,C,D) = cc+b+bdd+a+abdbd

111

11

111

111

00 01 11 10

00

01

11

10

00 01 11 10


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55--Variable KVariable K--MapMap

101198

13 141512

5 674

1 230

BC

DE

26272524

30312928

22232120

18191716

BC

DE

A=0

A=1

ABCDE

ABCDE


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ImplicantsImplicants andand

PrimePrime ImplicantsImplicants (PIs)(PIs)

An Implicant (P) of a function F is a product

term which implies F, i.e., F(P) = 1. An implicant (PI) of F is called a Prime

Implicant of F if any product term obtained

by deleting a literal of PI is NOT an implicantof F

Thus, a prime implicant is not contained inany larger implicant.


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ExampleExample Consider function f(a,b,c,d) whose K-

map is shown at right.

abis not a prime implicant becauseit is contained in b.

acdis not a prime implicant because

it is contained in ad. b, ad, and acd are prime implicants.

111

111

111

11

b

cd abad

acd

ab

acd


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Essential Prime Implicants (EPIs)

If a minterm of a function F is included in ONLY

one prime implicant p, then pis an essentialprime implicant of F.

An essential prime implicant MUST appear inall possible SOP expressions of a function

To find essential prime implicants: Generate all prime implicants of a function Select those prime implicants that contain at least one

1 that is not covered by any other prime implicant.

For the previous example, the PIs are b, ad,and acd; all of these are essential.

111

111

111

11

bad

acd


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Another ExampleAnother Example Consider f2(a,b,c,d), whose K-map

is shown below. The only essential PI is bd.

111

11

111

1cd

ab

S t ti P d fS t ti P d f


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Systematic Procedure forSystematic Procedure for

Simplifying Boolean FunctionsSimplifying Boolean Functions1. Generate all PIs of the function.

2. Include all essential PIs.3. For remaining minterms not included in

the essential PIs, select a set of other

PIs to cover them, with minimal overlapin the set.

4. The resulting simplified function is the

logical OR of the product terms selectedabove.


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ExampleExample f(a,b,c,d) =

m(0,1,2,3,4,5,7,14,15). Five grouped terms, not all

needed.

3 shaded cells covered by onlyone term

3 EPIs, since each shaded cell is

covered by a different term. F(a,b,c,d) = ab + ac + ad + abc

11

111

1111ab

cd


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Product of Sums SimplificationProduct of Sums Simplification Use sum-of-products simplification on the

zerosof the function in the K-map to getF. Find the complement of F, i.e. (F) = F

Recall that the complement of a booleanfunction can be obtained by (1) taking thedual and (2) complementing each literal.

OR, using DeMorgans Theorem.

POS E lPOS E l


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POS ExamplePOS Example

0000

1100

0111

1111

abcd

F(a,b,c,d) = ab + ac + abcd

Find dual of F, dual(F) = (a+b)(a+c)(a+b+c+d)

Complement of literals in dual(F) to get FF = (a+b)(a+c)(a+b+c+d)(verify that this is the same as in slide 60)


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Don't Care ConditionsDon't Care Conditions There may be a combination of input values which

will neveroccur

if they do occur, the output is of no concern. The function value for such combinations is called a

don't care.

They are usually denoted with x. Each x may bearbitrarily assigned the value 0 or 1 in animplementation.

Dont cares can be used to furthersimplify afunction


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Minimization using DonMinimization using Dont Carest Cares

Treat don't cares as if they are 1s to

generate PIs. Delete PI's that cover only don't care

minterms.

Treat the covering of remaining don't careminterms as optional in the selection

process (i.e. they may be, but need not be,covered).

cd


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ExampleExample Simplify the function f(a,b,c,d)

whose K-map is shown at the right.

f = acd+ab+cd+abc

or

f = acd+ab+cd+abd

The middle two terms are EPIs, whilethe first and last terms are selected tocover the minterms m1, m4, and m5.

(Theres a third solution!)

xx11

xx00

1011

1010

xx11

xx00

1011

1010

xx11

xx00

10111010

abcd

0001

11

10

00 01 11 10

cd


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Another ExampleAnother Example Simplify the function

g(a,b,c,d) whose K-mapis shown at right.

g = ac+ abor

g = ac+bd

0xx0

1xx1

x0x1

001x

0xx0

1xx1

x0x1

001x

0xx01xx1

x0x1

001x

abcd


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22-Oct-07 Chapter 2-i: Combinational Logic

Circuits (2.1-- 2.5)

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Algorithmic minimizationAlgorithmic minimization What do we do for functions with more

than 4-5 variables? You can code up a minimiser

(Computer-Aided Design, CAD)

Quine-McCluskey algorithm

Iterated consensus

We wont discuss these techniques here

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