Arquitectura de computadoras

Problem Solutions to Problems Marked With a * in Logic Computer Design Fundamentals, Ed. 2

CHAPTER 1
© 2000 by Prentice-Hall, Inc.

1-1.
Decimal, Binary, Octal and Hexadecimal Numbers from (16)10 to (31)10
Dec Bin Oct Hex 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 0000 1 0001 1 0010 1 0011 1 0100 1 0101 1 0110 1 0111 1 1000 1 1001 1 1010 1 1011 1 1100 1 1101 1 1110 1 1111 20 21 2223 24 25 26 27 30 31 32 33 34 35 36 37 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F

1-4.
( 1101001 ) 2 = 2 6 + 2 5 + 2 3 + 2 0 = 105 ( 10001011.011 )2 = 2 7 + 2 3 + 2 1 + 2 0 + 2 –2 + 2 – 3 = 139.375 ( 10011010 ) 2 = 2 7 + 2 4 + 2 3 + 2 1 = 154

1-7.
Decimal 369.3125 189.625 214.625 62407.625 Binary 101110001.0101 10111101.101 11010110.101 1111001111000111.101 Octal 561.24 275.5 326.5171707.5 Hexadecimal 171.5 BD.A D6.A F3C7.A

1-9.
a) 7562/8 945/8 118/8 14/8 1/8 0.45 × 8 0.60 × 8 0.80 × 8 0.20x8 (7562.45)10 b) c) (1938.257)10 (175.175)10 = = = = = = = = = = = = (16612.3463)8 (792.41CA)16 (10101111.001011)2 945 + 2/8 118 +1/8 14 + 6/8 1 + 6/8 1/8 3.6 4.8 6.4 3.2                   2 1 6 6 1 3 4 6 3

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Problem Solutions – Chapter 1

1-11.
a) b) c) (673.6)8 (E7C.B)16(310.2)4 = = = = = = (110 111 011.110)2 (1BB.C)16 (1110 0111 1100.1011)2 (7174.54)8 (11 01 00.10)2 (64.4)8

1-15.
a) (BEE)r = (2699)10 11 × r 2 + 14 × r 1 + 14 × r 0 = 2699 11 × r 2 + 14 × r – 2685 = 0 By the quadratic equation: r = 15 or r ≈ –16.27 ANSWER: r = 15 b) (365)r = (194)10 3 × r 2 + 6 × r 1 + 5 × r 0 = 194 3 × r 2 + 6 × r – 189 = 0 By the quadratic equation: r = -9 or 7 ANSWER: r = 71-17.
(694)10 (835)10 1 0110 +1000 1111 +0110 0001 0101 1001 +0011 1100 +0110 1 0010 0100 +0101 1001 +0000 1001 = = (0110 1001 0100)BCD (1000 0011 0101)BCD

1-20.
a) (0100 1000 0110 0111)BCD b) (0011 0111 1000.0111 0101)BCD = = = = (4867)10 (1001100000011)2 (378.75)10 (101111010.11)2

1-23.
a) b) c) (101101101)2 (0011 0110 0101)BCD 0011 0011 0011 0110 0011 0101ASCII

1-25.
BCD Digitswith Odd and Even Parity
Odd Even 0 1 0000 0 0000 1 0 0001 1 0001 2 0 0010 1 0010 3 1 0011 0 0011 4 0 0100 1 0100 5 1 0101 0 0101 6 1 0110 0 0110 7 0 0111 1 0111 8 0 1000 1 1000 9 1 1001 0 1001

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Problem Solutions to Problems Marked With a * in Logic Computer Design Fundamentals, Ed. 2

CHAPTER 2
© 2000 by Prentice-Hall, Inc.

2-1.
a) XYZ = X + Y + Z Verification of DeMorgan’s TheoremX 0 0 0 0 1 1 1 1 Y 0 0 1 1 0 0 1 1 Z 0 1 0 1 0 1 0 1 XYZ 0 0 0 0 0 0 0 1 XYZ 1 1 1 1 1 1 1 0 X+Y+Z 1 1 1 1 1 1 1 0

b)

X + YZ = ( X + Y ) ⋅ ( X + Z )

The Second Distributive Law
X 0 0 0 0 1 1 1 1 Y 0 0 1 1 0 0 1 1 Z 0 1 0 1 0 1 0 1 YZ 0 0 0 1 0 0 0 1 X+YZ 0 0 0 1 1 1 1 1 X+Y 0 0 1 1 1 1 1 1 X+Z 0 1 0 1 1 1 1 1 (X+Y)(X+Z) 0 0 0 1 1 1 1 1

c)
X 0 0 0 0 1 1 1 1 Y 0 0 1 1 0 0 1 1

XY+ YZ + XZ = XY + YZ + XZ
Z 0 1 0 1 0 1 0 1 XY 0 0 1 1 0 0 0 0 YZ 0 1 0 0 0 1 0 0 XZ XY+YZ+XZ XY 0 0 0 0 0 0 1 0 1 0 1 1 1 1 1 1 0 0 0 0 1 1 0 0 YZ 0 0 1 0 0 0 1 0 XZ XY+YZ+XZ 0 0 1 0 1 0 0 0 0 1 1 1 1 1 1 0

2-2.
a) X Y + XY + XY = (XY+ X Y ) + (X Y + XY) = X(Y + Y) + Y(X + X) + =X+Y = X+Y

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Problem Solutions – Chapter 2 b) A B+ B C + AB + B C = 1 = (A B+ AB) + (B C + B C) = B(A + A) +B(C + C) =B+B =1 c) Y+XZ+XY =Y+XY+XZ = (Y + X)(Y + Y) + X Z =Y+X+XZ = Y + (X + X)(X + Z) =X+Y+Z d) X Y + Y Z + XZ + XY + Y Z = X Y + Y Z(X + X) + XZ + XY + Y Z = X Y + X Y Z + X Y Z + XZ + XY + Y Z = X Y (1 + Z) + X Y Z +XZ + XY + Y Z = X Y + XZ(1 + Y) + XY + Y Z = X Y + XZ + XY (Z + Z)+ Y Z = X Y + XZ + XY Z +Y Z (1 + X) = X Y + XZ(1 + Y) + Y Z = X Y + XZ + Y Z = X Y + XZ + Y Z =X+Y+Z

2-7.
a)b) c) d) X Y + XYZ + XY = X + XYZ = (X + XY)(X + Z) = (X + X)(X + Y)(X + Z) = (X + Y)(X + Z) = X + YZ X + Y(Z + X Z) = X + YZ + X Y Z = X + (YZ + X)(YZ + YZ) = X + Y(X + YZ) = X + XY + YZ = (X + X)(X + Y) + YZ = X + Y + YZ = X + Y WX(Z + YZ) + X(W + W YZ) = WXZ + WXYZ + WX + WXYZ = WX + WXZ + WXZ = WX + WX = X ( AB + AB ) ( CD + CD ) + AC = ABCD + ABCD + ABCD + ABCD + A + C = A + C + ABCD = A +...