Skip to content
Textbook Guides

Solution Manual For Logic And Computer Design Fundamentals, 4th Edition

Get accurate solutions for your textbook problems with Solution Manual For Logic And Computer Design Fundamentals, 4th Edition, your go-to guide for textbook answers.

Preview 16 of 142 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    7 CHAPTER 1 1-1. (a) (1) Calm: (2) 10 mph (3) 100 mph (b) The microcomputer requires a table or equation for converting from rotations/second to miles/hour. The pulses pro- duced by the rotating disk must be counted over a known period of time, and the table or equation used to convert the binary count to miles per hour. 1-2. –34° quantizes to –30° => 1 V => 0001 +31° quantizes to +30° => 7 V => 0111 +77° quantizes to +80° => 12 V => 1100 +108° quantizes to +110° => 15 V => 1111 1-3.* 1-4. 1-5. 2 20 = (1,000,000 10 + d ) where d = 48,576 1Tb = 2 40 = (2 20 ) 2 = (1,000,000 + d ) 2 = (1,000,000) 2 + 2(1,000,000) d + d 2 = 1,000,000,000,000 + 97,152,000,000 + 2,359,627,776 = 1,099,511,627,776 1-6. Decimal, Binary, Octal and Hexadecimal Numbers from (16) 10 to (31) 10 Dec 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 Bin 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 Oct 20 21 22 23 24 25 26 27 30 31 32 33 34 35 36 37 Hex 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F or 96 K 96 2 10 × 98 304 Bits , = = 640 M 640 2 20 × 671 088 640 Bits , , = = 4 G 4 2 30 × 4 294 967 296 Bits , , , = = 11 1 Bits 2 11 1 – ⇒ 2047 = 25 1 Bits 2 25 1 – ⇒ 33 554 431 , , =

  2. Page 2
  3. Page 3
    Read page 3 as text

    8 Problem Solutions – Chapter 1 1-7.* 1-8. 2| 193 1 11000001 2| 751 1 1011101111 2| 96 0 2| 375 1 2| 48 0 2| 187 1 2| 24 0 2| 93 1 2| 12 0 2| 46 0 2| 6 1 2| 23 1 2| 1 1 2| 11 1 0 2| 5 1 2| 2 0 2| 1 1 0 2| 2007 1 11111010111 2| 19450 0 100101111111010 2| 1003 1 2| 9725 1 2| 501 1 2| 4862 0 2| 250 0 2| 2431 1 2| 125 1 2| 1215 1 2| 62 0 2| 607 1 2| 31 1 2| 303 1 2| 15 1 2| 151 1 2| 7 1 2| 75 1 2| 3 1 2| 37 1 2| 1 1 2| 18 0 0 2| 9 1 2| 4 0 2| 2 0 2|1 1 0 1-9.* 1-10.* a) 8| 7562 2 16612 0.45 × 8 = 3.6 => 3 8| 945 1 0.60 × 8 = 4.8 => 4 8| 118 6 0.80 × 8 = 6.4 => 6 8| 14 6 0.20 × 8 =3.2 => 3 3463 8| 1 1 0 (7 562.45) 10 = (16612.3463) 8 b) (1938.257) 10 = (792.41CB) 16 c) (175.175) 10 = (10101111.001011) 2 1-11.* a) (673.6) 8 = (110 111 011.110) 2 = (1BB.C) 16 b) (E7C.B) 16 = (1110 0111 1100.1011) 2 = (7174.54) 8 c) (310.2) 4 = (11 01 00.10) 2 = (64.4) 8 Decimal Binary Octal Hexadecimal 369.3125 101110001.0101 561.24 171.5 189.625 10111101.101 275.5 BD.A 214.625 11010110.101 326.5 D6.A 62407.625 1111001111000111.101 171707.5 F3C7.A 1001101 ( ) 2 2 6 2 3 2 2 2 0 + + + 77 = = 1010011.101 ( ) 2 2 6 2 4 2 1 2 0 2 1 – 2 3 – + + + + + 83.625 = = 10101110.1001 ( ) 2 2 7 2 5 2 3 2 2 2 1 2 1 – 2 4 – + + + + + + 174.5625 = =

  4. Page 4
    Read page 4 as text

    9 Problem Solutions – Chapter 1 1-12. a) 1101 b) 0101 c) 100111 × 1011 × 1010 × 011011 1101 0000 100111 1101 0101 100111 0000 0000 000000 1101 0101 100111 10001111 0110010 100111 000000 10000011101 1-13. + 10001 Quotient = 10001 101 )1010110 − 101 Remainder = 1 000 − 000 001 − 000 011 − 000 110 − 101 1 1-14. (a) 6 × 12 3 + 8 × 12 2 + 7 × 12 1 + 4 = 11608 (b) 12| 7569 9 4469 12 12| 630 6 12| 52 4 12| 4 4 0 1-15. a) 0 1 2 3 4 5 6 7 8 9 A B C D E F G H I J b) 20| 2007 7 507 20 20| 100 0 20| 5 5 0 c) 1-16.* a) (BEE) r = (2699) 10 By the quadratic equation: r = 15 or ≈ –16.27 ANSWER: r = 15 b) (365) r = (194) 10 BCI.G ( ) 20 11 20 2 × = 12 20 1 18 20 0 16 20 1 – × + × + × + 4658.8 ( ) 10 = 11 r 2 × 14 + r 1 14 r 0 × + × 2699 = 11 r 2 × 14 + r × 2685 – 0 = 3 r 2 × 6 + r 1 × 5 + r 0 × 194 =

  5. Page 5
    Read page 5 as text

    10 Problem Solutions – Chapter 1 By the quadratic equation: r = – 9 or 7 ANSWER: r = 7 1-17. Noting the order of operations, first add (34) r and (24) r Now, multiply the result by (21) r Next, set the result equal to (1480) r and reorganize. Finally, find the roots of this cubic polynomial. Solutions are: r = 8, – 1, – 1 ANSWER: The chicken has 4 toes on each foot (half of 8). 1-18.* a) (0100 1000 0110 0111) BCD = (4867) 10 = (1001100000011) 2 b) (0011 0111 1000.0111 0101) BCD = (378.75) 10 = (101111010.11) 2 1-19.* (694) 10 = (0110 1001 0100) BCD (835) 10 = (1000 0011 0101) BCD 1 0110 1001 0100 +1000 +0011 +0101 1111 1100 1001 +0110 +0110 +0000 0001 0101 1 0010 1001 1-20.* (a) 10 1 10 0 0111 1000 Move R 011 1100 0 10 0 column > 0111 Subtract 3 - 0011 011 1001 0 Subtract 3 -0011 01 1001 Move R 0 1100 110 10 0 column > 0111 Subtract 3 -0011 0 1001 110 Move R 0100 1110 Move R 010 01110 Move R 01 001110 Move R 0 1001110 Leftmost 1 in BCD number shifted out: Finished (b) 10 2 10 1 10 0 0011 1001 0111 Move R 001 1100 1011 1 10 1 and 10 0 columns > 0111 Subtract 3 -0011 -0011 001 1001 1000 1 Move R 00 1100 1100 01 10 1 and 10 0 columns > 0111 3 r 2 × 6 + r × 189 – 0 = 34 ( ) r 3 r 1 × 4 r 0 × + = 24 ( ) r 2 r 1 × 4 r 0 × + = 34 ( ) r 24 ( ) r + 5 r 1 8 r 0 × + × = 2 r 1 × 1 r 0 × + ( ) 5 r 1 × 8 r 0 × + ( ) × 10 r 2 23 r 1 8 + × + × = 10 r 2 23 r 1 8 + × + × 1 r 3 4 r 2 8 r 1 × + × + × = 1 r 3 6 r 2 15 r 1 × – 8 r 0 × – × – × 0 =

  6. Page 6
    Read page 6 as text

    11 Problem Solutions – Chapter 1 Subtract 3 -0011 -0011 00 1001 1001 01 Move R 0 0100 1100 101 10 0 column > 0111 Subtract 3 -0011 0 0100 1001 Move R 0010 0100 1101 Move R 001 0010 01101 Move R 00 1001 001101 100 column > 0111 Subtract 3 -0011 00 0110 001101 Move R 0 0011 0001101 Move R 0001 10001101 Move R 000 110001101Leftmost 1 in BCD number shifted out: Finished 1-21. (a) 10 2 10 1 10 0 1111000 1st Move L 1 111000 2nd Move L 11 11000 3rd Move L 111 1000 10 0 column > 100 Add 3 0011 1010 1000 4th Move L 1 0101 000 10 0 column > 100 Add 3 0011 1 1000 000 5th Move L 11 0000 00 6th Move L 110 0000 0 10 1 column > 100 Add 3 0011 1001 0000 0 7th Move L 1 0010 0000 Least significant bit in binary number moved in:Finished (b) 10 3 10 2 10 1 10 0 01110010111 1st Move L 0 1110010111 2nd Move L 01 110010111 3rd Move L 011 10010111 4th Move L 0111 0010111 10 0 column > 100 Add 3 0011 1010 0010111 5th Move L 1 0100 010111 6th Move L 10 1000 10111 10 0 column > 100 Add 3 0011 10 1011 10111 7th Move L 101 0111 0111 10 1 & 10 0 columns > 100 Add 3 0011 0011 1000 1010 0111 8th Move L 1 0001 0100 111 9th Move L 10 0010 1001 11 10 0 column > 100 Add 3 0011 10 0010 1100 11 10th Move L 100 0101 1001 1 10 1 &10 0 columns > 100 Add 3 0011 0011 100 1000 1100 1 11th Move L 1001 0001 1001 Least significant bit in binary number moved in: Finished 1-22. From Table 1-5, complementing the bit B 6 will switch an uppercase letter to a lower case letter and vice versa. 1-23. a) The name used is Brent M. Ledvina. An alternative answer: use both upper and lower case letters. 0100 0010 B 0101 0010 R 0100 0101 E

  7. Page 7
    Read page 7 as text

    12 Problem Solutions – Chapter 1 0100 1110 N 0101 0100 T 0010 0000 (SP) 0100 1101 M 0010 1110 . 0010 0000 (SP) 0100 1100 L 0100 0101 E 0100 0100 D 0101 0110 V 0100 1001 I 0100 1110 N 0100 0001 A b) 0100 0010 1101 0010 1100 0101 0100 1110 1101 0100 1010 0000 0100 1101 0010 1110 1010 0000 1100 1100 1100 0101 0100 0100 0101 0110 1100 1001 0100 1110 0100 0001 1-24. 1000111 G 1101111 o 0100000 1000010 B 1100001 a 1100100 d 1100111 g 1100101 e 1110010 r 1110011 s 0100001 ! 1-25.* a) (11111111) 2 b) (0010 0101 0101) BCD c) 011 0010 011 0101 011 0101 ASCII d) 0011 0010 1011 0101 1011 0101 ASCII with Odd Parity 1-26. 1-27. Binary Numbers from (32) 10 to (47) 10 with Odd and Even Parity Decimal 32 33 34 35 36 37 38 39 (a) Odd 100000 0 100001 1 100010 1 100011 0 100100 1 100101 0 100110 0 100111 1 (b) Even 100000 1 100001 0 100010 0 100011 1 100100 0 100101 1 100110 1 100111 0 Decimal 40 41 42 43 44 45 46 47 (a) Odd 101000 1 101001 0 101010 0 101011 1 101100 0 101101 1 101110 1 101111 0 (b) Even 101000 0 101001 1 101010 1 101011 0 101100 1 101101 0 101110 0 101111 1 Gray Code for Hexadecimal Digits Hex 0 1 2 3 4 5 6 7 8 9 A B C D E F Gray 0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000

  8. Page 8
    Read page 8 as text

    13 Problem Solutions – Chapter 1 1-28. As the wind direction changes, the codes change in the order of the rows of this table, assum - ing that the bottom row is “next to” the top row. From the table, the codes that result due to a wind direction change always change in a single bit. 1-29. + The percentage of power consumed by the Gray code counter compared to a binary code counter equals: Number of bit changes using Gray code Number of bit changes using binary code As shown in Table 1-6, and by definition, the number of bit changes per cycle of an n-bit Gray code counter is 1 per count = 2 n . Number of bit changes using Gray code = 2 n For a binary counter, notice that the least significant bit changes on every increment. The sec - ond least significant bit changes on every other increment. The third digit changes on every fourth increment of the counter, and so on. As shown in Table 1-6, the most significant digit changes twice per cycle of the binary counter. Number of bit changes using binary code % Power = (a) Wind Direction Gray Code Direction Code Word N 000 S 110 E 011 W 101 NW 100 NE 001 SW 111 SE 010 (b) Wind Direction Gray Code (directions in adjacent order) Direction Code Word N 000 NE 001 E 011 SE 010 S 110 SW 111 W 101 NW 100 2 n 2 n 1 – … 2 1 + + + 2 i i 1 = n ∑ = 2 i i 0 = n ∑ 1 – 2 n 1 + ( ) 1 – ( ) 1 – 2 n 1 + 2 – = = = 2 n 2 n 1 + ( ) 2 – ----------------------- - 100 ×

  9. Page 9
    Read page 9 as text

    14 CHAPTER 2 2-1.* a) b) c) 2-2.* a) = b) = 1 Verification of DeMorgan’s Theorem X Y Z XYZ XYZ X + Y + Z 0 0 0 0 1 1 0 0 1 0 1 1 0 1 0 0 1 1 0 1 1 0 1 1 1 0 0 0 1 1 1 0 1 0 1 1 1 1 0 0 1 1 1 1 1 1 0 0 The Second Distributive Law X Y Z YZ X+YZ X+Y X+Z ( X+Y )( X+Z ) 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 X Y Z X Y Y Z XZ X Y+Y Z+XZ XY YZ XZ XY +YZ +X Z 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 0 0 1 1 1 0 0 0 0 1 1 1 0 0 1 1 0 1 0 1 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 0 1 1 1 1 0 0 0 0 0 0 0 0 XYZ X Y Z + + = X YZ + X Y + ( ) X Z + ( ) ⋅ = XY YZ XZ + + XY YZ XZ + + = XY XY XY + + X Y + XY XY + ( ) XY XY + ( ) + = X Y Y + ( ) Y X X + ( ) + = X Y + = AB BC AB BC + + + AB AB + ( ) = BC BC + ( ) + B = A A + ( ) B C C + ( ) +

  10. Page 10
    Read page 10 as text

    15 Problem Solutions – Chapter 2 c) = d) = 2-3. + a) = b) = c) = = 2-4. + Given: Prove: = B B + 1 = Y XZ XY + + X Y Z + + Y = XY XZ + + Y X + ( ) = Y Y + ( ) XZ + Y = X XZ + + Y = X X + ( ) X Z + ( ) + X = Y Z + + XY YZ XZ XY YZ + + + + XY XZ YZ + + X = Y YZ X X + ( ) XZ XY YZ + + + + XY XYZ XYZ XZ XY YZ + + + + + = XY 1 Z + ( ) XYZ XZ XY YZ + + + + = XY XZ 1 Y + ( ) XY YZ + + + = XY XZ XY Z Z + ( ) YZ + + + = XY XZ XYZ YZ 1 X + ( ) + + + = XY XZ 1 Y + ( ) YZ + + = XY XZ YZ + + = ABC BCD BC CD + + + B CD + ABC ABC + BC BCD BCD CD + + + + = AB C C + ( ) BC D D + ( ) BC CD + + + = AB BC BC CD + + + = B AB CD + + = B CD + = WY WYZ WXZ WXY + + + WY WXZ XYZ XYZ + + + WY WXYZ + ( ) WXYZ WXYZ + ( ) WXYZ WXYZ + ( ) WXYZ WXYZ + ( ) + + + = WY WXYZ + ( ) WXYZ WXYZ + ( ) WXYZ WXYZ + ( ) WXYZ WXYZ + ( ) + + + = WY WXZ Y Y + ( ) XYZ W W + ( ) XYZ W W + ( ) + + + = WY WXZ XYZ XYZ + + + = AD AB CD BC + + + A B C D + + + ( ) A B C D + + + ( ) AD AB CD BC + + + = A D + ( ) A B + ( ) C D + ( ) B C + ( ) = AB AD BD + + ( ) BC BD CD + + ( ) = ABCD ABCD + = A B C D + + + ( ) = A B C D + + + ( ) A B C D + + + ( ) A B C D + + + ( ) A B ⋅ 0 A B + , 1 = = A C + ( ) A B + ( ) B C + ( ) BC AB AC BC + + ( ) B C + ( ) = AB AC BC + + =

  11. Page 11
    Read page 11 as text

    16 Problem Solutions – Chapter 2 2-5. + S tep 1: Define all elements of the algebra as four bit vectors such as A, B and C : A = ( A 3 , A 2 , A 1 , A 0 ) B = ( B 3 , B 2 , B 1 , B 0 ) C = ( C 3 , C 2 , C 1 , C 0 ) Step 2: Define OR 1 , AND 1 and NOT 1 so that they conform to the definitions of AND, OR and NOT presented in Table 2-1. a) A + B = C is defined such that for all i , i = 0, ... ,3, C i equals the OR 1 of A i and B i . b) A B = C is defined such that for all i , i = 0, ... ,3, C i equals the AND 1 of A i and B i . c) The element 0 is defined such that for A = “0”, for all i, i = 0, ... ,3, A i equals logical 0. d) The element 1 is defined such that for A = “1”, for all i, i = 0, ... ,3, A i equals logical 1. e) For any element A , A is defined such that for all i , i = 0, ... ,3, A i equals the NOT 1 of A i . 2-6. a) b) c) d) e) 2-7.* a) b) c) 0 C A B + ( ) + = C A B + ( ) 0 ( ) = C A B + ( ) A B + ( ) = C AB AB B + + ( ) = BC = AC ABC BC + + AC ABC A ( BC BC ) + + + = AC ABC ABC + ( ) BC + + = A ( C AC ) BC + + A BC + = = A B C + + ( ) ABC ( ) AABC ABBC ABCC + + = AA ( ) BC A BB ( ) C AB CC ( ) + + = ABC ABC ABC + + = ABC = ABC AC + A BC C + ( ) A B C + ( ) = = ABD ACD BD + + AB B AC + + ( ) D = A AC B + + ( ) D = A B + ( ) D = A B + ( ) A C + ( ) ABC ( ) AB ( ) = AC ( ) A B C + + ( ) ABC A B C + + ( ) = 0 = XY 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 Y Z XZ + ( ) + X Y Z X + ( ) Z Z + ( ) + X YZ XY + + = = = X X + ( ) X Y + ( ) YZ + = X Y YZ + + X Y + = = WX Z YZ + ( ) X W WYZ + ( ) + WXZ WXYZ WX WXYZ + + + =

  12. Page 12
    Read page 12 as text

    17 Problem Solutions – Chapter 2 d) 2-8. a) b) 2-9.* a) b) c) d) 2-10. * a) Sum of Minterms: Product of Maxterms: b) Sum of Minterms: Product of Maxterms: c) Sum of Minterms: Product of Maxterms: 2-11. a) , b) , Truth Tables a, b, c X Y Z a A B C b W X Y Z c 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 1 0 1 0 1 1 1 0 1 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 1 1 0 1 0 0 1 0 1 0 1 1 0 1 1 1 0 0 0 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 0 1 0 1 1 0 1 1 0 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 WXZ WXZ WX + + = WX WX + X = = AB AB + ( ) CD CD + ( ) AC + ABCD ABCD ABCD ABCD A C + + + + + = ABCD A C + + = A C A BCD ( ) + + A C C BD ( ) + + A C BD + + = = = F ABC AC AB + + = F ABC AC AB + + = A B C + + ( ) A C + ( ) A B + ( ) + + = ABC ( ) AC ( ) AB ( ) = F A B + ( ) A B + ( ) = F V W + ( ) X Y + ( ) Z = F W X + Y Z + ( ) Y Z + ( ) + [ ] W X + YZ YZ + + [ ] = F ABC A B + ( ) C A B C + ( ) + + = XYZ XYZ XYZ XYZ + + + X Y Z + + ( ) X Y Z + + ( ) X Y Z + + ( ) X Y Z + + ( ) ABC ABC ABC ABC + + + A B C + + ( ) A B C + + ( ) A B C + + ( ) A B C + + ( ) WXYZ WXYZ WXYZ WXYZ WXYZ WXYZ + + + + + W + XYZ 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 + + + ( ) W X Y Z + + + ( ) W X Y Z + + + ( ) E Σ m 1 2 4 6 , , , ( ) Π M 0 3 5 7 , , , ( ) = = F Σ m 0 2 4 7 , , , ( ) Π M 1 3 5 6 , , , ( ) = = E Σ m 0 3 5 7 , , , ( ) = F Σ m 1 3 5 6 , , , ( ) =

  13. Page 13
    Read page 13 as text

    18 Problem Solutions – Chapter 2 c) , d) , e) , 2-12.* a) s.o.p. p.o.s. b) p.o.s. s.o.p. c) p.o.s. s.o.p. 2-13. 2-14. 2-15. * E F + Σ m 0 1 2 4 6 7 , , , , , ( ) = E F ⋅ Σ m 2 4 , ( ) = E XYZ XYZ XYZ XYZ + + + = F XYZ XYZ XYZ XYZ + + + = E Z X Y + ( ) XYZ + = F Z X Y + ( ) XYZ + = AB C + ( ) B CD + ( ) AB ABCD BC + + AB BC + = = B A C + ( ) = X X X Y + ( ) Y Z + ( ) + X X + ( ) X X Y + ( ) Y Z + ( ) + ( ) = X X Y + + ( ) = X Y Z + + ( ) 1 Y + ( ) X Y Z + + ( ) = X Y Z + + = A BC CD + + ( ) B EF + ( ) A B C + + ( ) A B D + + ( ) A C D + + ( ) B EF + ( ) = A B C + + ( ) = A B D + + ( ) A C D + + ( ) B E + ( ) B F + ( ) A BC CD + + ( ) B EF + ( ) A B EF + ( ) BC B EF + ( ) CD B EF + ( ) + + = A = B AEF BCEF BCD CDEF + + + + a) b) c) C B A C X Z X Y Z D B A C Y W Y Y Z W X Z X Y A X Y Z A B C X Y Z A B C a) b) c) d) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 XZ XY + XY XZ YZ + + C AB + BC AB AC + + 1 or BC AB AC + + X Y Z A B C a) b) c) A B C 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 XZ XY + A CB + B C +

  14. Page 14
    Read page 14 as text

    19 Problem Solutions – Chapter 2 2-16. 2-17. 2-18. * 2-19.* a) b) c) 2-20. a) b) c) A B D W X Z A B D 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 AC AD ABC + + YZ XZ XYZ WXZ or WXY ( ) + + + BD ABC ACD + + W X Y Z A B C D a) b) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 F XZ WY WX XYZ + + + = F BD ABC ABC AD or CD ( ) + + + = 1 W X Y Z A B C D a) b) c) X Y Z Σ m 3 5 6 7 , , , ( ) Σ m 3 4 5 7 9 13 14 15 , , , , , , , ( ) Σ m 0 2 6 7 8 10 13 15 , , , , , , , ( ) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Prime XZ WX XZ WZ , , , = Prime CD AC BD ABD BC , , , , = Prime AB AC AD BC BD CD , , , , , = Essential XZ XZ , = Essential AC BD ABD , , = Essential AC BC BD , , = Prime XY XZ WYZ WXY XYZ WXZ , , , , , WYZ , = Prime ABC ACD ABC ACD BD , , , , = Essential XY XZ , = Essential ABC ACD ABC ACD , , , = F XY XZ WXY WXZ + + + = Redundant BD = F ABC ACD ABC ACD + + + = Prime YZ WY WZ WXZ XYZ WXY , , , , , = Essential WY WZ , = Redundant YZ = F WY WZ XYZ + + =

  15. Page 15
    Read page 15 as text

    20 Problem Solutions – Chapter 2 2-21. 2-22.* a) s.o.p. b) s.o.p. c) s.o.p. or p.o.s. p.o.s. p.o.s. 2-23. a) s.o.p. b) s.o.p. or p.o.s. p.o.s. or 2-24. 2-25. * W X Y Z A B C D a) b) F F 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 F Σ m 1 5 6 7 9 12 13 14 , , , , , , , ( ) = F Σ m 0 2 4 5 8 10 11 12 13 14 , , , , , , , , , ( ) = F YZ WXZ WXY + + = F BC BD AD ABC + + + = F Y Z + ( ) W X Z + + ( ) W X Y + + ( ) = F B C + ( ) B D + ( ) A D + ( ) A B C + + ( ) = CD AC BD + + AC BD AD + + BD ABD A ( BC + + ACD ) C D + ( ) A D + ( ) A B C + + ( ) C D + ( ) A D + ( ) A B C + + ( ) A B + ( ) B D + ( ) B C D + + ( ) ABC ABD ABC ABD + + + Z WX XY + + ACD BCD ACD BCD + + + W X Z + + ( ) X Y Z + + ( ) A B C + + ( ) A B D + + ( ) A B C + + ( ) A B D + + ( ) A C D + + ( ) B C D + + ( ) A C D + + ( ) B C D + + ( ) W X Y Z a) b) c) A B C 1 1 1 1 1 1 1 1 X X X X X F A C + = F XY WYZ WXZ + + ( WXY or WYZ ) + = A B C D 1 1 1 1 1 X X X X F BC BCD ABD + + = X X X X W X Y Z A B C D a) b) c) A B C 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 X X X X X X X X X X X Primes AB AC BC ABC , , , = F AB AC BC + + = Essential AB AC BC , , = Primes XZ XZ WXY WXY WYZ WYZ , , , , , = Essential XZ = F XZ WXY WXY + + = Primes AB C AD BD , , , = Essential C AD , = F C AD BD or AB ( ) + + =

  16. Page 16
    Read page 16 as text

    21 Problem Solutions – Chapter 2 2-26. 2-27. 2-28. + X A B C D a)(1) b)(1) X W X Y Z X X X 0 F BD = F B D + = X 1 X F BD = 0 X 0 0 A 0 X X B D C 1 X X X X W X Z X Y X X 1 F = W X +W Y + X Y Z F = ( WZ or X Z ) + WXY + WXY F = ((W + Z) or (X + Z)) X X 0 0 0 0 0 a)(2) b)(2) 0 0 X X X 1 X 1 X X X X X 1 1 1 1 1 1 + ( WXY or XYZ) + (W + X + Y) or (X + Y + Z) (W + X + Y)(W + X + Y) + W X Y A B C D a) F = A B C D + ABC + A C D + ABD + ABCD + BCD b) F = A C + AB + BD + AC + AB 1 1 1 1 1 1 1 1 1 1 A B C D 1 1 1 1 1 1 1 1 1 1 A B C D 1 1 1 1 1 1 1 1 1 1 A B C D 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 There are other solutions depending on how ties are resolved. There are other solutions depending on how ties are resolved. A B C D F = A B D + B C D + BC + AB + ACD 1 1 1 1 1 1 1 1 1 1 A B C D A B C D A B C D 1 1 1 1 F and F ; Cost = 18 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 EXPAND ESSENTIAL PRIMES X X X X X X X 1 IRREDUNDANT COVER; Cost = 17 A B C D 1 1 1 X X X X X X X 1 REDUCE A B C D 1 1 1 X X X X X X X 1 EXPAND A B D 1 1 1 X X X X X X X 1 IRREDUNDANT COVER; Cost = 13 REDUCE, EXPAND, IRREDUNDANT COVER, and LAST GASP produce no lasting changes. A B D 1 1 1 1 FINAL SOLUTION; Cost = 13 1 1 1 1 1 1 1

126 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Information Technology study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.