Question.8
Which of the following options represents an invalid (incorrect) decimal to hexadecimal conversion pair?
Digital electronics processes the information using discrete values or logic levels.
Digital circuits represent binary values using voltage ranges.
In a simplified 5 V logic example,
| 5 V | Logic 1 |
| 0 V or GND | Logic 0 |
Actual HIGH and LOW voltage thresholds depend on the logic family.
Digits: 0 to 9. (0,1,2,3,4,5,6,7,8,9)
Base = 10
Examples: - 3, 5, 87, 123, 543
Commonly used by humans for everyday numerical representation.
Digits: 0 and 1
Base = 2
Examples: - 0, 1, 10, 1001, 00110101, 1110001011010100
It is used in digital electronics because it enables –
Digits: 0 to 7 (0,1,2,3,4,5,6,7)
Base = 8
Examples: - 4, 7, 34, 23, 17 (Digits 8 and 9 do not appear in an octal number.)
Octal provides a compact representation of binary because one octal digit corresponds to three binary bits.
Digits: 0 to 9 and A to F (0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F)
Base = 16
Examples: - 2, 4, C, D, 3D, 2A, 23B, BD8, 45F
| Decimal | Binary | Octal | Hexadecimal |
|---|---|---|---|
| 0 | 0 | 0 | 0 |
| 1 | 1 | 1 | 1 |
| 2 | 10 | 2 | 2 |
| 3 | 11 | 3 | 3 |
| 4 | 100 | 4 | 4 |
| 5 | 101 | 5 | 5 |
| 6 | 110 | 6 | 6 |
| 7 | 111 | 7 | 7 |
| 8 | 1000 | 10 | 8 |
| 9 | 1001 | 11 | 9 |
| 10 | 1010 | 12 | A |
| 11 | 1011 | 13 | B |
| 12 | 1100 | 14 | C |
| 13 | 1101 | 15 | D |
| 14 | 1110 | 16 | E |
| 15 | 1111 | 17 | F |
| 16 | 10000 | 20 | 10 |
| 17 | 10001 | 21 | 11 |
| 18 | 10010 | 22 | 12 |
| 19 | 10011 | 23 | 13 |
| 20 | 10100 | 24 | 14 |
Same value can be represented in different number systems.
Decimal number 45 is converted into binary with following method:
| Division | Quotient | Remainder |
|---|---|---|
| 45 ÷ 2 | 22 | 1 (LSB) |
| 22 ÷ 2 | 11 | 0 |
| 11 ÷ 2 | 5 | 1 |
| 5 ÷ 2 | 2 | 1 |
| 2 ÷ 2 | 1 | 0 |
| 1 ÷ 2 | 0 | 1 (MSB) |
45₁₀ = 101101₂To convert binary number 10101 into decimal:
| Position (n) | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|
| Binary digits: (from MSB to LSB) | 1 | 0 | 1 | 0 | 1 |
| Positional Weights: 2ⁿ | 16 | 8 | 4 | 2 | 1 |
| Product of Weights and binary digit | 16 | 0 | 4 | 0 | 1 |
| Sum | 16 + 0 + 4 + 0 + 1 = 21 | ||||
Examples:
10101₂ = 21₁₀
110111₂ = 55₁₀
1001₂ = 9₁₀For an n-bit unsigned binary number, the maximum value is 2ⁿ − 1.
For 3-bit, highest value is 2³ – 1 = 8 – 1 = 7
For 4-bit, highest value is 2⁴ – 1 = 16 – 1 = 15Group 3 Binary Digits from LSB and write equivalent Octal Digit
Examples:
101100₂ = (101 100)₂ = 54₈
1011₂ = (001 011)₂ = 13₈Represent each Octal digit to its equivalent 3-digit Binary Number
Examples: -
34₈ = (011 100)₂ = 011100₂
65₈ = (110 101)₂ = 110101₂Binary Equivalent Octal Table
| Binary | Octal |
|---|---|
000 | 0 |
001 | 1 |
010 | 2 |
011 | 3 |
100 | 4 |
101 | 5 |
110 | 6 |
111 | 7 |
Group 4 Binary Digits from LSB and write equivalent Hexadecimal Digit
Examples:
00110111₂ = (0011 0111)₂ = 37₁₆
10100010₂ = (1010 0010)₂ = A2₁₆
11101011₂ = (1110 1011)₂ = EB₁₆Represent each Hexadecimal Digit to its equivalent 4-digit Binary or 4-bit Binary
5F₁₆ = (0101 1111)₂ = 01011111₂
3D9₁₆ = (0011 1101 1001)₂ = 001111011001₂
A54₁₆ = (1010 0101 0100)₂ = 101001010100₂Binary equivalent Hexadecimal
| Binary | Hexadecimal | Binary | Hexadecimal |
|---|---|---|---|
0000 | 0 | 1000 | 8 |
0001 | 1 | 1001 | 9 |
0010 | 2 | 1010 | A |
0011 | 3 | 1011 | B |
0100 | 4 | 1100 | C |
0101 | 5 | 1101 | D |
0110 | 6 | 1110 | E |
0111 | 7 | 1111 | F |
Convert the Decimal into Binary and convert the Binary into Hexadecimal
Example: Decimal number = 78₁₀
| Division | Quotient | Remainder |
|---|---|---|
| 78 ÷ 2 | 39 | 0 (LSB) |
| 39 ÷ 2 | 19 | 1 |
| 19 ÷ 2 | 9 | 1 |
| 9 ÷ 2 | 4 | 1 |
| 4 ÷ 2 | 2 | 0 |
| 2 ÷ 2 | 1 | 0 |
| 1 ÷ 2 | 0 | 1 (MSB) |
78₁₀ = 1001110₂ = (0100 1110)₂ = 4E₁₆
78₁₀ = 4E₁₆Convert the Hexadecimal into Binary and convert the Binary into Decimal
Example: - Hexadecimal number = D4₁₆
D4₁₆ = (1101 0100)₂ = 11010100₂| Position (n) | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| Binary digits: (from MSB to LSB) | 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
| Positional Weights: 2ⁿ | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| Product of Weights and binary digit | 128 | 64 | 0 | 16 | 0 | 4 | 0 | 0 |
| Sum | 128 + 64 + 0 + 16 + 0 + 4 + 0 + 0 = 212 | |||||||
D4₁₆ = 212₁₀Convert the Octal to Binary and then Binary to Decimal.
31₈ = (011 001)₂ = 11001₂ = 25₁₀
123₈ = (001 010 011)₂ = 1010011₂ = 83₁₀
234₈ = (010 011 100)₂ = 10011100₂ = 156₁₀Convert the Decimal to Binary and then Binary to Octal.
15₁₀ = 1111₂ = (001 111)₂ = 17₈
52₁₀ = 110100₂ = (110 100)₂ = 64₈
197₁₀ = 11000101₂ = (011 000 101)₂ = 305₈Convert Hexadecimal to Binary and then Binary to Octal.
Example: - Hexadecimal Number: - 72C₁₆
| Hexadecimal Number: | 7 | 2 | C | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Equivalent Binary Number: | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
| Equivalent Octal Number | 3 | 4 | 5 | 4 | ||||||||
4F0₁₆ = (0100 1111 0000)₂ = (010 011 110 000)₂ = 2360₈
AD₁₆ = (1010 1101)₂ = (010 101 101)₂ = 255₈Convert Octal to Binary and then Binary to Hexadecimal.
Examples: -
31₈ = (011 001)₂ = (0001 1001)₂ = 19₁₆
123₈ = (001 010 011)₂ = (0101 0011)₂ = 53₁₆
305₈ = (011 000 101)₂ = (1100 0101)₂ = C5₁₆Suppose A and B are two 1-bit binary numbers. The sum of A and B can be
| A | B | Sum | Carry |
|---|---|---|---|
0 | 0 | 0 | 0 |
0 | 1 | 1 | 0 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 1 |
Examples: -
Suppose A = 101 and B = 001
| A | 1 | 0 | 1 |
|---|---|---|---|
| B | 0 | 0 | 1 |
| Carry from previous bit | 1 | ||
| Sum | 1 | 1 | 0 |
011₂ + 011₂ = 110₂
1010₂ + 0110₂ = 10000₂ (Equivalently: 10 + 6 = 16)In common signed representations, the MSB indicates the sign.
MSB represents sign bit.
MSB = 0 | Positive Value |
MSB = 1 | Negative Value |
Remaining bits represents value depending on representation method.
Signed Binary Representation Methods:
Represent number as: Sign (MSB) and Magnitude (Remaining Bits).
Example for 4-bit signed magnitude representation.
0101 = +5
1101 = -5Limitations:
1000, 0000)Negative Numbers obtained by inverting all bits of positive number
Examples: -
| Numbers | Operations |
|---|---|
-3 | Given number |
011 | Represent equivalent binary of +3 |
100 | Take 1’s complement (invert 1 to 0 and 0 to 1) |
+3 = 011
-0 = 111Limitation: -
For Negative Number, after 1’s Complement, 1 is added to the result to generate 2’s Complement
Example for 4-bit binary number: -
| Number | Operation |
|---|---|
-5 | Given Number |
0101 | 4-bit binary representation of +5 |
1010 | 1s compliment |
1011 | 2s compliment (add 1 to the 1’s compliment) |
Range of n-bit signed binary number: −2ⁿ⁻¹ to +(2ⁿ⁻¹ − 1)3-bit signed binary equivalent decimal integer
| Signed Binary | Decimal |
|---|---|
100 | -4 |
101 | -3 |
110 | -2 |
111 | -1 |
000 | 0 |
001 | +1 |
010 | +2 |
011 | +3 |
Suppose A and B are 2 1-bit binary numbers.
| A | B | Subtraction | Borrow |
|---|---|---|---|
0 | 0 | 0 | 0 |
0 | 1 | 1 | 1 |
1 | 0 | 1 | 0 |
1 | 1 | 0 | 0 |
Examples: - Suppose A = 010 and B = 001
| A | 0 | 1 | 0 |
|---|---|---|---|
| B | 0 | 0 | 1 |
| Borrow from previous bit | 1 | ||
| Subtraction | 0 | 0 | 1 |
0110₂ – 0100₂ = 0010₂ (6 – 4 = 2)
0101₂ – 0111₂ = 1110₂ (5 – 7 = -2,
Here 1110 is 2s complement of 2 representing -2)A – B can also represent as: A + (-B)
Or it can be written as: A + (2s complement of B)
Example: - A = 0110₂, B = 0100₂ Find: A – B
| Operation | Values |
|---|---|
| Write A | 0110 (+6₁₀) |
| Write B | 0100 (+4₁₀) |
| 1s compliment of B | 1011 |
| 2s compliment of B | 1100 |
| A + 2s compliment of B | 0110 + 1100 = 1 0010 |
| Discard Carry Bit as it is out of 4-bit range | 0010 (+2₁₀) |
Examples: -
101₂ − 011₂ = 101₂ + 101₂ = 010₂
1011₂ − 0110₂ = 1011₂ + 1010₂ = 0101₂This method is used to perform addition and subtraction from single adder circuit
Predefined binary patterns used to represent different types of information in digital circuits.
Two consecutive Gray-code values differ by exactly one bit.
It is used to reduce errors or ambiguity during transitions between consecutive values.
| Decimal | Binary | Gray |
|---|---|---|
0 | 000 | 000 |
1 | 001 | 001 |
2 | 010 | 011 |
3 | 011 | 010 |
4 | 100 | 110 |
5 | 101 | 111 |
6 | 110 | 101 |
7 | 111 | 100 |
Applications
Each Decimal Digit is separately represented by 4-bit binary number.
Binary values: 1010 to 1111 are invalid in BCD Code
| Decimal | BCD | Decimal | BCD |
|---|---|---|---|
0 | 0000 | 5 | 0101 |
1 | 0001 | 6 | 0110 |
2 | 0010 | 7 | 0111 |
3 | 0011 | 8 | 1000 |
4 | 0100 | 9 | 1001 |
Examples: -
56 = 0101 0110
93 = 1001 0011
15 = 0001 0101Applications
Decimal measurement systems
Represent Decimal digit by adding 3 and converting result into binary.
Binary values 0000 to 0010 and 1101 to 1111 are invalid in Excess-3 Code
It is used to simplify decimal arithmetic and complement operations
| Decimal | Excess-3 | Decimal | Excess-3 |
|---|---|---|---|
0 | 0011 | 5 | 1000 |
1 | 0100 | 6 | 1001 |
2 | 0101 | 7 | 1010 |
3 | 0110 | 8 | 1011 |
4 | 0111 | 9 | 1100 |
Examples: -
23 = 0101 0110
85 = 1011 1000Applications
ASCII stands for American Standard Code for Information Interchange.
Represents – Numbers, Letters, Symbols, Control Characters
1000001 = A
1000010 = B
1100001 = a
0110000 = 0
0100000 = Space
0100001 = !
0111111 = ?Applications
| Code | Main Purpose | Type | Example | Main Application |
|---|---|---|---|---|
| Gray | Represent changing positions | Non-weighted | 7 = 100 | Rotary encoders |
| BCD | Represent decimal digits | Weighted 8421 | 59 = 0101 1001 | Displays and calculators |
| Excess-3 | Represent decimal digits with offset | Non-weighted | 12 = 0100 0101 | Decimal arithmetic |
| ASCII | Represent characters | Character code | A = 1000001 | Text communication |