basic_engineering_mathematics0

34 Basic Engineering Mathematics
(b) 3F 16 = 3 × 16 1 + F × 16 0 = 3 × 16 + 15 × 1
and 1BF 16 = 1 × 16 2 + B × 16 1 + F × 16 0
Thus 7A 16 = 122 10
= 1 × 16 2 + 11 × 16 1 + 15 × 16 0
= 256 + 176 + 15 = 447 10
= 48 + 15 = 63
Thus 3F 16 = 63 10
Table 5.2 compares decimal, binary, octal and hexadecimal
numbers and shows, for example, that
Problem 12. Convert the following hexadecimal numbers
23 10 = 10111 2 = 27 8 = 17 16
into their decimal equivalents: (a) C9 16 (b) BD 16
Table 5.2
Decimal Binary Octal Hexadecimal
(a) C9 16 = C × 16 1 + 9 × 16 0 = 12 × 16 + 9 × 1
= 192 + 9 = 201
Thus C9 16 = 201 10
0 0000 0 0
1 0001 1 1
(b) BD 16 = B × 16 1 + D × 16 0 = 11 × 16 + 13 × 1
2 0010 2 2
= 176 + 13 = 189
3 0011 3 3
4 0100 4 4
Thus BD 16 = 189 10
5 0101 5 5
6 0110 6 6
7 0111 7 7
Problem 13. Convert 1A4E 16 into a denary number.
8 1000 10 8
9 1001 11 9
10 1010 12 A
1A4E 16 = 1 × 16 3 + A × 16 2 + 4 × 16 1 + E × 16 0
11 1011 13 B
=1 × 16 3 + 10 × 16 2 + 4 × 16 1 + 14 × 16 0
12 1100 14 C
13 1101 15 D
= 1 × 4096 + 10 × 256 + 4 × 16 + 14 × 1
14 1110 16 E
= 4096 + 2560 + 64 + 14 = 6734
15 1111 17 F
16 10000 20 10
Thus 1A4E 16 = 6734 10
17 10001 21 11
18 10010 22 12
19 10011 23 13
20 10100 24 14
To convert from decimal to hexadecimal:
21 10101 25 15
This is achieved by repeatedly dividing by 16 and noting the
22 10110 26 16
remainder at each stage, as shown below for 26 10
23 10111 27 17
24 11000 30 18
16 26 Remainder
25 11001 31 19
26 11010 32 1A
16 1 10 A 16
27 11011 33 1B
28 11100 34 1C
0 1 1 16
29
31
11101
11111
35
37
1D
1F
30
32
11110
100000
36
40
1E
20
most significant bit → 1 A ← least significant bit
Hence 26 10 = 1A 16
Similarly, for 447 10
Problem 11. Convert the following hexadecimal numbers
16
16
447
27
Remainder
15 F 16
into their decimal equivalents: (a) 7A 16 (b) 3F 16
16 1 11 B 16
0 1 1 16
1 B F
(a) 7A 16 = 7 × 16 1 + A × 16 0 = 7 × 16 + 10 × 1
= 112 + 10 = 122
Thus 447 10 = 1BF 16