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### Chapter 3_Boolean Algebra _ Logic Gate (3).pptx

• 2. 2 Topics 3.1) Basic Operations of Boolean Algebra 3.2) Relationship between basic operation of Boolean and basic logic gate 3.3) Basic Theorems of Boolean Algebra 3.4) Relationship between Boolean Function and Logic Circuit
• 3. 3 3.5) Truth Table 3.6) Karnaugh Map 3.7) Example of Digital Problem 3.8) Create a Logic Circuit Using Only NAND Gates or NOR Gates
• 4. 4 3.1) Basic Operations of Boolean Algebra • Boolean Algebra was introduced by George Boole in the year 1854. • Like other algebras, it uses variables (called statements) and operations (called relations). • Variables in Boolean algebra is called logic variable that only has 2 possible values, either true (1) or false (0) whereas its operation is called logical operations.
• 5. 5 • There are 3 basic logical operations, i.e. AND (.), OR (+) and NOT ( ). • These operations are used to combine operands (logical constants and variables) to form logical expressions. • The following are a few logical expressions with X, Y and Z as the logical variables that can only have the value of FALSE (or 0) or TRUE (or 1). X = NOT X X.Y + Z = NOT( X AND Y) OR Z
• 6. 6 • Other than the basic logical operations of AND, OR and NOT, there are multiple of other logical operations, among them are NAND, NOR and XOR that are widely used in building logical circuits in computers. • These logical operations are combination of a few of the basic logical operations, e.g. : NAND is a combination of AND followed by NOT NOR is a combination of OR followed by NOT
• 7. 7 The following is a truth table for the logical operations of AND, OR, NOT, NAND, NOR and XOR. P Q NOT P P AND Q P OR Q P XOR Q P NAND Q P NOR Q 0 0 1 0 0 0 1 1 0 1 1 0 1 1 1 0 1 0 0 0 1 1 1 0 1 1 0 1 1 0 0 0 CONSIDER XAND – eXclusive AND whereby a positive output is only achieved if both inputs are equal.
• 8. XAND XAND- eXclusive AND whereby a positive output is only achieved if both inputs are equal. 8
• 9. 9 Summary 1. AND (symbol: .) ☞The AND operation only gives the TRUE (or value 1 in binary) result if and only if the value of all the variables are TRUE 2. OR (symbol : +) ☞The OR operation only gives the TRUE result if the value of one or all of the variables is/are TRUE. 3. NOT (symbol : ) ☞The NOT operation will change the value of the operand, i.e. TRUE to FALSE and vice-versa
• 10. 10 4. NAND ☞ The NAND operation only gives the TRUE result if one or both of the operands are FALSE 5. NOR ☞ The NOR operation only gives the TRUE result if and only if both operands are FALSE 6. XOR ☞ The XOR operation only gives the TRUE result if and only if only one of the operands is TRUE
• 11. 11 3.2) Relationship Between Basic Operation of Boolean and Basic Logic Gate • The basic construction of a logical circuit is gates. • The logical functions are presented through the combination of gates. • Gate is an electronic circuit that emits an output signal as a result of a simple Boolean operation on its inputs. • The basic gates that used in a digital logic is the same as the basic Boolean Algebra operations such AND, OR, NOT, NAND and NOR.
• 12. 12 • Table below shows the symbols, Boolean algebra and the truth table for the gates A B F 0 0 0 0 1 0 1 0 0 1 1 1 A B F 0 0 0 0 1 1 1 0 1 1 1 1 A B F 0 0 0 0 1 1 1 0 1 1 1 1 A B F 0 0 0 0 1 0 1 0 0 1 1 1 F F F F Name Graphic Symbol Boolean Algebra Truth Table A B A B A B A B A F AND OR NOT NAND NOR F = A . B Or F = AB F = A + B _____ F = A + B ____ F = A . B Or F = AB _ F = A B F 0 1 1 0
• 13. 13 3.3) Basic Theorems of Boolean Algebra 1. Identity Elements 2. Inverse Elements 1 . A = A A . A = 0 0 + A = A A + A = 1 3. Idempotent Laws 4. Boundess Laws A + A = A A + 1 = 1 A . A = A A . 0 = 0 5. Distributive Laws 6. Order Exchange Laws A . (B + C) = A.B + A.C A . B = B . A A + (B . C) = (A+B) . (A+C) A + B = B + A
• 14. 14 7. Absorption Laws 8. Associative Laws A + (A . B) = A A + (B + C) = (A + B) + C A . (A + B) = A A . (B . C) = (A . B) . C 9. Elimination Laws 10. De Morgan Theorem _ ______ _ _ A + (A . B) = A + B (A + B) = A . B _ _____ _ _ A . (A + B) = A . B (A . B) = A + B
• 15. 15 • All the theorems and laws can be proven by substituting the value of 0 or 1 for each variables A, B and C based on the Truth Table for each logical operation given. • These theorems and laws are extremely important in the computer environment because it is used to simplify logical expressions that are produced when designing the logical circuit of the computer
• 16. 16 3.5) Relationship Between Boolean Function and Logic Circuit • Any Boolean function can be implemented in electronic form as a network of gates called logic circuit. • The following are examples of Boolean functions and their logic circuit.
• 17. 17 1) F = A . B + C + D A B C D F A . B C + D
• 18. 18 2) G = A . (B + C + D) A B C D G
• 19. 19 3.5) Truth Table ⇨A truth table displays the relationship between the truth values of propositions. ⇨Eg: Construct a truth table for a Boolean function, X = A + B . C
• 20. 20 A B C A+B A+B A+B . C 0 0 0 0 1 0 0 0 1 0 1 1 0 1 0 1 0 0 0 1 1 1 0 0 1 0 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 1 1 0 0
• 21. 21 X Y X + Y X + Y X . Y 0 0 0 1 1 0 1 1 0 0 1 0 1 0 0 1 1 1 0 0 Logically equivalent : X + Y = X . Y Logical Equivalence: ⇨ Two logic expressions are called logically equivalent if and only if, they have identical values for each of the statements variables.
• 22. 22 3.6) Karnaugh Map ⇒ A graphical way of depicting the content of a truth table where the adjacent squares differ by only one variable. ⇒ For the purposes simplification, the Karnaugh map is a convenient way of representing a Boolean function of a small number (up to four) of variables. ⇒ The map is an array of 2n squares, representing all possible combination of values of n binary variables.
• 23. 23 ⇒ Example: Karnaugh Map with 2 variables, A and B 00 01 10 11 A B 0 0 1 1 _ _ A B _ A B _ A B A B B _ B B _ A A A
• 24. 24 A B F 0 0 0 1 1 0 1 1 00 01 10 11 B A 0 1 0 1 TRUTH TABLE input output Relationship between Truth Table and Karnaugh Map KARNAUGH MAP
• 25. 25 The number of squares in Karnaugh map depends on the number of variables: ⇨ if n variables, there are 2n squares in the Karnaugh Map Example : i) 2 Variables 2² = 4 squares in the Karnaugh Map ii) 3 Variables 2³ = 8 squares in the Karnaugh Map
• 26. 26 The arrangement of squares in the Karnaugh map is free as long as the adjacent squares that are next to each other only differ by one variable. BC BC 000 001 011 010 100 101 111 110 000 001 010 011 100 101 110 111 A _ _ B C _ B C B C _ B C _ A A 00 01 10 11 0 1 A Cannot!
• 27. 27 001 001 010 011 110 111 100 101 _ _ A B _ A B A B _ A B AB C _ C C 000 010 110 100 001 011 111 101 C _ _ _ _ A B A B A B A B AB Example: a) 3 variables _ C C
• 28. 28 0000 0001 0011 0010 0100 0101 0111 0110 1100 1101 1111 1110 1000 1001 1011 1010 00 01 11 10 00 01 11 10 AB CD or others as long as the value of adjacent squares differ by only 1 variable. b) 4 variables
• 29. 29 Getting an expression from a Karnaugh map Example : Given the following Truth Table with 3 inputs (A, B and C) and 1 output (F). A B C 0 0 0 0 0 1 0 1 0 0 0 1 1 1 0 0 1 0 1 1 1 0 0 1 1 1 0 F 1 0 1 1 1 Only 4 combinations of input that give output, F = 1
• 30. 30 Step 1: ⇒ Make a Karnaugh Map with 3 variables. Step 2: ⇒ For each output F = 1, write 1 in the square that has the same combination of input as in the truth table. 1 1 1 1 00 1 1 0 0 00 01 01 11 11 10 10 BC BC A A
• 31. 31 Step 3: ⇒ group the adjacent squares with the following steps : i) If a Karnaugh map has n variables, begin grouping with 2 n-1 ii) If there are no 2 n-1 adjacent squares (which the value is 1), proceed with 2 n-2 squares until 2 n-n OR until no squares which the value is 1 are not grouped 2n-1 = 23-1 = 22 = 4 2n-2 = 23-2 = 21 = 2 For n = 3 2n-3 = 23-3 = 20 = 1
• 32. 32 1 A BC ⇨only insert output which value is equal to 1 to the appropriate squares A B C 00 01 11 10 0 1 1 1 1 B C A B Answer : F = B C + A B + A B C
• 33. 33 A BC 3 variables => n = 3 00 01 11 10 0 1 1 1 1 1 Explaination: ⇨ begin grouping with 23-1 = 4 adjacent squares. (none in the above Karnaugh map) ⇨ proceed with 23-2 = 2 adjacent squares (there are 2 in the above Karnaugh map) ⇨ if there are still squares with value 1 that are not grouped yet, proceed with 2 3-3 = 1 square (there is one in the above Karnaugh map) ⇨ get the expression from each group of squares.
• 34. 34 SUMMARY (3 variables) n = 3 _ _ _ _ B C B C B C B C _ A A A BC _ A A ( 1) 4 adjacent squares _ C _ B B C
• 35. 35 _ _ _ _ B C B C B C B C _ A A _ A C A C _ _ A B _ A B _ A B A B A BC (2) 2 adjacent squares
• 36. 36 A BC A BC 1 1 1 1 1 0 1 00 01 11 10 1 1 1 1 1 00 01 11 10 0 1 _ C AB _ F = C + AB A _ B C _ F = A + B C Other examples
• 37. 37 _ _ _ _ _ _ F = C D + B D + AB C _ F = B + C 1 1 1 1 1 1 1 _ _ _ _ CD CD CD CD CD AB _ _ AB _ AB AB _ AB 00 01 11 10 0 1 BC A 1 1 1 1 1 1 _ C B
• 38. 38 3.7) Example of Digital Problem ⇨ To design a logic circuit of an Alarm System at the office (with one door and one window) that will be rang if the door or window is/are opened after working hours. ⇨ The followings are steps that are to be taken to design a logic circuit.
• 39. 39 1. Problem Determination Determine the problem that has to be solved ⇨ to design a logic circuit for an alarm system that will trigger the emergency bell if door or window is/are opened outside office hours. 2. Conceptualization Obtain the relevant logical variables and make a logical table and also a truth table. Obtain the logical expression from the truth table
• 40. 40 • The related variables are : Time T = 0 (work time) T = 1 (not work time) Doors D = 0 (closed) D = 1 (opened) Windows W = 0 (closed) W = 1 (opened) • Whether Bell B will ring (1) or will not ring (0) depends on all three logical variables (depending on the condition or problem given)
• 41. 41 Logic Table INPUT OUTPUT Time Door Window Bell Work closed closed Doesn’t ring Work closed opened Doesn’t ring Work opened closed Doesn’t ring Work opened opened Doesn’t ring Not Work closed closed Doesn’t ring Not Work closed opened Will Ring Not Work opened closed Will Ring Not Work opened opened Will Ring
• 42. 42 Truth Table (based on logical table above) INPUTS OUTPUT T D W B 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1
• 43. 43 • B is the output or the function that we have to find. The B function or expression can be obtained through many methods based on the truth table. One of the methods is by using a Karnaugh map. • Below is the Karnaugh Map for the Truth Table above 1 1 1 00 01 11 10 0 1 WD T TD TW From the Karnaugh Map above, B = TW + TD
• 44. 44 3. Solution or summary • The expression B above can be further summarized using theorem or the laws in Boolean algebra B = TW + TD = T(W + D) (from the Law of Distribution) 4. Execution • From the expression obtained, the following is the logic circuit for B.
• 46. 46 3.8) Building logical circuits using only NAND or only NOR gates • Most components in computers are built using only type of gate either the NAND or the NOR gates. This can further simplify the construction of such circuits (i.e. do not need to use various gates in a logic circuit) • To build a circuit that only uses NAND or NOR gates, firstly the expression for the circuit has to be changed into an expression that only has either the NAND or NOR operations only. To change it, the De Morgan and Involution theorems are used.
• 47. 47 The Involution theorem is as follow: Example: Using the logic expression in section 6.7 , B = T(W + D), draw a logic circuit by using: 1. Only NAND gates 2. Only NOR gates A = A =
• 48. 48 B = T.(W + D) = T.(W + D) Involution Theorem = T. (W . D) De Morgan theorem = T. (W . D) Involution Theorem 1. Using only the NAND gate To get an expression that only uses the NAND operations, eliminate the OR operation in the expression by using the Involution theorem and De Morgan theorem. ( W+D = W . D ) ( W+D = W+D )
• 49. 49 T W D Hence, the logic circuit for B that use only the NAND gates can be drawn as follow:
• 50. 50 2. Using only the NOR gate To get an expression that only uses the NOR operations, eliminate the AND operation in the expression by using the Involution theorem and the De Morgan Theorem. B = T.(W + D) = T . (W + D) Involution Theorem = T + (W + D) De Morgan theorem ( T.(W+D) = T.(W+D) )
• 51. 51 • The expression no longer has the AND operation and all the OR operation has the complement sign or NOT symbol (or the NOR operation). Hence, the logic circuit for B that only uses the NOR gates can be drawn as follow: T W D
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