1. 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

2. 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
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3. 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
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4. 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
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5. 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
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6. 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
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7. 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
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8. 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
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9. 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
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10. 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
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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
= {(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
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12. 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
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13. 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).
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14. 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
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15. 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)
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