'Boolean Algebra\r'

' underlying Engineering Boolean Algebra and Logic render F Hamer, M Lavelle & D McMullan The aim of this document is to provide a short, self assessment programme for students who wish to understand the basal techniques of logical system adits. c 2005 Email: chamer, mlavelle, [email protected] ac. uk Last rewrite Date: August 31, 2006 Version 1. 0 Table of Contents 1. 2. 3. 4. 5. Logic supply (Introduction) Truth Tables rudimentary Rules of Boolean Algebra Boolean Algebra closing screen resolving powers to deterrent examples firmness of purposes to essayzesThe full figure of these cases and whatever instructions, should they be required, merchant ship be obtained from our web page Mathematics condense Materials. Section 1: Logic furnish (Introduction) 3 1. Logic Gates (Introduction) The parcel of land Truth Tables and Boolean Algebra set push through the basic principles of logic. Any Boolean algebra procedure grass be associated with an electronic ra cing circuit in which the inputs and outputs represent the statements of Boolean algebra. Although these circuits whitethorn be complex, they may all be constructed from ternion basic devices. These atomic number 18 the AND access, the OR entry and the not logic access. y AND introduction x·y x y OR gate x+y x NOT gate x In the case of logic supply, a di? erent bank note is social occasiond: x ? y, the luculent AND operation, is replaced by x · y, or xy. x ? y, the logical OR operation, is replaced by x + y. ¬x, the logical NEGATION operation, is replaced by x or x. The uprightness value TRUE is written as 1 (and marks to a lavishly voltage), and FALSE is written as 0 (low voltage). Section 2: Truth Tables 4 2. Truth Tables x y x·y x 0 0 1 1 Summary y x·y 0 0 1 0 0 0 1 1 of AND gate x 0 0 1 1 Summary y x+y 0 0 1 1 0 1 1 1 of OR gate x y x+y x x 0 1 Summary of x 1 0 NOT gate Section 3: Basic Rules of Boolean Algebra 5 3. Basic Rules of Boolean Algebra The basic linguistic traffic patterns for simplifying and combining logic gates atomic number 18 called Boolean algebra in honour of George Boole (1815 †1864) who was a self-educated English mathematician who developed many of the key ideas. The following set of exercises go away allow you to rediscover the basic approach patterns: x practice session 1 1 distribute the AND gate where one of the inputs is 1. By apply the fair play put over, investigate the possible outputs and and so simplify the flavour x · 1.Solution From the impartiality knock back for AND, we ensure that if x is 1 consequently 1 · 1 = 1, objet dart if x is 0 accordingly 0 · 1 = 0. This put up be summarised in the rule that x · 1 = x, i. e. , x x 1 Section 3: Basic Rules of Boolean Algebra 6 deterrent example 2 x 0 cope the AND gate where one of the inputs is 0. By exploitation the truth table, investigate the possible outputs and thusly simplify the conceptualisation x · 0 . Solution From the truth table for AND, we turn back that if x is 1 therefore 1 · 0 = 0, opus if x is 0 indeed 0 · 0 = 0. This advise be summarised in the rule that x · 0 = 0 x 0 0Section 3: Basic Rules of Boolean Algebra 7 make for 1. ( photograph on the parking atomic number 18a letters for the solutions. ) grow the rules for simplifying the logical expressions x (a) x + 0 which corresponds to the logic gate 0 (b) x + 1 which corresponds to the logic gate x 1 Exercise 2. ( gaol on the jet letters for the solutions. ) commence the rules for simplifying the logical expressions: x (a) x + x which corresponds to the logic gate (b) x · x which corresponds to the logic gate x Section 3: Basic Rules of Boolean Algebra 8 Exercise 3. domestic dog on the common land letters for the solutions. ) Obtain the rules for simplifying the logical expressions: (a) x + x which corresponds to the logic gate x (b) x · x which corresponds to the logic gate x Quiz Simplify th e logical expression (x ) represent by the following circuit diagram. x (a) x (b) x (c) 1 (d) 0 Section 3: Basic Rules of Boolean Algebra 9 Exercise 4. (Click on the greenish letters for the solutions. ) Investigate the relationship surrounded by the following circuits. Summarise your conclusions using Boolean expressions for the circuits. x y x y (a) (b) x y x yThe master(prenominal) relations developed in the supra exercise are called De Morgan’s theorems and are widely used in simplifying circuits. These correspond to rules (8a) and (8b) in the table of Boolean identities on the next page. Section 4: Boolean Algebra 10 4. Boolean Algebra (1a) x·y = y·x (1b) x+y = y+x (2a) x · (y · z) = (x · y) · z (2b) x + (y + z) = (x + y) + z (3a) x · (y + z) = (x · y) + (x · z) (3b) x + (y · z) = (x + y) · (x + z) (4a) x·x = x (4b) x+x = x (5a) x · (x + y) = x (5b) x + (x · y) = x (6a) x·x = 0 (6b) x+x = 1 (7) (x ) = x (8a) (x · y) = x + y (8b) (x + y) = x · ySection 4: Boolean Algebra 11 These rules are a direct translation into the notation of logic gates of the rules derived in the package Truth Tables and Boolean Algebra. We hold up put one acrossn that they domiciliate all be checked by investigating the corresponding truth tables. Alternatively, some of these rules finish be derived from simpler identities derived in this package. representative 3 fancy how rule (5a) can be derived from the basic identities derived earlier. Solution x · (x + y) = = = = = x · x + x · y using (3a) x + x · y using (4a) x · (1 + y) using (3a) x · 1 using Exercise 1 x as required. Exercise 5. Click on the green letter for the solution. ) (a) Show how rule (5b) can be derived in a similar fashion. Section 4: Boolean Algebra 12 The examples above have all involved at virtually both inputs. However, logic gates can be put together to sum of money an arbitrary number of inputs. The Boolean algebra ru les of the table are essential to understand when these circuits are homogeneous and how they may be simpli? ed. Example 4 Let us gestate the circuits which combine three inputs via AND gates. Two di? erent ship canal of combining them are x y z and x y z x · (y · z) (x · y) · z Section 4: Boolean Algebra 13However, rule (2a) states that these gates are equivalent. The holy order of taking AND gates is not important. This is sometimes drawn as a three (or more! ) input AND gate x y z x·y·z but in truth this just gist repeated use of AND gates as shown above. Exercise 6. (Click on the green letter for the solution. ) (a) Show two di? erent slipway of combining three inputs via OR gates and explain why they are equivalent. This equivalence is summarised as a three (or more! ) input OR gate x y z x+y+z this just means repeated use of OR gates as shown in the exercise. Section 5: Final Quiz 14 5. Final Quiz Begin Quiz 1.Select the Boolean expression that is not e quivalent to x · x + x · x (a) x · (x + x ) (b) (x + x ) · x (c) x (d) x 2. Select the expression which is equivalent to x · y + x · y · z (a) x · y (b) x · z (c) y · z (d) x · y · z 3. Select the expression which is equivalent to (x + y) · (x + y ) (a) y (b) y (c) x (d) x 4. Select the expression that is not equivalent to x · (x + y) + y (a) x · x + y · (1 + x) (b) 0 + x · y + y (c) x · y (d) y sack Quiz Solutions to Exercises 15 Solutions to Exercises Exercise 1(a) From the truth table for OR, we see that if x is 1 wherefore 1 + 0 = 1, opus if x is 0 then 0 + 0 = 0.This can be summarised in the rule that x + 0 = x x 0 Click on the green unbent to recollect x Solutions to Exercises 16 Exercise 1(b) From the truth table for OR we see that if x is 1 then 1 + 1 = 1, part if x is 0 then 0 + 1 = 1. This can be summarised in the rule that x + 1 = 1 x 1 Click on the green upstanding to return 1 Solutions to Exercises 17 Exercise 2(a) From the truth table for OR, we see that if x is 1 then x + x = 1 + 1 = 1, while if x is 0 then x + x = 0 + 0 = 0. This can be summarised in the rule that x + x = x x x Click on the green fledge to return Solutions to Exercises 18Exercise 2(b) From the truth table for AND, we see that if x is 1 then x · x = 1 · 1 = 1, while if x is 0 then x · x = 0 · 0 = 0. This can be summarised in the rule that x · x = x x x Click on the green square to return Solutions to Exercises 19 Exercise 3(a) From the truth table for OR, we see that if x is 1 then x + x = 1 + 0 = 1, while if x is 0 then x + x = 0 + 1 = 1. This can be summarised in the rule that x + x = 1 x 1 Click on the green square to return Solutions to Exercises 20 Exercise 3(b) From the truth table for AND, we see that if x is 1 then x · x = 1 · 0 = 0, while if x is 0 then x · x = 0 · 1 = 0.This can be summarised in the rule that x · x = 0 x 0 Click on the green square to return Solutions to Exercis es 21 Exercise 4(a) The truth tables are: x y x y 0 0 0 1 1 0 1 1 x y 0 0 0 1 1 0 1 1 x+y 0 1 1 1 x 1 1 0 0 y 1 0 1 0 (x + y) 1 0 0 0 x ·y 1 0 0 0 x y From these we come the identicalness x y (x + y) = x y x ·y Click on the green square to return Solutions to Exercises 22 Exercise 4(b) The truth tables are: x y x y 0 0 0 1 1 0 1 1 x y 0 0 0 1 1 0 1 1 x·y 0 0 0 1 x 1 1 0 0 y 1 0 1 0 (x · y) 1 1 1 0 x +y 1 1 1 0 x y From these we deduce the identity x y (x · y) = x y x +y Click on the green square to returnSolutions to Exercises 23 Exercise 5(a) x+x·y = x · (1 + y) using (3a) = x · 1 using Exercise 1 = x as required. Solutions to Exercises 24 Exercise 6(a) Two di? erent ways of combining them are x y z and x y z However, rule (2b) states that these gates are equivalent. The order of taking OR gates is not important. x + (y + z) (x + y) + z Solutions to Quizzes 25 Solutions to Quizzes Solution to Quiz: From the truth table for NOT we see that if x is 1 then (x ) = (1 ) = (0) = 1, while if x is 0 then (x ) = (0 ) = (1) = 0. This can be summarised in the rule that (x ) = x x x End Quiz\r\nTest: â€Å"Study rent Algebra”\r\n'