Larger Kmaps. So far we have only discussed 2 and 3variable Kmaps. We can now create a 4variable map in the


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1 EET 3 Chapter 3 7/3/2 PAGE  23 Larger Kmaps The variable Kmap So ar we have only discussed 2 and 3variable Kmaps. We can now create a variable map in the same way that we created the 3variable Kmap. This time both the columns and rows are headed by two variables. Each side will be set up such that only one variable will change between adjacent cells. Examine the variable kmap to the right. Note that the cells are labeled with their minterm numbers (table row numbers). As the minterm count counts up, the count jumps rom row 2 to row then back to row 3. This can be veriied by noting the binary equivalent o each cell number as read rom the row and column headings. Cell 2 is the intersection o row and column creating 2, which is a binary 2. Cell is the intersection o row and column creating 2 which is a binary. ab cd The user should note that nothing else has changed. We will still group into groups o powers o two (, 2,, 8, 6, etc.) with the largest possible group in this variable map at 6. Note in the igure to the let that i we roll this map into a cylinder along the vertical axis, we see that the let and right edges ( abcd,,, ) bd are still adjacent, just as with the 3variable kmap. We can also roll the Kmap into a cylinder along the horizontal axis and we can then see that the top and bottom edges are adjacent as well. This adjacency brings up some interesting grouping potential as is demonstrated by the ollowing examples. ab cd ( abcd,,, ) b
2 EET 3 Chapter 3 7/3/2 PAGE  2 variable SOP Kmap examples Variable KMap Example (SOP): Problem Statement: Minimize the ollowing expression into an SOP expression: a, b,c,d m,,3,8,9, Due to the adjacency o the top and bottom edges, the ollowing groups: Groups (,,8,9) and (,3,9,) can be created and orm the minimized algebraic expression: ( a, b, c, d ) b c bd bd ab cd bc 2 3 It should be noted how one group shared part o the other group. Variable Kmap Example 2 (SOP): Problem Statement: Minimize the ollowing expression into an SOP expression: a, b, c, d m,2,6,8, ab cd bd 2 3 acd This example demonstrates that the cells in the our corners make up a group o our. The second group is obvious. The minimal answer is: ( abcd,,, ) bdacd
3 EET 3 Chapter 3 7/3/2 PAGE  2 Variable KMap Example 3 (SOP) Problem Statement: Minimize the ollowing expression into a SOP expression: a, b, c, d m,,2,3,, 8,, cd ab The minimal solution is: ( a, b, c, d ) b d a b a c ac 2 3 ab bd variable POS Kmap examples Variable KMap Example (POS) Problem Statement: Minimize the ollowing expression into a POS expression: a, b, c, d m, 6, 2, Note that the given switching unction was a minterm list and we need a maxterm list or a POS expression. So, the st thing to do is to convert the list to a maxterm list. ab cd b 2 3 a, b, c, d m,6,2, M,,2,3,,7, 8, 9,,, 3, This example plots the maxterms o an expression and results in the indicated expression below. Careully note d the logic level o the resulting terms and how they were obtained. ( abcd,,, ) b d
4 EET 3 Chapter 3 7/3/2 PAGE Variable KMap Example (POS) Problem Statement: Minimize the ollowing minterm list into a POS expression a,b,c,d,, m,2,, 7,9,,,3 Again, the provided unction was in minterm orm. It is necessary to convert it to a maxterm list. : a,b,c,d M, 3,,6,8,2 m,2,, 7,9,,,3,, Remember that when you read o a Max Term, you have to read it o like it was the inverse o a minterm. The minimum result is: d a c ( a, b, c, d ) a b a b d d a b d ab cd a c d 2 3 a b d
5 EET 3 Chapter 3 7/3/2 PAGE  27 The Implicant, Prime Implicant (PI), and Essential Prime Implicant (EPI) Let s discuss a couple deinitions which will come in handy later. Implicant: Any product term (element) or group o terms. Prime Implicant (PI): Largest possible term, or group o product terms, that cannot be combined with any other product term to generate a term with ewer literals than the original term. Essential Prime Implicant (EPI): A Prime Implicant which must be in the inal minimal answer. It is ESSENTIAL to the answer Now that we have these deinitions, let s revisit the solution to a previous example: A,B,C BC BC A A B C This solution has two dierent answers. The st two terms are in both answers. They are Essential Prime Implicants (EPI) because they are ESSENTIAL TO THE ANSWER. The second two terms are Prime Implicants (PI). While one or the other is in a solution, they are NOT ESSENTIAL to any single answer. Thereore, each minimal solution will consist o two EPI s and one PI. This topic will end up being used in the Tabular Reduction Method discussed at the end o the chapter. While this is not a KMap topic, these deinitions are best learned in a KMap environment. So, let s look at a ew more Kmaps and igure out what terms are PI s and what terms are EPI s.
6 EET 3 Chapter 3 7/3/2 PAGE  28 Essential Prime Implicant Example Problem Statement: Determine which terms in the Kmaps below are PI s and which are EPI s. abcd abcd epi epi's 2 3 ( abcd,,, ) bd abcd 2epi's (,7,3,) (,) pi 2 (,) 3 ( abcd,,, )bdabc ( abcd,,, ) bdacd Implicants (not Prime or Essential) It can be made part o a larger group abcd 2 3 EPI's I a '' were added here, it would be a IMPLICANT, but not 'PRIME' because it can be made "larger." I we were to group this term with minterm 2, then it would be an EPI.
7 EET 3 Chapter 3 7/3/2 PAGE  29 Essential Prime Implicant Example 2 Problem Statement: Determine which terms in the solution below are EPI s and PI s: Two EPI s (2,6) and (8,9) Four PI s (9,3), (7,), (3,), and (6,7) Thus, this example has THREE dierent answers. PI acd EPI ab c abcd 2 3 Solution EPI ' s PI # ( a, b, c, d ) acd ab c acd bcd PI ' s EPI acd bcd cd ab PI abd EPI ab c 2 3 Solution #2 EPI ' s PI abc ( a, b, c, d ) acd ab c abd abc EPI acd PI ' s cd ab PI abd EPI ab c 2 3 Solution #3 EPI ' s ( a, b, c, d ) acd ab c abd bcd PI ' s EPI acd PI bcd a, b, c, d abc acd bcd acd abc abd bcd abd
8 EET 3 Chapter 3 7/3/2 PAGE  3 Essential Prime Implicant Example 3 Problem Statement: Analyze the ollowing expression or PI s and EPI s. AB,C ABC, ABC ABC ABC A B C 3 A BC ( A, B, C ) AC AB BC A B C All our groups are Prime Implicant s because none o them can be combined with any other term to yield ewer literals. They are also all Essential because they each cover at least one " not covered by any other Prime Implicant. Since a term is always deined by its highest deinition, they are all designated as EPI s. They are ALL ESSENTIAL TO THE ANSWER.
9 EET 3 Chapter 3 7/3/2 PAGE  3 The Circular unction and the Prime Implicant Let s look at a special unction known as a CIRCULAR Function. A circular unction is made up o only PI s. There are NO EPI s. Circular Function Example Problem Statement: Examine the ollowing Kmaps to determine i the unction is CIRCULAR. AB CD 2 3 AB CD 2 3 This unction doesn t have any EPI s. Only 6 PI s: (,), (,), (,9),(,2), (9,3), (2,3)
10 EET 3 Chapter 3 7/3/2 PAGE  32 The variable Kmap The variable Kmap is three dimensional! The irst 6 cells are in ront and the 2 nd 6 cells are in the back. Obviously we can t group things in the back hal because we can t see it. So we slide out the back hal and place it to the right o the ront hal. Note that the ront hal has the MSB o (A in this example) while the back hal has the MSB o ( A in this example). O course, i this was to be used or mapping zeros instead o ones, the A would be swapped with the A. This will be a lot clearer ater a ew examples. DE BC 2 3 A A
11 EET 3 Chapter 3 7/3/2 PAGE Variable KMap Example (SOP): Problem Statement: Find the minimal SOP equation o the ollowing expression using a variable kmap. A,B,C,D,E 3,2,26, 29 m 2, 3, 7,8,2,3,8,9,2 DE BC 2 3 DE BC /A /A The only thing which remains is to turn the result into Boolean variables. The process below is a method that I ind helpul in the larger problems. A,B, C,D,E 2,3,8,9 B C D 3, 7,9,23 BDE 8,2 A B D E 3,29 BCDE 2,26 A B C E B C D B D E A B D E B C D E A B C E Just to review, there are 3 EPI s and 2 PI s in this list. Question: Can you pick them out? PI s are (8, 2) and (2, 26)
12 EET 3 Chapter 3 7/3/2 PAGE  3 Variable KMap Example 2 (POS): Problem Statement: Find the MINIMAL POS solution or the ollowing expression. A,B, C, D, E M,2,,8,,2,6,8,23 2,26,3, de bc 2 3 de bc /a /a,2,8,,6,8,2,26 C E 8, 2 A B D E 26,3 A B D E A B C D E 23 A B C D E A,B, C,D,E C EA B D EA B D E A B C D EA B C D E
13 EET 3 Chapter 3 7/3/2 PAGE  3 The 6 variable Kmap Just as there are variable Kmaps, there are also 6 variable Kmaps. They are xx three dimensional boxes and when expanded in the same manner as the variable Kmap create our boxes, two above the other two. However, I ind that it is very easy to make a mistake with 6 variable Kmaps. Thereore, I recommend that you use tabular reduction solution methods like QuineMcCluskey (discussed later) to simpliy expressions which contain 6 or more variables.
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