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Solution Manual For Digital Design, 5th Edition

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    2 CHAPTER 1 1.1 Base - 10: 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Octal: 20 21 22 23 24 25 26 27 30 31 32 3 3 34 35 36 37 40 Hex: 10 11 12 13 14 15 16 17 18 19 1A 1B 1C 1D 1E 1F 20 Base - 1 2 14 15 16 17 18 19 1A 1 B 20 21 22 23 24 25 26 27 28 1. 2 (a) 32,768 (b) 67,108,864 (c) 6,871,947,674 1. 3 (4310) 5 = 4 * 5 3 + 3 * 5 2 + 1 * 5 1 = 580 10 (198) 12 = 1 * 12 2 + 9 * 12 1 + 8 * 12 0 = 260 10 (4 35) 8 = 4 * 8 2 + 3 * 8 1 + 5 * 8 0 = 285 10 ( 345 ) 6 = 3 * 6 2 + 4 * 6 1 + 5 * 6 0 = 13 7 10 1.4 16 - bit binary: 11 11 _1111_1111_1111 Decimal equivalent : 2 1 6 - 1 = 65,535 10 Hexadecimal equivalent : F FFF 16 1. 5 Let b = bas e (a) 14/2 = (b + 4)/2 = 5, so b = 6 (b) 54/4 = (5*b + 4)/4 = b + 3, so 5 * b = 52 – 4, and b = 8 (c) (2 *b + 4) + (b + 7) = 4b, so b = 11 1. 6 (x – 3)(x – 6) = x 2 – (6 + 3)x + 6*3 = x 2 - 11x + 22 Therefore: 6 + 3 = b + 1m , so b = 8 Also, 6*3 = (18 ) 10 = (22) 8 1.7 64CD 16 = 0110_ 0100_1100 _11 01 2 = 110_010_011_001 _101 = ( 6 2315 ) 8 1.8 (a) Results of repeated division by 2 (quotients are followed by remainders): 431 10 = 215 (1); 107 ( 1); 5 3( 1 ); 2 6 (1); 1 3(0 ); 6 ( 1) 3 ( 0 ) 1(1) Answer: 1 111 _ 1010 2 = FA 16 (b) Results of repeated division by 16: 431 10 = 26(15); 1(10) (Faster) Answer: FA = 1111_1010 1. 9 (a) 10110.0101 2 = 16 + 4 + 2 + .25 + .0625 = 22.3125 (b) 16.5 16 = 16 + 6 + 5*(.0615) = 22.3125 (c) 26.24 8 = 2 * 8 + 6 + 2/8 + 4/64 = 2 2.3125 (d) DADA .B 16 = 1 4 *16 3 + 10*16 2 + 1 4 *16 + 10 + 11/16 = 60,138 .6875

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    3 (e) 1010 .1 101 2 = 8 + 2 + .5 + .25 + .0625 = 10.8125 1.10 (a) 1. 10010 2 = 0001 .1001 2 = 1.9 16 = 1 + 9/16 = 1.563 10 (b) 110.010 2 = 0110.0100 2 = 6.4 16 = 6 + 4/16 = 6.25 10 Reason: 110.010 2 is the same as 1.10010 2 shifted to the left by two places. 1011.11 1.11 101 | 1 1 1011.0000 101 01001 101 1001 101 1000 101 011 0 T he quotient is carried to two decimal places, giving 1011.11 Checking: 111011 2 / 101 2 = 59 10 / 5 10 ≅ 1011.11 2 = 58.75 10 1. 12 (a) 10000 and 110111 1011 1011 +101 x101 10000 = 16 10 1011 10 11 110111 = 55 10 (b) 62 h and 958 h 2E h 0010_1110 2E h + 34 h 0011_0100 x 34 h 62 h 0110_0010 = 98 10 B 3 8 8 2 A 9 5 8 h = 2392 10 1.13 (a) Convert 27.315 to binary: Integer Remainder Coefficient Quotient 27/2 = 13 + ½ a 0 = 1 13/2 6 + ½ a 1 = 1 6/2 3 + 0 a 2 = 0 3/2 1 + ½ a 3 = 1 ½ 0 + ½ a 4 = 1

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    4 27 10 = 11011 2 Integer Fraction Coefficient .315 x 2 = 0 + .630 a - 1 = 0 .630 x 2 = 1 + .26 a - 2 = 1 .26 x 2 = 0 + .52 a - 3 = 0 .52 x 2 = 1 + .04 a - 4 = 1 .315 10 ≅ .0101 2 = .25 + .0625 = .3125 27.315 ≅ 11011.0101 2 (b) 2/3 ≅ . 66 66666667 Integer Fraction Coefficient . 66 66 _ 6666 _ 67 x 2 = 1 + . 33 33 _ 3333 _ 34 a - 1 = 1 . 33 33333334 x 2 = 0 + . 66 66666668 a - 2 = 0 . 66 66666668 x 2 = 1 + . 33 33333336 a - 3 = 1 . 33 33333336 x 2 = 0 + . 66 66666672 a - 4 = 0 . 66 66666672 x 2 = 1 + . 33 33333344 a - 5 = 1 . 33 33333344 x 2 = 0 + . 66 66666688 a - 6 = 0 . 66 66666688 x 2 = 1 + .3 33 3333376 a - 7 = 1 . 33 33333376 x 2 = 0 + .6666666752 a - 8 = 0 . 6666666667 10 ≅ . 1 0101 010 2 = . 5 + .125 + .0313 + . .0078 = .6641 10 .101010102 = .1010_1010 2 = .AA 16 = 10/16 + 10/256 = .6641 10 (Same as (b)). 1.14 (a) 0 00 1 _ 0 000 (b) 0000_0000 (c) 1101_1010 1s comp: 1110 _1111 1s comp: 1111 _1111 1s comp: 0010_0101 2s comp: 1 111 _0000 2s comp: 0000_0000 2s comp: 0010_0110 (d) 1010_1010 (e) 1000_0101 (f) 1111_1111 1s comp: 0101_0101 1s comp: 0111_1010 1s comp: 0000_0000 2s comp: 0101_0110 2s comp: 0111_1011 2s co mp: 0000_0001 ` 1.15 (a) 25,478,036 (b) 63,325,600 9s comp: 74,521,963 9s comp: 36,674,399 10s comp: 74,521,964 10s comp: 36,674,400 (c) 25,000,000 ( d ) 00000000 9s comp: 74,999,999 9s comp: 99999999 10s comp: 75,000 ,000 10s comp: 100000000 1.16 C3DF C3DF : 1100_0011_1101_1111 15s comp: 3 C20 1s comp: 0011_1100_0010_0000 16s comp: 3 C21 2s comp: 0011_1100_0010_0001 = 3 C21 1.17 (a) 2,579 → 0 2,579 → 9 7,420 (9s comp) → 9 7,421 (10s comp) 4637 – 2,579 = 2,579 + 97,421 = 2058 10 (b) 1800 → 01800 → 9 8199 (9s comp) → 98200 (10 comp) 125 – 1800 = 00125 + 98200 = 98325 (negative) Magnitude: 1675 Result: 125 – 1800 = 1675

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    5 (c) 4,361 → 0 4361 → 95638 (9s comp) → 95639 (10s comp) 2043 – 4361 = 02043 + 9 5639 = 9 7682 (Negative) Magnitude: 2318 Result: 2043 – 6152 = - 2318 (d) 745 → 00745 → 99254 (9s comp) → 99255 (10s comp) 1631 - 745 = 01631 + 99255 = 0886 (Positive) Result: 1631 – 745 = 886 1.18 N ote: Consider sign extension with 2s complement arithmetic. (a) 0_ 100 10 ( b) 0_100110 1s comp : 1_ 011 01 1s comp: 1 _011001 with sign extension 2 s comp: 1_ 011 10 2s comp: 1_011010 0_ 10011 0_ 100010 Diff: 0_ 0000 1 ( Positive ) 1 _111100 sign bit indicates that the result is negative Check:19 - 18 = +1 0_000011 1 s complement 0_000100 2s complement 000100 magnitude Result: - 4 Check: 34 - 38 = - 4 (c) 0_110101 ( d ) 0_0 10 1 01 1s comp : 1_001010 1s comp: 1_ 10 1 01 0 with sign extension 2s comp: 1 _ 0 01011 2s comp: 1_ 101011 0_ 001001 0_101000 Diff: 1 _ 010100 (negative) 0 _ 0 10 011 sign bit indicates that the result is positive 0_ 101011 ( 1 s comp) Result: 19 10 0_10110 0 ( 2s complement) Check: 40 – 21 = 19 10 1 01100 (magnitude) - 4 4 10 (result) 1. 19 +9286 → 009286; +801 → 000801; - 928 6 → 990714; - 801 → 999199 (a) (+9286) + (_801) = 009286 + 000801 = 010 087 (b) (+9286) + ( - 801) = 009286 + 999199 = 00 848 5 (c) ( - 9286) + (+801) = 990714 + 000801 = 99151 5 (d) ( - 9286) + ( - 801) = 990714 + 999199 = 989 91 3 1.20 +49 → 0_ 110001 ( Needs leading zero extension to indicate + value); +29 → 0_011101 (Leading 0 indicates + value) - 4 9 → 1_ 00111 0 + 0_000001 → 1_ 001111 - 29 → 1_100011 (sign ex tension indicates negative value) (a) (+29) + ( - 49 ) = 0_011101 + 1_ 0 01111 = 1_ 101100 (1 indica tes negative value.) Magnitude = 0_ 010011 + 0_000001 = 0_01010 0 = 20 ; Result (+29) + ( - 49) = - 20 (b) ( - 29) + (+49) = 1_ 100011 + 0_110001 = 0_010100 (0 indicates positive value) ( - 29) + (+49) = +20

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    6 (c) Must increase word size by 1 (sign extension) to accomodate overflow of values: ( - 29) + ( - 49) = 1 1_ 100011 + 1 1_001111 = 10 _110010 (1 indicates negative result) Magnitude: 0 1 _001110 = 78 10 Result: ( - 29) + ( - 49) = - 78 10 1.21 +97 4 2 → 0097 4 2 → 9902 5 7 ( 9's comp ) → 9902 5 8 ( 10 s) comp +641 → 000641 → 999358 (9's comp) → 9993 59 (10s) comp (a) (+9742) + (+641) → 010383 (b) (+9742) + ( - 641) → 009742 + 9993 59 = 009102 Result: (+9742) + ( - 641) = 9102 (c) - 9742) + (+641) = 990258 + 000641 = 990899 (negative) Magnitude: 009101 Result: ( - 9742) + (641) = - 9101 (d) ( - 9742) + ( - 641) = 990258 + 999359 = 989617 (Negative) Magnitude: 10383 Result: ( - 9742) + ( - 641) = - 10383 1.22 6,514 BCD : 011 0_ 0101_0001_0100 ASCII: 0_011_ 0110_0_011 _ 0101 _ 1_011_0001_1_011_0100 ASCII: 0011_0110_0011_0101_1011_000 1_1011_0100 1.23 0111 1001 0001 ( 791 ) 0110 0101 1000 (+658 ) 1101 1110 10 0 1 0110 0110 0001 00 11 010 0 0001 0001 0001 0100 0100 1001 (1,449) 1.24 (a) (b) 6 3 1 1 Decimal 0 0 0 0 0 0 0 0 1 1 0 0 1 0 2 0 1 0 0 3 0 1 1 0 4 ( or 0101) 0 1 1 1 5 1 0 0 0 6 1 0 1 0 7 ( or 1001) 1 0 1 1 8 1 1 0 0 9 6 4 2 1 Decimal 0 0 0 0 0 0 0 0 1 1 0 0 1 0 2 0 0 1 1 3 0 1 0 0 4 0 1 0 1 5 1 0 0 0 6 ( or 0110) 1 0 0 1 7 1 0 1 0 8 1 0 1 1 9 1.25 (a) 6,248 1 0 BCD: 01 10 _0 010_0 100_1000 (b) Excess - 3 : 1001_0101 _01 11 _10 11 (c) 2421: 011 0 _ 001 0 _0 100 _ 1 11 0 (d) 6311: 1000 _ 0 010 _ 01 10 _10 1 1

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    7 1.26 6,248 9s Comp: 3,751 2421 code: 0 0 1 1 _ 011 1 _0101 _ 0001 1s comp c: 1001 _ 1101 _ 1011 _ 0001 (2421 code a lternative #1) 6,248 2421 0110_0010_0100_1110 (2421 code alternative #2) 1s comp c 1001_1101_1011_0001 Match

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    8 1.27 For a deck with 52 cards, we need 6 bits ( 2 5 = 32 < 52 < 64 = 2 6 ). Let the ms b's select the suit (e.g., diamonds, hearts, clubs, spades are encoded respectively as 00, 01, 10, and 11. The remaining four bits select the "number" of the card. Example: 0001 (ace) through 1011 (9), plus 101 through 1100 (jack, queen, king). This a jack of spades might be coded as 11_1010. (Note: only 52 out of 64 patterns are used.) 1.28 G (dot) (space) B o o l e 1 1000111_11101111_01101000_01101110_00100000_11000100_11101 111_11100101 1. 2 9 Steve Jobs 1.30 73 F4 E5 76 E5 4A EF 62 73 73: 0 _ 111 _0011 s F4: 1_111_0100 t E5: 1_110_0101 e 76: 0_111_0110 v E5: 1_110_0101 e 4A: 0_100_1010 j EF: 1_110_1111 o 62: 0_110_0010 b 73: 0_111_0011 s 1. 3 1 62 + 32 = 94 printing characters 1. 3 2 bit 6 from the right 1. 3 3 (a) 897 (b) 564 (c) 871 (d) 2,199 1.3 4 ASCII for decimal digits with even parity: (0): 0 0110000 (1) : 1 0110001 (2): 1 0110010 (3): 0 0110011 (4): 1 0110100 (5): 0 0110101 (6): 0 01 10110 (7): 1 0110111 (8): 1 0111000 (9): 0 0111001 1.35 (a) a b c f g a b c f g 1.36 a b f g a b f g

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    9 CHAPTER 2 2.1 (a) x y z 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 x + y + z 0 1 1 1 1 1 1 1 (x + y + z)' 1 0 0 0 0 0 0 0 x' 1 1 1 1 0 0 0 0 y' 1 1 0 0 1 1 0 0 z' 1 0 1 0 1 0 1 0 x' y' z' 1 0 0 0 0 0 0 0 x y z 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 (xyz) 0 0 0 0 0 0 0 1 (xyz)' 1 1 1 1 1 1 1 0 x' 1 1 1 1 0 0 0 0 y' 1 1 0 0 1 1 0 0 z' 1 0 1 0 1 0 1 0 x' + y' + z' 1 1 1 1 1 1 1 0 (b) (c) x y z 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 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 x y z 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 x(y + z) 0 0 0 0 0 1 1 1 xy 0 0 0 0 0 0 1 1 xz 0 0 0 0 0 1 0 1 xy + xz 0 0 0 0 0 1 1 1 (c) (d) x y z 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 x 0 0 0 0 1 1 1 1 y + z 0 1 1 1 0 1 1 1 x + (y + z) 0 1 1 1 1 1 1 1 (x + y) 0 0 1 1 1 1 1 1 x y z 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 yz 0 0 0 1 0 0 0 1 x(yz) 0 0 0 0 0 0 0 1 xy 0 0 0 0 0 0 1 1 (xy)z 0 0 0 0 0 0 0 1 (x + y) + z 0 1 1 1 1 1 1 1 2. 2 (a) xy + xy' = x(y + y') = x (b) (x + y)(x + y') = x + yy' = x(x +y') + y(x + y') = xx + xy' + xy + yy' = x (c) xyz + x'y + xyz' = xy(z + z') + x'y = xy + x'y = y (d) (A + B)'( A' + B') ' = (A'B')(A B ) = (A'B')(BA) = A'(B'B)A = 0 (e) (a + b + c')(a'b' + c) = aa'b' + ac + ba'b' + bc + c'a'b' + c'c = ac + bc +a'b'c' (f) a'bc + abc' + abc + a'bc' = a'b(c + c') + a b(c + c') = a'b + ab = (a' + a)b = b 2. 3 (a) ABC + A'B + ABC' = AB + A'B = B

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    10 (b ) x'yz + xz = (x'y + x)z = z(x + x')(x + y) = z(x + y) (c) (x + y)'(x' + y') = x'y'(x' + y') = x'y' (d) xy + x(wz + wz') = x(y +wz + wz') = x(w + y) (e ) (BC' + A'D)( AB' + CD') = BC'AB' + BC'CD' + A'DAB' + A'DCD' = 0 (f) (a' + c')(a + b' + c') = a'a + a'b' + a'c' + c'a + c'b' + c'c' = a'b' + a'c' + ac' + b'c' = c' + b'(a' + c') = c' + b'c' + a'b' = c' + a'b' 2. 4 (a) A'C' + ABC + AC' = C' + ABC = (C + C')(C' + AB) = AB + C' (b) (x'y' + z)' + z + xy + wz = (x'y')'z' + z + xy + wz = [ (x + y)z' + z ] + xy + wz = = (z + z')(z + x + y) + xy + wz = z + wz + x + xy + y = z(1 + w) + x(1 + y) + y = x + y + z (c) A'B( D ' + C'D) + B(A + A'CD) = B (A' D' + A'C'D + A + A'CD) = B(A'D' + A + A'D(C + C') = B(A + A'(D' + D)) = B(A + A') = B (d) (A' + C)(A' + C')(A + B + C'D) = (A' + CC')(A + B + C'D) = A'(A + B + C'D) = AA' + A'B + A'C'D = A'(B + C'D) (e) ABC'D + A'BD + ABCD = AB(C + C')D + A'BD = ABD + A'BD = BD 2.5 (a) x y F F simplified (b) x y F F simplified (c)

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    11 x y z F simplified F (d) B F F simplified A 0 ( e ) x y z F simplified F ( f )

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    12 x y z F F simplified 2. 6 (a) A B C F F simplified (b) x y z F F simplified (c) x y F F simplified

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    13 (d) w x y z F simplified F (e) A B C D F simplified = 0 F (f) w x y z F simplified F 2.7 (a) A B C D F simplified F

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    14 (b) w x y z F simplified F (c) A B C D F simplified F (d) A B C D F simplified F

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    15 (e) A B C D F F simplified 2. 8 F ' = (wx + yz)' = (wx)'(yz)' = (w' + x')(y' + z') FF' = wx(w' + x')(y' + z') + yz(w' + x' )(y' + z') = 0 F + F' = wx + yz + (wx + yz)' = A + A' = 1 with A = wx + yz 2. 9 (a) F' = ( x y' + x'y)' = (xy')'(x'y)' = (x' + y)(x + y') = xy + x'y' (b) F ' = [(a + c) (a + b')(a' + b + c') ]' = (a + c)' + (a + b')' + (a' + b + c')' =a'c' + a'b + ab'c (c) F ' = [ z + z'(v'w + xy) ]' = z'[z'(v'w + xy)]' = z'[z'v'w + xyz']' = z'[(z'v'w)'(xyz')'] = z'[(z + v + w') +( x' + y' + z)] = z'z + z'v + z'w' + z'x' + z'y' +z' z = z'(v + w' + x' + y') 2.10 (a) F 1 + F 2 = Σ m 1i + Σ m 2i = Σ (m 1i + m 2i ) (b) F1 F2 = Σ m i Σ m j where m i m j = 0 if i ≠ j and m i m j = 1 if i = j 2.11 (a) F(x, y, z) = Σ (1, 4, 5, 6, 7) (b) F( a, b, c ) = Σ ( 0, 2, 3, 7) x y z 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 F 0 1 0 0 1 1 1 1 F = xy + xy' + y'z a b c 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 F 1 0 1 1 0 0 0 1 F = bc + a'c' 2. 12 A = 1011 _0001 B = 1010_1100

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    16 (a) A AND B = 1010_0000 ( b ) A OR B = 1011_1 101 (c) A XOR B = 0001_1101 (d) NOT A = 010 0_1110 (e) NOT B = 0101_0011 2.13 (a) u x y Y = [(u + x')(y' + z)] z (u + x') (y' + z) (b) u x y Y = (u xor y)' + x (u xor y)' x (c) u x y Y = (u'+ x')(y + z') z (u'+ x') (y + z') (d) u x y Y = u(x xor z) + y' z u(x xor z) y' (e) u x y Y = u + yz +uxy z u yz uxy (f)

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