Boolean Algebra

staple fibre Engineering Boolean Algebra and logic Gates 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 basic techniques of logic adits. c 2005 Email chamer, mlavelle, emailprotected ac. uk Last edict Date August 31, 2006 Version 1. 0 Table of Contents 1. 2. 3. 4. 5. Logic Gates (Introduction) Truth Tables sanctioned Rules of Boolean Algebra Boolean Algebra Final Quiz Solutions to feats Solutions to QuizzesThe full range of these softwargons and some instructions, should they be required, freighter be obtained from our web page Mathematics Support Materials. naval division 1 Logic Gates (Introduction) 3 1. Logic Gates (Introduction) The package Truth Tables and Boolean Algebra set out the basic principles of logic. Any Boolean algebra cognitive process fucking be associated with an electronic circuit in which the in drifts and outputs represent the statements of Boolean alge bra. Although these circuits may be complex, they may all be constructed from three basic devices. These argon the AND access, the OR provide and the non gateway. y AND gate xy x y OR gate x+y x NOT gate x In the case of logic gates, a di? erent nonation is used x ? y, the uniform 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 loyalty value TRUE is written as 1 (and corresponds to a spunky voltage), and FALSE is written as 0 (low voltage). Section 2 Truth Tables 4 2. Truth Tables x y xy x 0 0 1 1 Summary y xy 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 bumps for simplifying and combining logic gates are called Boolean algebra in honour of George Boole (1815 1864) who was a self-educated English math ematician who developed many of the key ideas. The following set of exercises will allow you to discover the basic line ups x Example 1 1 ask the AND gate where one of the inputs is 1. By using the truth submit, investigate the possible outputs and hence simplify the verbal flavor x 1.Solution From the truth table for AND, we underwrite that if x is 1 whence 1 1 = 1, while if x is 0 then 0 1 = 0. This weed be summarised in the rule that x 1 = x, i. e. , x x 1 Section 3 Basic Rules of Boolean Algebra 6 Example 2 x 0 Consider the AND gate where one of the inputs is 0. By using the truth table, investigate the possible outputs and hence simplify the expression x 0. Solution From the truth table for AND, we command that if x is 1 then 1 0 = 0, while if x is 0 then 0 0 = 0. This thunder mug be summarised in the rule that x 0 = 0 x 0 0Section 3 Basic Rules of Boolean Algebra 7 Exercise 1. ( imbue on the yard letters for the solutions. ) Obtain the rules for simplifyin g 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. (Click on the green letters for the solutions. ) Obtain 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. Click on the green 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 the logical expression (x ) be 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 green letters for the solutions. ) Investigate the relationship amidst the following circuits. Summarise your conclusions using Boolean expressions for the circuits. x y x y (a) (b) x y x yThe important relations developed in t he above exercise are called De Morgans 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) xy = yx (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) xx = x (4b) x+x = x (5a) x (x + y) = x (5b) x + (x y) = x (6a) xx = 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 know seen that they can all be checked by investigating the corresponding truth tables. Alternatively, some of these rules can be derived from simpler identities derived in this package. Example 3 Show 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 most devil inputs. However, logic gates can be put together to join an arbitrary number of inputs. The Boolean algebra rules of the table are essential to understand when these circuits are equivalent and how they may be simpli? ed. Example 4 Let us consider the circuits which combine three inputs via AND gates. Two di? erent ways 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 order of taking AND gates is not important. This is sometimes drawn as a three (or more ) input AND gate x y z xyz but really this just message repeated use of AND gates as shown above. Exercise 6. (Click on th e green letter for the solution. ) (a) Show two di? erent ways 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 equivalent 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 End Quiz Solutions to Exercises 15 Solutions to Exercises Exercise 1(a) From the truth table for OR, we see that if x is 1 then 1 + 0 = 1, while 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 square to blow over 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, while 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 square 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 square 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 Solution s 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 Exercises 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 deduce the identity 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 xy 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+xy = 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 g ates 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 QuizTest Study Guide Algebra