Encoding Schemes and Number System - Class 11 Computer Science - Chapter 2 - Notes, NCERT Solutions & Extra Questions
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Notes - Encoding Schemes and Number System | Class 11 NCERT | Computer Science
Comprehensive Class 11 Notes on Encoding Schemes and Number Systems
Introduction to Encoding Schemes and Number Systems
Understanding Encoding Schemes
Encoding schemes are methodologies that convert data into codes, making it comprehensible for digital systems. These schemes play a pivotal role in enabling communication between different computer systems.
Importance of Number Systems in Computing
Number systems are integral to computing as they provide a structured method to represent and process numerical data. Various numbering systems like binary, octal, and hexadecimal are crucial for different computational operations.
Overview of Encoding Schemes
American Standard Code for Information Interchange (ASCII)
In the early days of computing, there was no standard for representing characters, which led to communication difficulties across different systems. ASCII, developed in the 1960s, standardised the representation of characters. Initially, it used 7 bits to represent 128 characters, sufficient for basic English text. For example, the ASCII code for the letter 'A' is 65, which translates to the binary 1000001.
Indian Script Code for Information Interchange (ISCII)
To facilitate the use of Indian languages, ISCII was introduced in the mid-1980s. This 8-bit code could represent 256 characters, including the standard ASCII set. ISCII enabled the encoding of various Indian scripts, making it a significant advancement in multilingual computing.
Unicode
Unicode was developed to overcome the limitations of previous encoding schemes by providing a unique number for every character in all written languages. Unicode supports encoding standards like UTF-8, UTF-16, and UTF-32, each varying in the number of bytes used. It ensures that text is displayed correctly across different systems and applications.
Introduction to Number Systems
Binary Number System
The binary number system, also known as base-2, uses only two digits—0 and 1. It's the fundamental language of computers due to its simplicity and direct correlation with electronic states (on and off). For example, the binary number 101 represents the decimal number 5.
Decimal Number System
The decimal system, or base-10, is used in everyday life and consists of ten digits (0-9). Each digit's value depends on its position and the power of 10. For example, the decimal number 237.25 can be broken down into:
$$ 2 \times 10^2 + 3 \times 10^1 + 7 \times 10^0 + 2 \times 10^{-1} + 5 \times 10^{-2} $$
Octal Number System
The octal system, or base-8, includes eight digits (0-7). It provides a more compact representation of binary numbers. For example, the binary number 110010 can be grouped into octal as 62.
Hexadecimal Number System
Hexadecimal, or base-16, uses sixteen symbols (0-9, A-F). It's widely used in computing to simplify binary code representation. Each hex digit represents four binary digits (bits). For example, the binary number 10101100 is equivalent to the hexadecimal number AC.
Conversion Between Number Systems
Conversion from Decimal to Binary
To convert a decimal number to binary, follow these steps:
Divide the decimal number by 2.
Record the remainder.
Repeat the process with the quotient until it becomes zero.
Write the remainders in reverse order.
For example, to convert 65 to binary:
$$ \begin{align*} 65 \div 2 & = 32 \quad \text{remainder} \quad 1 \\ 32 \div 2 & = 16 \quad \text{remainder} \quad 0 \\ 16 \div 2 & = 8 \quad \text{remainder} \quad 0 \\ 8 \div 2 & = 4 \quad \text{remainder} \quad 0 \\ 4 \div 2 & = 2 \quad \text{remainder} \quad 0 \\ 2 \div 2 & = 1 \quad \text{remainder} \quad 0 \\ 1 \div 2 & = 0 \quad \text{remainder} \quad 1 \\ \end{align*} $$
The binary equivalent is $1000001_2$.
Conversion from Decimal to Octal
To convert a decimal number to octal:
Divide the decimal number by 8.
Record the remainder.
Repeat with the quotient until it becomes zero.
Write the remainders in reverse order.
Conversion from Decimal to Hexadecimal
To convert a decimal number to hexadecimal:
Divide the decimal number by 16.
Record the remainder.
Repeat with the quotient until zero.
Write the remainders in reverse order, using A-F for values 10-15.
Conversion Between Binary, Octal, and Hexadecimal
Binary to Octal
Group the binary number in sets of three digits from right to left. Replace each group with the corresponding octal digit.
Binary to Hexadecimal
Group the binary number in sets of four digits from right to left and replace each group with the corresponding hexadecimal digit.
Real-world Applications of Encoding Schemes and Number Systems
Memory Addressing
Hexadecimal numbers simplify memory address representation. For instance, the 16-bit binary address 1100000011110001 becomes C0F1 in hexadecimal, making it easier to remember and use.
Web Colours
Web colours are represented in hexadecimal. For example, red is written as #FF0000, combining red, green, and blue components.
Frequently Asked Questions (FAQs)
Tools for Typing Indian Languages in Unicode
To type Indian languages using Unicode, one might need specific fonts and keyboard layouts designed for those languages.
Advantages of Using Unicode Fonts
Unicode fonts enable the display of diverse languages uniformly across different platforms and devices, removing compatibility issues.
Difference Between Various Number Systems
Binary, octal, and hexadecimal systems simplify the representation and processing of numbers in computing, each with unique benefits suited to different contexts.
Practical Examples of Encoding Schemes in Real Life
Encoding schemes are used in text processing, digital communications, web development, and data storage, ensuring accurate representation and interpretation of data across systems.
Conclusion
Encoding schemes and number systems form the backbone of digital communication and computation. Understanding these concepts is crucial for anyone delving into the field of computer science, ensuring a solid foundation for advanced studies and practical applications.
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Write base values of binary, octal and hexadecimal number system.
The base values are:
Binary Number System: Base 2
Octal Number System: Base 8
Hexadecimal Number System: Base 16
Give full form of ASCII and ISCII.
The full forms are:
ASCII: American Standard Code for Information Interchange
ISCII: Indian Script Code for Information Interchange
Try the following conversions.
(i) $(514)_{8}=\left(\right. ? {) }_{10}$
(ii) $(220)_{8}=(\text { ? })_{2}$
(iii) $(76 \mathrm{~F})_{16}=(\text { ? })_{10}$
(iv) $(4 \mathrm{D} 9)_{16}=(?)_{10}$
(v) $(11001010)_{2}=(?)_{10}$
(vi) $(1010111)_{2}=(\text { ? })_{10}$
Conversion (i) $ (514)_8 = (?)_{10} $
To convert from octal to decimal:
$ 514_8 = 5 \times 8^2 + 1 \times 8^1 + 4 \times 8^0 $
$ 5 \times 64 + 1 \times 8 + 4 \times 1 $
$ 320 + 8 + 4 = 332 $
So, $(514)_8 = (332)_{10}$
Conversion (ii) $ (220)_8 = (?)_{2} $
To convert from octal to binary:
Replace each octal digit with its 3-bit binary equivalent:
2 → 010
2 → 010
0 → 000
Combine the binary values: 0 100 100 000
So, $(220)_8 = (100100000)_2$
Conversion (iii) $ (76F)_{16} = (?)_{10} $
To convert from hexadecimal to decimal:
$ 76F_{16} = 7 \times 16^2 + 6 \times 16^1 + 15 \times 16^0 $
$ 7 \times 256 + 6 \times 16 + 15 $
$ 1792 + 96 + 15 = 1903 $
So, $(76F)_{16} = (1903)_{10}$
Conversion (iv) $ (4D9)_{16} = (?)_{10} $
To convert from hexadecimal to decimal:
$ 4D9_{16} = 4 \times 16^2 + 13 \times 16^1 + 9 \times 16^0 $
$ 4 \times 256 + 13 \times 16 + 9 $
$ 1024 + 208 + 9 = 1241 $
So, $(4D9)_{16} = (1241)_{10}$
Conversion (v) $ (11001010)_{2} = (?)_{10} $
To convert from binary to decimal:
Compute the positional value:
$ 1 \times 2^7 + 1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 $
$ 1 \times 128 + 1 \times 64 + 0 \times 32 + 0 \times 16 + 1 \times 8 + 0 \times 4 + 1 \times 2 + 0 \times 1 $
$ 128 + 64 + 8 + 2 = 202 $
So, $(11001010)_2 = (202)_{10}$
Conversion (vi) $ (1010111)_{2} = (?)_{10} $
To convert from binary to decimal:
Compute the positional value:
$ 1 \times 2^6 + 0 \times 2^5 + 1 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 $
$ 1 \times 64 + 0 \times 32 + 1 \times 16 + 0 \times 8 + 1 \times 4 + 1 \times 2 + 1 \times 1 $
$ 64 + 16 + 4 + 2 + 1 = 87 $
So, $(1010111)_2 = (87)_{10}$
Summary:
$(514)_8 = (332)_{10}$
$(4D9)_{16} = (1241)_{10}$
$(220)_8 = (100100000)_2$
$(11001010)_2 = (202)_{10}$
$(76F)_{16} = (1903)_{10}$
$(1010111)_2 = (87)_{10}$
Do the following conversions from decimal number to other number systems.
(i) $(54)_{10}=(\text { ? })_{2} \quad$
(ii) $(120)_{10}=(\text { ? })_{2} \quad$
(iii) $(76)_{10}=(\text { ?) })_{8} \quad$
(iv) $(889)_{10}=(\text { ?) })_{8} \quad$
(v) $(789)_{10}=(\text { ? })_{16}$
(vi) $(108)_{10}=\left(\right.$ (?) ${ }_{16}$
Let's perform the conversions step by step:
(i) $(54)_{10}$ to Binary $(?)_{2}$
To convert $54_{10}$ to binary:
Divide 54 by 2, note the remainder.
Continue dividing the quotient by 2 until the quotient is 0.
Collect the remainders from bottom to top to get the binary equivalent.
54 ÷ 2 = 27, remainder 0
27 ÷ 2 = 13, remainder 1
13 ÷ 2 = 6, remainder 1
6 ÷ 2 = 3, remainder 0
3 ÷ 2 = 1, remainder 1
1 ÷ 2 = 0, remainder 1
Therefore, (54)_{10} = (110110)_{2}
(ii) $(120)_{10}$ to Binary $(?)_{2}$
To convert $120_{10}$ to binary:
Divide 120 by 2, note the remainder.
Continue dividing the quotient by 2 until the quotient is 0.
Collect the remainders from bottom to top to get the binary equivalent.
120 ÷ 2 = 60, remainder 0
60 ÷ 2 = 30, remainder 0
30 ÷ 2 = 15, remainder 0
15 ÷ 2 = 7, remainder 1
7 ÷ 2 = 3, remainder 1
3 ÷ 2 = 1, remainder 1
1 ÷ 2 = 0, remainder 1
Therefore, (120)_{10} = (1111000)_{2}
(iii) $(76)_{10}$ to Octal $(?)_{8}$
To convert $76_{10}$ to octal:
Divide 76 by 8, note the remainder.
Continue dividing the quotient by 8 until the quotient is 0.
Collect the remainders from bottom to top to get the octal equivalent.
76 ÷ 8 = 9, remainder 4
9 ÷ 8 = 1, remainder 1
1 ÷ 8 = 0, remainder 1
Therefore, (76)_{10} = (114)_{8}
(iv) $(889)_{10}$ to Octal $(?)_{8}$
To convert $889_{10}$ to octal:
Divide 889 by 8, note the remainder.
Continue dividing the quotient by 8 until the quotient is 0.
Collect the remainders from bottom to top to get the octal equivalent.
889 ÷ 8 = 111, remainder 1
111 ÷ 8 = 13, remainder 7
13 ÷ 8 = 1, remainder 5
1 ÷ 8 = 0, remainder 1
Therefore, (889)_{10} = (1571)_{8}
(v) $(789)_{10}$ to Hexadecimal $(?)_{16}$
To convert $789_{10}$ to hexadecimal:
Divide 789 by 16, note the remainder.
Continue dividing the quotient by 16 until the quotient is 0.
Collect the remainders from bottom to top to get the hexadecimal equivalent.
789 ÷ 16 = 49, remainder 13 (D)
49 ÷ 16 = 3, remainder 1
3 ÷ 16 = 0, remainder 3
Therefore, (789)_{10} = (31D)_{16}
(vi) $(108)_{10}$ to Hexadecimal $(?)_{16}$
To convert $108_{10}$ to hexadecimal:
Divide 108 by 16, note the remainder.
Continue dividing the quotient by 16 until the quotient is 0.
Collect the remainders from bottom to top to get the hexadecimal equivalent.
108 ÷ 16 = 6, remainder 12 (C)
6 ÷ 16 = 0, remainder 6
Therefore, (108)_{10} = (6C)_{16}
Summary
$(54)_{10} = (110110)_{2}$
$(120)_{10} = (1111000)_{2}$
$(76)_{10} = (114)_{8}$
$(889)_{10} = (1571)_{8}$
$(789)_{10} = (31D)_{16}$
$(108)_{10} = (6C)_{16}$
Express the following octal numbers into their equivalent decimal numbers.
(i) 145
(ii) 6760
(iii) 455
(iv) 10.75
To convert numbers from octal to decimal, we use the positional value method. The base value of octal is 8.
Converting 145 from Octal to Decimal
Write down the positional values from right to left $8^0, 8^1, 8^2, ...$.
Multiply each digit by its positional value.
Sum the results.
$$ \begin{array}{cccc} 1 \cdot 8^2 & = & 1 \cdot 64 & = 64 \\ 4 \cdot 8^1 & = & 4 \cdot 8 & = 32 \\ 5 \cdot 8^0 & = & 5 \cdot 1 & = 5 \\ \end{array} $$
$$ 145_8 = 64 + 32 + 5 = 101 $$
Therefore, $(145)_{8} = (101)_{10}$
Converting 6760 from Octal to Decimal
Write down the positional values from right to left.
Multiply each digit by its positional value.
Sum the results.
$$ \begin{array}{cccc} 6 \cdot 8^3 & = & 6 \cdot 512 & = 3072 \\ 7 \cdot 8^2 & = & 7 \cdot 64 & = 448 \\ 6 \cdot 8^1 & = & 6 \cdot 8 & = 48 \\ 0 \cdot 8^0 & = & 0 \cdot 1 & = 0 \\ \end{array} $$
$$ 6760_8 = 3072 + 448 + 48 + 0 = 3568 $$
Therefore, $(6760)_{8} = (3568)_{10}$
Converting 455 from Octal to Decimal
Write down the positional values from right to left.
Multiply each digit by its positional value.
Sum the results.
$$ \begin{array}{cccc} 4 \cdot 8^2 & = & 4 \cdot 64 & = 256 \\ 5 \cdot 8^1 & = & 5 \cdot 8 & = 40 \\ 5 \cdot 8^0 & = & 5 \cdot 1 & = 5 \\ \end{array} $$
$$ 455_8 = 256 + 40 + 5 = 301 $$
Therefore, $(455)_{8} = (301)_{10}$
Converting 10.75 from Octal to Decimal
Write down the positional values from right to left for the integer part, and from left to right (negative exponents) for the fractional part.
Multiply each digit by its positional value.
Sum the results.
$$ \begin{array}{cccc} 1 \cdot 8^1 & = & 1 \cdot 8 & = 8 \\ 0 \cdot 8^0 & = & 0 \cdot 1 & = 0 \\ 7 \cdot 8^{-1} & = & 7 \cdot 0.125 & = 0.875 \\ 5 \cdot 8^{-2} & = & 5 \cdot 0.015625 & = 0.078125 \\ \end{array} $$
$$ 10.75_8 = 8 + 0 + 0.875 + 0.078125 = 8.953125 $$
Therefore, $(10.75)_{8} = (8.953125)_{10}$
Express the following decimal numbers into hexadecimal numbers.
(i) 548
(ii) 4052
(iii) 58
(iv) 100.25
To convert the given decimal numbers to their hexadecimal equivalents, we follow the steps outlined in the chapter. Below are the conversions for each number:
(i) $548_{10}$
First, divide the decimal number by 16 and note the remainder: $$ 548 \div 16 = 34 \text{ remainder } 4 $$
Then, divide the quotient (34) by 16: $ 34 \div 16 = 2 \text{ remainder } 2 $
Finally, the remainder after the final division is the quotient itself.
Collect the remainders from bottom to top: $$ (548_{10}) = \text{(2 2 4)}_{16} = 224_{16} $$
(ii) $4052_{10}$
First, divide the decimal number by 16 and note the remainder: $$ 4052 \div 16 = 253 \text{ remainder } 4 $$
Then, divide the quotient (253) by 16: $$ 253 \div 16 = 15 \text{ remainder } 13 $$ (Note that 13 in hexadecimal is D)
Finally, divide the quotient (15) by 16: $$ 15 \div 16 = 0 \text{ remainder } 15 $$ (Note that 15 in hexadecimal is F)
Collect the remainders from bottom to top:
$$ (4052_{10}) = \text{(F D 4)}_{16} = FD4_{16} $$
(iii) $58_{10}$
First, divide the decimal number by 16 and note the remainder: $$ 58 \div 16 = 3 \text{ remainder } 10 $$ (Note that 10 in hexadecimal is A) Finally, divide the quotient (3) by 16: $$ 3 \div 16 = 0 \text{ remainder } 3 $$
Collect the remainders from bottom to top: $$ (58_{10}) = \text{(3 A)}_{16} = 3A_{16} $$
(iv) $100.25_{10}$
First, convert the integer part (100) to hexadecimal: $$ 100 \div 16 = 6 \text{ remainder } 4 $$
Next, multiply the fractional part by 16 repeatedly: $$ 0.25 \times 16 = 4.00 \text{ (integer part is 4)} $$
Combine both integer and fractional parts: $$ (100.25_{10}) = (64.4)_{16} $$
Summary
(i) 548 → 224
(ii) 4052 → FD4
(iii) 58 → 3A
(iv) 100.25 → 64.4
Express the following hexadecimal numbers into equivalent decimal numbers.
(i) 4A2
(ii) 9E1A
(iii) 6BD
(iv) 6C.34
To convert hexadecimal numbers to their equivalent decimal values, we can use the method of positional values. Here are the conversions:
(i) 4A2
List the hexadecimal digits and their decimal equivalents:
(4) -> (4)
(A) -> (10)
(2) -> (2)
Multiply each digit with $16^{\text{position}}$, where position starts from 0 on the right:
$$ 4 \times 16^2 + 10 \times 16^1 + 2 \times 16^0 = 4 \times 256 + 10 \times 16 + 2 \times 1 = 1024 + 160 + 2 = 1186 $$
So, $ (4A2)_{16} = (1186)_{10} $
(ii) 9E1A
List the hexadecimal digits and their decimal equivalents:
(9) -> (9)
(E) -> (14)
(1) -> (1)
(A) -> (10)
Multiply each digit with $16^{\text{position}}$, where position starts from 0 on the right:
$$ 9 \times 16^3 + 14 \times 16^2 + 1 \times 16^1 + 10 \times 16^0 = 9 \times 4096 + 14 \times 256 + 1 \times 16 + 10 \times 1 = 36864 + 3584 + 16 + 10 = 40474 $$
So, $ (9E1A)_{16} = (40474)_{10} $
(iii) 6BD
List the hexadecimal digits and their decimal equivalents:
(6) -> (6)
(B) -> (11)
(D) -> (13)
Multiply each digit with $16^{\text{position}}$, where position starts from 0 on the right:
$$ 6 \times 16^2 + 11 \times 16^1 + 13 \times 16^0 = 6 \times 256 + 11 \times 16 + 13 \times 1 = 1536 + 176 + 13 = 1725 $$
So, $ (6BD)_{16} = (1725)_{10} $
(iv) 6C.34
List the hexadecimal digits and their decimal equivalents:
(6) -> (6)
(C) -> (12)
(.3) -> $3 \times 16^{-1} = 3 \times \frac{1}{16} = \frac{3}{16}$
(.4) -> $4 \times 16^{-2} = 4 \times \frac{1}{256} = \frac{4}{256}$
Convert each part separately:
$$ 6 \times 16^1 + 12 \times 16^0 + 3 \times 16^{-1} + 4 \times 16^{-2} = 6 \times 16 + 12 + 0.1875 + 0.015625 = 96 + 12 + 0.1875 + 0.015625 = 108.203125 $$
So, $$ (6C.34){16} = (108.203125){10} $$
Convert the following binary numbers into octal and hexadecimal numbers.
(i) 1110001000
(ii) 110110101
(iii) 1010100
(iv) 1010.1001
Let's convert the given binary numbers into both octal and hexadecimal systems:
(i) $ 1110001000 $
Binary to Octal:
Grouping bits from right to left in groups of 3: $ 1 ~\underline{110} ~\underline{001} ~\underline{000} $
Corresponding octal digits: $ 1 \ 6 \ 1 \ 0 $
Result: $ (1110001000)_2 = (1610)_8 $
Binary to Hexadecimal:
Grouping bits from right to left in groups of 4: $\underline{1110} \underline{0010} \ 000 $
Corresponding hex digits: $ E \ 2 \ 0 $
Result: $ (1110001000)_2 = (E20)_{16} $
(ii) $ 110110101 $
Binary to Octal:
Grouping bits from right to left in groups of 3: $ \underline{001} ~\underline{101} ~\underline{101} $
Corresponding octal digits: $ 1 \ 5 \ 5 $
Result: $ (110110101)_2 = (155)_8 $
Binary to Hexadecimal:
Grouping bits from right to left in groups of 4: $\underline{1101} ~\underline{1010} ~ 1 $
Corresponding hex digits: $ D \ A \ 1 $
Result: $ (110110101)_2 = (1DA)_{16} $
(iii) $ 1010100 $
Binary to Octal:
Grouping bits from right to left in groups of 3: $ \underline{001} ~\underline{010} ~\underline{100} $
Corresponding octal digits: $ 1 \ 2 \ 4 $
Result: $ (1010100)_2 = (124)_8 $
Binary to Hexadecimal:
Grouping bits from right to left in groups of 4: $\underline{1010}~ \underline{100} $
Corresponding hex digits: $ A \ 4 $
Result: $ (1010100)_2 = (54)_{16} $
(iv) $ 1010.1001 $
Binary to Octal:
Grouping integer part from right to left in groups of 3 and fractional part from left to right in groups of 3: $ 1 \underline{010} \cdot \underline{100} $
Corresponding octal digits: $ 1 \ 2 \cdot 4 $
Result: $ (1010.1001)_2 = (12.4)_8 $
Binary to Hexadecimal:
Grouping integer part from right to left in groups of 4 and fractional part from left to right in groups of 4: $ \underline{1010} \cdot \underline{1001} $
Corresponding hex digits: $ A \cdot 9 $
Result: $ (1010.1001)_2 = (A.9)_{16} $
Summary:
$ (1110001000)_2 = (1610)_8 = (E20)_{16} $
$ (110110101)_2 = (155)_8 = (1DA)_{16} $
$ (1010100)_2 = (124)_8 = (54)_{16} $
$ (1010.1001)_2 = (12.4)_8 = (A.9)_{16} $
Binary | Octal | Hexadecimal |
---|---|---|
1110001000 | 1610 | E20 |
110110101 | 155 | 1DA |
1010100 | 124 | 54 |
1010.1001 | 12.4 | A.9 |
Write binary equivalent of the following octal numbers.
(i) 2306
(ii) 5610
(iii) 742
(iv) 65.203
To write the binary equivalent of an octal number, each octal digit can be replaced with its corresponding 3-bit binary group. This method ensures accurate conversion.
(i) 2306
2: 010
3: 011
0: 000
6: 110
Therefore, $2306_8 = 010011000110_2$.
(ii) 5610
5: 101
6: 110
1: 001
0: 000
Therefore, $5610_8 = 101110001000_2$.
(iii) 742
7: 111
4: 100
2: 010
Therefore, $742_8 = 111100010_2$.
(iv) 65.203
6: 110
5: 101
.: .
2: 010
0: 000
3: 011
Therefore, $65.203_8 = 110101.010000011_2$.
Summary of Binary Equivalents:
2306 in binary: 010011000110
5610 in binary: 101110001000
742 in binary: 111100010
65.203 in binary: 110101.010000011
Write binary representation of the following hexadecimal numbers.
(i) 4026
(ii) BCA1
(iii) 98E
(iv) 132.45
To convert hexadecimal numbers to binary, each hexadecimal digit is substituted with its 4-bit binary equivalent. Let's convert the given hexadecimal numbers to binary:
Conversion Table:
Hexadecimal Symbol | Binary Equivalent |
---|---|
0 | 0000 |
1 | 0001 |
2 | 0010 |
3 | 0011 |
4 | 0100 |
5 | 0101 |
6 | 0110 |
7 | 0111 |
8 | 1000 |
9 | 1001 |
A | 1010 |
B | 1011 |
C | 1100 |
D | 1101 |
E | 1110 |
F | 1111 |
(i) 4026
Hexadecimal Digit | Binary Representation |
---|---|
4 | 0100 |
0 | 0000 |
2 | 0010 |
6 | 0110 |
Binary Representation of 4026: 0100 0000 0010 0110
(ii) BCA1
Hexadecimal Digit | Binary Representation |
---|---|
B | 1011 |
C | 1100 |
A | 1010 |
1 | 0001 |
Binary Representation of BCA1: 1011 1100 1010 0001
(iii) 98E
Hexadecimal Digit | Binary Representation |
---|---|
9 | 1001 |
8 | 1000 |
E | 1110 |
Binary Representation of 98E: 1001 1000 1110
(iv) 132.45
For the fractional part, each hexadecimal place value is treated separately before and after the point:
Hexadecimal Digit | Binary Representation |
---|---|
1 | 0001 |
3 | 0011 |
2 | 0010 |
. | . |
4 | 0100 |
5 | 0101 |
Binary Representation of 132.45: 0001 0011 0010.0100 0101
Summary
4026 (Hex) = 0100 0000 0010 0110 (Binary)
BCA1 (Hex) = 1011 1100 1010 0001 (Binary)
98E (Hex) = 1001 1000 1110 (Binary)
132.45 (Hex) = 0001 0011 0010.0100 0101 (Binary)
How does computer understand the following text? (hint: 7 bit ASCII code).
(i) HOTS
(ii) Main
(iii) CaSe
Computers understand text by converting each character into its 7-bit ASCII code, which is then translated into binary values for processing. Here is the conversion for each text:
(i) HOTS
H: ASCII = 72, Binary = 1001000
O: ASCII = 79, Binary = 1001111
T: ASCII = 84, Binary = 1010100
S: ASCII = 83, Binary = 1010011
(ii) Main
M: ASCII = 77, Binary = 1001101
a: ASCII = 97, Binary = 1100001
i: ASCII = 105, Binary = 1101001
n: ASCII = 110, Binary = 1101110
(iii) CaSe
C: ASCII = 67, Binary = 1000011
a: ASCII = 97, Binary = 1100001
S: ASCII = 83, Binary = 1010011
e: ASCII = 101, Binary = 1100101
Summary:
HOTS
Binary: 1001000 1001111 1010100 1010011
Main
Binary: 1001101 1100001 1101001 1101110
CaSe
Binary: 1000011 1100001 1010011 1100101
The hexadecimal number system uses 16 literals $(0-9$, A-F). Write down its base value.
The base value of the hexadecimal number system is 16, as it uses 16 unique symbols (0-9 and A-F).
Let $\mathrm{X}$ be a number system having B symbols only. Write down the base value of this number system.
The base value of a number system is equal to the number of unique symbols it has.
Therefore, if number system $\mathrm{X}$ has B symbols, the base value of this number system is B.
Write the equivalent hexadecimal and binary values for each character of the phrase given below.
हम सब एक
Characters and Their UNICODE Equivalent Hexadecimal Values:
ह: 0939
म: 092E
** **: 0020 (space)
स: 0938
ब: 092C
** **: 0020 (space)
ए: 090F
क: 0915
Hexadecimal and Binary Representation:
Character | Hexadecimal | Binary |
---|---|---|
ह | 0939 | 0000 1001 0011 1001 |
म | 092E | 0000 1001 0010 1110 |
Space | 0020 | 0000 0000 0010 0000 |
स | 0938 | 0000 1001 0011 1000 |
ब | 092C | 0000 1001 0010 1100 |
Space | 0020 | 0000 0000 0010 0000 |
ए | 090F | 0000 1001 0000 1111 |
क | 0915 | 0000 1001 0001 0101 |
Summary:
ह: 0939 → 0000100100111001
म: 092E → 0000100100101110
Space: 0020 → 0000000000100000
स: 0938 → 0000100100111000
ब: 092C → 0000100100101100
Space: 0020 → 0000000000100000
ए: 090F → 0000100100001111
क: 0915 → 0000100100010101
These are the UNICODE hexadecimal and binary equivalents for each character in the given phrase.
What is the advantage of preparing a digital content in Indian language using UNICODE font?
The advantage of preparing digital content in Indian language using UNICODE font includes:
Universality: Unicode provides a unique number for every character, ensuring that digital content can be read and interpreted correctly across different platforms and devices.
Compatibility: It ensures that text in Indian languages can be viewed and processed on various systems such as Linux, Windows, iOS, and different software applications.
Standardization: Unicode encompasses all characters of almost every written language, which ensures consistent representation and exchange of data globally.
This enhances interoperability and preserves the integrity of the content across different technological environments.
Explore and list the steps required to type in an Indian language using UNICODE.
To type in an Indian language using UNICODE, follow these steps:
Install Unicode Fonts: Ensure that you have installed the necessary Unicode fonts for the specific Indian language you want to use. These fonts can be downloaded from various sources online.
Enable Language Support:
Windows: Go to
Settings > Time & Language > Language
and add the desired Indian language. This will install the necessary language packs.Mac: Go to
System Preferences > Language & Region > Add a language
.
Switch Keyboard Layout:
Windows: Use
Windows Key + Spacebar
to switch between different keyboard layouts or click on the language icon in the taskbar.Mac: Use
Cmd + Space
to open Spotlight and type "Keyboard Preferences". Add the desired Indian input source and switch using the language icon in the menu bar.
Use Virtual Keyboard: If you have difficulty with the physical keyboard, you can use an on-screen virtual keyboard which can be enabled from the settings. This allows clicking characters directly.
Use Input Tools:
Google Input Tools: Install Google Input Tools which supports many Indian languages. It allows typing in transliterated format that converts phonetic English typing into the desired Indian language.
Other Software: Microsoft Indic Language Input Tool or other third-party software can be used for language-specific typing needs.
Type the Text: Open any text editor, document, or application where you want to type, switch to the desired language keyboard layout, and start typing using the appropriate Unicode characters.
By following these steps, you will be able to type seamlessly in any Indian language using Unicode encoding.
Encode the word 'COMPUTER' using ASCII and convert the encode value into binary values.
To encode the word "COMPUTER" using ASCII and convert the encoded values into their binary equivalents, follow these steps:
Determine the ASCII values for each character in "COMPUTER".
Convert each ASCII value to its binary equivalent.
ASCII Encoding:
Character | ASCII Code |
---|---|
C | 67 |
O | 79 |
M | 77 |
P | 80 |
U | 85 |
T | 84 |
E | 69 |
R | 82 |
Conversion to Binary:
Character | ASCII Code | Binary Code |
---|---|---|
C | 67 | 01000011 |
O | 79 | 01001111 |
M | 77 | 01001101 |
P | 80 | 01010000 |
U | 85 | 01010101 |
T | 84 | 01010100 |
E | 69 | 01000101 |
R | 82 | 01010010 |
Hence, the word "COMPUTER" in ASCII and its binary equivalents are as follows:
Character | Binary Code |
---|---|
C | 01000011 |
O | 01001111 |
M | 01001101 |
P | 01010000 |
U | 01010101 |
T | 01010100 |
E | 01000101 |
R | 01010010 |
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