Digital Design : BOOLEAN ALGEBRA AND THEOREMS, MINI TERMS AND MAX TERMS Part IV

Chapter 4

BOOLEAN ALGEBRA AND THEOREMS, MINI TERMS

AND MAX TERMS

Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006

2

Lesson 4

BOOLEAN EXPRESSION, TRUTH TABLE and

SUM OF THE PRODUCTS (SOPs) [MINITERMS]


3

Outline

• SOP two variables cases• SOP for three variable case • SOP for four variable case• Conversion of Boolean expression

into SOPs {Finding Miniterms]


4

Two variable Miniterms

0 0 mn0= A.B 0 0 0 1

Inputs Miniterms Outputs A B XOR AND OR NAND

1 0 mn1= A.B 1 0 1 10 1 mn2= A.B 1 0 1 11 1 mn3= A.B 0 1 1 0

A B S0 S1 S2 S3


5

XOR,AND, OR and NAND Outputs Two XOR,AND, OR and NAND Outputs Two Variable Cases Sum of Product Terms (SOPs)Variable Cases Sum of Product Terms (SOPs)

XOR: S0 = A.B + A.B =

ΣΣΣΣmn(1, 2)

AND: S1 = A.B =

mn(3)

OR: S2 = A.B + A.B + A.B =

ΣΣΣΣmn(1, 2, 3)

NAND: S3 = A.B + A.B + A.B =

ΣΣΣΣmn(0, 1, 2)


6

Advantage of using SOP form is that functions of any two input logic gate functions can be represented by maximum four ANDs at the inputs and four ORs at an output.

SOP form advantage


7

Outline

• SOP two variable cases•• SOP for three variable caseSOP for three variable case• SOP for four variable case• Conversion of Boolean expression



8

Three variable Miniterms 0 to 3

0 0 0 mn0= A.B.C 0 0 0 1

Inputs Miniterms Outputs A B C F1 F2 F3 F4

1 0 0 mn1= A.B. C 1 0 1 10 1 0 mn2= A.B .C 1 0 1 11 1 0 mn3= A.B .C 0 1 1 0

A B C S S’ S’’ S’’’


9

Three variable Miniterms 4 to 7

0 0 1 mn4= A.B.C 0 0 0 1

Inputs Miniterms Outputs A B C F1 F2 F3 F4

1 0 1 mn5= A.B. C 1 0 1 10 1 1 mn6= A.B .C 1 0 1 11 1 1 mn7= A.B .C 0 1 1 0

A B C S S’ S’’ S’’’


10

F1, F2, F3 and F4 outputs Three Variable F1, F2, F3 and F4 outputs Three Variable Cases Sum of Product Terms (SOPs)Cases Sum of Product Terms (SOPs)

F1 = S = A.B.C + A.B.C +A.B.C + A.B.C =

ΣΣΣΣmn(1, 2, 5, 6)F2 = S’= A.B.C + A.B.C =

ΣΣΣΣmn(3, 7)

F3 = S’’ = A.B .C + A.B .C + A.B .C + A.B .C + A.B .C + A.B .C =

ΣΣΣΣmn(1, 2, 3, 5, 6, 7)F4 = S’’’ = A.B .C + A.B .C + A.B .C + A.B .C + A.B .C + A.B .C =

ΣΣΣΣmn(0, 1, 2, 4, 5, 6)


11

Advantage of using SOP form is that functions of any three input logic gate functions can be represented by maximum eight ANDs at the inputs and eight ORs at an output.

SOP form advantage


12

Outline

• SOP two variable cases• SOP for three variable cases • SOP for four variable case• Conversion of Boolean expression



13

Four variable Miniterms 0 to 3

0 0 0 0 mn0= A.B.C .D 1

Inputs Miniterms Output A B C D F5

0 1 0 0 mn1= A.B. C .D 01 0 0 0 mn2 = A.B .C .D 11 1 0 0 mn3= A.B .C .D 0

A B C D S


14


0 0 1 0 mn4= A.B.C .D 0


0 1 1 0 mn5= A.B. C .D 01 0 1 0 mn6 = A.B .C .D 01 1 1 0 mn7= A.B .C .D 1

A B C D S


15


0 0 0 1 mn8= A.B.C .D 0


0 1 0 1 mn9= A.B. C .D 11 0 0 1 mn10 = A.B.C .D 01 1 0 1 mn11= A.B .C .D 0

A B C D S


16


0 0 1 1 mn12= A.B.C .D 1


0 1 1 1 mn13= A.B. C .D 01 0 1 1 mn14= A.B .C .D 01 1 1 1 mn15= A.B .C .D 0

A B C D S


17

F5 output Four Variable Case Sum of Product Terms (SOPs)

F5 = S = A.B.C.D + A.B.C .D +A.B.C .D + A.B.C .D + A.B.C .D= =

ΣΣΣΣmn(0, 2, 7, 9, 12)


18

Advantage of using SOP form is that functions of any four input logic gate functions can be represented by maximum sixteen ANDs at the inputs and sixteen ORs at an output.

SOP form advantage


19

Outline

• SOP two variable case• SOP for three variable case • SOP for four variable case• Conversion of Boolean expression

into SOPs [Finding Miniterms]


20

• Suppose in a four variable SOP, there is a term with two variables, only C.D. We perform AND operation with (A + A).(B + B). [Using OR rule Equation (4) that OR of a complement of a variable or term with itself is always 1.]

Example


21

C.D = (A + A).(B+B).C.D = (A + A)(B.C.D + B.C.D)= A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D= Σ mn (15, 11, 7, 3)

= Σ mn (3, 7, 11, 15) and obtain the SOP standard form.

Conversion


22

Summary


23

We learnt:• A Boolean expression output can be

written as an SOP expression• SOP expression has the miniterms• Each miniterm represent that row of

truth table in which output = 1• Each miniterm is implemented by

AND gate(s)• Miniterms after ORing gives the

output


24

We learnt:• Using OR rules, a Boolean

expression with lesser number of variables can be expanded into SOP form to get all the miniterms and obtain SOP standard form


25

End of Lesson 4

BOOLEAN EXPRESSION, TRUTH TABLE and

SUM OF THE PRODUCT


26

THANK YOU