1
Solution Manual
Digital Design With An Introduction To The Verilog
Hdl Vhdl And System Verilog
By Morris Mano, 6th Edition
, 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 33 34 35 36 37 40
Hex: 10 11 12 13 14 15 16 17 18 19 1a 1b 1c 1d 1e 1f 20
Base-12 14 15 16 17 18 19 1a 1b 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 * 53 + 3 * 52 + 1 * 51 = 58010
(198)12 = 1 * 122 + 9 * 121 + 8 * 120 = 26010
(435)8 = 4 * 82 + 3 * 81 + 5 * 80 = 28510
(345)6 = 3 * 62 + 4 * 61 + 5 * 60 = 13710
1.4 16-Bit Binary:
1111_1111_1111_1111
Decimal Equivalent: 216 -
1 = 65,53510
Hexadecimal Equivalent:
Ffff16
1.5 Let B = Base
(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) = X2 –(6 + 3)X + 6*3 = X2 -11x + 22
Therefore: 6 + 3 = B +
1m, So B = 8 Also,
6*3 = (18)10 = (22)8
, 3
1.7 64cd16 = 0110_0100_1100_11012 = 110_010_011_001 _101 = (62315 )8
1.8 (A) Results Of Repeated Division By 2 (Quotients Are Followed By Remainders):
43110 = 215(1); 107(1); 53(1); 26(1); 13(0); 6(1) 3(0) 1(1)
Answer: 1111_10102 = Fa16
(B) Results Of Repeated Division By 16:
43110 = 26(15); 1(10) (Faster)
Answer: Fa = 1111_1010
1.9 (A) 10110.01012 = 16 + 4 + 2 + .25 + .0625 = 22.3125
(b) 16.516 = 16 + 6 + 5*(.0615) = 22.3125
(c) 26.248 = 2 * 8 + 6 + 2/8 + 4/64 = 22.3125
(d) Dada.B16 = 14*163 + 10*162 + 14*16 + 10 + 11/16 = 60,138.6875
, 4
(e) 1010.11012 = 8 + 2 + .5 + .25 + .0625 = 10.8125
1.10 (A) 1.100102 = 0001.10012 = 1.916
= 1 + 9/16 = 1.56310
(B) 110.0102 = 0110.01002 = 6.416 = 6 +
4/16 = 6.2510
Reason: 110.0102 Is The Same As 1.100102 Shifted To The Left By Two Places.
1011.11
1.11 101 | 111011.0000
101
01001
101
100
1
10
1
10
00
101
0110
The Quotient Is Carried To Two
Decimal Places, Giving 1011.11
Checking: =
1011.112 = 58.7510
1.12 (A) 10000 And 110111
1011 1011
+101
Solution Manual
Digital Design With An Introduction To The Verilog
Hdl Vhdl And System Verilog
By Morris Mano, 6th Edition
, 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 33 34 35 36 37 40
Hex: 10 11 12 13 14 15 16 17 18 19 1a 1b 1c 1d 1e 1f 20
Base-12 14 15 16 17 18 19 1a 1b 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 * 53 + 3 * 52 + 1 * 51 = 58010
(198)12 = 1 * 122 + 9 * 121 + 8 * 120 = 26010
(435)8 = 4 * 82 + 3 * 81 + 5 * 80 = 28510
(345)6 = 3 * 62 + 4 * 61 + 5 * 60 = 13710
1.4 16-Bit Binary:
1111_1111_1111_1111
Decimal Equivalent: 216 -
1 = 65,53510
Hexadecimal Equivalent:
Ffff16
1.5 Let B = Base
(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) = X2 –(6 + 3)X + 6*3 = X2 -11x + 22
Therefore: 6 + 3 = B +
1m, So B = 8 Also,
6*3 = (18)10 = (22)8
, 3
1.7 64cd16 = 0110_0100_1100_11012 = 110_010_011_001 _101 = (62315 )8
1.8 (A) Results Of Repeated Division By 2 (Quotients Are Followed By Remainders):
43110 = 215(1); 107(1); 53(1); 26(1); 13(0); 6(1) 3(0) 1(1)
Answer: 1111_10102 = Fa16
(B) Results Of Repeated Division By 16:
43110 = 26(15); 1(10) (Faster)
Answer: Fa = 1111_1010
1.9 (A) 10110.01012 = 16 + 4 + 2 + .25 + .0625 = 22.3125
(b) 16.516 = 16 + 6 + 5*(.0615) = 22.3125
(c) 26.248 = 2 * 8 + 6 + 2/8 + 4/64 = 22.3125
(d) Dada.B16 = 14*163 + 10*162 + 14*16 + 10 + 11/16 = 60,138.6875
, 4
(e) 1010.11012 = 8 + 2 + .5 + .25 + .0625 = 10.8125
1.10 (A) 1.100102 = 0001.10012 = 1.916
= 1 + 9/16 = 1.56310
(B) 110.0102 = 0110.01002 = 6.416 = 6 +
4/16 = 6.2510
Reason: 110.0102 Is The Same As 1.100102 Shifted To The Left By Two Places.
1011.11
1.11 101 | 111011.0000
101
01001
101
100
1
10
1
10
00
101
0110
The Quotient Is Carried To Two
Decimal Places, Giving 1011.11
Checking: =
1011.112 = 58.7510
1.12 (A) 10000 And 110111
1011 1011
+101
