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its about different types of gates
Typology: Exercises
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Basic Engineering
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 gates.
©c 2005 Email: chamer, mlavelle, [email protected] Last Revision Date: August 31, 2006 Version 1.
The full range of these packages and some instructions, should they be required, can be obtained from our web page Mathematics Support Materials.
Section 2: Truth Tables 4
x y
x · y
x y x · y 0 0 0 0 1 0 1 0 0 1 1 1 Summary of AND gate x y x + y 0 0 0 0 1 1 1 0 1 1 1 1 Summary of OR gate
x y
x + y
x x′
x x′ 0 1 1 0 Summary of NOT gate
Section 3: Basic Rules of Boolean Algebra 5
The basic rules for simplifying and combining logic gates are 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 will allow you to rediscover the basic rules:
Example 1 x 1
Consider the AND gate where one of the inputs is 1. By using the truth table, investigate the possible outputs and hence simplify the expression x · 1.
Solution From the truth table for AND, we see that if x is 1 then 1 · 1 = 1, while if x is 0 then 0 · 1 = 0. This can be summarised in the rule that x · 1 = x, i.e., x 1
x
Section 3: Basic Rules of Boolean Algebra 7
Exercise 1. (Click on the green letters for the solutions.) Obtain the rules for simplifying the logical expressions
(a) x + 0 which corresponds to the logic gate
x 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:
(a) x + x which corresponds to the logic gate
x
(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′)′^ represented by the following circuit diagram.
x
(a) x (b) x′^ (c) 1 (d) 0
Section 4: Boolean Algebra 10
(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′^ · y′
Section 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 have 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 iden- tities 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 13
However, 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
x · y · z
but really this just means repeated use of AND gates as shown above.
Exercise 6. (Click on the green letter for the solution.)
(a) Show two different 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
Begin Quiz
End Quiz
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
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 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
