Logic Equation Simplification

Logic equation simplification.

Digital Logic and Software Applications

© University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

The following presentation is a part of the level 4 module -- Digital Logic and Signal Principles. This resources is a part of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1 st year undergraduate programme.

The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond. Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This course has been designed to provide you with knowledge, skills and practical experience encountered in everyday engineering environments.

Contents Introduction Karnaugh Maps 2 Input Karnaugh Maps The Process Steps Example 3 Input Karnaugh Maps Example A function F has the truth table shown below. De... Example Three judges A, B and C vote: 1 guilty and 0 not ... Four Judge example Credits

In addition to the resource below, there are supporting documents which should be used in combination with this resource. Please see: Holdsworth B, Digital Logic Design, Newnes 2002 Crisp J, Introduction to Digital Systems, Newnes 2001

Logic Equation Simplification

Often we are given a Boolean expression which could be written in a simplified way or we are presented with a truth table where we have no expression. The process of simplification is the way in which we generated the simplest (shortest) expression for the function. The reason that this is important is that when implementing the expression, the simpler it is the less number of gates, therefore the less number of I.C.s and the smaller the space required.

e.g. Consider the OR gate truth table

A B Y

0 0 0

0 1 1

1 0 1

1 1 1

We have 3 combinations which generate an output:

BABABAY We know that this can be written as:

but how do we get from the first equation to the second?

BAY

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

We will look at two methods.1.Boolean Algebra simplification rules

1 AA AAA AAA 0AA

CABACBA )( AA 0

11 A AA 100 A ABBA

ABBA BAAB )(

ABAA ABAA )(

BABAA BABAA )(

BABA BABA

Examples.

1.

2.

BABABAY

)()()( CBACBACBAF

Boolean Algebra requires the user to:have a good understanding of the rules, experience of reducing equations, a knowledge of where to start a knowledge of when we have reached the simplest

solution.


2. Karnaugh Maps

This method coverts the truth table information into a two-dimensional map. It then converts areas of 1’s on the map into groups. These groups are then identified and this gives us the simplest expression.


11

101

10

000

Y10B AYBA

2-input Karnaugh MapThis has 4 entries on the Truth Table and so the

Karnaugh Map has 4 squares

The top left square is where A = 0 and where B = 0 and so the value of Y for A = 0 and B = 0 would be placed in here.Each entry in the Truth Table has one square in the Karnaugh Map.


STEP ONE

of the simplification process would be to fill in the Karnaugh Map

(Note: we normally only transfer 1’s onto the map.)


STEP TWOis to group the 1’s. Groups are formed using the following rules:

1. Group sizes must be powers of 2 – 1, 2, 4, 8, 16, etc – no other size groups are allowed.

2. Groups must be square of rectangles (1 x 4 or 2 x 2 etc)

3. Groups must be as large as possible (never group 2 groups of 2 if a group of 4 can be made.)

4. All 1’s must be grouped.5. A 1 may be grouped more than once.6. Do not include redundant groups – a redundant

group is a group that contains 1’s which have all been previously grouped.


STEP THREEidentifies the expression for each group. The groups are examined one at a time. For a group the following question is asked for each input one at a time:

For the group:Is the input logic state for every square in the group:

Always 1 – if it is then the input appears in the expression

Always 0 - if it is then the not input appears in the expression

Both 1 and 0 - if it is then the input does not appears in the expression

After each input has been checked, the expression is the AND of the inputs states identified.

STEP FOUR

identifies the complete expression for the function. The individual group expressions are OR-ed together to give the simplified expression.


Example

The Truth Table and Karnaugh Map are shown below:

BABABAY

A B Y B A 0 1 Y

0 0 0

0 1

1 0 1

1 1


111

1101

010

0100

Y10B AYBA


1111

1101

01011

0100

Y10B AYBA

STEP 1


A B Y B A 0 1 Y

0 0 1 01 1

0 1 0

1 0 1 11

1 1 1

STEP 2


A B Y B A 0 1 Y

0 0 1 01 1

0 1 0

1 0 1 11

1 1 1

STEP 3A always 1 so AB 1 and 0 so no BExpression

A 1 and 0 so no AB always 0 so not BExpression AB


A B Y B A 0 1 Y

0 0 1 01 1

0 1 0

1 0 1 11

1 1 1

STEP 4

Complete expression: BAY


0

3-input Karnaugh MapThis has 8 entries on the Truth Table and so the

Karnaugh Map has 8 squares

111

011

101

001

1110

010

100

Y0110C B000

1100AYCBA


Note there is one additional rule for grouping 1’s on this map and larger maps:

Rule: 1’s may be grouped between the left hand column and the right hand column.


ExampleA function F has the truth table shown below. Determine the simplest Boolean Expression for the function.

A B C F A 0 0 1 1

0 0 0 1 C B

0 1 1 0 F

0 0 1 0 0

0 1 0 0

0 1 1 1 1

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1Logic Equation Simplification

1111

1011

1101

1001111

11110

0010111

00100

F0110C B

1000

1100AFCBA


1111

1011

1101

1001111

11110

0010111

00100

F0110C B

1000

1100AFCBA

A always 1 so AB 1 and 0 so no BC 1 and 0 so no CExpression

A 1 and 0 so no AB always 1 so BC always 1 so CExpression

A

CB

A 1 and 0 so no AB always 0 so not BC always 0 so not CExpression

CB

1111

1011

1101

1001111

11110

0010111

00100

F0110C B

1000

1100AFCBA

Complete expression

CBCBAF


ExampleThree judges A, B and C vote: 1 guilty and 0 not guilty. Design a logic circuit using NAND only which will allow a majority decision (F) to be found. e.g. A = 1, B = 0, C = 0 gives an output of 0 (not guilty)

A B C F A 0 0 1 1

0 0 0 C B

0 1 1 0 F

0 0 1 0

0 1 0

0 1 1 1

1 0 0

1 0 1

1 1 0

1 1 1Logic Equation Simplification

A B C D Y A 0 0 1 1

0 0 0 0 C D B 0 1 1 0 Y

0 0 0 1 0 00 0 1 0

0 0 1 1 0 1

0 1 0 0

0 1 0 1 1 1

0 1 1 0

0 1 1 1 1 0

1 0 0 0

1 0 0 1

1 0 1 0

1 0 1 1

1 1 0 0

1 1 0 1

1 1 1 0

1 1 1 1

4-inputKarnaugh Map

This has 16 entries on the Truth Table and so the Karnaugh Map has 16 squares


1111

0111

1011

0011

1101

0101

1001

0001

1 01110

0110

1 11010

0010

0 11100

0100

0 0

1000

Y0110C D B0000

1100AYDCBA

Note there is one additional rule for grouping 1’s on this map and larger maps:Rule: 1’s may be grouped between the top row and the bottom row.


ExampleFour judges A, B, C and D vote: 1 guilty and 0 not

guilty. Obtain a Boolean Expression that will allow a majority decision to be found. In the case of a split decision the vote of A determines the outcome Y.


11111

10111

11011

10011

11101

10101

11001

0000111

1 011110

00110111

1 101010

0001011

0 101100

001001

0 0

01000

Y0110C D B00000

1100AYDCBA


11111

10111

11011

10011

11101

10101

11001

0000111

1 011110

00110111

1 101010

0001011

0 101100

001001

0 0

01000

Y0110C D B00000

1100AYDCBA

Now form groups


11111

10111

11011

10011

11101

10101

11001

0000111

1 011110

00110111

1 101010

0001011

0 101100

001001

0 0

01000

Y0110C D B00000

1100AYDCBA

Identify groups

A always 1 so AB 1 and 0 so no BC always 1 so CD 1 and 0 so no DExpression

CA

CA

A always 1 so AB 1 and 0 so no BC 1 and 0 so no CD always 1 so DExpression DA

A always 1 so AB always 1 so BC 1 and 0 so no CD 1 and 0 so no DExpression

BA

A 1 and 0 so no AB always 1 so BC always 1 so CD always 1 so DExpression

DCB

11111

10111

11011

10011

11101

10101

11001

0000111

1 011110

00110111

1 101010

0001011

0 101100

001001

0 0

01000

Y0110C D B00000

1100AYDCBA

CA

Boolean Expression

DCBDACABAY


ExampleTwo 2-bit numbers (A,B) and (C,D) are to be

compared.

If (A,B) > (C,D) then the G (greater than) output is to equal 1

If (A,B) < (C,D) then the L (less than) output is to equal 1

If (A,B) = (C,D) then the E (Equal to) output is to equal 1

e.g. A = 1, B = 0, C = 1, D = 1 10 (2) is less than 11 (3) so L = 1


1111

0111

1011

0011

1101

0101

1001

0001

1 01110

0110

1 11010

0010

0 11100

0100

0 0

1000

G0110C D B0000

1100AELGDCBA


A 0 0 1 1

C D B

0 1 1 0 L

0 0

0 1

1 1

1 0

A 0 0 1 1

C D B 0 1 1 0 E

0 0

0 1

1 1

1 0

Expression G

Expression L

Expression E


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© 2009 University of Wales Newport

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Logic Equation Simplification

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