logic equation simplification
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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 1st 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.TRANSCRIPT
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.
Logic Equation Simplification
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.
Logic Equation Simplification
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.
Logic Equation Simplification
STEP ONE
of the simplification process would be to fill in the Karnaugh Map
(Note: we normally only transfer 1’s onto the map.)
Logic Equation Simplification
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.
Logic Equation Simplification
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.
Logic Equation Simplification
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
Logic Equation Simplification
111
1101
010
0100
Y10B AYBA
Logic Equation Simplification
1111
1101
01011
0100
Y10B AYBA
STEP 1
Logic Equation Simplification
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
Logic Equation Simplification
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
Logic Equation Simplification
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
Logic Equation Simplification
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
Logic Equation Simplification
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.
Logic Equation Simplification
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
Logic Equation Simplification
1111
1011
1101
1001111
11110
0010111
00100
F0110C B
1000
1100AFCBA
Logic Equation Simplification
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
Logic Equation Simplification
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
Logic Equation Simplification
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.
Logic Equation Simplification
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.
Logic Equation Simplification
11111
10111
11011
10011
11101
10101
11001
0000111
1 011110
00110111
1 101010
0001011
0 101100
001001
0 0
01000
Y0110C D B00000
1100AYDCBA
Logic Equation Simplification
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
Logic Equation Simplification
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
Logic Equation Simplification
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
Logic Equation Simplification
1111
0111
1011
0011
1101
0101
1001
0001
1 01110
0110
1 11010
0010
0 11100
0100
0 0
1000
G0110C D B0000
1100AELGDCBA
Logic Equation Simplification
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
Logic Equation Simplification
This resource was created by the University of Wales Newport and released as an open educational resource through the Open Engineering Resources project of the HE Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme.
© 2009 University of Wales Newport
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Logic Equation Simplification