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### logic

Course: ECE 200, Spring 2012
School: Drexel
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Word Count: 897

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Topic EC262 7: Advanced Topics in Karnaugh Maps Don't Cares If a function has don't cares in its truth table, we can treat those K-map cells as holding a 1 or a 0, depending on which values (1 or 0) allow us to construct the simplest Boolean expression. When considering the implicants and prime implicants of a K-map, it is most beneficial to treat all of the don't care values (usually denoted as X) as holding the...

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Topic EC262 7: Advanced Topics in Karnaugh Maps Don't Cares If a function has don't cares in its truth table, we can treat those K-map cells as holding a 1 or a 0, depending on which values (1 or 0) allow us to construct the simplest Boolean expression. When considering the implicants and prime implicants of a K-map, it is most beneficial to treat all of the don't care values (usually denoted as X) as holding the value: Larger groupings The Boolean expression corresponding to a larger When considering the essential prime implicants of a K-map, it is most beneficial to treat all of the don't care values (denoted as X) as holding the value: ____ . Why? Example (from text) Determine the minimal Boolean expression for a truth table that has the SOP expression: = f ( A, B, C , D ) m (1, 7,10,11,13) + d ( 5,8,15) m (1, 7,10,11,13) Recall that the m in the expression above stands for _________, so the 11 in the expression means that: The d in the expression above stands for _____________, so the 5 in the expression means that: What is the K-map for this canonical SOP expression? AB CD 00 01 11 10 Circle the prime implicants: AB CD 00 01 11 10 00 01 11 10 00 01 11 10 d ( 5,8,15) 1 Now...which of these prime implicants are essential? Remember, when looking for 1's not covered by any other prime implicants, you do not consider the cells holding X's. In other words, a cell holding an X will never force a prime implicants to be essential. Draw arrows pointing to the 1's that are covered by only one prime implicant: AB CD 00 01 11 10 00 01 11 10 So, the essential prime implicants are: AB CD 00 01 11 10 How would you sweep up all the additional 1's? AB CD 00 01 11 10 Thus, the minimal SOP expression is: Now...think about the benefit of using don't cares. What would have been the minimal SOP expression for the original K-map if we decided to treat all don't cares as 0's when forming the prime implicants? AB CD 00 01 11 10 00 01 11 10 00 01 11 10 00 01 11 10 So, by considering the don't cares as 1's when forming prime implicants we have have reduced the SOP expression from four terms with 14 literals down to three terms with 8 literals. What would have happened if we had treated don't cares as 1's when examining essential prime implicants? 2 Example (from text) Determine the minimal SOP expression for the K-map shown below. wx yz 00 01 11 10 Let's first circle the prime implicants. wx yz 00 01 11 10 00 X X X X 01 1 1 11 1 1 10 1 1 00 X X X X 01 1 1 11 1 1 10 1 1 Draw arrows pointing to the 1's that are covered by only one prime implicants: wx yz 00 01 11 10 00 X X X X 01 1 1 11 1 1 10 1 1 So, the essential prime implicants are: wx yz 00 01 11 10 00 X X X X 01 1 1 11 1 1 10 1 1 How would you sweep up the remaining 1's? wx yz 00 01 11 10 Thus, the minimal SOP expression is: Note that there are two other equally good ways that we could have swept up remaining the 1's (after circling the essential prime implicants), so there are three different correct answers to this question. 3 00 X X X X 01 1 1 11 1 1 10 1 1 Product of Sums Sometimes we might want the minimum POS expression for a digital logic circuit. Recall how we found a POS expression for a function f using its truth table : We formed the SOP expression for f '(the complement of f) using the lines of the truth table that had an output of 0. We then complemented f 'to obtain f. If we use DeMorgan's Law, the expression we obtain for f will be in POS form. How do you think we can obtain a minimal POS expression using a K-map? Example (from text) Find a minimal POS expression for a digital logic circuit whose K-map is shown below. AB CD 00 01 11 10 00 1 1 01 1 1 1 1 1 11 10 AB CD 00 01 11 10 00 01 11 10 We can readily see two essential prime implicants: but we have two equally good choices for covering the remaining 1: Let's flip a coin and go with Then our minimal SOP expression for the complemented K-map is and therefore the minimal POS expression for the original K-map is: 4 Example (from text) Find a minimal POS solution = for g ( w, x, y, z ) m (1,3, 4, 6,11) + d ( 0,8,10,12,13) wx yz 00 01 11 10 00 01 11 10 wx yz 00 01 11 10 00 01 11 10 wx yz 00 01 11 10 00 01 11 10 Now, are there any 1's that are covered only by one prime implicant? (These will become essential prime implicants.) wx 00 01 11 10 yz 00 01 11 10 Let's circle the essential prime implicants: wx yz 00 01 11 10 5 00 01 11 10 Now there is only one remaining 1 still to be covered, and it can be covered in two equally good ways. Let's go with the third column. Thus the final SOP expression for g' is and therefore the minimal POS expression for g is The Five-Variable Karnough Map The five-input K-map has 32 possible input arrangements, where the outputs for each possible minterms are arranged in two layers as shown below. Squares directly above and below each other are adjacent. Examples of adjacencies are shown in the figure below : So, now we must recognize groups of 2, 4, 8, 16, etc., where the groups can be arranged as rectangular solids. 6 Example (from text) Consider the five-variable K-map shown below. From Introduction to Logic Design, Alan Marcovitz, McGraw Hill, 2010 It is helpful to search for 1's on a layer that have a 0 in the corresponding position in the other layer. The 1 in the box for m4 meets this criteria. Thus A'B'C is an essential prime implicant. Circling essential prime implicants on one-layer yields : From Introduction to Logic Design, Alan Marcovitz, McGraw Hill, 2010 The two remaining 1's are covered by an essential prime implicant that spans both layers, as shown below : From Introduction to Logic Design, Alan Marcovitz, McGraw Hill, 2010 The overall solution is : A'B'C + A'BE + AB'C'E' + ABCD'E' + BDE 7
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