Programable Logic Controller questions. PLCS

MODULE TITLE: PROGRAMMABLE LOGIC CONTROLLERS

TOPIC TITLE: INTRODUCTORY BASIC THEORY

LESSON 2: NUMBER SYSTEMS

PLC - 1 - 2

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________________________________________________________________________________________

INTRODUCTION ________________________________________________________________________________________

In this lesson we shall learn what we mean by a number ‘base’ and consider

number systems other than the one that we use in everyday life, the denary or

decimal number system. Central processing units of computers require digital

signals, i.e. two-state signals, which make the binary system of numbers

convenient for processing taks. The binary numbering system, however, is

unwieldly and the ‘hexadecimal’ system is a much more compact system and

more convenient to use for tasks associated with computers. As programmable

logic controllers are essentially computers, it will benefit us to understand

these different numbering systems.

________________________________________________________________________________________

YOUR AIMS ________________________________________________________________________________________

On completing this lesson you should be able to:

• explain what is meant by positional notation

• define the numer bases denary, binary and hex

• convert between one number base and another

• perform binary addition and subtraction

• understand what is meant by the complement of a number

• understand binary addition involving negative binary numbers

• understand BCD systems of binary coding.

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________________________________________________________________________________________

NUMBER SYSTEMS ________________________________________________________________________________________

We are used to counting things in batches of ten. That is to say, we normally

work to a base of ten. Habitual use of this number base may, mistakenly, lead

us to the conclusion that the number ten has some special property not to be

found in other numbers. This is not so and the choice of ten as our ‘everyday’

number base is due only to the biological fact that we have a total of ten digits

on our two hands. The number ten has no special mathematical significance.

The numbering system to a base of ten is called the DECIMAL or DENARY

system. In the denary system the symbols 0, 1, 2 ......9 are used to represent

zero, one, two ......nine objects. If we count beyond nine then POSITIONAL

NOTATION is used, so that ten objects are represented by the two symbols

‘10’, eleven by ‘11’, etc. In positional notation the numerical weighting of

each digit increases by a factor of ten each time we move one position to the

left. This gives us the ‘units, tens, hundreds and thousands’ of the denary

system. Thus the symbols ‘1011’ in denary are taken to represent:-

[(1 × 103) + (0 × 102) + (1 × 101) + (1 × 100)] Thousands Hundreds Tens Units

(Remember, any number raised to the power of zero is equal to 1.

Thus 100 = 1, 80 = 1, x0 = 1)

Although what we have outlined above might seem obvious, it is important to

remember that not all numbering systems have used the same symbols and

notation as used in denary. Consider, for example, Roman numerals which use

the symbols I, V, X, L, C, D and M for the numbers one, five, ten, fifty,

one hundred, five hundred and one thousand respectively. Thus 1999 is

represented by MCMXCIX!

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We can also have denary numbers with a decimal point, such as

So, 101.101 represents (1 × 102) + (0 × 101) + (1 × 100) + (1 × 10–1) + (0 × 10–2) + (1 × 10–3)

OTHER NUMBER BASES

When we wish to work to bases other than ten, then it makes sense to use, as

far as possible, the same symbols as used in denary and also to employ

positional notation. Thus, if working to a base of 8 (OCTAL) then we will use

the symbols 0 – 7 and the numerical weighting of each digit will increase by a

factor of eight for each move to the left.

Thus the number ‘1011’ in OCTAL represents

[(1 × 83) + (0 × 82) + (1 × 81) + (1 × 80)].

You may care to check that this is equivalent to the denary number ‘521’.

(1 × 83 = 512, 0 × 82 = 0, 1 × 81 = 8, 1 × 80 = 1.) To avoid confusion about which number base is being used, a subscript may be added to the

number to indicate its base. Thus:

10118 = 52110 base of 8 base of 10

(Note that the subscript is always given in denary. The notation

101110 = 52112 (giving the bases in octal) really could cause confusion!)

0 1 1

10 1

10 10

0 01 1

100 1

10 10

1 1

2

.

.

–= = =

= = =and –– ,2 etc.

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BINARY NUMBERS

Digital computers use two-state devices (the states represented by ‘0’ and ‘1’)

to store and process data. The binary counting system, using a base of two,

must therefore be used. In binary, only the symbols ‘0’ and ‘1’ are required

and each position in a binary number represents a power of two.

Thus the binary number ‘1011’ represents:

[(1 × 23) + (0 × 22) + (1 × 21) + (1 × 20)] giving 10112 = 1110.

Similarly, the binary number ‘11000111’ represents:

[(1 × 27) + (1 × 26) + (0 × 25) + (0 × 24) + (0 × 23) + (1 × 22) + (1 × 21) + (1 × 20)]

= 128 + 64 + 4 + 2 + 1 = 199 in denary

i.e. 110001112 = 19910

Most significant bit Least significant bit

27

128

0

26

64

0

25

32

0

24

16

1

23

8

1

22

4

0

21

2

0

20

1

0

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BINAY NUMBERS WITH ‘DECIMAL’ POINTS

As with decimal numbers, fractional binary numbers can be represented with a

binary point. For example:

and

As we can have only 0 or 1 in binary then

0.5 in decimal becomes 0.1 binary

3.5 decimal would become 11.1 in binary

3.75 decimal would become 11.11 (i.e. 3 + 1/2 + 1/4).

These rules also apply to other bases.

binary 0.1 = 1 2

=

binary 0.01 = 1

2 = ,

2

2

2

1

2

–

– eetc

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CONVERTING BINARY NUMBERS TO DECIMAL

Example 1

Convert 1100011102 to denary.

Solution

i.e. 1100011102 = 39810

It is immediately apparent that whilst the binary system requires only two

symbols (and is therefore ideally suited to use with two-state devices) its

number length becomes very cumbersome when representing large numbers.

For example, ‘one thousand’ in binary is ‘1111101000’. To avoid having to

handle very long binary numbers, the number is often split into groups of four

bits – each pattern of four bits having its own symbol. This gives rise to a new

number system – the HEXADECIMAL system (or simply ‘HEX’).

110001110 1 2 1 2 0 2 0 2

0

2 8 7 6 5= ×( ) + ×( ) + ×( ) + ×( )⎡⎣

+ × 22 1 2 1 2 1 2

0 2

256 128

4 3 2 1

0

( ) + ×( ) + ×( ) + ×( )

+ ×( )⎤⎦ = + ++ + + =8 4 2 398.

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HEXADECIMAL NUMBER SYSTEM

Although a computer works in binary, the data may be entered into the

machine using ‘HEX’, as HEX is more concise and easier for us to handle.

The machine will then convert the HEX to binary. The HEX system uses a

base of sixteen – but what symbols shall we use in a system that has more than

ten digits? We could invent a completely new set of symbols but these would

have to be memorised; it is best that we stick with the symbols we are already

familiar with. In fact, in HEX, the numerical symbols 0 – 9 and the

alphabetical symbols A – F have been chosen. The table below and overleaf

gives the HEX representation of the denary numbers 0 to 32. The binary

equivalents are also given (expressed as 8 bit numbers).

DENARY HEX BINARY

0 0 0000 0000

1 1 0000 0001

2 2 0000 0010

3 3 0000 0011

4 4 0000 0100

5 5 0000 0101

6 6 0000 0110

7 7 0000 0111

8 8 0000 1000

9 9 0000 1001

10 A 0000 1010

11 B 0000 1011

12 C 0000 1100

13 D 0000 1101

14 E 0000 1110

15 F 0000 1111

16 10 0001 0000

17 11 0001 0001

18 12 0001 0010

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19 13 0001 0011

20 14 0001 0100

21 15 0001 0101

22 16 0001 0110

23 17 0001 0111

24 18 0001 1000

25 19 0001 1001

26 1A 0001 1010

27 1B 0001 1011

28 1C 0001 1100

29 1D 0001 1101

30 1E 0001 1110

31 1F 0001 1111

32 20 0010 0000

• • •

• • •

• • •

48 30 0011 0000

• • •

• • •

• • •

96 60 0110 0000

• • •

• • •

• • •

255 FF 1111 1111

This ‘alphanumeric’ system of symbols in a counting system may seem a little

strange at first, but with a little practice you will soon get used to it! Note that

in the HEX system, positional notation is still used as illustrated by the

following example.

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CONVERSION FROM HEX TO DENARY

Example 2

Convert the HEX number 7A4E into denary.

Solution

From the preceding table A = 1010 and E = 1410. By means of positional

notation we can write:

7A4E = [(7 × 163) + (10 × 162) + (4 × 161) + (14 × 160) = 31310

163 162 161 160

Should there be any risk of confusion over which number base is being used

then the convention is that the HEX number is followed by a capital ‘H’.

Thus 13H = 1910, 21H = 3310, 1000H = 409610.

In this module we shall only be using binary, denary or HEX systems and

therefore any number with alphabetical symbols will automatically be HEX.

OTHER BASE CONVERSIONS

Hex to Binary

This is particularly easy as the HEX base of 16 is equal to 24. All we need do

is convert each HEX symbol into its four bit binary equivalent using our

conversion table.

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Example 3

Convert ABCD to binary.

Solution

A = 1010, B = 1011, C = 1100 and D = 1101

Thus ABCD = 1010 1011 1100 11012

It would be a useful exercise for you to check this result by converting both

numbers independently to denary.

Binary to Hex

This is simply the reverse of the above process. The binary number is split

into groups of four starting with the least significant bit (LSB):

Example 4

Convert 1010 0000 1111 00112 to HEX

Solution

1010 0000 1111 0011 = A0F3

| | | |

A 0 F 3

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________________________________________________________________________________________

ADDITION OF BINARY NUMBERS ________________________________________________________________________________________

In principle, binary numbers are added in exactly the same way as in denary

addition – but remember we are working to a base of two! This means that

(1 + 1) will generate a ‘carry’. The rules of binary addition are summarised

below:

0 0 1 1

+ 0 + 1 + 0 + 1

0 1 1 0 carry 1

You may find it worthwhile to write these rules down several times so that you

remember them.

Then study the following examples.

Example 5

Add the binary numbers 1010 and 0011.

Solution

1 0 1 0

+ 0 0 1 1

1 1 0 1

carry

The bits in each column are added together and any ‘carry bit’ generated is

added into the column on the left.

∪

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Example 6

Add the binary numbers 1011 and 0011.

Solution

* 1 0 1 1

+ 0 0 1 1

1 1 1 0

carry

In this example the column marked with an asterisk represents the addition

(1 + 1 + 1) = 11. This means that 1 is entered into this column and one is

carried over to the next most significant column.

∪∪

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________________________________________________________________________________________

BINARY SUBTRACTION ________________________________________________________________________________________

There is no theoretical reason why binary numbers cannot be subtracted using

the same rules as for denary subtraction, i.e. starting with the least significant

column we subtract one bit from another. The rules for binary subtraction will

then be:

0 1 1 0

– 0 – 0 – 1 – 1 borrow 1 from left hand

0 1 0 1 column

In the fourth rule, ‘1’ from ‘0’ ‘won’t go’ and we have to borrow a bit from the

adjacent column. This now forms the equivalent of (2 – 1) = 1.

Example 7

Subtract 0111 from 1111.

Solution

1 1 1 1

– 0 1 1 1

1 0 0 0

Note that the answer can be checked by adding the result to the number

subtracted.

0 1 1 1

+ 1 0 0 0

1 1 1 1

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Example 8

Subtract 0011 from 1101.

Solution

1 1 0 1 In column 2 we have to borrow

– 0 0 1 1 from column 3.

1 0 1 0

borrow

Good fun as it might be(!), this is as far as we shall go in binary subtraction.

The reason for this is that computers do not perform subtraction in this way.

There are two reasons why they do not:

(a) Binary addition is carried out in the arithmetic logic unit (ALU) built into

the CPU. The addition is done by special ‘adder ’ hardware. If

subtraction were to be done in the ‘conventional’ way then ‘subtractor’

hardware would also be required.

(b) Subtraction can give rise to negative results. Thus we need a means of

representing positive and negative numbers. Conventionally we do this

by using the symbols ‘+’ and ‘–’. We must remember, however, that a

binary machine will only recognise two symbols (namely ‘0’ and ‘1’).

We cannot therefore use extra symbols to denote a number’s sign.

∪

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________________________________________________________________________________________

REPRESENTATION OF NEGATIVE NUMBERS ________________________________________________________________________________________

Two methods will be considered. The first of these is widely used in data

communications systems.

(a) THE SIGN-MAGNITUDE METHOD

The magnitude of a number is its value, ignoring its sign. Thus the magnitude

of ‘+3’ is ‘3’ and of ‘–8’ is ‘8’.

The magnitude of ‘minus one hundred’ is equal to the magnitude of ‘plus one

hundred’.

In the sign magnitude method, the sign of a binary number is given by the

MSB. The remaining bits give the magnitude of the number. If positive

numbers are represented by the MSB being set to ‘0’, then negative numbers

are represented by the MSB being set to ‘1’. To illustrate the idea, the binary

equivalents of –5 to +5 are listed below (using 1 byte numbers):

SIGN BIT MAGNITUDE ___________________________

–5 1 0000101 ___________________________

–4 1 0000100 ___________________________

–3 1 0000011 ___________________________

–2 1 0000010 ___________________________

–1 1 0000001 ___________________________

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–0 1 0000000 ___________________________

+0 0 0000000 ___________________________

+1 0 0000001 ___________________________

+2 0 0000010 ___________________________

+3 0 0000011 ___________________________

+4 0 0000100 ___________________________

+5 0 0000101 ___________________________

As a matter of curiosity, note that this method gives two zeros, 10000000

(minus zero) and 00000000 (plus zero)! This, however, does not cause any

problems.

Also, there is, as a matter of fact, no hard and fast rule as to whether a ‘1’ or a

‘0’ is used to represent a positive number. Which convention is adopted will

depend upon the particular system.

The advantage of this method is the simplicity with which it allows the

magnitude of the number to be quickly established. A serious disadvantage

though, is that the method cannot easily be used for binary arithmetic.

Consider the following examples.

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Example 9

Add (+3) to (–3) using binary signed numbers.

Solution

+3 = 00000011 and –3 = 10000011

0 0 0 0 0 0 1 1

+ 1 0 0 0 0 0 1 1

1 0 0 0 0 1 1 0

Clearly the result should have been zero ((+3) + (–3) = 0) but the addition

has come up with –6.

Example 10

Add (–5) to (+2).

Solution

–5 = 10000101 and +2 = 00000010

1 0 0 0 0 1 0 1

0 0 0 0 0 0 1 0

1 0 0 0 0 1 1 1

This time the answer is –7 and not –3 as it should be.

This system is clearly no good at arimethic. Nevertheless it does have the

value of simplicity and is used in systems where the data does not have to be

processed arithmetically.

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(b) THE TWO’S COMPLEMENT METHOD

The ‘one’s complement’ of a binary number is obtained by changing all the

‘1’s to ‘0’s and vice versa. (This process is known as INVERTING the

number.) A few examples of binary numbers and their ‘one’s complements’

are given below.

NUMBER ONE’S COMPLEMENT

0 0 0 0 1 1 1 1

1 0 1 0 0 1 0 1

0 0 1 1 1 1 0 0

1 1 1 1 0 0 0 0

The two’s complement of a binary number is formed by taking the one’s

complement and adding 1 to it.

Example 11

Find the ‘two’s complement’ of 14. Express the answer as an 8 bit number.

Solution

1410 = 00001110

‘One’s complement’ = 11110001

‘Two’s complement’ = (‘one’s complement’ + 1)

= (11110001 + 1)

= 11110010

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The ‘two’s complement’ form can be used to represent the negative of a binary

number which can be used in arithmetic operations. It may be a bit tricky for

us to use and recognise the ‘two’s complement’ form, but relatively simple

arithmetic circuits in the CPU are quite at home with them. To convince

yourself that the ‘two’s complement’ representation of binary numbers works,

study the following examples.

Example 12

Use the ‘addition of the two’s complement’ method perform the operation

(25 – 14).

Solution

(25 – 14) may be rewritten as (25 + (–14)).

The procedure is to represent (–14) in two’s complement form and then add

this to 25.

2510 = 0 0 0 1 1 0 0 1

add (–1410) = 1 1 1 1 0 0 1 0

[1] 0 0 0 0 1 0 1 1

↑ carry

As we are working in 8 bit numbers, the ‘carry one’ is ignored. The result is,

therefore, 000010112 = 1110, the correct answer.

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Example 13

Use the ‘two’s complement’ representation show that 25 – 25 = 0.

Solution

25 = 0 0 0 1 1 0 0 1

add (–25) = 1 1 1 0 0 1 1 1

[1] 0 0 0 0 0 0 0 0

↑ carry

Again, the carry bit is ignored and a zero result is obtained.

Example 14

Use the ‘two’s complement’ representation evaluate (14 – 25).

Solution

14 = 0 0 0 0 1 1 1 0

add (–25) = 1 1 1 0 0 1 1 1

1 1 1 1 0 1 0 1

This answer highlights one of the difficulties of the ‘two’s complement’ form.

In the ‘complement’ system all negative numbers have a MSB of 1, so we

know our result is negative, but the magnitude of the number is by no means

obvious!

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Before attempting to evaluate, in denary, the solution to example 14, let us

investigate our new system a little more closely. Initially, for simplicity, we

shall use 4 bit signed numbers. This gives a numbering range of (–8) to (+7)

as indicated on the ‘number line’ shown below.

The clue to evaluating the magnitude of a negative number represented in

‘two’s complement’ form is to note that, for example, (–3) = (–8 + 5).

Looking at the number here we see that (–3) = 1101. Thus the ‘two’s

complement’ form can be regarded as consisting of a mixture of two numbers:

(a) a negative number represented by the MSB and equal to (–2n–1) where n

is the bit length of the number

(b) a positive number evaluated from the remaining bits.

The result is, of course, obtained by adding the two numbers together.

[Those students who have studied logarithms will perhaps recall that a

similar technique was used to represent negative logarithms by means of

‘bar’ numbers. Thus represents (–1 + 0.1234).]1 1234.

NEGATIVE NUMBER POSITIVE NUMBER

(–2n–1) = (–23) = –8 (22 + 20) = 5

1 1 0 1

–8 –7 –6 –5 –4 –3 –2 –1 0 +1 +2 +3 +4 +5 +6 +7

1000

(–8+0)

1001

(–8+1)

1010

(–8+2)

1011

(–8+3)

1100

(–8+4)

1101

(–8+5)

1110

(–8+6)

1111

(–8+7)

0000 0001 0010 0011 0100 0101 0110 0111

DENARY

BINARY

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We are now in a position to return to Example 14. The result is an 8 bit signed

number and thus the number range will be from (–128) to (+127). This range

was evaluated using the formulae:

(a) lower limit = (–2n–1) giving –128 for n = 8

(b) upper limit = (2n–1 – 1) giving +127 for n = 8

Splitting (11110101) into two parts yields:

(–28–1) = –128 (26 + 25 + 24 + 22 + 20) = 117

The answer, in denary is, therefore, (–128 + 117) = –11.

Alternative method

An alternative way to find the 2’s complement, instead of firstly finding the 1’s

complement is:

(a) Write down the binary number, starting with the LSB, up to and including

the first ‘1’.

(b) Invert the remaining bits.

The following example shows how this is done.

Example 15

Write down the ‘two’s complement’ of 01110010.

11110101 � �� ��

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Solution

invert these bits to give 'up to and including the first '1" gives

'1 0 0 0 1 1' '1 0'

Thus ‘two’s complement of 01110010 is 10001110.

Example 16

Perform the arithmetic operation (17 – 97) in binary. Express your answer in

‘two’s complement’ form and then convert it to denary. Work in 8 bit

numbers.

Solution

17 = 00010001

97 = 01100001

∴ (–97) = 10011111 (in ‘two’s complement)

Adding (–97) to (+17):

0 0 0 1 0 0 0 1

+ 1 0 0 1 1 1 1 1

1 0 1 1 0 0 0 0

Thus in ‘two’s complement’ the result is 10110000 (which is clearly negative

as the MSB is set to ‘1’).

01110 010 ��� �� �

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Converting this to denary:

10110000 = (–28–1) + (25 + 24)

= (–128 + 48)

= –80

This technique can be used in reverse for easy identification of the magnitude

of negative binary numbers if the binary number is known to be negative.

Consider the result to Example 16. Applying the technique in reverse we:

(a) write down the number up to and including the first ‘1’ (yielding – 10000)

(b) invert all the other bits (giving ‘01010000’).

The magnitude is thus ‘01010000’ = 80

As we know we are dealing with a negative number the answer is –80.

Which method you use (the more formal ‘[–2n -1 + (positive number)]’ or the

‘trick’) is entirely up to you – there is little to choose between them in terms of

speed of conversion.

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________________________________________________________________________________________

BINARY CODED DECIMAL ________________________________________________________________________________________

This is a system of representing each denary digit by its binary equivalent.

The system is based on a weighting system of 8421. This simply means that

each decimal digit is represented by 4 binary digits and that the binary LSB is

represented by 20 (i.e. 1) through to the MSB which is 23 (i.e. 8). The

weighting is therefore 8421.

The number 457 will be represented, therefore, with four binary digits for each

denary or decimal number as:

0100 0101 0111

4 5 7

BCD is commonly used with input and output devices. A thumbwheel switch is

one example of an input device that uses BCD. The binary numbers are broken

into groups of four bits, each group representing a decimal equivalent. A four-

digit thumbwell switch, like the one shown here, would control 16 (4 × 4) PLC inputs.

Decimal numbers

0 1 2 3 4 5 6 7 8 9

0 2 0 5

0000 0010 0000 0101

BCD numbers

0000 0001 0010 0011 0100 0101 0110 0111 1000 1001

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BINARY CODED DECIMAL (BCD) AND THE TEN'S COMPLEMENT

Two's complement notation is used to represent negative binary numbers. For

example, in two's complement 1111 1111 represents minus one. The

complement of a number, however, is not peculiar to the binary system.

The complement notation can be used to represent negative numbers in any

number base. Here we shall consider the representation of signed BCD

numbers using the ten's complement.

Quickly, mentally subtract 7 from 6 and then compare your result with that

below :

6

– 7

(1) 9

This result, at first sight, might seem a little strange, for, no doubt, you

expected an answer of – 1! But how did you arrive at your answer? I suspect

you looked at the two numbers, decided that 7 was the larger and then

subtracted 6 from it. You then called the result negative. This technique is fine

for humans but it does require a little intuition, a property for which computers

are not particularly renowned. In subtracting two numbers they will perform

the operation in the order in which the numbers are given. To carry out the

above calculation, therefore, a computer will borrow when attempting to

subtract 7 from 6 so that the calculation becomes 7 from 16 to give 9. The 1

that has been borrowed is shown in brackets in the working above. If the

machine had a two-digit display, then 99 would be displayed.

As a further illustration imagine an odometer (‘mileometer’) set to ...0006 miles

and then wound back 7 places. The resulting display would be ...9999, i.e. ‘9’

followed by a long succession of nines. This is, in fact, the ten’s complement

representation of minus one.

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The number line below shows the numbers –10 to +9, the negative numbers

being represented by their two-digit, ten’s complement form. Note that in such

representation all negative numbers are prefixed by a nine and all positive

numbers by a zero.

When working in BCD, negative numbers will be represented by the computer

in their ten's complement form. So for example, 17 – 39 will be displayed as

78 on a two-digit display. The ten’s complement of a single decimal digit can

be found by subtracting it from ten.

(where N is the number and its ten’s

complement)

i.e. a denary number and its ten’s complement

equals 10. Generally, in any base, a number

plus its complement equals the base.

However, it is easier to program a machine to first form the nine’s complement

of the number and then add 1 to the result. (The nine’s complement is formed

by subtracting the number from 9). This process avoids having to ‘borrow’.

The above equation can be re-written as:

But (9 – N) is the nine’s complement of N.

N N

N N

10

10

9 1

9 1

′

′

= +( )

∴ = ( ) +

–

–

or N N10 10 ′ + =

N10 ′N N10 10

′ = –

Ten's complement Signed numbers

90 91 92 93 94 95 96 97 98 99 00 01 02 03 04 05 06 07 08 09

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 +1 +2 +3 +4 +5 +6 +7 +8 +9

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Hence

This process should be compared with the formation of the two’s complement

in natural binary. You'll recall that to form the two’s complement of a binary

number, the number is first complemented by inverting the bits and then

adding 1. The table below gives the nine’s and ten’s complement in BCD of

the numbers 0 to 9.

Decimal Number BCD nine’s complement ten’s complement

0 0000 1001 1010

1 0001 1000 1001

2 0010 0111 1000

3 0011 0110 0111

4 0100 0101 0110

5 0101 0100 0101

6 0110 0011 0100

7 0111 0010 0011

8 1000 0001 0010

9 1001 0000 0001

What is the result of adding a BCD number to its ten’s complement?

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....................................................................................................................................................

__________________________________________________________________________

The result is always ten (1010).

where represents the nine's complementN9 ′( )).

N N10 9 1 ′ ′= +

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Can the nine's complement be formed by bit inversion?

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__________________________________________________________________________

No, it is one of the disadvantages of this particular BCD code that the nine’s complement

cannot be formed by simple bit inversion. This does make arithmetic operations more

difficult.

We shall now look at examples of other BCD codes in which bit inversion can

be used to form the nine’s complement. Codes which possess this property are

said to be self-complementary.

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________________________________________________________________________________________

OTHER BCD CODES ________________________________________________________________________________________

The BCD above is known as 8-4-2-1 BCD, the 8, 4, 2 and 1 referring to the

weighting given to each bit position. Various other BCD codes have been

devised, usually for processors which do not have decimal adjust instructions,

to make arithmetic operations easier. We shall briefly acquaint ourselves with

one such code.

2-4-2-1 BCD

The most significant bit of 2-4-2-1 BCD code has a weighting of only two,

instead of the customary eight. This is done so that the code automatically

compensates for the difference of six between the HEX and decimal number

bases.

The code is listed below. Note that a disadvantage of 2-4-2-1 BCD is that

some of the numbers have two code words. For example, five can be

represented by 1011 or by 0101.

Decimal (2-4-2-1)BCD nine’s complement ten’s complement

0 0000 1111 0000

1 0001 1110 1111

2 0010 (1000) 1101 1110

3 0011 (1001) 1100 1101

4 0100 (1010) 1011 1100

5 1011 (0101) 0100 0101

6 1100 (0110) 0100 0100

7 1101 (0111) 0011 0011

8 1110 0001 0010

9 1111 0000 0001

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The code is self-complementary, which means that the ten’s complement can

be formed by an ALU (Arithmetic and Logic Unit) designed to process natural

binary, that is by bit inversion and adding one to the result.

An example will show how 2-4-2-1 BCD works.

Example 17

Add eight to nine using 2-4-2-1 BCD.

Solution

decimal 2-4-2-1 BCD

8 1110

add 9 1111

17 1 1101

The 1 carried over in the BCD sum gives the correct result of seventeen

without the need for any adjustment.

That completes our look at number systems. Now try the Self-Assessment

Questions that follow to test your understanding of the lesson.

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________________________________________________________________________________________

SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. Convert the following binary numbers to denary:

(a) 100000

(b) 1010101

(c) 000101

2. Convert the following binary numbers to HEX:

(a) 1010

(b) 11101001

(c) 1011100001111111

3. Convert the following HEX numbers to binary:

(a) A0

(b) C2

(c) BEF

4. Find the ‘two’s complement’ of 25.

5. Add seven to eight using 2-4-2-1 BCD.

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________________________________________________________________________________________

NOTES ________________________________________________________________________________________

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________________________________________________________________________________________

ANSWERS TO SELF-ASSESSMENT QUESTIONS ________________________________________________________________________________________

1. (a)

(b)

(c)

2. (a) 10102 = A

(b) 1110 10012 = E9

(c) 1011 1000 0111 11112 = B87F.

3. (a) A0 = 10100000

(b) C2 = 11000010

(c) BEF = 101111101111.

000101 1 2 0 2 1 2

5

2 2 1 0

10

= ×( ) + ×( ) + ×( )⎡⎣ ⎤⎦ =

1010101 1 2 0 2 1 2 0 2

1 2

2 6 5 4 3

2

= ×( ) + ×( )⎡⎣ + ×( ) + ×( )

+ ×(( ) + ×( ) + ×( )⎤⎦ = + + + + + +

=

0 2 1 2

64 0 16 0 4 0 1

85

1 0

10

100000 1 2 0 2 0 2 0 2

0 2

2 5 4 3 1

0

= ×( ) + ×( ) + ×( )⎡⎣ + ×( )

+ ×( ))⎤⎦ = 3210

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4.

5. Decimal 2-4-2-1 BCD

7 1101

8 1110

15 1 1011

25 00011001

11100

10 2=

∴ = 'two's complement' 1110 1 11100111+( ) =

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________________________________________________________________________________________

SUMMARY ________________________________________________________________________________________

We have encountered what is meant by a number system and should now be

aware of the following.

(a) Numbers consist of an ordered set of symbols, the numerical weighting of

each symbol increasing by a constant factor (called the BASE) as we

move from right to left. Such a numbering system uses POSITIONAL

NOTATION.

(b) In the denary (or decimal) system a base of TEN is used.

e.g. 3061 = [(3 × 103) + (0 × 102) + (6 × 101) + (1 × 100)]

(c) In the binary system a base of TWO is used.

e.g. 10112 = [(1 × 22) + (1 × 21) + (1 × 20)]

(d) In HEXADECIMAL a base of 16 is used.

e.g. ABC = [(10 × 162) + (11 × 161) + (12 × 160)]

(e) Conversion from binary and HEX to denary is by straight application of

positional notation.

We have also encountered other forms of binary coding. BCD is convenient

for representing decimal numbers in a binary form.

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