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1 Gate-level Minimization Although truth tables representation of a function is unique, it can be expressed algebraically in different forms The procedure of simplifying Boolean expressions (in 2-4) is difficult since it lacks specific rules to predict the successive steps in the simplification process. Alternative: Karnaugh Map (K-map) Method. Straight forward procedure for minimizing Boolean Function Fact: Any function can be expressed as sum of minterms K-map method can be seen as a pictorial form of the truth table. m0m0 m1m1 m2m2 m3m3 x y y x Two-variable map

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2 x y y x Two-variable K-MAP x y y x x y y x

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3 x y y x The three squares can be determined from the intersection of variable x in the second row and variable y in the second column. x y y x Two-variable K-MAP

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4 Any two adjacent squares differ by only one variable. M5 is row 1 column 01. 101= xy’z=m5 Since adjacent squares differ by one variable (1 primed, 1 unprimed) From the postulates of Boolean algebra, the sum of two minterms in adjacent squares can be simplified to a simple AND For example m5+m7=xy’z+xyz=xz(y’+y)=xz Three-Variable K-Map

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5 Example 1 Three-Variable K-Map

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6 Example 2 Three-Variable K-Map Simplify: m0m0 m1m1 m3m3 m2m2 m4m4 m5m5 m7m7 m6m6

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7 Three-Variable K-Map Example 3 Simplify: m0m0 m1m1 m3m3 m2m2 m4m4 m5m5 m7m7 m6m6

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8 Three-Variable K-Map Example 3 Simplify: m0m0 m1m1 m3m3 m2m2 m4m4 m5m5 m7m7 m6m6

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9 Example 4 Three-Variable K-Map Given: (a) Express F in sum of minterms. (b) Find the minimal sum of products using K-Map (a)

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10 Three-Variable K-Map Example 4 (continued) m0m0 m1m1 m3m3 m2m2 m4m4 m5m5 m7m7 m6m6

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11 Three-variable K-Map: Observations One square represents one minterm a term of 3 literals Two adjacent squares a term of 2 literals Four adjacent squares a term of 1 literal Eight adjacent squares the function equals to 1

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12 Four-Variable K-Map

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13 Four-Variable K-Map Example 5 Simplify F(w,x,y,z) = (0,1,2,4,5,6,8,9,12,13,14) 1

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14 Four-Variable K-Map Example 6 Simplify F(A,B,C,D) = Represented by 0001 or 0000

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16 Need to ensure that all Minterms of function are covered But avoid any redundant terms whose minterms are already covered Prime Implicant is product Term obtained by combining maximum possible number of adjacent squares If a minterm in a square is covered by only prime implicant then ESSENTIAL PRIME IMPLICANT Prime Implicants Essential prime implicant BD and B’D’ Non Essential prime implicant CD, B’C, AD and AB’

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17 Four-variable K-Map: Observations One square represents one minterm a term of 4 literals Two adjacent squares a term of 3 literals Four adjacent squares a term of 2 literal Eight adjacent squares a term of 1 literal sixteen adjacent squares the function equals to 1

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19 Simplify the following Boolean function in: (a) sum of products (b) product of sums Combining the one’s: Combining the zero’s: Taking the the complement: SUM of PRODUCT and PRODUCT OF SUM (a) (b)

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20 SUM OF PRODUCT (SOP) PRODUCT OF SUM (POS) SOP and POS gate implementation

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21 Draw the logic diagram for the following function: F = (a.b)+(b.c) a b c F Implementation of Boolean Functions

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22 Implement a circuit –2 Level –More than two level –SOP –POS Implement a circuit using OR and Inverter Gates only Implement a circuit using AND and Inverter Gates only Implement a circuit using NAND Gates only Implement a circuit using NOR Gates only

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23 NAND IMPLEMENTATION

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25 TWO LEVEL IMPLEMENT- ATION F=AB+CD F=(A’B’)’+(C’D’)’ F=[(AB)’.(CD)’]’=AB+CD

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26 F(X,Y,Z)=(1,2,3,4,5,7) SUM OF PRODUCT

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27 COVERT AND TO NAND WITH AND INVER. CONVERT OR TO NAND WITH INVERT OR. SINGLE BUBBLE WITH INVERTER CHAPTER 4

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42 SIMPLIFICATION WITH TABULATION METHOD DO IT ON BOARD

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