DKT 122/3 DIGITAL SYSTEM 1


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1 Company LOGO DKT 122/3 DIGITAL SYSTEM 1 BOOLEAN ALGEBRA (PART 2)
2 Boolean Algebra Contents Boolean Operations & Expression Laws & Rules of Boolean algebra DeMorgan s Theorems Boolean analysis of logic circuits Simplification using Boolean Algebra Standard forms of Boolean Expressions Boolean Expressions & truth tables
3 Boolean Algebra (Cont.) The Karnaugh Map Karnaugh Map SOP minimization Karnaugh Map POS minimization Programmable Logic
4 Solve this.. A B X C (a) Write the equivalent Boolean expression of circuit shown above. (b) Simplify the Boolean expression found in (a).
5 Solve this.. Simplify the following Boolean expressions: (AB(C + BD) + AB)C ABC + ABC + ABC + ABC + ABC Write the Boolean expression of the following circuit. RULE: Break the longest bar first!
6 Standard Forms of Boolean Expressions Sumofproducts (SOP) Refer to two or more product terms, summed (added) by Boolean addition Productofsums (POS) Refer to multiplication of two or more sum terms
7 Sumofproducts (SOP) In an SOP expression, a single overbar cannot extend over more than one variable
8 Sumofproducts (SOP) The standard SOP form Refer to expression in which all the variables in the domain appear in each product term in the expression Example: ABCD + ABCD + ABCD (standard SOP expression) ABC + ABD + ABCD (nonstandard SOP expression)
9 Sumofproducts (SOP) The standard SOP form All variables appear in each product term. Each of the product term in the expression is called as minterm. Example f ( A, B, C ) = ABC + ABC + ABC In compact form, f(a,b,c) may be written as f ( A, B, C ) = m2 + m3 + m f ( A, B, C) = Σm(2,3,6) 6
10 Productofsums (POS) In a POS expression, a single overbar cannot extend over more than one variable
11 Standard Forms of Boolean Expressions The standard POS form Refer to expression in which all the variables in the domain appear in each sum term in the expression Example: (A + B + C)(B + C + D)(A + B + C + D) (nonstandard POS expression) (A + B + C + D)(A + B + C + D) (standard POS expression)
12 All variables appear in each product term. Each of the product term in the expression is called as maxterm. Example: ) ( ) ( ) ( ),, ( C B A C B A C B A C B A f = Standard Forms of Boolean Expressions The standard POS form Example: ) ( ) ( ) ( ),, ( C B A C B A C B A C B A f = In compact form, f(a,b,c) may be written as ),, ( M M M C B A f = (1,4,5) ),, ( M C B A f π =
13 Solve this.. Identify each of the following expressions as SOP, standard SOP, POS or standard POS: (i) AB + ABD + ACD (ii) (A + B + C)(A + B + C) (iii) ABC + ABC (iv) A(A + C)(A + B) Convert the following Boolean expressions to SOP form: (i) (A + B)(B + C + D) (ii) AB + B(CD + EF)
14 Example (Standard SOP) Convert the following Boolean expressions into standard SOP form: Solution AB + ABCD 1 st step: AB = AB (C + C) = ABC + ABC 2 nd step: ABC (D + D) + ABC (D + D) = ABCD + ABCD + ABCD + ABCD 3 rd step: ABCD + ABCD + ABCD + ABCD + ABCD
15 Example (Standard POS) Convert the following Boolean expressions into standard POS form: Solution (A + B + C)(B + C + D) 1 st step: A + B + C = A + B + C + DD = (A + B + C + D) (A + B + C + D) 2 nd step: B + C + D = B + C + D + AA = (A + B + C + D)(A + B + C + D) Final answer: Combine answer for 1 st and 2 nd step
16 Boolean Expressions & Truth Tables Converting SOP to Truth Table Examine each of the products to determine where the product is equal to a 1. Set the remaining row outputs to 0.
17 Boolean Expressions & Truth Tables Converting POS to Truth Table Opposite process from the SOP expressions. Each sum term results in a 0. Set the remaining row outputs to 1.
18 Boolean Expressions & Truth Tables Determining Standard Expressions from Truth Table INPUT OUTPUT A B C X SOP Expressions X = ABC + ABC + ABC + ABC POS Expressions X = (A + B + C)(A + B + C) (A + B + C)(A + B + C)
19 Solve this.. Convert the following SOP expression to an equivalent POS expression: f ( A, B, C)= ABC+ ABC + ABC + ABC Develop a truth table for the expression: f ( A, BC, ) = ( A+ B+ C) ( A+ B+ C) ( A+ B+ C) ( A+ B+ C)
20 The Karnaugh Map (KMap) Karnaugh Map (Kmap) is an array of cells in which each cell represents a binary value of the input variables KMapping is used to minimize the number of logic gates that are required in a digital circuit. This will replace Boolean reduction when the circuit is large. The number of cells in a Kmap is equal to the total number of possible input variable combinations KMap is similar to Truth Table because it present all possible values of input variables and the resulting output for each value
21 The Kmap The map is made up of a table of every possible SOP using the number of variables that are being used. If 2 variables are used, then a 2X2 map is used, If 3 variables are used, then a 4X2 map is used, If 4 variables are used, then a 4X4 map is used, If 5 variables are used, then a 8X4 map is used
22 Kmap SOP Minimization Mapping Standard SOP Expression For an SOP in standard form, a 1 is placed on the Kmap for each product term in the expression Gray Code Example of standard SOP form
23 Kmap SOP Minimization Mapping Nonstandard SOP Expression Expand the nonstandard expression A + AB + ABC Question How to expand this expression?
24 Kmap SOP Minimization Mapping directly from a truth table to a K map Thomas L. Floyd Digital Fundamentals, 9e Copyright 2006 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved.
25 Kmap SOP Minimization 2 variables Kmap A A B B The upper right hand cell is A B if X= A B then put an X in that cell Notice that the map is going false to true, left to right and top to bottom A A B B 1 This show the expression true when A = 0 and B = 0
26 Kmap SOP Minimization 2 variables Kmap If X=AB + AB then put an X in both of these cells B B A 1 A 1 From Boolean reduction we know that A B + A B = B From the Karnaugh map we can circle adjacent cell and find that X = B A A B B 1 1
27 Kmap SOP Minimization 3 variables Kmap Gray Code 0 1 C C 00 A B 01 A B 11 A B 10 A B
28 Kmap SOP Minimization 3 variables Kmap X = A B C + A B C + A B C + A B C 00 A B 01 A B 11 A B 10 A B 0 1 C C Each 3 variable term is one cell on a 4 X 2 Karnaugh map
29 Kmap SOP Minimization 3 variables Kmap X = A B C + A B C + A B C + A B C 00 A B 01 A B 11 A B 10 A B 0 1 C C One simplification could be X = A B + A B
30 Kmap SOP Minimization X = A B C + A B C + A B C + A B C 3 variables Kmap X = A B C + A B C + A B C + A B C 00 A B 01 A B 11 A B 10 A B 0 1 C C Another simplification could be X = B C + B C A Karnaugh Map does wrap around
31 Kmap SOP Minimization X = A B C + A B C + A B C + A B C 3 variables Kmap X = A B C + A B C + A B C + A B C 00 A B 01 A B 11 A B 10 A B 0 1 C C The best simplification would be X = B
32 Kmap SOP Minimization 3 variables Kmap Conclusions One cell requires 3 variables Two adjacent cells require 2 variables Four adjacent cells require 1 variable Eight adjacent cells is a 1
33 Kmap SOP Minimization 4 variables Kmap Gray Code A B 01 A B 11 A B 10 A B C D C D C D C D
34 Simplify: Kmap SOP Minimization X = A B C D + A B C D + A B C D + A B C D + A B C D + A B C D Gray Code A B 01 A B 11 A B 10 A B C D C D C D C D X = ABD + ABC + CD 1 Now, try it with Boolean reductions..
35 Kmap SOP Minimization 4 variables Kmap Conclusions One Cell requires 4 variables Two adjacent cells require 3 variables Four adjacent cells require 2 variables Eight adjacent cells require 1 variable Sixteen adjacent cells give a 1 or true
36 Kmap SOP Minimization Simplify Z = B C D + B C D + C D + B C D + A B C Gray Code C D C D C D C D 00 A B 01 A B 11 A B 10 A B Z = BD + C
37 Example Simplify the following circuit using Kmap method.
38 Example (Cont.) Y = A + B + B C + ( A + B ) ( C + D) Y = A B + B C + A + B + ( C + D ) Y = A B + B C + (A + B ) + C D Simplified SOP expression
39 Example (Cont.) Then, map the SOP expression into the Kmap & simplify the equation Gray Code C D C D C D C D 00 A B A B A B A B Y = 1
40 Kmap POS Minimization Assume A, B, C, and D are variables. 3 variables 4 variables
41 3 variables Kmap Kmap POS Minimization
42 4 variables Kmap Kmap POS Minimization
43 4 variables Kmap Kmap POS Minimization
44 Kmap Minimization Don t Cares 3 variables with output don t cares (X) Input Output
45 Kmap Minimization Don t Cares 4 variables with output don t cares (X)
46 Example Determine the minimal SOP using KMap: F(A, B, C, D) = πm(0,2,6,8, 9,10) D(5,12,13, 14,15) D refers to don t cares What is this?
47 Example (Cont.) F(A, B, C, D) = πm(0,2,6,8, 9,10) D(5,12,13, 14,15) CD AB BC X X X X X Minimum SOP expression is: F ( A, B, C, D) = CD+ BC + AD AD CD
48 Solve this.. Reduce (a), (b) and (c) using Kmap: (a) (b) (c) f(a, B, C, D) = m (1, 3, 5, 7, 9) + d(6,12, 13)
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