9 Digital computation and intelligent devices

 Octal/hexadecimal to binary conversion

Octal and hexadecimal conversion is very simple. Each octal/hexadecimal digit is taken in turn and converted to its binary representation according to the table above.

Example 9.3

Convert the octal number 7654 to binary.

Solution

Using the table above, write down the binary equivalent of each octal digit:

                                      |  7  |   6  |   5  |  4  |

                                      |111|110|101|100|

Thus, the binary code is 111110101100.

Example 9.4

Convert the hexadecimal number ABCD to binary.

Solution

Using the table above, write down the binary equivalent of each hexadecimal digit:

                                         |     A    |     B    |     C    |    D   |

                                          | 1010 | 1011 | 1100 | 1101 |

Thus, the binary code is 1010101111001101.

Binary to octal/hexadecimal conversion

Conversion from binary to octal or hexadecimal is also simple. The binary digits are taken in groups of three at a time (for octal) or four at a time (for hexadecimal), starting at the least significant end of the number (right-hand side) and writing down the appropriate octal or hexadecimal digit for each group.

Example 9.5

Convert the binary number 010111011001 into octal and hexadecimal.

Solution

                                             | 010 | 111 | 011 | 001 |

                                             |   2    |   7   |   3   |    1   |

                                                   = 2731 octal

                                              | 0101 | 1101 | 1001 |

                                              |     5    |     D   |     9    |

                                                     = 5D9 hexadecimal

Example 9.6

The 24-bit binary number 011111001001001101011010 is to be entered into a computer. How would it be entered using (a) octal code and (b) hexadecimal code?

Solution

(a) Divide the 24-bit number into groups of three, starting at the right-hand side:

011 | 111 | 001 | 001 | 001 | 101 | 011 | 010 | = 37111532 octal

Thus, the number would be entered as 37111532 using octal code.

(b) Divide the 24-bit number into groups of four, starting at the right-hand side:

| 0111 | 1100 | 1001 | 0011 | 0101 | 1010 | = 7C935A hexadecimal

In carrying out such conversions, it is essential that the groupings of binary digits start from the right-hand side. Groupings starting at the left-hand side give completely wrong values unless the number of binary digits happens to be an integer multiple of the grouping size. Consider a 10-digit binary number: 1011100011.

Grouping digits starting at the right-hand side gives the values 1343 octal and 2E3 hexadecimal.

Grouping digits starting at the left gives the (incorrect) values of 5611 octal and B83 hexadecimal.

When converting a binary number to octal or hexadecimal representation, a check must also be made that all of the binary digits represent data. In some systems, the first (left-hand) digit is used as a sign bit and the last (right-hand) digit is used as a parity bit.

Example 9.7

In a system that uses the first bit as a sign bit and the last bit as a parity bit, what is the octal and hexadecimal representation of the binary code: 110111000111?

Solution

The ten data bits are 1011100011. This converts to 1343 octal and 2E3 hexadecimal.

 

Programming and program execution

In most modes of usage, including use as part of intelligent devices, computers are involved in manipulating data. This requires data values to be input, processed and output according to a sequence of operations defined by the computer program. However, in practice, programming the microprocessor within an intelligent device is not normally the province of the instrument user, indeed, there is rarely any provision for the user to create or modify operating programs even if he/she wished to do so. There are several reasons for this. Firstly, the signal processing needed within an intelligent device is usually well defined, and therefore it is more efficient for a manufacturer to produce this rather than to have each individual user produce near identical programs separately. Secondly, better program integrity and instrument operation is achieved if a standard program produced by the instrument manufacturer is used. Finally, use of a standard program allows it to be burnt into ROM, thereby protecting it from any failure of the instrument power supply. This also facilitates software maintenance and updates, by the mechanism of the manufacturer providing a new ROM that simply plugs into the slot previously occupied by the old ROM.

However, even though it is not normally a task undertaken by the user, some appreciation of microprocessor programming for an intelligent device is useful background knowledge. To illustrate the techniques involved in programming, consider a very simple program that reads in a value from a sensor, adds a pre-stored value to it to compensate for a bias in the sensor measurement, and outputs a corrected reading to a display device.

Let us assume that the addresses of the sensor and output display device are 00C0 and 00C1 respectively, and that the required scaling value has already been stored in memory address 0100. The instructions below are formed from the instruction set for a Z80Ł microprocessor and make use of CPU registers A and B.

                                                        IN A,C0

                                                        IN B,100

                                                        ADD A,B

                                                        OUT C1,A

This list of four instructions constitutes the computer program that is necessary to execute the required task. The CPU normally executes the instructions one at a time, starting at the top of the list and working downwards (though jump and branch instructions change this order). The first instruction (IN A,C0) reads in a value from the sensor at address C0 and places the value in CPU register A (often called the accumulator). The mechanics of the execution of this instruction consist of the CPU putting the required address C0 on the address bus and then putting a command on the control bus that causes the contents of the target address (C0) to be copied onto the data bus and subsequently transferred into the A register. The next instruction (IN B,100) reads in a value from address 100 (the pre-stored biasing value) and stores it in register B. The following instruction (ADD A,B) adds together the contents of registers A and B and stores the result in register A. Register A now contains the measurement read from the sensor but corrected for bias. The final instruction (OUT C1,A) transfers the contents of register A to the output device on address C1.