Lecture 6
Download as PPT, PDF1 like2,670 views
This document discusses Karnaugh maps and their use in simplifying Boolean functions. It begins with an introduction to Karnaugh maps, including how they are constructed for 2, 3, and 4 variable functions. It then discusses how to use Karnaugh maps to find the simplest sum-of-products expression for a Boolean function by grouping adjacent 1's in the map. Examples are provided to demonstrate algebraic simplification and simplification using Karnaugh maps.
Lecture 6
- 1. Computer & Network Technology Chamila Fernando 02/03/14 Information Representation BSc(Eng) Hons,MBA,MIEEE 1
- 2. Lecture 6: Karnaugh Maps 02/03/14 Function Simplification Algebraic Simplification Half Adder Introduction to K-maps Venn Diagrams 2-variable K-maps 3-variable K-maps 4-variable K-maps 5-variable and larger K-maps Quick Review Questions (3) 2
- 3. Lecture 5: Karnaugh Maps 02/03/14 Simplification using K-maps Converting to Minterms Form Simplest SOP Expressions Getting POS Expressions Don’t-care Conditions Review Examples Quick Review Questions (3) 3
- 4. Function Simplification Why simplify? Simpler expression uses less logic gates. Thus: cheaper, less power, faster (sometimes). Simplification techniques: Algebraic Simplification. Simplification simplify symbolically using theorems/postulates. requires skill but extremely open-ended. Karnaugh Maps. Maps diagrammatic technique using ‘Venn-like diagram’. easy for humans (pattern-matching skills). simplified standard forms. limited to not more than 6 variables. 02/03/14 Quick Review Questions (3) 4
- 5. Function Simplification Simplification techniques: Quine-McCluskey tabulation technique. technique tabulation technique based on certain ‘cancellation theorems’. simplified standard forms. tedious, repetitive step-by-step technique. boring to humans BUT suitable for computers. larger variables possible, but computationally intensive. 02/03/14 Quick Review Questions (3) 5
- 6. Algebraic Simplification Algebraic simplification aims to minimise (i) number of literals, and (ii) number of terms But sometimes conflicting. Let’s aim at reducing the number of literals. 02/03/14 Quick Review Questions (3) 6
- 7. Algebraic Simplification Example: (x+y).(x+y').(x'+z) (6 literals) (x+y).(x+y'). = (x.x+x.y'+x.y+y.y').(x'+z) (assoc.) x.y'+x.y y.y' = (x+x.(y'+y)+0).(x'+z) (idemp.,assoc., compl.) (y'+y) = (x+x.(1)+0).(x'+z) (complement) x.(1) = (x+x+0).(x'+z) (identity 1) x+x+0 = (x).(x'+z) (idemp, identity 0) = (x.x'+x.z) (assoc.) x.x' = (0+x.z) (complement) = x.z (identity 0) Number of literals reduced from 6 to 2. 02/03/14 Quick Review Questions (3) 7
- 8. Algebraic Simplification Find minimal SOP and POS expressions of f(x,y,z) = x'.y.(z + y'.x) + y'.z x'.y.(z+y'.x) + y'.z = x'.y.z + x'.y.y'.x + y'.z (distributivity) = x'.y.z + 0 + y'.z (complement, null element 0) = x'.y.z + y'.z (identity 0) = x'.z + y’.z (absorption) = (x' + y').z (distributivity) Minimal SOP of f = x'.z + y'.z (2 2-input AND gates and 1 2-input OR gate) Minimal POS of f = (x' + y').z (1 2-input OR gate and 1 2-input AND gate) 02/03/14 Quick Review Questions (3) 8
- 9. Algebraic Simplification Find minimal SOP expression of f(a,b,c,d) = a.b.c + a.b.d + a'.b.c' + c.d + b.d' a.b.c + a.b.d + a'.b.c' + c.d + b.d' = a.b.c + a.b + a'.b.c' + c.d + b.d' (absorption) = a.b.c + a.b + b.c' + c.d + b.d' (absorption) = a.b + b.c' + c.d + b.d' (absorption) = a.b + c.d + b.(c' + d') (distributivity) = a.b + c.d + b.(c.d)' (DeMorgan) = a.b + c.d + b (absorption) = b + c.d (absorption) Number of literals reduced form 13 to 3. 02/03/14 Quick Review Questions (3) 9
- 10. Algebraic Simplification Difficulty – needs good algebraic manipulation skills. Advantage – very open-ended (to your desired form!) 02/03/14 Quick Review Questions (3) 10
- 11. Half Adder Half-Adder is a circuit which adds two single bits (called X,Y) together, to produce a result of two bits (called C, S). A black-box representation of this circuit is: Truth table representation is: X Y 02/03/14 Half adder S C (X+Y) Quick Review Questions (3) X 0 0 1 1 Y 0 1 0 1 C 0 0 0 1 S 0 1 1 0 11
- 12. Half Adder X 0 0 1 1 In sum-of-minterms forms: C = X.Y S = X'.Y + X.Y' Y 0 1 0 1 C 0 0 0 1 S 0 1 1 0 Algebraic simplification could simplify S to: S = X'.Y + X.Y' = X⊕Y Giving: X Y S C 02/03/14 Quick Review Questions (3) 12
- 13. Introduction to K-maps Systematic method to obtain simplified sum-of-products (SOPs) Boolean expressions. Objective: Fewest possible terms/literals. Diagrammatic technique based on a special form of Venn diagram. Advantage: Easy with visual aid. Disadvantage: Limited to 5 or 6 variables. 02/03/14 Quick Review Questions (3) 13
- 14. Venn Diagrams Venn diagram to represent the space of minterms. Example of 2 variables (4 minterms): a'b' ab' ab a'b a b 02/03/14 Quick Review Questions (3) 14
- 15. Venn Diagrams Each set of minterms represents a Boolean function. Examples: { a.b, a.b' } a.b + a.b' = a.(b+b') = a { a‘.b, a.b } a‘.b + a.b = (a'+a).b = b { a.b } a.b { a.b, a.b', a‘.b } {} a.b + a.b' + a‘.b = a + b 0 { a‘.b',a.b,a.b',a‘.b } 1 02/03/14 a'b' a Quick Review Questions (3) ab' ab a'b b 15
- 16. 2-variable K-maps Karnaugh-map (K-map) is an abstract form of Venn diagram, organised as a matrix of squares, where each square represents a minterm adjacent squares always differ by just one literal (so that the unifying theorem may apply: a + a' = 1) 1 For 2-variable case (e.g.: variables a,b), the map can be drawn as: 02/03/14 Quick Review Questions (3) 16
- 17. 2-variable K-maps Alternative layouts of a 2-variable (a, b) K-map Alternative 1: b a'b ' a b OR m0 a'b ab' Alternative 2: ab a m1 m2 m3 a a'b' b ab OR m0 ab' a'b a b m2 m1 m3 Alternative 3: a 02/03/14 ab a'b ab' b a OR a'b' b m3 m1 m2 and others… m0 Quick Review Questions (3) 17
- 18. 2-variable K-maps Equivalent labeling: b equivalent to: a 0 1 1 0 0 1 a a b equivalent to: b 02/03/14 b a 0 1 Quick Review Questions (3) 18
- 19. 2-variable K-maps The K-map for a function is specified by putting a ‘1’ in the square corresponding to a minterm a ‘0’ otherwise For example: Carry and Sum of a half adder. b 0 a 0 0 1 C = ab 02/03/14 b 0 a 1 1 0 S = ab' + a'b Quick Review Questions (3) 19
- 20. 3-variable K-maps There are 8 minterms for 3 variables (a, b, c). Therefore, there are 8 cells in a 3-variable K-map. b b a bc 01 0 a 00 11 a'b'c' a'b'c a'bc a'bc' 1 ab'c' ab'c abc abc' a a 00 01 11 10 0 10 OR bc m0 m1 m3 m2 1 m4 m5 m7 m6 c c Above arrangement ensures that minterms of adjacent cells differ by only ONE literal. (Other arrangements which satisfy this criterion may also be used.) 02/03/14 Quick Review Questions (3) Note Gray code sequence 20
- 21. 3-variable K-maps There is wrap-around in the K-map: a'.b'.c' (m0) is adjacent to a'.b.c' (m2) a.b'.c' (m4) is adjacent to a.b.c' (m6) a bc 00 01 11 10 0 m0 m1 m3 m2 1 m4 m5 m7 m6 Each cell in a 3-variable K-map has 3 adjacent neighbours. In general, each cell in an n-variable K-map has n adjacent neighbours. For example, m0 has 3 adjacent neighbours: m1, m2 and m4. 02/03/14 Quick Review Questions (3) 21
- 22. Quick Review Questions (1) Textbook page 104. 5-1. The K-map of a 3-variable function F is shown below. What is the sum-of-minterms expression of F? b a bc 01 0 a 00 11 10 1 0 0 1 1 0 1 0 0 c 5-2. Draw the K-map for this function A: A(x, y, z) = x.y + y.z’ + x’.y’.z 02/03/14 Quick Review Questions (3) 22
- 23. 4-variable K-maps There are 16 cells in a 4-variable (w, x, y, z) K-map. y wx yz 01 11 10 00 m0 m1 m3 m2 01 w 00 m4 m5 m7 m6 11 10 x m12 m13 m15 m14 m8 m9 m11 m10 z 02/03/14 Quick Review Questions (3) 23
- 24. 4-variable K-maps There are 2 wrap-arounds: a horizontal wrap-around and a vertical wrap-around. Every cell thus has 4 neighbours. For example, the cell corresponding to minterm m0 has neighbours m1, m2, m4 and m8. wx y yz m0 m3 m2 m4 w m1 m5 m7 m6 m12 m13 m15 m14 m8 m9 m11 m10 x z 02/03/14 Quick Review Questions (3) 24
- 25. 5-variable K-maps Maps of more than 4 variables are more difficult to use because the geometry (hyper-cube configurations) for combining adjacent squares becomes more involved. For 5 variables, e.g. vwxyz, need 25 = 32 squares. 02/03/14 Quick Review Questions (3) 25
- 26. 5-variable K-maps Organised as two 4-variable K-maps: v' wx v y yz 01 11 m0 m1 m3 m2 01 m4 m5 m7 wx 10 00 w 00 m6 11 10 m13 m15 w m8 m9 m11 m10 z 01 11 10 m16 m17 m19 m18 01 m14 00 00 x m12 y yz m20 m21 m23 m22 11 10 x m28 m29 m31 m30 m24 m25 m27 m26 z Corresponding squares of each map are adjacent. Can visualise this as being one 4-variable map on TOP of the other 4-variable map. 02/03/14 Quick Review Questions (3) 26
- 27. Larger K-maps 6-variable K-map is pushing the limit of human “patternrecognition” capability. K-maps larger than 6 variables are practically unheard of! Normally, a 6-variable K-map is organised as four 4- variable K-maps, which are mirrored along two axes. 02/03/14 Quick Review Questions (3) 27
- 28. Larger K-maps w b ef a‘.b' 10 a‘.b 11 01 00 00 00 01 11 10 m0 m1 m3 m2 m18 m19 m17 m16 m4 cd m5 m7 m6 ef m22 m23 m21 m20 00 11 10 10 a 11 cd 01 ef 00 m30 m31 m29 m28 01 m11 m10 m26 m27 m25 m24 11 m40 m41 m43 m42 m58 m59 m57 m56 10 10 m44 m45 m47 m46 01 m12 m13 m15 m14 cd m62 m63 m61 m60 m36 m37 m39 m38 m54 m55 m53 m52 m32 m33 m35 m34 m50 m51 m49 m48 m8 00 m9 01 11 a.b' 10 10 11 01 a.b 00 11 01 cd ef 00 Try stretch your recognition capability by finding simplest sum-of-products expression for Σ m(6,8,14,18,23,25,27,29,41,45,57,61). 02/03/14 Quick Review Questions (3) 28
- 29. Simplification Using K-maps Based on the Unifying Theorem: Theorem A + A' = 1 In a K-map, each cell containing a ‘1’ corresponds to a minterm of a given function F. Each group of adjacent cells containing ‘1’ (group must have size in powers of twos: 1, 2, 4, 8, …) then corresponds to a twos simpler product term of F. Grouping 2 adjacent squares eliminates 1 variable, grouping 4 squares eliminates 2 variables, grouping 8 squares eliminates 3 variables, and so on. In general, grouping 2 n squares eliminates n variables. 02/03/14 Quick Review Questions (3) 29
- 30. Simplification Using K-maps Group as many squares as possible. The larger the group is, the fewer the number of literals in the resulting product term. Select as few groups as possible to cover all the squares (minterms) of the function. The fewer the groups, the fewer the number of product terms in the minimized function. 02/03/14 Quick Review Questions (3) 30
- 31. Simplification Using K-maps Example: F (w,x,y,z) = w’.x.y'.z' + w'.x.y'.z + w.x'.y.z' + w.x'.y.z + w.x.y.z' + w.x.y.z = Σ m(4, 5, 10, 11, 14, 15) y wx yz 00 01 1 11 10 1 00 01 w x 1 10 1 1 11 (cells with ‘0’ are not shown for clarity) 1 z 02/03/14 Quick Review Questions (3) 31
- 32. Simplification Using K-maps Each group of adjacent minterms (group size in powers of twos) corresponds to a possible product term of the given function. y yz wx 00 01 1 11 10 1 00 A 01 x 1 1 10 w 11 1 1 B z 02/03/14 Quick Review Questions (3) 32
- 33. Simplification Using K-maps There are 2 groups of minterms: A and B, where: A = = = w'.x.y'.z' + w‘.x.y'.z w'.x.y'.(z' + z) w'.x.y' B = = = = = w.x'.y.z' + w.x'.y.z + w.x.y.z' + w.x.y.z yz w.x'.y.(z' + z) + w.x.y.(z' + z) wx w.x'.y + w.x.y 00 w.(x'+x).y A 01 w.y y 00 01 1 11 10 1 x 1 1 10 w 11 1 1 B z 02/03/14 Quick Review Questions (3) 33
- 34. Simplification Using K-maps Each product term of a group, w'.x.y' and w.y, represents w.y the sum of minterms in that group. Boolean function is therefore the sum of product terms (SOP) which represent all groups of the minterms of the function. F(w,x,y,z) = A + B = w'.x.y' + w.y 02/03/14 Quick Review Questions (3) 34
- 35. Simplification Using K-maps Larger groups correspond to product terms of fewer literals. In the case of a 4-variable K-map: 1 cell = 4 literals, e.g.: w.x.y.z, w'.x.y'.z 2 cells = 3 literals, e.g.: w.x.y, w.y'.z' 4 cells = 2 literals, e.g.: w.x, x'.y 8 cells = 1 literal, e.g.: w, y', z 16 cells 02/03/14 = no literal, e.g.: 1 Quick Review Questions (3) 35
- 36. Simplification Using K-maps Other possible valid groupings of a 4-variable K-map include: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 02/03/14 1 1 1 Quick Review Questions (3) 1 1 36
- 37. Simplification Using K-maps Groups of minterms must be (1) rectangular, and (2) have size in powers of 2’s. Otherwise they are invalid groups. Some examples of invalid groups: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 02/03/14 Quick Review Questions (3) 1 1 37
- 38. Converting to Minterms Form The K-map of a function is easily drawn when the function is given in canonical sum-of-products, or sum-of-minterms form. What if the function is not in sum-of-minterms? Convert it to sum-of-products (SOP) form. Expand the SOP expression into sum-of-minterms expression, or fill in the K-map directly based on the SOP expression. 02/03/14 Quick Review Questions (3) 38
- 39. Converting to Minterms Form Example: f(A,B,C,D) = A(C+D)'(B'+D') + C(B+C'+A'D) = A(C'D')(B'+D') + BC + CC' + A'CD = AB'C'D' + AC'D' + BC + A'CD AB'C'D' + AC'D' + BC + A'CD = AB'C'D' + AC'D'(B+B') + BC + A'CD = AB'C'D' + ABC'D' + AB'C'D' + BC(A+A') + A'CD = AB'C'D' + ABC'D' + ABC + A'BC + A'CD = AB'C'D' + ABC'D' + ABC(D+D') + A'BC(D+D') + A'CD(B+B') = AB'C'D' + ABC'D' + ABCD + ABCD' + A'BCD + A'BCD' + A'B'CD 02/03/14 Quick Review Questions (3) A AB CD 00 01 11 1 00 10 1 01 C 11 10 1 1 1 1 D 1 B 39
- 40. Simplest SOP Expressions To find the simplest possible sum of products (SOP) expression from a K-map, you need to obtain: minimum number of literals per product term; and minimum number of product terms This is achieved in K-map using bigger groupings of minterms (prime implicants) where possible; implicants and no redundant groupings (look for essential prime implicants) implicants Implicant: a product term that could be used Implicant to cover minterms of the function. 02/03/14 Quick Review Questions (3) 40
- 41. Simplest SOP Expressions A prime implicant is a product term obtained by combining the maximum possible number of minterms from adjacent squares in the map. Use bigger groupings (prime implicants) where possible. 1 1 1 1 1 1 02/03/14 1 1 1 1 1 1 Quick Review Questions (3) 41
- 42. Simplest SOP Expressions No redundant groups: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Essential prime implicants An essential prime implicant is a prime implicant that includes at least one minterm that is not covered by any other prime implicant. 02/03/14 Quick Review Questions (3) 42
- 43. Quick Review Questions (2) Textbook page 104. 5-3. Identify the prime implicants and the essential prime implicants of the two K-maps below. b a bc CD 00 01 0 a 11 1 1 0 1 1 0 1 0 0 c A AB 10 00 00 01 1 11 1 1 1 01 1 10 C 11 10 1 1 1 1 1 D 1 B 02/03/14 Quick Review Questions (3) 43
- 44. Simplest SOP Expressions Algorithm 1 (non optimal): 1. Count the number of adjacencies for each minterm on the K-map. 2. Select an uncovered minterm with the fewest number of adjacencies. Make an arbitrary choice if more than one choice is possible. 3. Generate a prime implicant for this minterm and put it in the cover. If this minterm is covered by more than one prime implicant, select the one that covers the most uncovered minterms. 4. Repeat steps 2 and 3 until all the minterms have been covered. 02/03/14 Quick Review Questions (3) 44
- 45. Simplest SOP Expressions Algorithm 2 (non optimal): 1. Circle all prime implicants on the K-map. 2. Identify and select all essential prime implicants for the cover. 3. Select a minimum subset of the remaining prime implicants to complete the cover, that is, to cover those minterms not covered by the essential prime implicants. 02/03/14 Quick Review Questions (3) 45
- 46. Simplest SOP Expressions Example: f(A,B,C,D) = ∑ m(2,3,4,5,7,8,10,13,15) A AB CD 00 01 11 00 1 01 1 1 1 1 11 1 10 C 10 1 1 All prime implicants D 1 B 02/03/14 Quick Review Questions (3) 46
- 47. Simplest SOP Expressions CD AB A 00 01 00 1 1 11 1 10 C CD 1 A AB 10 1 01 11 00 00 D 1 1 1 1 1 1 C 1 11 1 00 01 11 00 1 01 1 D 1 B 10 1 1 Essential prime implicants 1 1 A AB 10 1 10 B Minimum cover 1 11 1 10 C 11 01 1 CD 01 1 1 D 1 B 02/03/14 Quick Review Questions (3) 47
- 48. Simplest SOP Expressions A AB CD A'BC' 00 01 11 00 1 01 1 1 1 AB'D' 1 11 1 10 C 10 1 1 D 1 BD B A'B'C f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C' 02/03/14 Quick Review Questions (3) 48
- 49. Quick Review Questions (3) Textbook page 104. 5-4. Find the simplified expression for G(A,B,C,D). A AB CD 00 01 11 10 1 00 01 C 1 1 1 1 1 11 1 1 10 5-5 to 5-7. 02/03/14 D B Quick Review Questions (3) 49
- 50. Getting POS Expressions Simplified POS expression can be obtained by grouping the maxterms (i.e. 0s) of given function. Example: Given F=∑m(0,1,2,3,5,7,8,9,10,11), we first draw K-map, then group the maxterms together: A AB CD 00 01 11 10 00 1 0 0 1 01 1 1 0 1 11 1 1 0 1 10 C the 1 0 0 1 D B 02/03/14 Quick Review Questions (3) 50
- 51. Getting POS Expressions CD A AB 11 1 0 0 1 1 1 0 1 11 1 1 0 1 10 1 0 0 CD 10 01 C 01 00 K-map of F 00 1 A AB B 11 10 0 1 1 0 01 C 01 00 D 00 0 0 1 0 11 0 0 1 0 10 0 1 1 0 D K-map of F' B This gives the SOP of F' to be: F' = B.D' + A.B To get POS of F, we have: F = (B.D' + A.B)' = (B.D')'.(A.B)' = (B'+D).(A'+B') 02/03/14 DeMorgan DeMorgan Quick Review Questions (3) 51
- 52. Don’t-care Conditions In certain problems, some outputs are not specified. These outputs can be either ‘1’ or ‘0’. They are called don’t-care conditions, denoted by X (or conditions sometimes, d). Example: An odd parity generator for BCD code which has 6 unused combinations. 02/03/14 Quick Review Questions (3) No. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 P 1 0 0 1 0 1 1 0 0 1 X X X X X X 52
- 53. Don’t-care Conditions Don’t-care conditions can be used to help simplify Boolean expression further in K-maps. They could be chosen to be either ‘1’ or ‘0’, depending on which gives the simpler expression. 02/03/14 Quick Review Questions (3) 53
- 54. Don’t-care Conditions For comparison: WITHOUT Don’t-cares: AB 00 00 P = A'.B'.C'.D’ + A'.B'.C.D + A'.B.C'.D + A'.B.C.D' + A.B'.C'.D C CD 01 11 1 01 10 1 1 1 11 A B 1 10 D WITH Don’t-cares: P = A'.B'.C'.D' + B'.C.D + B.C'.D + B.C.D' + A.D AB C CD 00 00 01 11 1 10 1 01 A 1 11 X X X X 10 1 X X 1 B D 02/03/14 Quick Review Questions (3) 54
- 55. Review – The Techniques Algebraic Simplification. requires skill but extremely open-ended. Karnaugh Maps. can obtain simplified standard forms. easy for humans (pattern-matching skills). limited to not more than 6 variables. Other computer-aided techniques such as QuineMcCluskey method (not covered in this course). 02/03/14 Quick Review Questions (3) 55
- 56. Review – K-maps Characteristics of K-map layouts: (i) each minterm in one square/cell (ii) adjacent/neighbouring minterms differ by only 1 literal (iii) n-literal minterm has n neighbours/adjacent cells Valid 2-, 3-, 4-variable K-maps b a'b ' a 02/03/14 a'b ab' b ab m0 OR Quick Review Questions (3) a m1 m2 m3 56
- 57. Review – K-maps b a b bc a 00 01 0 a 11 bc 10 a'b'c' a'b'c a'bc a'bc' 1 ab'c' ab'c abc abc' 00 11 10 0 m0 m1 m3 m2 1 a 01 m4 m5 m7 m6 c wx c y yz 00 11 10 m0 m1 m3 m2 m4 00 01 m5 m7 m6 x m12 m13 m15 m14 11 w 01 m8 m9 m11 m10 10 02/03/14 z Quick Review Questions (3) 57
- 58. Review – K-maps Groupings to select product-terms must be: (i) rectangular in shape (ii) in powers of twos (1, 2, 4, 8, etc.) (iii) always select largest possible groupings of minterms (i.e. prime implicants) (iv) eliminate redundant groupings Sum-of-products (SOP) form obtained by selecting groupings of minterms (corresponding to product terms). 02/03/14 Quick Review Questions (3) 58
- 59. Review – K-maps Product-of-sums (POS) form obtained by selecting groupings of maxterms (corresponding to sum terms) and by applying DeMorgan’s theorem. Don’t cares, marked by X (or d), can denote either 1 or 0. They could therefore be selected as 1 or 0 to further simplify expressions. 02/03/14 Quick Review Questions (3) 59
- 60. Examples Example #1: f(A,B,C,D) = ∑ m(2,3,4,5,7,8,10,13,15) A AB CD 00 01 11 00 1 01 1 1 1 1 11 1 10 C 10 1 1 Fill in the 1’s. D 1 B 02/03/14 Quick Review Questions (3) 60
- 61. Examples Example #1: f(A,B,C,D) = ∑ m(2,3,4,5,7,8,10,13,15) A AB CD 00 01 11 00 1 01 1 1 1 1 11 1 10 C 10 1 1 D These are all the prime implicants; but do we need them all? 1 B 02/03/14 Quick Review Questions (3) 61
- 62. Examples Example #1: f(A,B,C,D) = ∑ m(2,3,4,5,7,8,10,13,15) A AB CD 00 01 11 00 1 01 1 1 1 1 11 1 10 C 10 1 1 Essential prime implicants: D 1 B 02/03/14 Quick Review Questions (3) B.D A'.B.C' A.B'.D' 62
- 63. Examples Example #1: f(A,B,C,D) = ∑ m(2,3,4,5,7,8,10,13,15) A AB CD 00 01 11 00 1 01 1 1 1 1 11 1 10 C 10 1 1 Minimum cover. D 1 B EPIs: B.D, A'.B.C', A.B'.D' + A'.B'.C f(A,B,C,D) = B.D + A'.B.C' + A.B'.D' + A'.B'.C 02/03/14 Quick Review Questions (3) 63
- 64. CD AB A 00 01 00 11 1 01 1 1 11 1 10 C Examples CD 1 A AB 10 00 00 D 1 1 1 1 1 C 1 01 11 00 1 01 1 1 1 D 1 SUMMARY 10 1 11 1 1 1 Minimum cover D 1 B 02/03/14 1 Essential prime implicants 1 B 1 10 C 11 A AB 10 1 10 B 00 11 01 1 1 CD 01 f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C' Quick Review Questions (3) 64
- 65. Examples Example #2: f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D' A AB CD 00 00 01 11 1 10 1 01 C 1 1 11 1 1 1 Fill in the 1’s. 1 10 1 D B 02/03/14 Quick Review Questions (3) 65
- 66. Examples Example #2: f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D' A AB CD 00 00 01 11 1 10 1 01 C 1 1 11 1 1 1 Find all PIs: 1 10 1 D B A.D A.C B'.D' Are all ‘1’s covered by the PIs? Yes, so the answer is: f(A,B,C,D) = A.D + A.C + B'.D' 02/03/14 Quick Review Questions (3) 66
- 67. Examples Example #3 (with don’t cares): f(A,B,C,D) = ∑ m(2,8,10,15) + ∑ d(0,1,3,7) A AB CD 00 00 10 C X 1 10 X 11 11 X 01 01 1 X Fill in the 1’s and X’s. D 1 1 B 02/03/14 Quick Review Questions (3) 67
- 68. Examples Example #3 (with don’t cares): f(A,B,C,D) = ∑ m(2,8,10,15) + ∑ d(0,1,3,7) A AB CD 00 00 10 C X 1 Do we need to have an additional term A'.B' to cover the 2 remaining x’s? 10 X 11 11 X 01 01 1 X D 1 1 No, because all the 1’s (minterms) have been covered. B f(A,B,C,D) = B'.D' + B.C.D 02/03/14 Quick Review Questions (3) 68
- 69. Examples To find simplest POS expression for example #2: f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D' Draw the K-map of the complement of f, f '. AB CD 00 01 11 1 00 1 01 C 1 1 11 1 1 10 From K-map, A 10 f ' = A'.B + A'.D + B.C'.D' Using DeMorgan’s theorem, D f = (A'.B + A'.D + B.C'.D')' = (A+B').(A+D').(B'+C+D) 1 B 02/03/14 Quick Review Questions (3) 69
- 70. Examples • To find simplest POS expression for example #3: f(A,B,C,D) = ∑ m(2,8,10,15) + ∑ d(0,1,3,7) • Draw the K-map of the complement of f, f '. f '(A,B,C,D) = ∑ m(4,5,6,9,11,12,13,14) + ∑ d(0,1,3,7) A AB CD 01 11 00 X 1 X 1 1 11 X X 10 10 f ' = B.C' + B.D' + B'.D 1 1 1 1 B 02/03/14 From K-map, 1 01 C 00 D Using DeMorgan’s theorem, f = (B.C' + B.D' + B'.D)' = (B'+C).(B'+D).(B+D') Quick Review Questions (3) 70
- 71. End of file 02/03/14 Quick Review Questions (3) 71