chapter 3: principles of combinational logic (sections 3.1 – 3.5)

CHAPTER 3: PRINCIPLES CHAPTER 3: PRINCIPLES

OF COMBINATIONAL LOGICOF COMBINATIONAL LOGIC

(Sections 3.1 – 3.5)(Sections 3.1 – 3.5)

3.1 DEFINITION OF 3.1 DEFINITION OF COMBINATIONAL LOGICCOMBINATIONAL LOGIC

Logic circuits without feedback from output to the Logic circuits without feedback from output to the input, constructed from a functionally complete input, constructed from a functionally complete gate set, are said to be combinational.gate set, are said to be combinational.

Inputs OutputsCombinatioCombinational Logic nal Logic FunctionsFunctions

· ·

····

· ·

····

3.1.1 PROBLEM STATEMENTS TO 3.1.1 PROBLEM STATEMENTS TO TRUTH TABLESTRUTH TABLES

E.g : Conveyor system(P85, E.g : Conveyor system(P85, Ex. 3.1) Ex. 3.1)

M=a’bms+ab’ms+abmM=a’bms+ab’ms+abmss


E.g. A NASA E.g. A NASA system(P87, Ex. 3.2)system(P87, Ex. 3.2)

C1C1

C2C2C3C3


E.g. Conveyor system(P88, E.g. Conveyor system(P88, ex. 3.4)ex. 3.4)

S3

S1

S2

m3

m1

m2 m4

The process of converting a verbal problem The process of converting a verbal problem statement into a truth table :statement into a truth table : Determine the input variables and output variables that

are involved Assign mnemonic or letter or number symbols to each

variable determine the size of the truth table; how many input

combinations exist: 2x=y where x=number of input variables and y=number of combinations.

Construct a truth table containing all of the input variable combinations

By careful reading of the problem statement determine the combinations of inputs that cause a given output to be true


Exe: Design a truth table to indicate a Exe: Design a truth table to indicate a majority of three inputs is truemajority of three inputs is true


II33 I I2 2 I I11 O O11

0 0 0 00 0 0 0

0 0 1 00 0 1 0

0 1 0 00 1 0 0

0 1 1 10 1 1 1

1 0 0 01 0 0 0

1 0 1 11 0 1 1

1 1 0 11 1 0 1

1 1 1 11 1 1 1

3.1.2 Deriving Switching Equations3.1.2 Deriving Switching Equations

DefinitionsDefinitions Product term: literals connected by • Sum term: literals connected by + Minterm: a product term in which all the variables

appear exactly once, either complemented or uncomplemented

Maxterm: a sum term in which all the variables appear exactly once, either complemented or uncomplemented

MintermMinterm

Represents exactly one combination in the Represents exactly one combination in the truth table.truth table.

Denoted by Denoted by mmjj, where , where j j is the decimal is the decimal equivalent of the minterm’s corresponding equivalent of the minterm’s corresponding binary combination binary combination (b(bjj))..

A variable in A variable in mmjj is complemented if its value is complemented if its value in in bbjj is 0, otherwise is uncomplemented. is 0, otherwise is uncomplemented.

M(a,b,m,s)=a’bms+ab’ms+abmsM(a,b,m,s)=a’bms+ab’ms+abms (m(m77) (m) (m1111) (m) (m1515))

MaxtermMaxterm

Represents exactly one combination in the truth Represents exactly one combination in the truth table.table.

Denoted by Denoted by MMjj, where , where j j is the decimal equivalent is the decimal equivalent of the maxterm’s corresponding binary of the maxterm’s corresponding binary combination combination (b(bjj))..

A variable in A variable in MMjj is complemented if its value in is complemented if its value in bbjj is 1, otherwise is uncomplemented.is 1, otherwise is uncomplemented.

F(a,b,c)=(a’+b+c’)(a+b+c’)(a+b’+c’)F(a,b,c)=(a’+b+c’)(a+b+c’)(a+b’+c’) 101 001 011101 001 011

MM55 M M11 M M33

Truth Table notation for Minterms Truth Table notation for Minterms and Maxtermsand Maxterms

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

3.2 Canonical(3.2 Canonical( 规范规范 ) Forms) Forms Any Boolean function F( ) can be expressed Any Boolean function F( ) can be expressed

as a as a uniqueunique sumsum of of minminterms and a unique terms and a unique productproduct of of maxmaxterms (under a fixed variable terms (under a fixed variable ordering).ordering).

every function F() has two canonical forms:every function F() has two canonical forms: Canonical Sum-Of-Products (sum of minterms) Canonical Product-Of-Sums (product of maxterms)

Expand Expand non-canonicalnon-canonical terms by inserting terms by inserting equivalent of 1 in each missing variable x:equivalent of 1 in each missing variable x: (x + x’) = 1 (x + x’) = 1

Remove duplicate mintermsRemove duplicate minterms ff11(a,b,c) = a’b’c + bc’ + ac’(a,b,c) = a’b’c + bc’ + ac’

= a’b’c + = a’b’c + (a+a’)(a+a’)bc’ + abc’ + a(b+b’)(b+b’)c’c’ = a’b’c + = a’b’c + abc’abc’ + a’bc’ + + a’bc’ + abc’abc’ + ab’c’ + ab’c’ = a’b’c + abc’ + a’bc + ab’c’ = a’b’c + abc’ + a’bc + ab’c’

Convert switching equations to Convert switching equations to canonical formcanonical form


SOPSOP

E.g. P=f(a,b,c)=ab’+ac’+bcE.g. P=f(a,b,c)=ab’+ac’+bc

=ab’(c+c’)+ac’ (b+b’)+bc(a+a’)=ab’(c+c’)+ac’ (b+b’)+bc(a+a’)

=ab’c+ab’c’+abc’+ab’c’+abc+a’bc=ab’c+ab’c’+abc’+ab’c’+abc+a’bc

=ab’c+ab’c’+abc’+abc+a’bc=ab’c+ab’c’+abc’+abc+a’bc

Expand noncanonical terms by adding 0 in terms of Expand noncanonical terms by adding 0 in terms of missing variables (missing variables (e.g.e.g., xx’ = 0) and using the , xx’ = 0) and using the distributive lawdistributive law

Remove duplicate maxtermsRemove duplicate maxterms ff11(a,b,c) = (a+b+c)•(b’+c’)•(a’+c’)(a,b,c) = (a+b+c)•(b’+c’)•(a’+c’)

= (a+b+c)•( = (a+b+c)•(aa’aa’+b’+c’)•(a’++b’+c’)•(a’+bb’bb’+c’)+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’)(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’)



POSPOS

E.g. T=f(a,b,c)=(a+b’)(b’+c)E.g. T=f(a,b,c)=(a+b’)(b’+c)

=(a+b’+cc’)(b’+c+aa’)=(a+b’+cc’)(b’+c+aa’)

=(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’)(a’+b’+c)=(a+b’+c) (a+b’+c’)(a’+b’+c)

Exe:Exe:

P=f(w,x,y,z)=w’x+yz’P=f(w,x,y,z)=w’x+yz’

T=f(a,b,c,d)=(a+b’+c)(a’+d)T=f(a,b,c,d)=(a+b’+c)(a’+d)


Ans:Ans:

P=f(w,x,y,z)=wxyz’+wx’yz’+w’xyz+w’xyz’+w’xy’zP=f(w,x,y,z)=wxyz’+wx’yz’+w’xyz+w’xyz’+w’xy’z+w’xy’z’+w’x’yz’+w’xy’z’+w’x’yz’

T=f(a,b,c,d)= (a+b’+c+d) (a+b’+c+d’) (a’+b+c+d) T=f(a,b,c,d)= (a+b’+c+d) (a+b’+c+d’) (a’+b+c+d) (a’+b’+c+d) (a’+b+c’+d) (a’+b’+c’+d)(a’+b’+c+d) (a’+b+c’+d) (a’+b’+c’+d)

3.3 GENERATION OF SWITCHING 3.3 GENERATION OF SWITCHING EQUATIONS FROM TRUTH TABLESEQUATIONS FROM TRUTH TABLES

Canonical Sum-Of-Products:Canonical Sum-Of-Products:The minterms included are those mThe minterms included are those mjj such such

that F( ) = 1 in row that F( ) = 1 in row jj of the truth table for F( of the truth table for F( ).).

Canonical Product-Of-Sums:Canonical Product-Of-Sums:The maxterms included are those MThe maxterms included are those Mjj such such

that F( ) = 0 in row that F( ) = 0 in row jj of the truth table for F( of the truth table for F( ).).

ExampleExample

Truth table for fTruth table for f11(a,b,c) at right(a,b,c) at right The canonical sum-of-products form for The canonical sum-of-products form for

ff11 is isff11(a,b,c) = m(a,b,c) = m11 + m + m22 + m + m44 + m + m66

= a’b’c + a’bc’ + ab’c’ + abc’ = a’b’c + a’bc’ + ab’c’ + abc’ The canonical product-of-sums form for The canonical product-of-sums form for

ff1 1 isisff11(a,b,c) = M(a,b,c) = M00 •• M M33 • • MM55 • • MM77 = (a+b+c) = (a+b+c)••(a+b’+c’)(a+b’+c’)• •

(a’+b+c’)(a’+b+c’)••(a’+b’+c’).(a’+b’+c’). Observe that: mObserve that: mjj = M = Mjj’’

00111111

11001111

00110011

11000011

00111100

11001100

11110000

00000000

ff11ccbbaa

Shorthand: Shorthand: ∑∑ and ∏ and ∏ ff11(a,b,c) = (a,b,c) = ∑∑ mm(1,2,4,6), where(1,2,4,6), where ∑ indicates that this ∑ indicates that this

is a sum-of-products form, and m(1,2,4,6) is a sum-of-products form, and m(1,2,4,6) indicates that the minterms to be included are mindicates that the minterms to be included are m11, ,

mm22, m, m44, and m, and m66..

ff11(a,b,c) = ∏(a,b,c) = ∏ M(0,3,5,7), where ∏ indicates that M(0,3,5,7), where ∏ indicates that

this is a product-of-sums form, and M(0,3,5,7) this is a product-of-sums form, and M(0,3,5,7) indicates that the maxterms to be included are Mindicates that the maxterms to be included are M00, ,

MM33, M, M55, and M, and M77..

∑∑ m(1,2,4,6) = ∏m(1,2,4,6) = ∏ M(0,3,5,7) = fM(0,3,5,7) = f11(a,b,c) (a,b,c)

Conversion Between Canonical FormsConversion Between Canonical Forms

Replace Replace ∑∑ with with ∏∏ (or (or vice versavice versa) and replace those ) and replace those j’j’s that s that appeared in the original form with those that do not.appeared in the original form with those that do not.

Example:Example:ff11(a,b,c)(a,b,c) = a’b’c + a’bc’ + ab’c’ + abc’ = a’b’c + a’bc’ + ab’c’ + abc’

= m= m11 + m + m22 + m + m44 + m + m66

= = ∑∑((1,2,4,61,2,4,6))

= = ∏∏((0,3,5,70,3,5,7)) = (a+b+c)•(a+b’+c’)•(a’+b+c’)•(a’+b’+c’) = (a+b+c)•(a+b’+c’)•(a’+b+c’)•(a’+b’+c’)

Karnaugh MapsKarnaugh Maps

Why Karnaugh mapsWhy Karnaugh maps ？？to simplify boolean equationsto simplify boolean equations

Karnaugh maps (K-maps) are Karnaugh maps (K-maps) are graphicalgraphical representations of boolean functions.representations of boolean functions.

One One map cellmap cell corresponds to a row in the corresponds to a row in the truth table.truth table.

one map cell corresponds to a minterm or a one map cell corresponds to a minterm or a maxterm in the boolean expressionmaxterm in the boolean expression

Karnaugh MapsKarnaugh Maps

mm

00mm

33

mm

11

mm

22

0 1

1

0

AB

mm

00mm

11

mm

33

mm

77

mm

55

mm

22

mm

66

mm

44

00 01 11 10

0

1

AB

C

mm

00mm

11mm

33

mm

55

mm11

33

mm

99

mm

44

mm11

22

mm

88

mm

66

mm1414 mm11

00

mm

77

mm11

55

mm11

11mm

22

ABCD 00 01 11 10

00

01

11

10

Logic adjacencyLogic adjacency : Any two : Any two adjacent cells in the map adjacent cells in the map differ by ONLY one. differ by ONLY one.

Example:Example:mm0 0 (=x(=x11’x’x22’) is adjacent to m’) is adjacent to m1 1

(=x(=x11’x’x22) and m) and m2 2 (=x(=x11xx22’) but ’) but

NOT mNOT m3 3 (=x(=x11xx22) )

2-2-Variable Map -- Example Variable Map -- Example E.g. E.g.

f(xf(x11,x,x22) = x) = x11’x’x22’+ x’+ x11’x’x22 + x + x11xx22’ ’ = m = m00 + m + m11 + m + m22

What (simpler) function is What (simpler) function is

represented by each dashed represented by each dashed rectangle?rectangle? x1’x2’+ x1’x2 =x1’(x2’+x2)=x1’ x1’x2’+ x1x2’=x2’(x1’+x1) =x2’

f(x1,x2) =x1’+x2’

001111

111100

1100xx22

x1

0 2

1 3

Minimization as SOP using K-mapMinimization as SOP using K-map

Enter 1s in the K-map for each product Enter 1s in the K-map for each product term in the functionterm in the function

Group Group adjacentadjacent K-map cells containing 1s to K-map cells containing 1s to obtain a product with fewer variables. obtain a product with fewer variables. Groups must be in power of 2 (2, 4, 8, …)Groups must be in power of 2 (2, 4, 8, …)

Handle “Handle “boundary wrapboundary wrap” for K-maps of 3 ” for K-maps of 3 or more variables.or more variables.

Three-Variable MapThree-Variable Map

mm66mm77mm55mm4411

mm22mm33mm11mm0000

1010111101010000

yzyz

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 three other cells (left or right or top or bottom or edge wrap)

Three-Variable Map (cont.)Three-Variable Map (cont.)

The types of structures that are The types of structures that are either minterms or are generated either minterms or are generated by repeated application of the by repeated application of the minimization theorem on a three minimization theorem on a three variable map are shown at right. variable map are shown at right. Groups of 1, 2, 4, 8 are possible.Groups of 1, 2, 4, 8 are possible.

minterm

group of 2 terms

group of 4 terms

SimplificationSimplification

Example: f(a,b,c) = ac’ + abc + bc’Example: f(a,b,c) = ac’ + abc + bc’

=a=a(b+b’)(b+b’)c’+abc+c’+abc+(a+a’)(a+a’)bc’bc’

=abc’+ab’c’+abc+abc’+a’bc’=abc’+ab’c’+abc+abc’+a’bc’

=abc’+ab’c’+abc+a’bc’=abc’+ab’c’+abc+a’bc’ Result: f(a,b,c) = Result: f(a,b,c) = ac’ac’++abab++bc’bc’

11

111111

abc

1010111101010000

11

00

More ExamplesMore Examples

ff11(x, y, z) = (x, y, z) = ∑∑ m(2,3,5,7) m(2,3,5,7)

ff22(x, y, z) = ∑ m (0,1,2,3,6)(x, y, z) = ∑ m (0,1,2,3,6)

ff11(x, y, z) = (x, y, z) = x’yx’y + + xzxz

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

1010111101010000

xyxy

zz

11

00

11 11

11

11

1111

111111

xyxy

zz

1010111101010000

11

00

1010

1111

0101

0000

Four-Variable MapsFour-Variable Maps

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

Note variable ordering (WXYZ).Note variable ordering (WXYZ).

mm1010mm1111mm99mm88

mm1414mm1515mm1313mm1212

mm66mm77mm55mm44

mm22mm33mm11mm00

1010 1111 0101 0000WXWX

YZ

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

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

of 2 literals.of 2 literals. A rectangle of 8 squares represents a product term A rectangle of 8 squares represents a product term

of 1 literal.of 1 literal. A rectangle of 16 squares produces a function that A rectangle of 16 squares produces a function that

is equal to logic 1.is equal to logic 1.

ExampleExample Simplify the following Boolean function Simplify the following Boolean function

(A,B,C,D) = (A,B,C,D) = ∑∑m(0,1,2,3,4,5,6,7,8,10,13).m(0,1,2,3,4,5,6,7,8,10,13). First put the function g( ) into the map, and then First put the function g( ) into the map, and then

group as many 1s as possible.group as many 1s as possible.abab

cdcd

11 1111

1111

11 1111

11 1111

g(A,B,C,D) = g(A,B,C,D) = a’a’++b’d’b’d’++bc’dbc’d

111111

1111

111111

111111

1010

1111

0101

0000 1010 1111 0101 0000

Implicants ,Prime Implicants (PIs) and Implicants ,Prime Implicants (PIs) and Essential Prime ImplicantEssential Prime Implicant (EPI)(EPI)

Implicant: Implicant: any single minterm

Prime Implicant Prime Implicant (PI)(PI): a group of minterms that cannot be combined with any other minterms or groups

Essential Prime Implicant Essential Prime Implicant (EPI): (EPI): a PI in which one or more miniterms are unique

How to find PI and EPI?How to find PI and EPI?

1

1 1

1

1 1

A’BC’DBCD

BDABC’

AC’D’

00 01 11 10

00

01

11

10

ABCD

1?1?

ExampleExample

a’b’a’b’ is is notnot a prime implicant a prime implicant because it is contained in because it is contained in b’.b’.

acdacd is is notnot a prime implicant a prime implicant because it is contained in because it is contained in adad..

b’, ad, and a’cd’ are prime b’, ad, and a’cd’ are prime implicants.implicants.

b’, ad, and a’cd’ are essentialb’, ad, and a’cd’ are essential

prime implicants.prime implicants.111111

111111

111111

1111

b’

cd ab

ad

a’cd’

a’b’

acd

Another ExampleAnother Example Consider fConsider f22(a,b,c,d), whose K-map (a,b,c,d), whose K-map

is shown below.is shown below. The only essential PI isThe only essential PI is b’d.b’d.

111111

1111

111111

11cd

ab

Systematic Procedure for Systematic Procedure for Simplifying Boolean FunctionsSimplifying Boolean Functions

1. Load the minterms into the K-map by placing a 1 in the appropriate square.

2. Look for groups of minterms (PI)1. The group size must be a power of 22. Find the largest groups of minterms first, then smaller

collections until all groups are found.3. Find all essential PIs.4. For remaining min-terms not included in the

essential PIs, select a set of other PIs to cover them.

5. The resulting simplified function is the logical OR of the product terms selected above.

ExampleExample

f(a,b,c,d) = f(a,b,c,d) = ∑∑m(0,1,4,5,8,11,12,13,15).m(0,1,4,5,8,11,12,13,15).

F(a,b,c,d) = c’d’ + a’c’ + bc’ F(a,b,c,d) = c’d’ + a’c’ + bc’ + acd+ acd

1111

111111

11111111cd

ab

HomeworkHomework

Simplifying Boolean FunctionsSimplifying Boolean Functions F(x,y,z)=F(x,y,z)=∑(0,2,3,4,5,7)∑(0,2,3,4,5,7) F(a,b,c,d)=F(a,b,c,d)=∑(0,3,4,5,7,11,13,15)∑(0,3,4,5,7,11,13,15) F(w,x,y,z)=∑(0,1,4,5,9,11,13,15)F(w,x,y,z)=∑(0,1,4,5,9,11,13,15) F(a,b,c,d)=∑(0,1,2,4,5,6,8,9,12,13,14)F(a,b,c,d)=∑(0,1,2,4,5,6,8,9,12,13,14) F(a,b,c,d)=∑(1,3,4,5,7,8,9,11,15)F(a,b,c,d)=∑(1,3,4,5,7,8,9,11,15)

F(w,x,y,z)=∑(1,5,7,8,9,10,11,13,15)F(w,x,y,z)=∑(1,5,7,8,9,10,11,13,15) P151: 4-c, 4-eP151: 4-c, 4-e

chapter 3: principles of combinational logic (sections 3.1 – 3.5)

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