# Logic Computer Design

Topics: Binary numeral system, Hexadecimal, Bit Pages: 37 (1967 words) Published: December 12, 2012
C HAPTER 1

1-1.

(a)
(1) Calm:

or

(2) 10 mph

(3) 100 mph
(b) The microcomputer requires a table or equation for converting from rotations/second to miles/hour. The pulses produced 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.*
Decimal, Binary, Octal and Hexadecimal Numbers from (16)10 to (31)10 Dec
Bin
Oct
Hex

1-4.

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
22
23
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

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

1-5.

220 = (1,000,00010 + d) where d = 48,576
1Tb = 240 = (220)2 = (1,000,000 + d)2
= (1,000,000)2 + 2(1,000,000) d + d2
= 1,000,000,000,000
+

97,152,000,000

+

2,359,627,776

= 1,099,511,627,776
11 1 Bits
25 1 Bits

1-6.

2 11 – 1 = 2047

2 25 – 1 = 33, 554, 431

7

Problem Solutions – Chapter 1

1-7.*

( 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

1-8.

2|193
2|96
2|48
2|24
2|12
2|6
2|1
0

1
0
0
0
0
1
1

11000001

2|751
2|375
2|187
2|93
2|46
2|23
2|11
2|5
2|2
2|1
0

2|2007
2|1003
2|501
2|250
2|125
2|62
2|31
2|15
2|7
2|3
2|1
0

1
1
1
0
1
0
1
1
1
1
1

11111010111

1
1
1
1
0
1
1
1
0
1

2|19450 0
2|9725 1
2|4862 0
2|2431 1
2|1215 1
2|607 1
2|303 1
2|151 1
2|75 1
2|37 1
2|18 0
2|9 1
2|4 0
2|2 0
2|1 1
0

1011101111

100101111111010

1-9.*
Decimal

Binary

Octal

369.3125

101110001.0101

561.24

171.5

189.625

10111101.101

275.5

BD.A

214.625

1-10.*

11010110.101

326.5

D6.A

62407.625

1111001111000111.101

171707.5

F3C7.A

a)
8|7562
8|945
8|118
8|14
8|1
0
b)
c)

1-11.*

2
1
6
6
1

0.45 × 8 = 3.6 => 3
0.60 × 8 = 4.8 => 4
0.80 × 8 = 6.4 => 6
0.20 × 8 =3.2 => 3

16612

(7562.45)10 = (16612.3463)8
(1938.257)10 = (792.41CB)16
(175.175)10 = (10101111.001011)2

a)

(673.6)8

b)

(E7C.B)16

c)

(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

8

3463

Problem Solutions – Chapter 1

1-12.

a)

1101

b)

0101

c)

100111

×1011

×1010

×011011

1101