西安电子科技大学通信工程学院信息论与编码理论（研究型）课件Chapter3 More Algorithmic Simplification Techniques.pptVIP

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西安电子科技大学通信工程学院信息论与编码理论（研究型）课件Chapter3 More Algorithmic Simplification Techniques.ppt

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Problems with the algebraic methods to simplify logic functions No formal method Totally heuristic(试探性),depending heavily on experience Cannot be sure whether or not it is a minimum Furthermore, rather difficult to do algebraic simplification with more than 4 or 5 variables § 3.1The Karnaugh Map(K-Map) We have introduced the Karnaugh map in chapter 2, and we can plot any function on the map, either sum of minterms form or sum of products form. Below is an example: Map Terminology related to K-map Implicant(蕴涵项):a product term that can be used in a sum of products expression; An implicant is a rectangle of 1、2、4、8…(any power of 2)1’s, which may not include any 0’s. Terminology related to K-map(Con.) Prime implicant(质蕴涵项):an implicant that is not fully contained in any one other implicant; Terminology related to K-map(Con.) The purpose of the K-map is to help us find minimum sum of products expressions. Terminology related to K-map(Con.) Essential prime implicant(实质蕴涵项): a prime implicant that includes at least one 1 that is not included in any other prime implicant; § 3.1.1 Minimum Sum of Product Expressions Using the K-Map In this section,we will describe two methods for finding minimum sum of products expressions (最简与或式) using the Karnaugh map. In the process of finding prime implicants,we will be considering each of the 1’s on the map starting with the most isolated 1’s. Note: In an n-variable map, each square has n adjacent square. Examples for 3- and 4-variable maps are shown in Map 3.3 Map Method 1 1．Find all essential prime implicants. It’s usually quickest to start with the most isolated 1’s, that is, those that have the fewest adjacent squares with 1’s in them. 2．Find enough other prime implicants to cover the function. Do this using two criteria: Example 3.3 Example 3.4 Example 3.5 Example 3.6 (3.7、3.8、3.9、3.10)An example with multiple minimum solutions Map Method 2 Example 3.11 Example 3.12 This is a case with more 1’s left uncovered afte