simplification of boolean algebra

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simplification of boolean algebra

  1. 2. <ul><li>The map method provides a simple straight forward procedure for minimizing a Boolean function. This method maybe regarded either as a pictorial form or as an extension of Venn Diagram. The map method, first proposed by Vietch and modified by Karnaugh, is also known as “Vietch Diagram” or “Karnaugh Map”. </li></ul>
  2. 3. <ul><li>The map is a diagram made up of squares. Each square represents one minterm. Since any Boolean function can be a sum of minterms, it follows that a Boolean function can be graphically represented in a map by the areas function. </li></ul><ul><li>The map presents a visual diagram of all possible ways a function may be expressed in a standard form. </li></ul>
  3. 4. <ul><li>Here’s how to make the Karnaugh map of this truth table: </li></ul><ul><li>Begin by drawing the blank map: </li></ul>A B A B F 0 0 0 0 1 1 1 0 1 1 1 1 0 1 0 A’B’ A’B 1 AB’ AB
  4. 5. <ul><li>The first 1 output to appear is for the input of A=1 and B=0. The fundamental product for this is AB’. </li></ul><ul><li>Similarly, this table has an output 1 appearing for input of A=1 and B=1, and A=0 and B=1. </li></ul><ul><li>Now, enter 1’s in the Karnaugh map to represent AB and A’B. </li></ul>B A 0 1 0 1 1 1 1
  5. 6. <ul><li>The final step is to enter 0’s in the remaining spaces. </li></ul><ul><li>This will how the Karnaugh map looks in its final form. </li></ul>B A 0 1 0 0 1 1 1 1
  6. 7. <ul><li>The truth table: The Karnaugh Map: </li></ul><ul><li>Begin by drawing a map consist of 8 squares. </li></ul>C AB 0 1 00 A’B’C’ A’B’C 01 A’BC’ A’BC 11 ABC ’ ABC 10 AB’C’ AB’C
  7. 8. <ul><li>NOTE: The binary combinations are not arranged in sequence. This order is not a binary progression; instead it follows the order of 00, 01, 11, and 10. The reason for this is, so that the map will show only one variable changing from complemented to uncomplemented form in adjacent rows. </li></ul>
  8. 9. <ul><li>Next, look for the outputs of 1. The fundamental products for these 1 outputs are A’BC’, ABC’, ABC. Enter these 1’s in the map. </li></ul><ul><li>The last step is to put 0’s in the remaining </li></ul><ul><li>spaces. </li></ul>
  9. 10. <ul><li>Many MSI (Marketing Science Institute) circuits process binary words of 4 bits each. For this reason, logic circuits are often designed to handle 4 variables. This is the reason why four-variable map is the most important . </li></ul>
  10. 11. <ul><li>The truth table: </li></ul>A B C D F 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 1 0 0 1 0 0 0 0 1 0 1 0 0 1 1 0 1 0 1 1 1 1 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 0 1 0 1 1 1 0 1 1 1 1 1 0
  11. 12. <ul><li>Begin again by drawing the blank map. </li></ul><ul><li>In the truth table, the output 1 corresponds to these fundamental products: A’B’C’D, A’BCD’, A’BCD, and ABCD’. Enter the output 1 to the map. </li></ul>AB CD CD AB 00 01 11 10 00 A’B’C’D’ A’B’C’D A’B’CD A’B’CD’ 01 A’BC’D’ A’BC’D A’BCD A’BCD’ 11 ABC’D’ ABC’D ABCD ABCD’ 10 AB’C’D’ AB’C’D AB’CD AB’CD’ 00 01 11 10 00 1 01 1 1 11 1 10
  12. 13. <ul><li>A five-variable map needs 32 squares and a six-variable map needs 64 squares. When the number of variables becomes large, the number of squares becomes excessively large and the geometry for combining adjacent squares becomes more involved. </li></ul>
  13. 14. <ul><li>Truth table for five-variable map. </li></ul>A B C D E F 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 1 0 0 1 1 1 0 0 1 0 0 0 0 0 1 0 0 1 1 0 1 0 1 0 0 0 1 0 1 1 0 0 1 1 0 0 0 0 1 1 0 1 1 0 1 1 1 0 0 0 1 1 1 1 0 A B C D E F 1 0 0 0 0 0 1 0 0 0 1 0 1 0 0 1 0 0 1 0 0 1 1 0 1 0 1 0 0 0 1 0 1 0 1 1 1 0 1 1 0 0 1 0 1 1 1 1 1 1 0 0 0 0 1 1 0 0 1 1 1 1 0 1 0 0 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 0
  14. 15. <ul><li>It consists of 2 four-variable maps with variables A, B, C, D, and E. The leftmost variable distinguishes between the two maps. </li></ul>BC DE DE BC A=1 A=0 00 01 11 10 00 1 0 0 1 01 1 0 0 1 11 0 1 0 0 10 0 1 0 0 00 01 11 10 00 0 0 0 0 01 0 1 1 0 11 0 1 1 0 10 0 1 0 0
  15. 17. AB CD CD AB AB CD CD CD AB AB B A C D E 00 01 11 10 00 0 0 0 0 01 0 0 1 0 11 0 0 1 0 10 0 0 0 0 00 01 11 10 00 0 0 0 0 01 0 0 1 0 11 0 1 1 0 10 0 1 0 0 00 01 11 10 00 0 0 0 0 01 0 0 1 1 11 0 1 0 0 10 0 1 0 0 00 01 11 10 00 0 0 0 0 01 0 0 1 1 11 0 0 0 0 10 0 0 0 0 00 01 11 10 00 0 0 0 0 01 0 0 0 0 11 1 0 0 1 10 0 0 0 0
  16. 18. <ul><li>As we move from the first 1 to the second 1, only one variable changes (e.g. A to A’) and the other three variables remain unchanged. Whenever this happens, eliminating the variable that changes is possible, therefore: </li></ul><ul><li>F=BCD </li></ul>
  17. 19. <ul><li>It is customary to encircle a pair of adjacent 1s for easy identification. This way, when you look at the map, you can tell at a glance that one variable and its complement can be dropped from the Boolean equation. </li></ul><ul><li>Figure B shows a pair of adjacent 1s that are horizontally adjacent. These 1s corresponds to the products A’BCD and A’BCD’. Notice that the only one variable changes from not complemented form – D to D’. All the other variables remain unchanged. Thus D can be dropped. The final product of </li></ul>
  18. 20. <ul><li>the encircled pairs is: </li></ul><ul><li>F = A’BC </li></ul><ul><li>If more than one pair exist on a Karnaugh map as in Figure 3.5 (c) and (d), you can OR the simplified products to get the Boolean equation. Therefore: </li></ul><ul><li>for (c) F = AC’D + A’BC </li></ul><ul><li>And </li></ul><ul><li>for F = BC’D + A’CD </li></ul>
  19. 21. CD AB 00 01 11 10 00 0 0 1 0 01 0 0 1 0 11 0 0 1 0 10 0 0 1 0 CD AB 00 01 11 10 00 0 0 0 0 01 0 0 1 1 11 0 0 1 1 10 0 0 0 0 a b
  20. 22. <ul><li>A Quad is a group of four 1s that are end to end as shown in Figure A or in the form of a square as shown in figure B. When you see a quad, always encircle it because it leads to a simpler product. In fact, a quad means that two variables and their complements can be dropped from the Boolean equation. </li></ul>
  21. 23. <ul><li>To get the simplified equation… </li></ul><ul><li>“ A” </li></ul><ul><li>F=A’B’CD+A’BCD+ABCD+AB’CD </li></ul><ul><li>Therefore, </li></ul><ul><li>F=CD </li></ul><ul><li>“ B” </li></ul><ul><li>F=A’BCD+ABCD+A’BCD’+ABCD’ </li></ul><ul><li>Therefore, </li></ul><ul><li>F=BC </li></ul>
  22. 24. <ul><li>An octet is a group of eight adjacent 1s like this: </li></ul><ul><li>Notice that in this map the three variables changed (ABC) and only A retains its value. </li></ul>CD AB 00 01 11 10 00 0 0 1 1 01 0 0 1 1 11 0 0 1 1 10 0 0 1 1
  23. 25. <ul><li>F = A’B’CD </li></ul><ul><li>A’BCD </li></ul><ul><li>ABCD </li></ul><ul><li>AB’CD </li></ul><ul><li>+ A’B’CD’ </li></ul><ul><li>A’BCD’ </li></ul><ul><li>ABCD’ </li></ul><ul><li>AB’CD’ </li></ul><ul><li>Therefore, </li></ul><ul><li>F=C </li></ul>

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