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- 1. 2.2 PERFORM OPERATION WITH BOOLEAN ALGEBRA
- 2. 2.2.1 Define logic gates • A logic gate performs a logical operation on one or more logic inputs and produces a single logic output and most commonly found at digital circuits.
- 3. 2.2.2 Explain the operation of logic gates. 2.2.3 Draw logic symbols for gates. 2.2.4 Construct truth table of logic gates.
- 4. AND Gate • Logic Symbol, Truth Table And Logic Expression YXZX 0 0 10 Y 01 11 0 0 0 1 Logic Symbol Truth Table Logic Expression
- 5. OR Gate • Logic Symbol, Truth Table And Logic Expression YXZX 0 0 10 Y 01 11 0 1 1 1 Logic Symbol Truth Table Logic Expression
- 6. Inverter/Not Gate • Logic Symbol, Truth Table And Logic Expression XZX 0 1 01 Logic Symbol Truth Table Logic Expression
- 7. NOR • Logic Symbol, Truth Table And Logic Expression YXZX 0 0 10 Y 01 11 1 0 0 0 YXZ X Y Logic Symbol Truth Table Logic Expression
- 8. NAND • Logic Symbol, Truth Table And Logic Expression YXZX 0 0 10 Y 01 11 1 1 1 0 YXZ X Y Logic Symbol Truth Table Logic Expression
- 9. XOR • Logic Symbol, Truth Table And Logic Expression YXZX 0 0 10 Y 01 11 0 1 1 0 YXZ X Y 1) Result is „1‟ when exactly one input is „1‟ 2) The output is always 1 when we have a different set of input Logic Symbol Truth Table Logic Expression
- 10. XNOR • Logic Symbol, Truth Table And Logic Expression YXZX 0 0 10 Y 01 11 1 0 0 1 YXZ X Y Result is „1‟ when both inputs are the same logic Logic Symbol Truth Table Logic Expression
- 11. 2.3 Build sequential logic circuit • Circuits whose outputs depends not only on the present input value but also the past input value are known as sequential logic circuits. • Are circuits that contain memory element. • Example: flip-flop 2.3.1 Define sequential logic circuit.
- 12. 2.3.2 Differentiate between combinational logic circuit and sequential logic circuit • Combinational Logic Circuit – refers to circuits whose output is strictly depended on the present value of the inputs. Example: logic gates • Sequential Logic Circuit- Circuits whose outputs depends not only on the present input value but also the past input value are known as sequential logic circuits. Example: flip-flop
- 13. 2.3.3 Describe Flip Flop • Is a logic circuit that has two stable states or memory where one state is compliment with other state. • can be divided into common types either synchronous(clock) or asynchronous (no clock):
- 14. 2.3.4 List the types of flip-flop: a. SR flip – flop (SR- set reset) b. Clocked SR flip – flop c. JK flip – flop d. T flip flop (Toggle) e. D flip flop (Delay or Data)
- 15. 2.3.5 Build SR, JK, T and D flip flop using logic gates. 2.3.6 Draw the symbol and truth table of SR, JK, T and D flip –flop.
- 16. 1.SR FLIP FLOP • Can build from NOR or NAND gate. From NOR gate From NAND gate S R Q Q S R Q Q S R Keluaran (Q) 0 0 Tak logik 0 1 1 (set) 1 0 0(reset) 1 1 Tak ubah S R Keluaran (Q) 0 0 Tak ubah 0 1 0 (reset) 1 0 1 (set) 1 1 Tak logik symbol
- 17. Con‟t Timing digram for Flip-Flop SR-get NOR Timing digram for Flip-Flop SR-get NAND S R Q T1 T2 T3 T4 T5 T6
- 18. 2) CLOCKED SR FLIP FLOP From NOR gate From NAND gate Timing diagram for SR flip flop with clock S KLOK R Q Q S KLOK R Q Q S R klok Q keadaanawal set takubah reset reset takubah set
- 19. 3) JK FLIP FLOP Truth table Timing Digram nQ J K clock Q t.ubah set t.ubah toggle reset t.ubah Klok J K Qn+1 1 0 0 Qn 1 0 1 0 1 1 0 1 1 1 1 nQ symbol
- 20. 4) T FLIP FLOP JAM T Qn Qn+1 CATATA N 1 0 0 0 Tak Ubah 1 0 1 1 Tak Ubah 1 1 1 0 Toggle 1 1 0 1 Toggle JAM T Qn+1 1 0 Qn 1 1 nQ T clock Q Logic Symbol Logic circuit Truth table Truth table Timing diagram
- 21. 5) D Flip flop Jam D Qn+1 0 0 0 1 1 0 0 1 1 1 nQ nQ D Qn+1 0 0 1 1 D clock Q Symbol Circuit Truth table Timing diagram
- 22. COMBINATIONAL LOGIC CIRCUIT • refers to circuits whose output is strictly depended on the present value of the inputs • Are made of logic gates with no feedback. • To design combinational logic circuit, we need to know about basic logic equation : – If sign “+” between two or more variables, it means all variables using OR gate. For example : A + B + C – If sign “.” between two or more variables, it means all variables using AND gate operation. For example : A.B.C
- 23. Example : • Given logic equation Y = A . B + A . B. Draw the logic diagram base on the equation. Solution • the equation has 2 variables A and B. • reference A . B used AND gate and A used NOT gate • reference A . B used AND gate • Finally, both reference used OR gate to form equation of Y
- 24. Boolean Theorem • Basic Rules 1. A + 0 = A 2. A + 1 = 1 5. A . 0 = 0 6. A . 1 = A 3. A + A = A 7. A . A = A 4. A + A = 1 8. A . A = 0 9. A = A = 10. A + AB = A 12. (A + B)(A + C) = A + BC 11. A + AB = A + B
- 25. BOOLEAN THEOREMS XXX X XX XX XXX X X .7 11.6 0.5 0.4 .3 X1.2 00.1 XZWZXYWYZYXWb XZXYZYXa XYZZXYYZX ZYXZYXZYX XYYX XYYX XX .13 .13 .12 .11 .10 .9 1.8 AA YXXY YXYX YXYXX XXYX .18 .17 .16 .15 .14
- 26. Boolean Simplification - Example • Using Boolean theorem, Simplify the expression: )()( CBBCBAAB • Apply distributive law, BCBBACABAB • Apply rule 7 (BB = B), and rule 5 (AB + AB = AB) BACAB • Apply rule 10 (B + BC = B) BCBACAB
- 27. Boolean Simplification - Example BACAB • Apply rule 10 (AB + B = B) ACB At this point, the expression is simplified as much as possible Original expression is )()( CBBCBAAB Which is logically equal to ACB In terms of design, what is the advantage of Boolean simplification?
- 28. Boolean Simplification - Example Original expression is )()( CBBCBAAB Which is logically equal to ACB Faster Compact design Lower cost A B C A B C
- 29. DeMorgan‟s Theorem • The complement of a product of variables is equal to the sum of the complemented variables AB = A + B A + B A B AB A B NAND Negative-OR BAA 0 0 10 B 01 11 1 1 1 0 BA 1 1 1 0 Theorem 1
- 30. DeMorgan‟s Theorem BAA 0 0 10 B 01 11 1 0 0 0 BA 1 0 0 0 Theorem 2 AB A B A + B A B NOR Negative-AND A + B = A . B
- 31. Example 1: • Given Z = A + B . C .Simplified the equation below using De‟ Morgan Theorem. Solution; Z = A + B.C = A . B.C = A .( B+C) = A . (B+C)
- 32. Example 2: • Given Z = (A + C).(B+D) .Simplified the equation below using De‟ Morgan Theorem. Solution : Z = (A + C) . (B + D) = (A + C) + (B + D) = (A . C) + (B . D) = AC + BD
- 33. Sum-of-Products • SOP expressions consist of two or more AND terms (products) that are ORed together • In SOP an inversion cannot cover more than one variable in a term Example: • ABC + ABC • A B + A B + A B • A B C + A B C • A B + A B C + C D + C
- 34. Product-of-Sums • POS expressions consist of two or more OR terms (sums) that are ANDed together • Example: – X = (A + B + C)(A + C) – X = (A + B)(C + D)F – X = ( A + B ) . ( B + C ) – X = ( B + C + D ) . ( B C + E ) – X = ( A + C ) . ( B + E ). ( C + B )
- 35. Karnaugh Map Method • A graphical method of simplifying logic equations or truth tables. • Also called a K map. • Theoretically can be used for any number of input variables, but practically limited to 5 or 6 variables.
- 36. Karnaugh Map Method • The truth table values are placed in the K map. • Adjacent K map square differ in only one variable both horizontally and vertically. • The pattern from top to bottom and left to right must be in the form • A SOP expression can be obtained by ORing all squares that contain a 1.
- 37. Karnaugh Map Method • Looping adjacent groups of 2, 4, or 8 1s will result in further simplification. • When the largest possible groups have been looped, only the common terms are placed in the final expression. • Looping may also be wrapped between top, bottom, and sides. • Looping a pair (or quad or octet and so on) of adjacent 1s in a K map eliminates the variable that appears in complemented and uncomplemented form.
- 38. Karnaugh maps and truth tables for (a) two, (b) three, and (c) four variables.
- 39. Examples of looping pairs of adjacent 1s.
- 40. Examples of looping groups of fours 1s (quads).
- 41. Examples of looping groups of eight 1s (octets).
- 42. Karnaugh Map Method • Complete K map simplification process – Construct the K map, place 1s as indicated in the truth table. – Loop 1s that are not adjacent to any other 1s. (Isolated 1s) – Loop 1s that are in pairs – Loop 1s in octets even if they have already been looped. – Loop quads that have one or more 1s not already looped. (Use minimum number of loops) – Loop any pairs necessary to include 1s not already looped. – Form the OR sum of terms generated by each loop.
- 43. Examples :
- 44. Example : The same K map with two equally good solutions.
- 45. Example : • Use a K map to simplify: Y = C(ABD + D) + ABC + D
- 46. SOLUTION :

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