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CS1104: Computer Organisation http://www.comp.nus.edu.sg/~cs1104 Lecture 5: Karnaugh Maps Lecture 5: Karnaugh Maps 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 CS1104-5 Lecture 5: Karnaugh Maps 2 Lecture 5: Karnaugh Maps Simplification using K-maps Converting to Minterms Form Simplest SOP Expressions Getting POS Expressions Don‟t-care Conditions Review Examples CS1104-5 Lecture 5: Karnaugh Maps 3 Function Simplification (1/2) Why simplify? Simpler expression uses less logic gates. Thus: cheaper, less power, faster (sometimes). Simplification techniques: Algebraic Simplification. simplify symbolically using theorems/postulates. requires skill but extremely open-ended. Karnaugh Maps. diagrammatic technique using „Venn-like diagram‟. easy for humans (pattern-matching skills). simplified standard forms. limited to not more than 6 variables. CS1104-5 Function Simplification 4 Function Simplification (2/2) Simplification techniques: Quine-McCluskey tabulation 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. CS1104-5 Function Simplification 5 Algebraic Simplification (1/5) 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. CS1104-5 Algebraic Simplification 6 Algebraic Simplification (2/5) Example: (x+y).(x+y').(x'+z) (6 literals) = (x.x+x.y'+x.y+y.y').(x'+z) (assoc.) = (x+x.(y'+y)+0).(x'+z) (idemp.,assoc., compl.) = (x+x.(1)+0).(x'+z) (complement) = (x+x+0).(x'+z) (identity 1) = (x).(x'+z) (idemp, identity 0) = (x.x'+x.z) (assoc.) = (0+x.z) (complement) = x.z (identity 0) Number of literals reduced from 6 to 2. CS1104-5 Algebraic Simplification 7 Algebraic Simplification (3/5) 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) CS1104-5 Algebraic Simplification 8 Algebraic Simplification (4/5) 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. CS1104-5 Algebraic Simplification 9 Algebraic Simplification (5/5) Difficulty – needs good algebraic manipulation skills. Advantage – very open-ended (to your desired form!) CS1104-5 Algebraic Simplification 10 Half Adder (1/2) 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 Half S X Y C S adder (X+Y) 0 0 0 0 Y C 0 1 0 1 1 0 0 1 1 1 1 0 CS1104-5 Half Adder 11 Half Adder (2/2) X Y C S In sum-of-minterms forms: 0 0 0 0 C = X.Y 0 1 0 1 1 0 0 1 S = X'.Y + X.Y' 1 1 1 0 Algebraic simplification could simplify S to: S = X'.Y + X.Y' = XY X Giving: Y S C CS1104-5 Half Adder 12 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. CS1104-5 Introduction to K-maps 13 Venn Diagrams (1/2) Venn diagram to represent the space of minterms. Example of 2 variables (4 minterms): a'.b' a.b' a.b a'.b a b CS1104-5 Venn Diagrams 14 Venn Diagrams (2/2) 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 = a + b {} 0 a.b' a.b a'.b a { a'.b',a.b,a.b',a'.b } 1 b CS1104-5 Venn Diagrams 15 2-variable K-maps (1/4) 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) For 2-variable case (e.g.: variables a,b), the map can be drawn as: CS1104-5 2-variable K-maps 16 2-variable K-maps (2/4) Alternative layouts of a 2-variable (a, b) K-map Alternative 1: Alternative 2: a b b a OR OR a'b' a'b m0 m1 a'b' ab' m0 m2 a ab a m3 b b m1 m3 ab' m2 a'b ab Alternative 3: a a OR b ab a'b b m3 m1 and others… ab' a'b' m2 m0 CS1104-5 2-variable K-maps 17 2-variable K-maps (3/4) Equivalent labeling: b b a 0 1 0 equivalent to: a 1 a a b 1 0 equivalent to: 0 b 1 CS1104-5 2-variable K-maps 18 2-variable K-maps (4/4) 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 b 0 0 0 1 a 0 1 a 1 0 C = a.b S = a.b' + a'.b CS1104-5 2-variable K-maps 19 3-variable K-maps (1/2) There are 8 minterms for 3 variables (a, b, c). Therefore, there are 8 cells in a 3-variable K-map. b b bc bc a a 00 01 11 10 00 01 11 10 a'b'c' a'b'c a'bc a'bc' 0 m0 m1 m3 m2 0 OR a m4 m5 m7 m6 a ab'c' ab'c abc abc' 1 1 c c Above arrangement ensures that minterms Note Gray code sequence of adjacent cells differ by only ONE literal. (Other arrangements which satisfy this criterion may also be used.) CS1104-5 3-variable K-maps 20 3-variable K-maps (2/2) 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) bc a 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. CS1104-5 3-variable K-maps 21 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 bc a 00 01 11 10 0 1 0 0 1 a 0 1 0 0 1 c 5-2. Draw the K-map for this function A: A(x, y, z) = x.y + y.z' + x'.y'.z CS1104-5 Quick Review Questions (1) 22 4-variable K-maps (1/2) There are 16 cells in a 4-variable (w, x, y, z) K-map. y yz wx 00 01 11 10 00 m0 m1 m3 m2 01 m4 m5 m7 m6 x m12 m13 m15 m14 11 w m8 m9 m11 m10 10 z CS1104-5 4-variable K-maps 23 4-variable K-maps (2/2) 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. yz y wx m0 m1 m3 m2 m4 m5 m7 m6 x m12 m13 m15 m14 w m8 m9 m11 m10 z CS1104-5 4-variable K-maps 24 5-variable K-maps (1/2) 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. CS1104-5 5-variable K-maps 25 5-variable K-maps (2/2) Organised as two 4-variable K-maps: v' v y y yz yz wx 00 01 11 10 wx 00 01 11 10 00 m0 m1 m3 m2 00 m16 m17 m19 m18 01 m4 m5 m7 m6 01 m20 m21 m23 m22 x x m12 m13 m15 m14 m28 m29 m31 m30 11 11 w w m8 m9 m11 m10 m24 m25 m27 m26 10 10 z 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. CS1104-5 5-variable K-maps 26 Larger K-maps (1/2) 6-variable K-map is pushing the limit of human “pattern-recognition” 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. CS1104-5 Larger K-maps 27 Larger K-maps (2/2) b ef a'.b' a'.b ef cd 00 01 11 10 10 11 01 00 cd 00 m0 m1 m3 m2 m18 m19 m17 m16 00 01 m4 m5 m7 m6 m22 m23 m21 m20 01 11 m12 m13 m15 m14 m30 m31 m29 m28 11 10 m8 m9 m11 m10 m26 m27 m25 m24 10 10 m40 m41 m43 m42 m58 m59 m57 m56 10 11 m44 m45 m47 m46 m62 m63 m61 m60 11 a 01 m36 m37 m39 m38 m54 m55 m53 m52 01 00 m32 m33 m35 m34 m50 m51 m49 m48 00 cd 00 01 11 10 10 11 01 00 cd ef ef a.b' a.b Try stretch your recognition capability by finding simplest sum-of-products expression for S m(6,8,14,18,23,25,27,29,41,45,57,61). CS1104-5 Larger K-maps 28 Simplification Using K-maps (1/9) Based on the Unifying 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 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 2n squares eliminates n variables. CS1104-5 Simplification Using K-maps 29 Simplification Using K-maps (2/9) 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. CS1104-5 Simplification Using K-maps 30 Simplification Using K-maps (3/9) 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 = S m(4, 5, 10, 11, 14, 15) y yz wx 00 01 11 10 00 01 1 1 x (cells with „0‟ are not 1 1 w 11 shown for clarity) 1 1 10 z CS1104-5 Simplification Using K-maps 31 Simplification Using K-maps (4/9) 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 11 10 00 A 01 1 1 x 11 1 1 w 10 1 1 B z CS1104-5 Simplification Using K-maps 32 Simplification Using K-maps (5/9) 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 y = w.x'.y.(z' + z) + w.x.y.(z' + z) yz wx 00 01 11 10 = w.x'.y + w.x.y 00 = w.(x'+x).y A = w.y 01 1 1 x 11 1 1 w 10 1 1 B z CS1104-5 Simplification Using K-maps 33 Simplification Using K-maps (6/9) Each product term of a group, w'.x.y' and w.y, represents 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 CS1104-5 Simplification Using K-maps 34 Simplification Using K-maps (7/9) 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 = no literal, e.g.: 1 CS1104-5 Simplification Using K-maps 35 Simplification Using K-maps (8/9) 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 1 1 1 1 1 P P P CS1104-5 Simplification Using K-maps 36 Simplification Using K-maps (9/9) 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 1 1 O O CS1104-5 Simplification Using K-maps 37 Converting to Minterms Form (1/2) 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. CS1104-5 Converting to Minterms Form 38 Converting to Minterms Form (2/2) Example: f(A,B,C,D) = A.(C+D)'.(B'+D') + C.(B+C'+A'.D) = A.(C'.D').(B'+D') + B.C + C.C' + A'.C.D = A.B'.C'.D' + A.C'.D' + B.C + A'.C.D A A.B'.C'.D' + A.C'.D' + B.C + A'.C.D AB CD 00 01 11 10 = A.B'.C'.D' + A.C'.D'.(B+B') + B.C + A'.C.D 00 1 1 = A.B'.C'.D' + A.B.C'.D' + A.B'.C'.D' + B.C.(A+A') + A'.C.D 01 D = A.B'.C'.D' + A.B.C'.D' + A.B.C + A'.B.C + 11 1 1 1 A'.C.D C 10 1 1 = A.B'.C'.D' + A.B.C'.D' + A.B.C.(D+D') + A'.B.C.(D+D') + A'.C.D.(B+B') B = A.B'.C'.D' + A.B.C'.D' + A.B.C.D + A.B.C.D' + A'.B.C.D + A'.B.C.D' + A'.B„.C.D CS1104-5 Converting to Minterms Form 39 Simplest SOP Expressions (1/3) 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; and no redundant groupings (look for essential prime implicants) Implicant: a product term that could be used to cover minterms of the function. CS1104-5 Simplest SOP Expressions 40 Simplest SOP Expressions (2/3) 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 1 1 1 O 1 1 1 P CS1104-5 Simplest SOP Expressions 41 Simplest SOP Expressions (3/3) No redundant groups: 1 1 1 1 1 1 1 1 O 1 1 1 1 P 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. CS1104-5 Simplest SOP Expressions 42 Quick Review Questions (2) Textbook page 104. 5-3. Identify the prime implicants and the essential prime implicants of the two K-maps below. A b AB bc CD 00 01 11 10 a 00 01 11 10 00 1 1 1 0 1 1 0 1 01 1 1 a D 1 0 1 0 0 11 1 1 1 C c 10 1 1 1 B CS1104-5 Quick Review Questions (2) 43 Simplest SOP Expressions (1/5) 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. CS1104-5 Simplest SOP Expressions 44 Simplest SOP Expressions (2/5) 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. CS1104-5 Simplest SOP Expressions 45 Simplest SOP Expressions (3/5) Example: f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15) A AB CD 00 01 11 10 00 1 1 All prime implicants 01 1 1 D 11 1 1 1 C 10 1 1 B CS1104-5 Simplest SOP Expressions 46 Simplest SOP Expressions (4/5) A AB A CD 00 01 11 10 AB CD 00 01 11 10 00 1 1 01 1 1 00 1 1 Essential prime D C 11 1 1 1 01 1 1 implicants 10 1 1 D 11 1 1 1 B C 10 1 1 B A AB CD 00 01 11 10 00 1 1 Minimum cover 01 1 1 D 11 1 1 1 C 10 1 1 B CS1104-5 Simplest SOP Expressions 47 Simplest SOP Expressions (5/5) A AB CD 00 01 11 10 A'BC' 00 1 1 AB'D' 01 1 1 D 11 1 1 1 C 10 1 1 B.D B A'B'C f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C' CS1104-5 Simplest SOP Expressions 48 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 00 1 01 1 1 1 D 11 1 1 1 C 10 1 B 5-5 to 5-7. CS1104-5 Quick Review Questions (3) 49 Getting POS Expressions (1/2) 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 the K-map, then group the maxterms together: A AB CD 00 01 11 10 00 1 0 0 1 01 1 1 0 1 D 11 1 1 0 1 C 10 1 0 0 1 B CS1104-5 Getting POS Expressions 50 Getting POS Expressions (2/2) A A AB AB CD 00 01 11 10 CD 00 01 11 10 K-map 00 1 0 0 1 00 0 1 1 0 K-map of F 01 1 1 0 1 01 0 0 1 0 of F' D D 11 1 1 0 1 11 0 0 1 0 C C 10 1 0 0 1 10 0 1 1 0 B 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)' DeMorgan = (B'+D).(A'+B') DeMorgan CS1104-5 Getting POS Expressions 51 Don’t-care Conditions (1/3) In certain problems, some No. A B C D P 0 0 0 0 0 1 outputs are not specified. 1 0 0 0 1 0 These outputs can be either „1‟ or 2 3 0 0 0 0 1 1 0 1 0 1 „0‟. 4 0 1 0 0 0 5 0 1 0 1 1 They are called don‟t-care 6 0 1 1 0 1 conditions, denoted by X (or 7 0 1 1 1 0 8 1 0 0 0 0 sometimes, d). 9 1 0 0 1 1 Example: An odd parity generator 10 11 1 1 0 0 1 1 0 1 X X for BCD code which has 6 12 1 1 0 0 X unused combinations. 13 1 1 0 1 X 14 1 1 1 0 X 15 1 1 1 1 X CS1104-5 Don‟t-care Conditions 52 Don’t-care Conditions (2/3) 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. We usually use the notation Sd to denote the set of don‟t-care minterms. For example, the function P in the odd-parity generator for BCD can be written as: P = Sm(0, 3, 5, 6, 9) + Sd(10, 11, 12, 13, 14, 15) CS1104-5 Don‟t-care Conditions 53 Don’t-care Conditions (3/3) For comparison: CD C WITHOUT don‟t-cares: AB 00 01 11 10 00 P = A'.B'.C'.D‟ + A'.B'.C.D + A'.B.C'.D 1 1 + A'.B.C.D' + A.B'.C'.D 01 1 1 B 11 A 10 1 WITH don‟t-cares: D P = A'.B'.C'.D' + B'.C.D + B.C'.D CD C + B.C.D' + A.D AB 00 01 11 10 00 1 1 01 1 1 B 11 X X X X A 10 1 X X D CS1104-5 Don‟t-care Conditions 54 Review – The Techniques Algebraic Simplification. requires skills 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 Quine- McCluskey method (not covered in this course). CS1104-5 Review 55 Review – K-maps (1/4) 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 b a'b' a'b m0 m1 OR a ab' ab a m2 m3 CS1104-5 Review 56 Review – K-maps (2/4) b b bc bc a a 00 01 11 10 00 01 11 10 0 a'b'c' a'b'c a'bc a'bc' 0 m0 m1 m3 m2 a ab'c' ab'c abc abc' a m4 m5 m7 m6 1 1 c c y yz wx 00 01 11 10 00 m0 m1 m3 m2 01 m4 m5 m7 m6 x 11 m12 m13 m15 m14 w 10 m8 m9 m11 m10 z CS1104-5 Review 57 Review – K-maps (3/4) 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). CS1104-5 Review 58 Review – K-maps (4/4) 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. CS1104-5 Review 59 Examples (1/11) Example #1: f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15) A AB CD 00 01 11 10 00 1 1 Fill in the 1‟s. 01 1 1 D 11 1 1 1 C 10 1 1 B CS1104-5 Examples 60 Examples (2/11) Example #1: f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15) A AB CD 00 01 11 10 These are all the 00 1 1 prime implicants; but 01 1 1 do we need them D 11 1 1 1 all? C 10 1 1 B CS1104-5 Examples 61 Examples (3/11) Example #1: f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15) A AB CD 00 01 11 10 00 1 1 Essential prime implicants: 01 1 1 D 11 1 1 1 B.D C 10 1 1 A'.B.C' B A.B'.D' CS1104-5 Examples 62 Examples (4/11) Example #1: f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15) A AB CD 00 01 11 10 00 1 1 Minimum cover. 01 1 1 D EPIs: B.D, A'.B.C', A.B'.D' 11 1 1 1 C 1 1 + 10 B A'.B'.C f(A,B,C,D) = B.D + A'.B.C' + A.B'.D' + A'.B'.C CS1104-5 Examples 63 Examples (5/11) A AB A CD 00 01 11 10 AB CD 00 01 11 10 00 1 1 01 1 1 00 1 1 Essential prime D C 11 1 1 1 01 1 1 implicants 10 1 1 D 11 1 1 1 B C 10 1 1 SUMMARY A B AB CD 00 01 11 10 00 1 1 Minimum cover 01 1 1 D 11 1 1 1 C 10 1 1 f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C' B CS1104-5 Examples 64 Examples (6/11) Example #2: f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D' A AB CD 00 01 11 10 00 1 1 Fill in the 1‟s. 01 1 1 D 11 1 1 C 10 1 1 1 B CS1104-5 Examples 65 Examples (7/11) Example #2: f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D' A AB CD 00 01 11 10 00 1 1 Find all PIs: 01 1 1 D A.D 11 1 1 C A.C 10 1 1 1 B'.D' B A.B' A.D, A.C and B'.D' are EPIs, and they cover all the minterms. So the answer is: f(A,B,C,D) = A.D + A.C + B'.D' CS1104-5 Examples 66 Examples (8/11) 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 01 11 10 00 X 1 Fill in the 1‟s and X‟s. 01 X D 11 X X 1 C 10 1 1 B CS1104-5 Examples 67 Examples (9/11) 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 01 11 10 Do we need to have an 00 X 1 additional term A'.B' to 01 X cover the 2 remaining x‟s? D 11 X X 1 No, because all the 1‟s C 1 1 (minterms) have been 10 covered. B f(A,B,C,D) = B'.D' + B.C.D CS1104-5 Examples 68 Examples (10/11) 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 '. A From K-map, AB CD 00 01 11 10 f ' = A'.B + A'.D + B.C'.D' 00 1 1 Using DeMorgan‟s theorem, 01 1 1 D 11 1 1 f = (A'.B + A'.D + B.C'.D')' C 10 1 = (A+B').(A+D').(B'+C+D) B CS1104-5 Examples 69 Examples (11/11) 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 00 01 11 10 From K-map, 00 X 1 1 f ' = B.C' + B.D' + B'.D 01 X 1 1 1 Using DeMorgan‟s theorem, D 11 X X 1 C f = (B.C' + B.D' + B'.D)' 10 1 1 = (B'+C).(B'+D).(B+D') B CS1104-5 Examples 70 End of file