Question.8
Which of the following options represents an invalid (incorrect) decimal to hexadecimal conversion pair?
Digital electronics processes information using separate or discrete values.
The word digital comes from digit, which means a number symbol.
Digital circuits normally use two logic levels:
0 – LOW1 – HIGHPositive and Negative Logic
Logic defines how voltage levels represent binary values.
Table 1: Logic Conventions
| Logic Type | HIGH Voltage | LOW Voltage |
|---|---|---|
| Positive logic | 1 | 0 |
| Negative logic | 0 | 1 |
Positive logic is used more widely in digital circuits.
Importance of Number Systems
Number systems are used to:
A number system uses a fixed set of digits and a base.
The value of each digit depends on its position:
Value = Digit × BaseᵖᵒˢⁱᵗⁱᵒⁿTable 2: Common Number Systems
| Number System | Base | Digits | Sequence Example | Data Example |
|---|---|---|---|---|
| Decimal | 10 | 0 to 9 | 8, 9, 10, 11 | 45₁₀ |
| Binary | 2 | 0, 1 | 0, 1, 10, 11, 100 | 101101₂ |
| Octal | 8 | 0 to 7 | 6, 7, 10, 11 | 55₈ |
| Hexadecimal | 16 | 0 to 9, A to F | E, F, 10, 11 | 2D₁₆ |
In hexadecimal:
A = 10B = 11C = 12D = 13E = 14F = 15
The same value will be used in all examples:
45₁₀ = 101101₂ = 55₈ = 2D₁₆
Decimal to Binary
Repeatedly divide the decimal number by 2.
Write the remainders from bottom to top.
Example: Convert 45₁₀ to Binary
| Division | Quotient | Remainder |
|---|---|---|
| 45 ÷ 2 | 22 | 1 |
| 22 ÷ 2 | 11 | 0 |
| 11 ÷ 2 | 5 | 1 |
| 5 ÷ 2 | 2 | 1 |
| 2 ÷ 2 | 1 | 0 |
| 1 ÷ 2 | 0 | 1 |
Reading upward:
45₁₀ = 101101₂
Decimal to Octal
Repeatedly divide the decimal number by 8.
45 ÷ 8 = 5, remainder 5
5 ÷ 8 = 0, remainder 5
Therefore:
45₁₀ = 55₈
Decimal to Hexadecimal
Repeatedly divide the decimal number by 16.
45 ÷ 16 = 2, remainder 13
13 = D
Therefore:
45₁₀ = 2D₁₆
Table 3: Decimal Conversion Summary
| Required System | Division Base | Result for 45₁₀ |
|---|---|---|
| Binary | 2 | 101101₂ |
| Octal | 8 | 55₈ |
| Hexadecimal | 16 | 2D₁₆ |
Binary to Decimal
Multiply every binary bit by its positional weight.
101101₂
= 1×2⁵ + 0×2⁴ + 1×2³ + 1×2² + 0×2¹ + 1×2⁰
= 32 + 8 + 4 + 1
= 45₁₀Binary to Octal
Group binary bits into sets of 3, starting from the right.
101 101
101₂ = 5₈101₂ = 5₈Therefore:
101101₂ = 55₈
Binary to Hexadecimal
Group binary bits into sets of 4, starting from the right.
Add leading zeros when required:
0010 1101
0010₂ = 2₁₆1101₂ = D₁₆Therefore:
101101₂ = 2D₁₆
Table 4: Binary Conversion Summary
| Required System | Method | Result for 101101₂ |
|---|---|---|
| Decimal | Positional weights | 45₁₀ |
| Octal | Group into 3 bits | 55₈ |
| Hexadecimal | Group into 4 bits | 2D₁₆ |
Octal to Binary
Replace each octal digit with its 3-bit binary value.
5₈ = 101₂
Therefore:
55₈ = 101 101₂
55₈ = 101101₂
Octal to Decimal
Multiply every digit by its positional weight.
55₈
= 5×8¹ + 5×8⁰
= 40 + 5
= 45₁₀Octal to Hexadecimal
First convert octal to binary:
55₈ = 101101₂
Group the binary value into 4 bits:
0010 1101
Therefore:
55₈ = 2D₁₆
Table 5: Octal Conversion Summary
| Required System | Method | Result for 55₈ |
|---|---|---|
| Binary | Replace each digit with 3 bits | 101101₂ |
| Decimal | Positional weights | 45₁₀ |
| Hexadecimal | Octal → Binary → Hexadecimal | 2D₁₆ |
Hexadecimal to Binary
Replace each hexadecimal digit with its 4-bit binary value.
2₁₆ = 0010₂D₁₆ = 1101₂Therefore:
2D₁₆ = 0010 1101₂
Removing leading zeros:
2D₁₆ = 101101₂
Hexadecimal to Decimal
Multiply every digit by its positional weight.
2D₁₆
= 2×16¹ + 13×16⁰
= 32 + 13
= 45₁₀Hexadecimal to Octal
First convert hexadecimal to binary:
2D₁₆ = 00101101₂
Group the bits into sets of 3:
00 101 101
101₂ = 5₈
Therefore:
2D₁₆ = 55₈
Table 6: Hexadecimal Conversion Summary
| Required System | Method | Result for 2D₁₆ |
|---|---|---|
| Binary | Replace each digit with 4 bits | 101101₂ |
| Decimal | Positional weights | 45₁₀ |
| Octal | Hexadecimal → Binary → Octal | 55₈ |
Table 7: Number-System Conversion Methods
| From | To | Method |
|---|---|---|
| Decimal | Binary, Octal or Hexadecimal | Repeated division by required base |
| Binary | Decimal | Add positional weights |
| Binary | Octal | Group into 3 bits |
| Binary | Hexadecimal | Group into 4 bits |
| Octal | Binary | Replace each digit with 3 bits |
| Hexadecimal | Binary | Replace each digit with 4 bits |
| Octal | Hexadecimal | Convert through binary |
| Hexadecimal | Octal | Convert through binary |
Signed binary numbers represent both positive and negative values.
The most significant bit is used as the sign bit:
0: Positive1: NegativeThe remaining bits represent the value.
Sign-Magnitude Range
For an n-bit sign-magnitude number:
Range = −(2ⁿ⁻¹ − 1) to +(2ⁿ⁻¹ − 1)
For 4 bits:
Range = −7 to +7
Sign-magnitude has two zero values:
0000 = +01000 = −0
Binary addition follows four basic rules.
Table 8: Binary Addition Rules
| Addition | Sum | Carry |
|---|---|---|
0 + 0 | 0 | 0 |
0 + 1 | 1 | 0 |
1 + 0 | 1 | 0 |
1 + 1 | 0 | 1 |
1 + 1 + 1 | 1 | 1 |
Example
1011
+ 0110
-------
100011011₂ = 11₁₀
0110₂ = 6₁₀
10001₂ = 17₁₀
The 1's complement of a binary number is found by inverting every bit.
0 to 1.1 to 0.Example
Original number:
0101
Invert all bits:
1010
Therefore, in 4-bit 1's complement:
+5 = 0101−5 = 1010Limitation
1's complement has two representations of zero:
00001111Addition may also require an end-around carry.
The 2's complement is found in two steps:
1.Example: Find −5
Positive 5:
0101
1's complement:
1010
Add 1:
1010
+ 0001
------
1011Therefore:
−5 = 1011
Range
For an n-bit 2's complement number:
Range = −2ⁿ⁻¹ to +(2ⁿ⁻¹ − 1)
For 4 bits:
Range = −8 to +7
Benefits
Table 9: Signed Binary Representations
| Representation | +5 | −5 | 4-Bit Range | Zero Values |
|---|---|---|---|---|
| Sign-magnitude | 0101 | 1101 | −7 to +7 | Two |
| 1's complement | 0101 | 1010 | −7 to +7 | Two |
| 2's complement | 0101 | 1011 | −8 to +7 | One |
Binary subtraction uses borrowing, similar to decimal subtraction.
Basic rules:
0 − 0 = 01 − 0 = 11 − 1 = 00 − 1 requires a borrowAfter borrowing:
10₂ − 1₂ = 1₂
Example
10110
- 00101
--------
1000110110₂ = 22₁₀
00101₂ = 5₁₀
10001₂ = 17₁₀
Subtraction Using 2's Complement
Binary subtraction can be changed into addition:
A − B = A + 2's complement of B
Example: 10110 − 00101
Step 1: Find 2's Complement of 00101
1's complement:
11010
Add 1:
11011
Step 2: Add It to 10110
10110
+ 11011
--------
1 10001Discard the final carry:
10001
Therefore:
10110₂ − 00101₂ = 10001₂
Result Rule
Binary codes represent numbers, characters or changing states using binary bits.
Gray code is a binary code in which two consecutive values differ by only one bit.
It is also called a unit-distance code.
Example
| Decimal | Binary | Gray |
|---|---|---|
| 0 | 000 | 000 |
| 1 | 001 | 001 |
| 2 | 010 | 011 |
| 3 | 011 | 010 |
Properties
Applications
BCD means Binary-Coded Decimal.
Each decimal digit is represented separately using four binary bits.
BCD normally uses the 8421 weighted code.
Example
Decimal value:
59
5 = 01019 = 1001Therefore:
59₁₀ = 0101 1001 in BCD
BCD 1010 to 1111 are invalid decimal-digit codes.
Properties
Applications
Excess-3 represents each decimal digit after adding 3 to it.
Example
Convert decimal 59:
For digit 5:
5 + 3 = 8 = 1000₂
For digit 9:
9 + 3 = 12 = 1100₂
Therefore:
59₁₀ = 1000 1100 in Excess-3
Properties
0011.Applications
ASCII means American Standard Code for Information Interchange.
It represents:
Standard ASCII uses 7 bits.
Examples
A = 65₁₀ = 1000001₂a = 97₁₀ = 1100001₂0 = 48₁₀ = 0110000₂Properties
Applications
Table 10: Binary Code Comparison
| Code | Main Purpose | Type | Example | Main Application |
|---|---|---|---|---|
| Gray | Represent changing positions | Non-weighted | Decimal 2 = 011 | Rotary encoders |
| BCD | Represent decimal digits | Weighted 8421 | Decimal 59 = 0101 1001 | Displays and calculators |
| Excess-3 | Represent decimal digits with offset | Non-weighted | Decimal 59 = 1000 1100 | Decimal arithmetic |
| ASCII | Represent characters | Character code | A = 1000001 | Text communication |