Page 23 - computer science (868) class 11
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Example 1: Convert (A6D9)  to decimal.
                                         16
                 Answer:     3      2      1       0            Place value

                             A      6      D       9            Digits

                                          2
                                  3
                         =   10 × 16  + 6 × 16  + 13 × 16  + 9 × 16 0
                                                    1
                         =   40960 + 1536 + 208 + 9
                         =   (42713)
                                  10
                 Example 2: Convert (0.C82)  to decimal.
                                         16
                 Answer:     0                -1      -2       -3
                             0        .       C       8        2

                                           -2
                         =  12 × 16  + 8 × 16  + 2 × 16 -3
                                  -1
                         =  3/4 + 1/32 + 1/2048
                         =   (0.7817)
                                   10

                 1.2.7 Binary to Octal
                 The steps to be followed are:
                 1.  Group all the 1s and 0s in the binary number in sets of three, starting from the left of the binary point for the integer
                   part and right of the point for the fractional part.
                 2.  For the integer part, add necessary zeros to the left of the MSB if it contains less than 3 bits to make a set of three.
                   Similarly, for the fractional part, add 0s to the right if necessary, to make a set of three.
                                             0
                            2
                 3.  Write 4 (2 ), 2 (2 ), and 1 (2 )  above each set of three bits, multiply it with the corresponding bit and add the
                                   1
                   products to get the octal equivalent of each set.
                 4.  Repeat the process for the other sets.
                 5.  Arrange the newly converted octal digits together to form the final result.
                 Example 1: Convert (1101011)  to octal.
                                            2
                 Answer: On grouping 3 bits from LSB  1     101      011
                                                                                2
                                                                                                0
                                                                                        1
                 In leftmost 1 adding two 0s, we get  001   101      011     4 (2 )  2 (2 )  1 (2 )         Octal
                                                     ↓       ↓       ↓         0       0       1     0 × 4 + 0 × 2 + 1 × 1 = 1
                                                    001     101      011       1       0       1     1 × 4 + 0 × 2 + 1 × 1 = 5
                                                   2  2  2 0  2  2  2 0  2  2  2 0  0  1       1     0 × 4 + 1 × 2 + 1 × 1 = 3
                                                      1
                                                    2
                                                                    2
                                                                      1
                                                            2
                                                              1
                                                        001        101         011
                                                         1          5           3
                 (1101011)  = (153)
                          2      8
                 Example 2: Convert (10101.0101)  to octal.
                                               2
                 Answer: Making sets of 3 bits    10   101 . 010      1      4 (2 )  2 (2 )  1 (2 )         Octal
                                                                                        1
                                                                                                0
                                                                                2
                 Adding one 0 before 10 (integer   010  101 . 010    100       0       1       0     0 × 4 + 1 × 2 + 0 × 1 = 2
                 part), and two 0s after 1       ↓      ↓      ↓               1       0       1     1 × 4 + 0 × 2 + 1 × 1 = 5
                 (fractional part), we get      010    101     010   100
                                                                     2 1 0
                                               2  2  2 2  2  2  2  2  2 2  2  2  0     1       0     0 × 4 + 1 × 2 + 0 × 1 = 2
                                                2 1 0
                                                              2 1 0
                                                       2 1 0
                                                                               1       0       0     1 × 4 + 0 × 2 + 0 × 1 = 4
                                                  010      101       .      010     100
                                                   2        5        .       2       4
                 (10101.0101)  = (25.24) 8
                             2
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