Speaker: Fuw-Yi Yang 楊伏夷 伏夷非征番, 道德經 察政章(Chapter 58) 伏者潛藏也 數位系統  Digital Systems  Department of Computer Science and Information Engineering, Chaoyang University of Technology 朝陽科技大學資工系 Speaker: Fuw-Yi Yang 楊伏夷 伏夷非征番, 道德經 察政章(Chapter 58) 伏者潛藏也 道紀章(Chapter 14) 道無形象, 視之不可見者曰夷 Fuw-Yi Yang

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.1 Demonstrate the validity of the following identities by means of truth tables: (a) DeMorgan’s theorem for three variables: (x + y + z) = x y z and (xyz) = (x+ y+ z) (b) The distributive law: x(y + z) = xy + xz and x + yz = (x + y)(x + z) (c) The associative law: x + (y + z) = (x + y) + z and x(yz) = (xy)z Fuw-Yi Yang 2

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.1 Solution: (x + y + z) = x y z x y z x+y+z (x+y+z) xyz 1 Fuw-Yi Yang 3

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.2 Simplify the following Boolean expressions to a minimum number of literals: (a) xy + xy (b) (x + y) (x + y) (c) xyz + xyz +yz (d) (A + B)(A + B) (e) (a + b + c)(ab + c) (f) abc + abc + abc + abc Fuw-Yi Yang 4

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.2 Solution: (a) xy + xy = x(y + y) = x; The distributive law x(y + z) = xy + xz (b) (x + y)(x + y) = x + (yy) = x; The distributive law x + yz = (x + y)(x + z) (c) xyz + xyz +yz = yz(x + x) +yz = yz + yz = z (c) xyz + xyz +yz = yz(x + 1) + xyz = yz + xyz = y(z + xz ) = y(z + x)(z + z ) =y(z + x) (d) (A + B)(A + B) = (AB)(AB) = 0 (e) (a + b + c)(ab + c) = abc + ac + bc (f) abc + abc + abc + abc = c(ab + ab + ab + ab) = c Fuw-Yi Yang 5

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.14 Implement the Boolean function F = xy + xz + xy (a) With AND, OR, and inverter gates (b) With OR and inverter gates (c) With AND and inverter gates (d) With NAND and inverter gates (e) With NOR and inverter gates Fuw-Yi Yang 6

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.14 Solution: With AND, OR, and inverter gates F = xy + xz + xy = { (x)(y) + (x)(z) + (x)(y) } Two inverters (NOT gates), cost 2 * 2 = 4 Three 2-input AND gates, cost 3 * 3 = 9 One 3-input OR gates, cost 1 * 4 = 4 Total cost: 17 Fuw-Yi Yang 7

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.14 Solution: (b) With OR and inverter gates F = xy + xz + xy = (x + y) + (x + (z)) + ((x) + (y)) Six inverters (NOT gates), cost 6 * 2 = 12 Three 2-input OR gates, cost 3 * 3 = 9 One 3-input OR gates, cost 1 * 4 = 4 Total cost: 25 Fuw-Yi Yang 8

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.14 Solution: (c) With AND and inverter gates F = xy + xz + xy = ((x)(y)) + (x)z + xy = { [(x)(y)] [(x)z] [xy]} Six inverters (NOT gates), cost 6 * 2 = 12 Three 2-input AND gates, cost 3 * 3 = 9 One 3-input AND gates, cost 1 * 4 = 4 Total cost: 25 Fuw-Yi Yang 9

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.14 Solution: (d) With NAND and inverter gates F = xy + xz + xy Sum of Product = 2-levels NAND = {(xy + xz + xy)} = {(xy) (xz) (xy)} = { [(x)(y)] [(x)z] (xy)} Two inverters (NOT gates), cost 2 * 2 = 4 Three 2-input NAND gates, cost 3 * 3 = 9 One 3-input NAND gates, cost 1 * 4 = 4 Total cost: 17 Fuw-Yi Yang 10

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.14 Solution: (e) With NOR and inverter gates F = xy + xz + xy = {(xy + xz + xy)} = { (x + y) (x +z) (x + y) } Product of Sum = 2-levels NOR = { (x + y) + [x +(z)] + [(x) + (y)] } Four inverters (NOT gates), cost 4 * 2 = 8 Three 2-input NOR gates, cost 3 * 3 = 9 One 3-input NOR gates, cost 1 * 4 = 4 Total cost: 21 Fuw-Yi Yang 11

Text Book: Digital Design 5th ed. Chapter 2 Problems 2.14 Solution: (e) With NOR and inverter gates F = xy + xz + xy = (0, 1, 3, 6, 7) F = (2, 4, 5) = xyz + xyz + xyz = xyz + xy F = (x+ y + z)(x + y) Product of Sum = 2-levels NOR Two inverters (NOT gates), cost 2 * 2 = 4 Two 2-input NOR gates, cost 2 * 3 = 6 One 3-input NOR gates, cost 1 * 4 = 4 Total cost: 14 Fuw-Yi Yang 12

Text Book: Digital Design 5th ed. Chapter 2 Problems Mid-Term Assume that both the normal and complement inputs are available. Implement the Boolean function F = xyz + xy (a) With AND, OR, and inverter gates (b) With NAND gates (c) With NOR gates (d) Compare the cost of implementations (a), (b), and (c). Fuw-Yi Yang 13

Text Book: Digital Design 5th ed. Chapter 2 Problems F = xyz + xy With AND, OR, and inverter gates (b) With NAND gates Sol (a), (b) (a) AND gates: One 3-input and one 2-input OR gate: one 2-input (b) NAND gates: One 3-input and two 2-input Fuw-Yi Yang 14

Text Book: Digital Design 5th ed. Chapter 2 Problems F = xyz + xy (c) With NOR gates Sol.1 (c) F = xyz + xy = xyz + xyz + xyz = (2, 4, 5) F = (0, 1, 3, 6, 7) = xy + xy + xz or = xy + xy + yz F = (x + y)(x + z)(x + y) or = (x + y)(y + z)(x + y) Sol.2 (c) F = xyz + xy = ((x + y + z)(x + y)) = (xy + xy + xz + yz) = (x + y)(x + y)(x + z)(y + z) Fuw-Yi Yang 15