Problem Solutions – Chapter 1
1
CHAPTER 1
1-1. (a)
(1) Calm:
(2) 10 mph
or
1-2.
–34° quantizes to –30° => 1 V => 0001
1-3.*
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
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 1111
Oct
20
21
22
23
24
25
26
27
30
31
32
33
34
35
37
1-4.
10
128 128 2 131,072
 
K Bits
Problem Solutions – Chapter 1
2
1-5.
220 = (1,000,00010 + d) where d = 48,576
1-6.
1-7.*
6 3 2 0
2
(1001101) 2 2 2 2 77
 
1-8.
2|187 1 10111011 2|891 1 1101111011
2| 93 1 2|445 1
2|2014 0 11111011110 2|20486 0 101 0000 0000 0110
2|1007 1 2|10243 1
2|503 1 2|5121 1
2|251 1 2|2560 0
Problem Solutions – Chapter 1
1-9.*
Decimal
Binary
Octal
Hexadecimal
369.3125
101110001.0101
561.24
171.5
275.5
326.5
D6.A
F3C7.A
1-10.*
a) 8|7562 2 16612 0.45 × 8 = 3.6 => 3
8|945 1 0.60 × 8 = 4.8 => 4
1-11.*
a) (673.6)8 = (110 111 011.110)2
1-12.
a) 1010 b) 0110 c) 1111001
×1100 ×1001 ×011101
0000 0110 1111001
0000 0000 000000
Problem Solutions – Chapter 1
4
1-13.+
10001
Quotient = 10001
1-14.
(a) 6 × 123 + 8 × 122 + 7 × 121 + 4 = 11608
0
1-15.
a) 0 1 2 3 4 5 6 7 8 9
20 10
1-16.*
a) (BEE)r = (2699)10
2 1 0
11 14 14 2699
r r r
    
Problem Solutions – Chapter 1
1-17.
Errata: The text has an error: 1480 should be 1460. This will be corrected in future printings.
Noting the order of operations, first add (34)r and (24)r
10
(34) 3 4
rr
 
1-18.*
a) (0100 1000 0110 0111)BCD = (4867)10
1-19.*
(694)10 = (0110 1001 0100)BCD
(835)10 = (1000 0011 0101)BCD
1
0110 1001 0100
1-20.*
(a) 101 100
0111 1000
Move R 011 1100 0 100 column > 0111
Subtract 3 −0011
011 1001 0
(b) 102 101 100
0011 1001 0111
Move R 001 1100 1011 1 101 and 100 columns > 0111
Subtract 3 −0011 -0011
001 1001 1000 1
1-21.
(a) 102 101 100
1111000
1st Move L 1 111000
2nd Move L 11 11000
3rd Move L 111 1000 100 column > 100
(b) 103 102 101 100
01110010111
1st Move L 0 1110010111
2nd Move L 01 110010111
Problem Solutions – Chapter 1
1010 0010111
5th Move L 1 0100 010111
1-22.
From Table 1-5, complementing the bit B6 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
0100 1110 N 0101 0100 T 0010 0000 (SP)
1-24. 1000111 G
1101111 o
0100000
1000011 C
011
101
100
001
111
1-25.*
a) (11111111)2
1-26.
a) U+0040 = 01000000
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
(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
101111 1
1-28.
Gray Code for Hexadecimal Digits
1-29.
(a) Wind Direction Gray Code
Direction
Code Word
N
000
S
110
Problem Solutions – Chapter 1
9
(b) Wind Direction Gray Code (directions in adjacent order)
Direction
Code Word
N
000
1-30.+
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 =
2n.
001
E
011
010
110
111
101
100