Number System & Binary Codes - 2 - Free MCQ Practice Test with solutions,

Binary code for 5 → 0101
The procedure to convert a gray code to a binary one is as shown:

A₃ → Copy MSB → 0
A₂ → 0 ⊕ 1 → 1
A₁ → 1 ⊕ 0 → 1
A₀ → 0 ⊕ 1 → 1
Grey code → 0111

Formula:
If we covert x decimal to binary, divide x successively by 2 until the quotient is 0.

Calculation:
Divide 23 successively by 2 until the quotient is 0:

23/2 = 11, remainder is 1 (LSB)
11/2 = 5, remainder is 1
5/2 = 2, remainder is 1
2/2 = 1, remainder is 0
1/2 = 0, remainder is 1 (MSB)

Read from the bottom (MSB) to top (LSB) as 10111
∴ 10111 is the binary equivalent of decimal number 23

Concept:

1’s complement:

• 1’s complement simply inverted every bit of input.
• Can be implemented using only NOT gate for each bit of binary number input.  

Concept:
A number system includes the number of independent digits used in the number system (the base), the place values of the different digits constituting the number, and the maximum numbers that can be written with the given number of digits.
Octal numbers: These numbers use digits from 0 to 7, total of 8 digits, and hence they are called octal number system. Octal numbers have base 8.
Hexadecimal numbers: The numbers which have base 16. It uses 16 different digits to represent the numbers. It is denoted as h16, where h is a hexadecimal number. It may be a combination of alphabets and numbers. Thus, it includes numbers from 0 to 9 and alphabets A to F.

Calculation:
Hexadecimal to Octal Conversion:
Given, B34₁₆ is a hexadecimal number.
B → 1011, 3 → 0011, 4 → 0100
1011 0011 0100
Now group them from right to left, each having 3 digits.
101, 100, 110, 100
101 → 5, 100 → 4, 110 → 6, 100 → 4
Hence, B34₁₆ = 5464₈

Concept:

• 8421 is also known as BCD code
• BCD is a weighted code.
• In weighted codes, each successive digit from right to left represents weights equal to some specified value, and to get the equivalent decimal number to add the products of the weights by the corresponding binary digit.

The following represents the 4-bit binary representation of decimal values: 

Concept:

1. Signed magnitude representation uses the most significant bit (MSB) a sign bit.

• If the sign bit is ‘0’ then the number is positive.
• If the sign bit is ‘1’ then the number is negative.

The remaining bits represent the magnitude of the binary number.

2. 1’s complement representation:
It is a representation of a binary number obtained by toggling all bits in it i.e. transforming the 0 bit to 1 and the 1 bit to 0.

3. 2’s complement representation:
It is obtained by simply adding 1 to the 1’s complement of that binary number.

Calculation:
The binary form of 7 ↔ 0111
1's complement of -7 = 1(for sign) (000)(1's complement of 7) = 1000
2's complement of -7 = 1's complement + 1 = 1000 + 1 = 1001

Concept:
Hexadecimal number: In this, value of the base is 16. Each digit is represented by 4-bit binary no.
Octal number: For octal number, value of base is 8. Each digit of an octal number is represented by 3-bit binary no.

Explanation:
Octal number = 657
Binary representation for this number (each digit of a octal number is converted into 3 binary bits) 
So, 657 in binary is equivalent to 110 101 111
Now group this binary number into 4 bits starting from right to left. 
i.e. 0001 1010 1111
Hexadecimal representation for this number is : 1AF

Concept:

Another number system to Decimal
In each and every representation of numbers with different bases, the maximum value in a number system with the base ‘r’ is r – 1. Since numbers vary from 0 to r – 1.
To convert any number which is in the different base to decimal number system we use binary-weighted representation.
Eg: let the number be ( abc⋯ ⋯ yz)_r
Now to convert the above number into the decimal number system

If we convert all numbers into decimal then we can perform normal addition and subtraction etc.

Application:
Given:
(110)_x = (132)₄,
The decimal equivalent of this number will be:
On solving this quadratic equation we'll get:
x = 5, -6
Base can't be a negative so;
x = 5

In BCD each decimal digit is represented by a 4-bit binary number.
The binary representation of 8 → 1000
The binary representation of 1 → 0001
(81)₁₀ = 10000001

Binary Coded Decimal (BCD) code:

• BCD is a way to express each of the decimal digits with a binary code.
• In this code, each decimal digit is represented by its 4-bit binary equivalent.
• Also, with four bits we can represent sixteen numbers (0000 to 1111)
• But as there are 10 decimal digits from 0 to 9, BCD code uses only the first ten of these (0000 to 1001). The remaining six code combinations i.e. 1010 to 1111 are invalid in BCD.

Octal to Decimal Conversion:

• Step 1: Since an octal number only uses digits from 0 to 7, we first arrange the octal number with the power of 8.
• Step 2: We evaluate all the power of 8 values such as 80 is 1, 81 is 8, etc., and write down the value of each octal number.
• Step 3: Final step is to add the product of all the numbers to obtain the decimal number.

 
Application:

Step 1: Write 23 with the power of 8. Start from the right-hand side.
2 × 81 + 3 × 80

Step 2: Evaluate the power of 8 values for each octal number.
16 + 3 = 19 (Decimal Number)

Convert the given b-radix into a decimal system.



 

b = 7 Here b value should be greater than 5 because it maximum digit in b radix.
Hence the correct answer is 7.

Concept:

A number system includes the number of independent digits used in the number system (the base), the place values of the different digits constituting the number, and the maximum numbers that can be written with the given number of digits.
Octal numbers: These numbers use digits from 0 to 7, total of 8 digits, and hence they are called octal number system. Octal numbers have base 8.
Hexadecimal numbers: The numbers which have base 16. It uses 16 different digits to represent the numbers. It is denoted as h16, where h is a hexadecimal number. It may be a combination of alphabets and numbers. Thus, it includes numbers from 0 to 9 and alphabets A to F.

Calculation:
Hexadecimal to Octal Conversion:
Given, (FB2)₁₆ is a hexadecimal number.
F → 1111, B → 1011, 2 → 0010 1111 1011 0010
Now group them from right to left, each having 3 digits.
111, 110, 110, 010
111 → 7, 110 → 6, 110 → 6, 010 → 2
Hence, FB2₁₆ = 766₂₈

• Excess 3 code is a 4-bit code.
• It can be derived from the BCD code by adding ‘3’ to each coded number. Thus, in excess 3 code 0011 is added to each BCD number.
• It is an un-weighted code.
• It is a self-complimenting code i.e. the 1’s complement of an excess three number is the excess 3 code for the 9’s complement of the corresponding decimal number.

Given BCD number = 0001 0010 0110
BCD is converted to a decimal by taking a pair of 4 and representing it in its equivalent decimal, i.e.
The decimal equivalent of the given BCD will be:
Decimal equivalent = 126
Converting this to an equivalent binary through successive division, we get:
Binary equivalent = 1111110