Pune Vidyarthi Griha’s · Block Schematic of ALU IC 74181 IC 74381 8/29/2017 26. Karnaugh Map...

Pune Vidyarthi Griha’s

COLLEGE OF ENGINEERING, NASIK - 4

“Minimization techniques”

Prof. Anand N. Gharu

Assistant Professor

Computer Department

Combinational Logic Circuits

Introduction Standard representation of canonical forms (SOP &

POS), Maxterm and Minterm , Conversion between SOP and POS forms

K-map reduction techniques upto 4 variables (SOP & POS form), Design of Half Adder, Full Adder, Half Subtractor & Full Subtractor using k-Map

Code Converter using K-map: Gray to Binary Binary to Gray Code Converter (upto 4 bit)

IC 7447 as BCD to 7- Segment decoder driver IC 7483 as Adder & Subtractor, 1 Digit BCD Adder Block Schematic of ALU IC 74181 IC 74381

8/29/2017 2

Standard Representation

Any logical expression can be expressed in the

following two forms:

Sum of Product (SOP) Form

Product of Sum (POS) Form

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SOP Form

For Example, logical expression given is;

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Product

. . .Y A B B C AC

POS Form

For Example, logical expression given is;

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Product

( ).( ).( )Y A B B C A C

Standard or Canonical SOP & POS Forms

We can say that a logic expression is said to be in the standard (or canonical) SOP or POS form if each product term (for SOP) and sum term (for POS) consists of all the literals in their complemented or uncomplemented form.

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Standard SOP

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Each product term consists all the literals

Y ABC ABC ABC

Standard POS

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Each sum term consists all the literals

( ).( ).( )Y A B C A B C A B C

Sr. No. Expression Type

1 Non Standard SOP

2 Standard SOP

3 Standard POS

4 Non Standard POS

Examples

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Y AB ABC ABC

Y AB AB AB

( ).( ).( )Y A B A B A B

( ).( )Y A B A B C

Conversion of SOP form to Standard SOP

Procedure: 1. Write down all the terms. 2. If one or more variables are missing in any

product term, expand the term by multiplying it with the sum of each one of the missing variable and its complement .

3. Drop out the redundant terms

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Example 1

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Convert given expression into its standard SOP form

Missing literal is A

Missing literal is B

Missing literal is C

Term formed by ORing of missing literal & its complement

Y AB AC BC

.( ) .( ) .( )Y AB C C AC B B BC A A

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Example 1 Continue….

Standard SOP form Each product term consists all the literals

.( ) .( ) .( )Y AB C C AC B B BC A A

Y ABC ABC ABC ABC ABC ABC

Y ABC ABC ABC ABC

Conversion of POS form to Standard POS

Procedure: 1. Write down all the terms. 2. If one or more variables are missing in any

sum term, expand the term by adding the products of each one of the missing variable and its complement .

3. Drop out the redundant terms

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Example 2

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Convert given expression into its standard SOP form

Missing literal is A

Missing literal is B

Missing literal is C

Term formed by ANDing of missing literal & its complement

( ).( ) ( )Y A B A C B C

( ).( ).( )Y A B CC A C BB B C AA

( ).( ) ( )Y A B A C B C

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Example 2 Continue….

Standard POS form Each sum term consists all the literals

( ).( ).( )Y A B CC A C BB B C AA

( )( ).( )( ).( )( )Y A B C A B C A B C A B C A B C A B C

( )( )( )( )Y A B C A B C A B C A B C

Concept of Minterm and Maxterm

Minterm: Each individual term in the standard SOP form is called as “Minterm”.

Maxterm: Each individual term in the standard

POS form is called as “Maxterm”.

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The concept of minterm and max term allows us to introduce a very convenient shorthand notation to express logic functions

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Minterms & Maxterms for 3 variable/literal logic function

8/29/2017 Amit Nevase 18

Variables Minterms Maxterms

A B C mi Mi

1ABC m

2ABC m

3ABC m

4ABC m

5ABC m

6ABC m

7ABC m

4A B C M

3A B C M

2A B C M

1A B C M

0A B C M 0ABC m

5A B C M

6A B C M

7A B C M

Minterms and maxterms

Each minterm is represented by mi where i=0,1,2,3,…….,2n-1

Each maxterm is represented by Mi where i=0,1,2,3,…….,2n-1

If ‘n’ number of variables forms the function, then number of minterms or maxterms will be 2n • i.e. for 3 variables function f(A,B,C), the number of

minterms or maxterms are 23=8

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Minterms & Maxterms for 2 variable/literal logic function

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Variables Minterms Maxterms

A B mi Mi

3AB m 3A B M

2A B M

1A B M

0A B M

Representation of Logical expression using minterm

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m7 m3 m4 m5

Logical Expression

Corresponding minterms

where denotes sum of products

Y ABC ABC ABC ABC

7 3 4 5Y m m m m

(3,4,5,7)Y m

( , , ) (3,4,5,7)Y f A B C m

Representation of Logical expression using maxterm

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M2 M0 M6

Logical Expression

Corresponding maxterms

where denotes product of sum

( ).( ).( )Y A B C A B C A B C

2 0 6. .Y M M M

(0,2,6)Y M

( , , ) (0,2,6)Y f A B C M

Conversion from SOP to POS & Vice versa

The relationship between the expressions using minters and maxterms is complementary.

We can exploit this complementary

relationship to write the expressions in terms of maxterms if the expression in terms of minterms is known and vice versa

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Conversion from SOP to POS & Vice versa

For example, if a SOP expression for 4 variable is given by,

Then we can get the equivalent POS

expression using the complementary relationship as follows,

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(0,1,3,5,6,7,11,12,15)Y m

(2,4,8,9,10,13,14)Y M

Examples

1. Convert the given expression into standard form

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2. Convert the given expression into standard form

Y A BC ABC

( ).( )Y A B A C

Combinational Logic Circuits

Introduction Standard representation of canonical forms (SOP & POS),

Maxterm and Minterm , Conversion between SOP and POS forms

K-map reduction techniques upto 4 variables (SOP & POS form), Design of Half Adder, Full Adder, Half Subtractor & Full Subtractor using k-Map

Code Converter using K-map: Gray to Binary, Binary to Gray Code Converter (upto 4 bit)

IC 7447 as BCD to 7- Segment decoder driver IC 7483 as Adder & Subtractor, 1 Digit BCD Adder Block Schematic of ALU IC 74181 IC 74381

8/29/2017 26

Karnaugh Map (K-map)

In the algebraic method of simplification, we need to write lengthy equations, find the common terms, manipulate the expressions etc., so it is time consuming work.

Thus “K-map” is another simplification

technique to reduce the Boolean equation.

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It overcomes all the disadvantages of algebraic simplification techniques.

The information contained in a truth table or

available in the SOP or POS form is represented on K-map.

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K-map Structure - 2 Variable A & B are variables or inputs 0 & 1 are values of A & B 2 variable k-map consists of 4 boxes i.e. 22=4

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K-map Structure - 2 Variable Inside 4 boxes we have enter values of Y i.e.

output

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K-map & its associated minterms

Relationship between Truth Table & K-map

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K-map Structure - 3 Variable A, B & C are variables or inputs 3 variable k-map consists of 8 boxes i.e. 23=8

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01 11 10

A BC 0 1

3 Variable K-map & its associated product terms

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01 11 10

A BC 0 1

00 01 11 10

ABC ABC ABC ABC

ABC ABC ABC

3 Variable K-map & its associated minterms

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01 11 10

A BC 0 1

00 01 11 10

1m 3m0m 4m

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01 11 10

K-map Structure - 4 Variable A, B, C & D are variables or inputs 4 variable k-map consists of 16 boxes i.e. 24=16

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01 11 10

4 Variable K-map and its associated product terms

ABCD ABCD ABCD ABCD ABCD

ABCD ABCD ABCD ABCD

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m0 m4 m12 m8

m1 m5 m13 m9

m3 m7 m15 m11

m2 m6 m11 m10

01 11 10

m0 m1 m3 m2

m4 m5 m7 m6

m12 m13 m15 m14

m8 m9 m11 m10

01 11 10

4 Variable K-map and its associated minterms

Representation of Standard SOP form expression on K-map

For example, SOP equation is given as

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The given expression is in the standard SOP form. Each term represents a minterm. We have to enter ‘1’ in the boxes corresponding to each

minterm as below

1 1 0 0

1 0 1 1

01 11 10

Y ABC ABC ABC ABC ABC

ABC ABC

BC BC BC BC

Simplification of K-map

Once we plot the logic function or truth table on K-map, we have to use the grouping technique for simplifying the logic function.

Grouping means the combining the terms in adjacent cells.

The grouping of either 1’s or 0’s results in the simplification of boolean expression.

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If we group the adjacent 1’s then the result of simplification is SOP form

If we group the adjacent 0’s then the result of simplification is POS form

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Simplification of K-map

Grouping

While grouping, we should group most number of 1’s.

The grouping follows the binary rule i.e we can group 1,2,4,8,16,32,…..…number of 1’s.

We cannot group 3,5,7,………number of 1’s Pair: A group of two adjacent 1’s is called as

Pair Quad: A group of four adjacent 1’s is called as

Quad Octet: A group of eight adjacent 1’s is called as

Octet 8/29/2017 41

Grouping of Two Adjacent 1’s : Pair

A pair eliminates 1 variable

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0 0 1 1

0 0 0 0

01 11 10

BC BC BC BC

ABCABC

Y ABC ABC

( )Y AB C C

( 1)Y AB C C

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0 0 0 0

1 0 0 1

01 11 10

0 1 0 0

01 11 10

0 1 1 1

0 0 1 0

01 11 10

BC BC BC BC

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0 1 0 0

0 0 0 0

0 1 0 0

AB 00 01 11 10

CD CD CD CD

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Possible Grouping of Four Adjacent 1’s : Quad

0 0 0 0

1 1 1 1

AB 00 01 11 10

0 1 0 0

AB 00 01 11 10

A Quad eliminates 2 variable

CD CD CD CD

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0 0 0 0

1 1 0 0

0 0 0 0

AB 00 01 11 10

0 1 1 0

0 0 0 0

0 1 1 0

AB 00 01 11 10

CD CD CD CD

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1 0 0 1

0 0 0 0

1 0 0 1

AB 00 01 11 10

0 0 0 0

1 0 0 1

0 0 0 0

AB 00 01 11 10

CD CD CD CD

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0 0 0 0

0 1 1 1

0 0 0 0

AB 00 01 11 10

0 0 0 0

0 1 1 0

AB 00 01 11 10

CD CD CD CD

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Possible Grouping of Eight Adjacent 1’s : Octet

0 0 0 0

1 1 1 1

AB 00 01 11 10

0 1 1 0

AB 00 01 11 10

A Octet eliminates 3 variable

CD CD CD CD

8/29/2017 50

Possible Grouping of Eight Adjacent 1’s : Octet

1 1 1 1

0 0 0 0

1 1 1 1

AB 00 01 11 10

1 0 0 1

AB 00 01 11 10

A Octet eliminates 3 variable

CD CD CD CD

Rules for K-map simplification

1. Groups may not include any cell containing a zero.

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Not Accepted Accepted

2. Groups may be horizontal or vertical, but may not be diagonal

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3. Groups must contain 1,2,4,8 or in general 2n cells

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0 1 1 1

0 0 0 0

01 11 10

0 1 1 1

0 0 0 0

01 11 10

BC BC BC BC

4. Each group should be as large as possible

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1 1 1 1

0 0 1 1

01 11 10

1 1 1 1

0 0 1 1

01 11 10 BC BC BC BC

BC BC BC BC

5. Each cell containing a one must be in at least one group

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0 0 0 1

0 0 1 0

01 11 10

BC BC BC BC

6. Groups may be overlap

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1 1 1 1

0 0 1 1

01 11 10

BC BC BC BC

7. Groups may wrap around the table. The leftmost cell in

a row may be grouped with rightmost cell and the top cell in a column may be grouped with bottom cell

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1 0 0 1

01 11 10 1 1 1 1

0 0 0 0

1 1 1 1

AB 00 01 11 10

BC BC BC BC

CD CD CD CD

8. There should be as few groups as possible, as long as this does not contradict any of the previous rules.

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1 1 1 1

0 0 1 1

01 11 10

1 1 1 1

0 0 1 1

01 11 10 BC BC BC BC

BC BC BC BC

9. A pair eliminates one variable. 10. A Quad eliminates two variables. 11. A octet eliminates three variables

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Example 1

For the given K-map write simplified Boolean expression

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0 1 1 1

0 0 1 0

01 11 10

AB AB AB AB

Example 1 continue…..

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0 1 1 1

0 0 1 0

01 11 10

Simplified Boolean expression

AB AB AB AB

Y BC AB AC

Example 2

For the given K-map write simplified Boolean expression

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1 1 0 1

1 0 0 1

01 11 10

AB AB AB AB

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1 1 0 1

1 0 0 1

01 11 10

AB AB AB AB

Y B AC

Example 3

A logical expression in the standard SOP form is as follows; Minimize it with using the K-map technique

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Y ABC ABC ABC ABC

Example 3 continue……

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1 0 1 1

0 1 0 0

01 11 10

Y ABC ABC ABC ABC

BC BC BC BC

Y AC AB ABC

Example 4

A logical expression representing a logic circuit is; Draw the K-map and find the minimized logical expression

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(0,1,2,5,13,15)Y m

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1 1 0 1

0 1 0 0

0 1 1 0

0 0 0 0

AB 00 01 11 10

0 1 3 2

4 5 7 6

8 9 11 10

12 13 15 14 Simplified Boolean expression

(0,1,2,5,13,15)Y m

CD CD CD CD

ABDACD

Y ABD ACD ABD

Example 5

Minimize the following Boolean expression using K-map ;

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( , , , ) (1,3,5,9,11,13)f A B C D m

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0 1 1 0

0 1 0 0

0 1 1 0

AB 00 01 11 10

0 1 3 2

4 5 7 6

8 9 11 10

CD CD CD CD

f BD CD

( , , , ) (1,3,5,9,11,13)f A B C D m

( )f D B C

Example 6

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( , , , ) (4,5,8,9,11,12,13,15)f A B C D m

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0 0 0 0

1 1 0 0

1 1 1 0

AB 00 01 11 10

0 1 3 2

4 5 7 6

8 9 11 10

CD CD CD CD

f BC AC AD

( , , , ) (4,5,8,9,11,12,13,15)f A B C D m

Example 7

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2( , , , ) (0,1,2,3,11,12,14,15)f A B C D m

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1 1 1 1

0 0 0 0

1 0 1 1

0 0 1 0

AB 00 01 11 10

0 1 3 2

4 5 7 6

8 9 11 10

CD CD CD CD

ACDABD

2f AB ABD ACD

2( , , , ) (0,1,2,3,11,12,14,15)f A B C D m

Example 8

Solve the following expression with K-maps; 1. 2.

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1( , , ) (0,1,3,4,5)f A B C m

2( , , ) (0,1,2,3,6,7)f A B C m

8/29/2017 75

1 1 1 0

1 1 0 0

01 11 10

0 1 3 2

4 5 7 6

1 1 1 1

0 0 1 1

01 11 10

0 1 3 2

4 5 7 6

1( , , ) (0,1,3,4,5)f A B C m 2( , , ) (0,1,2,3,6,7)f A B C m

BC BC BC BC

2f A B 1f AC B

Example 9

Simplify ;

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( , , , ) (0,1,4,5,7,8,9,12,13,15)f A B C D m

8/29/2017 77

1 1 0 0

1 1 1 0

1 1 0 0

AB 00 01 11 10

0 1 3 2

4 5 7 6

8 9 11 10

CD CD CD CD

f C BD

( , , , ) (0,1,4,5,7,8,9,12,13,15)f A B C D m

Example 10

Solve the following expression with K-maps; 1. 2.

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1( , , , ) (0,1,3,4,5,7)f A B C D m

2( , , ) (0,1,3,4,5,7)f A B C m

8/29/2017 79

1 1 1 0

01 11 10

0 1 3 2

4 5 7 6

1 1 0 0

1 1 1 0

0 0 0 0

AB 00 01 11 10

0 1 3 2

4 5 7 6

8 9 11 10

12 13 15 14

BC BC BC BC

2f B C

1f AC AD

1( , , , ) (0,1,3,4,5,7)f A B C D m 2( , , ) (0,1,3,4,5,7)f A B C m

CD CD CD CD

K-map and don’t care conditions

For SOP form we enter 1’s corresponding to the combinations of input variables which produce a high output and we enter 0’s in the remaining cells of the K-map.

For POS form we enter 0’s corresponding to the combinations of input variables which produce a high output and we enter 1’s in the remaining cells of the K-map.

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But it is not always true that the cells not containing 1’s (in SOP) will contain 0’s, because some combinations of input variable do not occur.

Also for some functions the outputs corresponding to certain combinations of input variables do not matter.

8/29/2017 81

In such situations we have a freedom to assume a 0 or 1 as output for each of these combinations.

These conditions are known as the “Don’t Care Conditions” and in the K-map it is represented as ‘X’, in the corresponding cell.

The don’t care conditions may be assumed to be 0 or 1 as per the need for simplification

8/29/2017 82

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K-map and don’t care conditions - Example

Simplify ;

( , , , ) (1,3,7,11,15) (0,2,5)f A B C D m d

8/29/2017 Amit Nevase 84

X 1 1 X

0 X 1 0

0 0 1 0

AB 00 01 11 10

0 1 3 2

4 5 7 6

8 9 11 10

K-map and don’t care conditions - Example

CD CD CD CD

f CD AB AD

( , , , ) (1,3,7,11,15) (0,2,5)f A B C D m d

Pune Vidyarthi Griha’s · Block Schematic of ALU IC 74181 IC 74381 8/29/2017 26. Karnaugh Map...

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