Test: Number System & Binary Codes - 1 - Electrical Engineering (EE) MCQ

Hexadecimal Number System:

• It has a base of 16. Hence, it uses sixteen distinct counting digits 0 through 9 and A through F
• Place value (or weight) for each digit is in ascending powers of 16 for integers and descending powers of 16 for fractions.
• The chief use of this system is in connection with byte-organized machines.
• It is used for specifying addresses of different binary numbers stored in computer memory.

 
Complement of a Number: In digital work, two types of complements of a binary number are used for complemental sub-traction:

1’s complement:

• The 1’s complement of a binary number is obtained by changing 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 1002 is 0112, and 0112 is 00012.

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
• It is also known as a true complement.

 
Octal Number System:

• It has a base of 8 which means that it has eight distinct counting digits: 0, 1, 2, 3, 4, 5, 6, and 7
• These digits 0 through 7, have precisely the same physical meaning as in the decimal system.
• For counting beyond 7, 2-digit combinations are formed taking the second digit followed by the first, then the second followed by the second, and so on.
• Hence, after 7, the next octal number is 10 (second digit followed by first), then 11 (second digit followed by second), and so on.

Excess-3 Code:

• It is an unweighted code and is a modified form of BCD.
• It is widely used to represent numerical data in digital equipment.
• It is abbreviated as XS-3. As its name implies, each coded number in XS-3 is three larger than in the BCD code. 

Concept:

Conversion of decimal to binary:
Step 1: Divide the number by 2 keeping notice of the quotient and the remainder. Continue dividing the quotient by 2 until you get a quotient of zero.
Step 2: Write out the remainders in the reverse order to get the equivalent binary number.

For converting decimal fractions to binary numbers, follow these steps:
For converting decimal fractions to a binary numbers, follow these steps:
Then write out the integer parts from the results of each multiplication to get the equivalent binary number.

Conversion of binary to octal:
Make pair of three binary number which forms an octal number.

Conversion of binary to hexadecimal:
Make pair of four binary number which forms a hexadecimal number.

Calculation:
Given, that the decimal number = (98.75)10 
98 / 2 = 49 with remainder 0
49 / 2 = 24 with remainder 1
24 / 2 = 12 with remainder 0
12 / 2 = 6 with remainder 0
6 / 2 = 3 with remainder 0
3 / 2 = 1 with remainder 1
1 / 2 = 0 with remainder 1
Write in reverse order:
98 = 1100010
0.75 × 2 = 1 + 0.5
0.5 × 2 = 1 + 0
.75 = .11
(98.75)₁₀ = (1100010.11)₂ 
(98.75)₁₀ = (001  100  010.  110)₂ = (142.6)₈ ...........(from table 1)
(98.75)₁₀ = (0110  0010.  1100)₂ = (62.C)₁₆...........(from table 2)

Write the remainder from bottom to top i.e. in the reverse chronological order.
This will give the binary equivalent of 127. 
Therefore, the binary equivalent of decimal number 127 is 1111111.

Tips and Tricks:

Steps to writing 2’s complement to any binary number:

• Start from right to left and search for the first ‘1’
• Write down the bits until that first ‘1’ as it is.
• Write down the remaining left bits with their respective complement.

Step 1: Divide (21)₁₀ successively by 2 until the quotient is 0.
21/2 = 10, remainder is 1
10/2 = 5, remainder is 0
5/2 = 2, remainder is 1
2/2 = 1, remainder is 0
1/2 = 0, remainder is 1

Step 2: Read from bottom (MS2) to top (LS2) as 10101
This is the binary equivalent of decimal number 21
Step 3:Binary equivalent of 0.125 is, multiplying by 2 until we get 1 and writing down the integer after each multiplication,
⇒ 0.125 × 2 = 0.25
⇒ 0.25 × 2 = 0.5
⇒ 0.5 × 2 = 1
⇒ Binary equivalent of 0.125 = 001
∴ The binary code of (21.125)₁₀ is,
(21.1250)₁₀ = 10101.001

The correct answer is 13.

Step by step solution:

• Step 1: Write down the binary number: 1101
• Step 2: Multiply each digit of the binary number by the corresponding power of two: 1x2³ + 1x2² + 0x2¹ + 1x2⁰
• Step 3: Solve the powers: 1x8 + 1x4 + 0x2 + 1x1 = 8 + 4 + 0 + 1
• Step 4: Add up the numbers written above: 8 + 4 + 0 + 1 = 13
• So, (1101)₂ = (13)₁₀

The circuit is redrawn as:

y₁ = x₁
y₂ = x₁ ⊕ x₂
y₃ = x₂ ⊕ x₃
y₄ = x₃ ⊕ x₄

Example:
Let the input to the circuit be 1010
∴ For an input 1010, we get the output as 1111.
Similarly, let the input be 0110.
The output for 0110 input is 0101
 

Observations:

∴ The given circuit converts 4 bit binary to Grey code converter.

The decimal equivalent of the binary number 101.101 is,

= 1 x 2² + 0 x 2¹ + 1 x 2⁰ + 1 x 2^-1 + 0 x 2^-2 + 1 × 2^-3
= 4 + 0 + 1 + 0.5 + 0 + 0.125  
= 5.625

Concept:

Decimal to binary:

• Take decimal number as dividend.
• Divide the number by 2.
• Get the integer quotient for the next iteration.
• Get the remainder (it will be either 0 or 1 because of divisor 2).
• Repeat the steps until the quotient is equal to 0
• Write the remainders in reverse order (which will be equivalent binary number of given decimal number).

Decimal to binary: (fractional part)

• Take decimal number as multiplicand.
• Multiple this number by 2 (2 is base of binary so multiplier here).
• Store the value of integer part of result in an array (it will be either 0 or 1 because of multiplier 2).
• Repeat the above two steps until the number became zero.
• Write these resultant integer part

Calculation:
Binary of  57:

Now, write remainder from bottom to up (in reverse order), this will be 111001 which is equivalent binary number of decimal integer 57.
Convert decimal fractional number 0.375 into binary number.
Here, decimal fraction: 0.375

Now, write these resultant integer part, this will be 0.0110 which is equivalent binary fractional number of decimal fractional 0.375.
∴ 57.375 can be written as 111001.011 in binary
Hence, option (1) is correct.

The range of n bit word in 2’s complement representation is,

(n-1) is used here because out of n bits 1 bit is used as a sign bit 
There is one extra negative number because "0" has only a single representation in 2's complement form
For 8 bit word, the range will be -128 to 127.