There are two basic types of circuits that may be considered as digital devices: Logic
Gates, and Flip-flop. Electronic calculator is an example of a digital circuits, where information is processed in binary form, and the output are displayed as decimal
numbers. The gradual turning of the potentiometer shaft is the analog input in a circuit
consisting of a battery, potentiometer and ammeter all in series.
Examples of analog signals include: sine wave, audio and video signals, while the
square wave is an example of a digital signal. A digital signal has two distinct am-
plitudes such as 0 and +5 V. The pulse is either all on or all off i.e. High or Low.
Activity 4.1
Number Systems
(i) Manipulation of numbers in different bases
A number system is a set of numbers, together with one or more operations, such as
addition or multiplication.
Examples of number systems include: natural numbers, integers, rational numbers,
algebraic numbers, real numbers, complex numbers, p-adic numbers, surreal num-
bers, and hyperreal numbers.
Numeral
The numerals used when writing numbers with digits or symbols can be divided into
two types that might be called the arithmetic numerals 0,1,2,3,4,5,6,7,8,9 and the
geometric numerals 1,10,100,1000,10000... respectively.
There are four systems of arithmetic numerals which are often used in digital circuits.
These systems are:
1. Decimal; it has a base (or radix) of 10 i.e. it uses 10 different symbols to
represent numbers.
2. Binary; it has a base of two i.e. it uses only two different symbols.
3. Octal; it has a base of 8 i.e. it uses eight different symbols.
4. Hexadecimal; it has a base of 16, i.e. it uses sixteen different symbols.
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All these systems use the same type of positional notation except that
a) decimal system which uses powers of 10 are used to represent quantities which
are outside the digital system.
b) binary system which uses power of 2, is extensively used by digital systems
like digital computers which operate on binary information.
c) octal system which uses power of 8, has certain advantages in digital work
because it requires less circuitry to get information into and out of a digital
system. Moreover, it is easier to read, record and print out octal numbers than
binary numbers.
d) hexadecimal system which uses power of 16 is particularly suited for micro-
computers
Activity 4.2
The decimal Number System
(i) Base or Radix
The decimal number system has a base of 10 meaning that it contains ten unique
symbols (or digits). These are: 0,1,2,3,4,5,6,7,8,9. Any one of these may be used in
each position of the number.
(ii) Position Value
Position value of a number 2573 given by:
2573 = 2*103 + 5*102 + 7*101 + 3*100
The number, 3 is the least significant digit (LSD) whereas 2 is the most significant
digit (MSD).
Again, the number 2573.469 can be written as
2573.469 = 2*103 +5*102 + 7*101 +3*100 +4*10-1 +6*10-2 +9*10-3
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Activity 4.3
Binary Number System
Like decimal number (or denary) system, it has a radix and it also uses the same type
of position value system.
Radix
Its base or radix is two because it uses only two digits 0 and 1 (the word ‘binary digit’
is contracted to bit). All binary numbers consist of a string of 0s and 1s. Examples
are 10, 101, and 1011 which are read as one-zero, one-zero-one and one-zero-one-
one. To avoid confusion subscripts of 10 for decimal and of 2 for binary are added
as shown below.
10 , 101 , 6574 ___ decimal number and
10
10
10
10 , 101 , 110001 ____ binary numbers.
2
2
2
Position Value
The position value of each bit corresponds to some power of 2. A 7-bit binary number
1101.011 is as illustrated below
MSD
LSD
1
1
0
1
•
0
1
1
23
22
21
20
2−1
2−2
2−3
Binary Point
The decimal equivalent is
1101.011 = 1× 23
(
)+ 1× 22
(
)+ 0× 21
(
)+ 1× 20
(
)+ 0× 2−1
(
)+ 1× 2−2
(
)+ 1× 2−3
(
)
2
1 1
= 8 + 4 + 0 + 1+ 0 + = 8 + 4 + 0 + 1+ 0 + + = 13.375
2 8
10
Binary numbers are used extensively by all digital systems primarily due to the nature
of electronics itself. The bit 1 may be represented by a saturated (fully-conducting)
transistor, a light turned ON, a relay energised or a magnet magnetized in a particular
direction. The bit 0, on the other hand, can be represented as a cut-off transistor, a light
turned OFF, a relay de-energised or a magnet magnetised in the opposite direction.
In such cases, there are only two values which a device can assume.
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Activity 4.4
Binary to Decimal Conversion
The following procedures should be adopted for converting a given binary integer
(whole number) into its equivalent decimal number:
Step 1. Write the binary number i.e. all its bits in a row.
Step 2. Directly under the bits, write 1, 2, 4, 8, 16, …Starting from right to left.
Step 3. Cross out the decimal weights which lie under the0 bits.
Step 4. Add the remaining weights to get the decimal equivalent.
Example 4.1 Convert 1101 to its equivalent decimal number
2
Solution The four steps involved in the conversion are given as follow.
Step 1.
1
1
0
0
1
Step 2
16
8
4
2
1
Step 3
16
8
4
2
1
Step 4
16+8+1 = 25.
∴1101 = 25
2
10
Activity 4.5
Binary Fractions
The procedure is the same as binary integers except that the following weights are
used for different bit positions.
Example 4.2. Convert the binary fraction 0.101 into its decimal equivalent.
Solution. The following four steps will be used for this purpose.
Step 1.
0
1
0
1
Step 2.
½
¼
1/8
Step 3.
½
¼
1/8
Step 4.
½ +1/8 = 0.625
∴0.101 = 0.625
2
10
Example 4.3 Find the decimal equivalent of the 6-bit binary number 101.101 2
Solution
1
0
1
1
0
1
4
2
1
½
¼
1/8
4
2
1
½
¼
1/8
= 5 + ½ +1/8 = 5.625
∴101.101 = 5.625
2
10
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Activity 4.6
Double-Dadd Method
This method of converting binary integers into decimal equivalents is much simpler
and quicker than the method so far given especially in the case of large numbers.
The three steps are involved:
1. Double the first bit to the extreme left and add this doubled value to the next
bit on the right.
2. Double the sum obtained and add the doubled value to the next bit.
3. Continue step 2 until the last bit has been added to the previously-doubled
sum.
The conversion of 11001 is as follows. It is seen that 11001 =25
2
2
10
1
1
0 0
1
2 × 1 = 2,2 + 1 = 3, 2 × 3 = 6,6 + 0 = 6, 2 × 6 = 12,12 + 0 = 12, 2 × 12 = 24,24 + 1 = 25
(i)Using double-dadd method, let us convert
into its binary equivalent.
1110102
1. 2 × 1 = 2 ,
add next bit 1 so that 2+1 =3
2. 2 × 3 = 6 ,
add next bit 1 so that 6+1 =7
3. 2 × 7 = 14 ,
add next bit 0 so that 14+0 =14
4. 2 × 14 = 28,
add next bit 1 so that 28+1 =29
5. 2 × 29 = 58 ,
add next bit 0 so that 58+0 =58
Therefore 111010 = 58 .
2
10
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4.7
Decimal to Binary Conversion
(a) Integers
Such conversions can be achieved by using the so-called double-dadded method.
Or by divide-by-two methods. As an example, let us convert 25 into its binary
10
equivalent.
25 ÷ 2 =12+ remainder of 1
TOP
12 ÷ 2 = 6 + remainder of 0
6 ÷ 2 = 3 + remainder of 0
3 ÷ 2 = 1 + remainder of 1
1 ÷ 2 = 0 + remainder of 1
BOTTOM
Therefore
25 = 11001
10
2
(b) Fractions
In this case, Multiply-by-two rule is used we multiply each bit by 2 and record the
carry in the integer position.. The steps below shows how 0.8125 is converted into
10
its binary equivalent.
0.8125 × 2 = 1.625 = 0.625
with a carry of 1
0.625 × 2 = 1.25
= 0.25
with a carry of 1
0.25 × 2 = 0.5
= 0.5
with a carry of 0
0.5 × 2 = 1.5
= 0.0
with a carry of 1
∴0.8125 = 0.1101
10
2
Activity 4.9
Binary Operations
The following four binary operations are considered:
1. addition
2. subtraction
3. multiplication
4. division
(a) Use your school knowledge of mathematics to carry out binary addition.
(b)
Binary subtraction.
Binary subtraction requires more borrowing operations than decimal subtrac-
tionThe four rules for binary subtraction are provided as follows:
1. 0 – 0 = 0,
2. 1 - 0 = 1,
3. 1 – 1 = 0,
4. 0 – 1 = 1 with a borrow of 1 from the next column of the minuend
or 10 – 1 = 1
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Example 4.4
Let us subtract 0101 from 1110 . The various steps are explained below:
2
2
1110
-0101
1001
Explanation
1. In the first column, since we can not subtract 1 from 0, we borrow 1 from the
next column to the left. Hence, we put down 1 in the answer and change the
1 of the next left column to a 0.
2. We apply Rule 1 to the next column i.e. 0 – 0 = 0.
3. We apply Rule 3 to the third column i.e. 1 – 1 = 0.
4. Finally, we apply Rule 2 to the last i.e. fourth column it. 1 – 0 = 1
As a check, it may be noted that talking in terms of decimal numbers, we have sub-
tracted 5 from 14. Obviously, the answer has to be 9 (1001 ).
2
Activity 4.10
Complement of a Number.
In digital work, two types of compliments of binary number are used for comple-
mental subtraction:
(a) 1’s complement
The 1’s complement of a binary number is obtained by changing its each 0
into a 1 and each 1 into a 0. It is also called radix-minus-one-complement.
For example, 1’s complement of 100 is 011 and of 1110 is 0001 .
2
2
2
2
(b) 2’s complement
The 2’s complement of a binary number is obtained by adding 1 to its 1’s
complement.
2’s complement = 1’s complement + 1
This is also known as true complement.
Example: 4.5
The 2’s complement of 1011 is found as follows:
2
Step 1: Its 1’s complement is 0100
2 .
Step 2: Add 1 to 0100 in order to get 0101 .
2
2
Step 3: Thus, 2’s complement of 1011 is 0101 .
2
2
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The complement method of subtraction reduces subtraction to an addition process.
In digital computers this method is popular because
1. Only adder circuits are needed thus simplifying the circuitry.
2. It is easy with digital circuits to get the complements.
Activity 4.10
1’s Complemental Subtraction.
Subtracting of a number is done by adding its 1’s complement to the minuend, using
the following rules:
1. Compute the 1’s complement of the subtrahend by changing all its 1 to 0
s
s
and all its 0 to 1 .
s
s.
2. Add this complement to the minuend.
3. Perform the end-around carry of the last 1 or 0.
4. If there is no end-around carry (i.e. 0 carry), then the answer must be recom-
plemented and a negative sign attached to it.
5. No recomplementing is necessary if the end-around carry is 1.
Example 4.6
Subtracting 101 from 111 is performed as follows:
2
2
Solution
111
+ 010 ← 1’s complement of subtrahend 101
1001
1 ← end-round carry
010
The final answer is obtained by removing from the addition sum carry in the last
position and added it onto the remainder. This is called end-round carry.
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Task 4.1 Using the rules stated under 4.13 in solving problems
Use the rules for subtraction and carry out the following subtractions.
a. Subtract 1101 from 1010 . The final answer is -0011. Explain how this is
2
2
obtained.
b. Subtract 1110 from 0110 . The answer is -000 .
2
2
2
c. Subtract 01101 from 11011 . The answer is 00 .
2
2
2
Activity 4.11
2’s Complemental Subtraction
The essential steps are given as follows:
1. Find the 2’s complement of the subtrahend,
2. Add this complement to the minuend,
3. Drop the final carry,
4. If the carry is 1, the answer is positive and needs no recomplementing,
5. Recomplement the answer and attach minus sign if there is no carry.
Example 4.7 Using 2’s complemental subtraction
Subtract 1010 from 1101
2
2
Solution.
The 1’s complement of 1010 is 0101
The 2’s complement is 0101 + 1 = 0110
Adding this to 1101 we get
1101
+0110
10011
Dropping the final carry gives the final answer as 0011 .2
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Activity 4.12
Binary Multiplication and Division
The procedure for multiplication and division are the same as for decimal multipli-
cation and division.
(a) The four simple rules for multiplication are:
i. 0 × 0 = 0
ii. 0 × 1 = 0
iii. 1 × 0 = 0
iv. 1 × 1 = 1
(b)Rules for division are:
i. 0 ÷1=0
ii. 1 ÷ 1= 1
iii. Division of 1 by 0 is meaningless
Task 4.2 Further Reading and note making
a) Use available references and read about multiplication and division of binary
numbers.
b) Do as many examples as possible in order to come to grip with such pro-
blems.
c) Multiply 1101 by 1100 (Answer is 1001100 ).
2
2
2
d) Multiply 111 by 111 (Answer is 101001 ).
2
2
e) Divide 11001 by 101 (Answer is 101 ).
2
2
2
f) Divide 110011 by 100 (Answer is 110.11 ).
2
2
2
Activity 4.13
Octal Number System.
(i) Radix or Base
It has a base of 8 which means that it has eight distinct counting digits:
0,1,2,3,4,5,6, and 7
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These digits 0 through 7 have exactly the same physical meaning as in decimal system.
Beyond 7, the counting becomes
0, 1,
2,
3,
4,
5,
6,
7,
→ 10, 11,
12,
13,
14,
15,
16,
17,
20, 21,
22,
23,
24,
25,
26,
27,
30, 31,
32,
…
…
…
…
…
(ii) Position value
The position value (or weight) for each digit is given by different powers of 8 as
shown below:
← 8 4
8 3
8 2
8 1
8 0
•
8 −1
8 −2
→
↑
octal point
Example 4.8 Conversion of octal numbers to decimal
453
8 ≡ 4 × 82 + 5 × 81 + 3 × 80 = 4 × 64 + 5 × 8 + 3 × 1 = 29910
1
1
453.27 ≡ 4 × 82 + 5× 81 + 3× 80 = 4 × 64 + 5× 8 + 3×1+ 2 × + 7 ×
8
8
64
; 299.359410
Example 4.9 Conversion of decimal numbers to octal
Here 8 acts as a multiplying factor for integers and as a dividing factor for frac-
tions.
256 .43 → octal
10
256 ÷ 8 = 32
with 0 remainder
32 ÷ 8 = 4
with 0 remainder
4 ÷ 8 = 0
with 4 remainder
⇒ 256 ≡ 400
10
8
Similarly,
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0.43 → octal is worked out as follows
10
0.43 × 8 = 3.44 = 0.44
with a carry 3
0.44 × 8 = 3.52 = 0.52