Unit 2 Session - 7 Combinational Logic Circuits
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1 Objectives Unit 2 Session - 7 Combinational Logic Circuits Use 5 Variable Karnaugh Map and Entered Variable Map to simplify Boolean expressions Use Quine-McClusky tabular method for simplification of Boolean expressions having more than 4 variables 5 Variable Karnaugh Map A given 5-variable minterm will have 5 adjacent terms. One way of drawing Karnaugh map for 5 variables is as shown below. The visual advantage of identifying the adjacent terms is not there. Example: The simplified expression is Y = B D + BCDE + ABE Entered Variable Map In entered variable map one of the input variables is placed inside Karnaugh map. This is done separately noting how the input variable is related with output. This reduces the Karnaugh map size by one degree. This technique is particularly useful for mapping problems with more than four input variables. B. S. Umashankar, BNMIT Page 1
2 Example: Consider the 3 - variable truth table as shown below. The output Y is rewritten in terms of variable C. A B C Y Y C 0 1 The 3 variable truth table reduces to 2 variable truth table as shown below: A B Y C The 2 variable Karnaugh map is drawn as shown below: The Karnaugh map is now called an entered variable map. The simplification of entered variable map is as illustrated next: B. S. Umashankar, BNMIT Page 2
3 The product term representing each group is obtained by including map entered variable in the group as an additional ANDed term. Group 1 gives B.(C ) and group 2 gives AB.1. Therefore, the simplified expression is obtained as Y = BC + AB. Disadvantage of Karnaugh Map Simplifications Karnaugh map depends on the user s ability to identify patterns. It becomes difficult to adapt for simplification of 5 or more variables. Simplification by Quine-McClusky Method Quine-McClusky method is a systematic approach for logic simplification. It was developed by W. V. Quine and Edward J. McCluskey. It is functionally identical to Karnaugh map. It does not have the limitation of number of variables. The tabular form makes it more efficient for use in computer algorithms. It is also called as the tabulation method. Main Steps The 2 main steps are: 1. Find all prime implicants of the given Boolean equation. 2. Use the prime implicants in a prime implicant chart to find the essential prime implicants and other prime implicants that are required to completely cover the given equation. Example 1: Simplify the following function: Solution: Step 1 Y = f(a, B, C, D) = Σ m(0, 1, 2, 3, 10, 11, 12, 13, 14, 15) We find all Prime Implicants. B. S. Umashankar, BNMIT Page 3
4 In stage 1 of this step, the minterms are put in to different groups depending on the number of 1 s they have in their binary equivalents. In each group the minterms are ordered in increasing values of their decimal equivalent. Finding Number of 1 s Minterms Binary rep. No. of 1 s Minterms Binary rep. No. of 1 s Stage 1 Stage 1 No. of 1 s Minterm Binary Representation In stage 2, we form pairs, i.e. compare minterms of group n and group n + 1 ( n varies from 0 to 3). Where there is only one bit change in the binary representations, the corresponding variable is eliminated using the adjacency theorem and in its position a dash ( - ) is written. After the stage 2 process the table is obtained as shown: B. S. Umashankar, BNMIT Page 4
5 Next we go to stage 3 and same process is repeated i.e. pair of pairs is formed. After the stage 3 process we stop since further comparison is not possible. Next, we check off all minterms or group of minterms that are covered by larger groups. The unchecked minterms are the prime implicants as shown below: B. S. Umashankar, BNMIT Page 5
6 Step 2. We find essential prime implicants using the prime implicant chart. The chart is drawn as shown below. Prime implicants are shown as rows and minterms as columns. An X is placed to indicate the covering of the minterm by the prime implicant. For example, minterms 0, 1, 2, and 3 are covered by prime implicant A B. Find the column that has a single X and circle it. The associated prime implicant is an essential prime implicant since it covers at least one minterm not covered by other prime implicants. An asterisk (*) is placed to indicate that the prime implicant is essential. Then check off all the minterms covered by the essential prime implicants as shown below: Since all the minterms are not checked off, we have to find secondary essential prime implicants by drawing a reduced prime implicant chart. The reduced chart is obtained by removing the essential prime implicants that are already found and included in the solution and also removing the checked off minterms. B. S. Umashankar, BNMIT Page 6
7 The reduced prime Implicant chart is as shown below: PIs Minterms B C X X **AC X X We note that column with single X is not there. The complexity or cost of both the prime implicants are same. So we select one of them as secondary essential prime implicants (say, AC). It covers all the minterms. So we have the solution: Example 2: Y = A B + AB + AC Minimize f(a, B, C, D) = Σ (0, 1, 2, 8, 9, 15, 17, 21, 24, 25, 27, 31). Step 1: Say we have obtained the following prime implicants: P1: BC D P4: ABDE P7: ABC D P10: A B C E P2: C D E P5: BCDE P8: A B D E P11: A B C D P3: A C D P6: ABC E P9: B C D E Step 2: Prime Implicant Chart B. S. Umashankar, BNMIT Page 7
8 The prime implicants P5, P7, and P10 are essential. They are included in the solution. They do not cover all the minterms. So secondary essential prime implicants have to be found by using the reduced prime implicant chart. Reduced Prime Implicant Chart (Essential prime implicants removed) P1 X X X X P2 X X X P3 X X X P4 X P6 X X P7 X X P9 P11 X X We are not able to find columns with single X. Now we find the dominance relations. Column Dominance: 9 > 8 25 > 24 Row Dominance: P1 > P7 P6 > P4 P2 > P9 P2 > P11 Prime Implicant Chart Reduction Steps: 1. All the dominating columns and dominated rows of a prime-implicant chart can be removed without affecting the table for obtaining a minimal solution. 2. Dominating column is guaranteed to be covered by the row that covers its dominated column. 3. The columns of the dominated row are guaranteed to be covered by its dominating row. B. S. Umashankar, BNMIT Page 8
9 Finding Secondary Essential PIs PI Chart after the dominating columns and the dominated rows are deleted: Final Solution Minterm 1 can be covered by P2 or P3. If we select P2, we have the solution: Questions Y = P1 + P2 + P5 + P6 + P8 + P What are the drawbacks of Karnaugh map? 2. Define prime implicant and essential prime implicant. Find prime implicant and essential prime implicants for the following function using Quine-McClusky method: f(a, b, c, d) = m(0, 2, 3, 6, 7, 8, 10, 12, 13) 3. Find the prime implicants for the following Boolean expressions using Quine Mc Clusky's method. i. f(w, x, y, z) = m(1, 3, 6, 7, 8, 9, 10, 12, 13, 14) ii. f(a, b, c, d) = m(1, 2, 8, 9, 10, 12, 13, 14) ii. f(a, B, C, D) = m(1, 3, 6, 7, 8, 9, 10, 12, 14, 15) + d(11, 13) B. S. Umashankar, BNMIT Page 9
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