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Digital Logic Design Lecture 14 based on video from: https://www.youtube.com/watch?v=pQ3MfzqGlrc

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Announcements HW5 due on 10/21 Quiz during recitation on Monday, 10/20. Upcoming: Midterm on Tuesday, 10/28.

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Agenda Last time: – Minimal expressions for incomplete Boolean functions (4.6) – 5 and 6 variable K-Maps (4.7) – Petrick’s method of determining irredundant expressions (4.9) This time: – Quine-McCluskey method (4.8) – Prime Implicant Table Reductions (4.10) – The Multiple Output Simplification Problem (4.12)

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Tabular Representations 0001 1111 0111 0100 00011110 00 01 11 10

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Prime Implicants

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0001 1111 00011110 0 1

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Prime Implicants 0001 1111 00011110 0 1 MintermXYZF 00000 10010 20101 30110 41001 51011 61101 71111

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Finding Prime Implicants Step 1 2010 4100 5101 6110 7111 Step 2 (2,6)10 (4,5)10 (4,6)10 (5,7)11 (6,7)11 Step 3 (4,5,6,7)1 (4,6,5,7)1 All unchecked entries are Prime Implicants

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Prime Implicants 0001 1111 00011110 0 1 MintermXYZF 00000 10010 20101 30110 41001 51011 61101 71111

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Essential Prime Implicants 1111 0110 0011 1001 00011110 00 01 11 10 Find the Essential Prime Implicants using Quine- McClusky

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Essential Prime Implicants 1111 0110 0011 1001 00011110 00 01 11 10 00000 10001 20010 81000 30011 50101 1010 70111 141110 151111 minterms

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Finding Prime Implicants 00000 10001 20010 81000 30011 50101 101010 70111 141110 151111 (0,1)000 (0,2)000 (0,8)000 (1,3)001 (1,5)001 (2,3)001 (2,10)010 (8,10)100 (3,7)011 (5,7)011 (10,14)110 (7,15)111 (14,15)111 (0,1,2,3)00 (0,2,1,3)00 (0,2,8,10)00 (0,8,2,10)00 (1,3,5,7)01 (1,5,3,7)01

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Find Essential Prime Implicants Prime Implicant Covered Minterms0123578101415 10,14XX 7,15XX 14,15XX 0,1,2,3XXXX 0,2,8,10XXXX 1,3,5,7XXXX Minterms

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3 Prime Implicants 1111 0110 0011 1001 00011110 00 01 11 10

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3 Prime Implicants 1111 0110 0011 1001 00011110 00 01 11 10

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Column and Row Reductions

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Example AX BXXX CXXX DXXX EXX FX GXX HXX IXX

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Cost AX4 BXXX4 CXXX4 DXXX4 EXX4 FX5 GXX5 HXX5 IXX5 Cost is 1 plus number of literals in the term

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Example Cost AX4 BXXX4 CXXX4 DXXX4 EXX4 FX5 GXX5 HXX5 IXX5

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Example Cost AX4 BXXX4 CXXX4 DXXX4 EXX4 FX5 GXX5 HXX5 IXX5

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Example Cost AX4 BXX4 CXX4 DXXX4 EX4 FX5 GXX5 HXX5 IXX5

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Example Cost AX4 BXX4 CXX4 DXXX4 EX4 FX5 GXX5 HXX5 IXX5 Dominated rows: A is dominated by B since B has X’s in all columns in which A has X’s and B has at least one more X. Which rows dominate E and F?

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Example Cost BXX4 CXX4 DXXX4 GXX5 HXX5 IXX5 Delete rows A, E, F since dominated row has cost equal to its dominating row. Why is this ok?

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Example Cost **BXX4 CXX4 **DXXX4 **GXX5 HXX5 IXX5

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Example Cost HX5 IX5 Can select either H or I since they both have the same cost. Final minimal cover is either: B,D,G,H B,D,G,I Note: Unlike Petrick’s method, not all minimal covers are necessarily obtained.

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Summary: Prime Implicant Selection Procedure 1.Find all essential prime implicants. Rule a line through the essential rows and all columns which have an X in an essential row. 2.Rule a line through all dominating columns and dominated rows, keeping in mind the cost restriction for deleting rows. 3.Check to see if any unruled column has a single X. If there are no such columns, then the table is cyclic. If there are some columns with a single X, place a double asterisk next to the rows in which these X’s appear. These are called secondary essential rows. Rule a line through each secondary essential row and each column in which an X appears in a secondary essential row. 4.If all columns are ruled out, then the minimal sum is given by the sum of all the prime implicants which are associated with rows that have asterisks next to them. If all columns are not ruled out, then repeat Steps 2 and 3 until either there are no columns to be ruled or a cyclic table results. 5.If a cyclic table results, then Petrick’s method is applied to the cyclic table and a minimal cover is obtained for it. The sum of all prime implicants that are marked with asterisks plus the prime implicants for the minimal cover of the cyclic table as determined by Petrick’s method is a minimal sum.

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Multiple Output Minimal Sums and Products

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The Multiple-Output Simplification Problem

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Naïve Approach Construct a minimal expression for each output function independently of the others. Example: XYZ 00010 00110 01000 01111 10000 10100 11001 11111

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Naïve Approach

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A More Economical Realization

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Pitfalls of Naïve Approach

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Naïve Approach: Better Approach:

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ENEE244-02xx Digital Logic Design Lecture 7. Announcements Homework 3 due on Thursday. Review session will be held by Shang during class on Thursday.

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