chapter 1 number and code systems

8/12/2019 Chapter 1 Number and Code Systems

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EE202 [DIGITAL ELECTRONIC]NUMBER AND CODE SYSTEMS

Prepared by: WAN HAMIDAH BINTI WAN ABAS [PKK\JKE]_version01_Dis2012 Page 1-1

CHAPTER 1

NUMBER AND CODE SYSTEMS

There are many types of number system is used in our life. The binary number system is

the most important one in digital systems, but several others are also important. Thedecimal system is important because it is universally used to represent quantities outside a

digital system. This means that there will be situations where decimal values must be

converted to binary values before they are entered into the digital system. Likewise, there

will be situations where the binary values at the outputs of a digital system must be

converted to decimal values for presentation to the outside world. Two other number

systems that also important are octal and hexadecimal system.

1.1 Number SystemsThere are 4 types of numbering system used in digital system.

a. Decimal system (N10)0, 1, 2, 3, 4, 5, 6, 7, 8, 9

b. Binary system (N2)0, 1

c. Octal system (N8)0, 1, 2, 3, 4, 5, 6, 7

d. Hexadecimal system (N16)0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

Example: 23410

OUTCOMES:

Upon completion of this topic, students should be able to:

1. illustrate the knowledge of digital number systems: decimal, binary, octal andhexadecimal

2. illustrate the knowledge of code systems: BCD 8421 and ASCII code

Basic

number

Base

number


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Table 1: Numbering Systems

Decimal

System

Binary

System

Octal

System

Hexadecimal

System

0 0000 0 0

1 0001 1 1

2 0010 2 23 0011 3 3

4 0100 4 4

5 0101 5 5

6 0110 6 6

7 0111 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 D14 1110 16 E

15 1111 17 F

Binary Number System:

System Digits: 0 and 1

Bit (short for binary digit): A single binary digit

LSB (least significant bit): The rightmost bit

MSB (most significant bit): The leftmost bit

1.2 Conversion Number Systems1.2.1 Decimal to Binary Conversions

The decimal number will divide by 2 until obtain minimum values. If the decimal

number have the decimal point, so each the decimal point value must be multiply by 2.

Example:

Convert 12310to it binary equivalent.

Ans: 2 123 remainder of 1

2 61 remainder of 12 30 remainder of 0

2 15 remainder of 1

2 7 remainder of 1

2 3 remainder of 1

1

12310= 1 1 1 1 0 1 12MSB LSB


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Example:

Convert 97.5410to it binary equivalent.

Ans: 2 97 - 1 0.54 x 2 = 1 .08

2 48 - 0 0.08 x 2 = 0 .16

2 24 - 0 0.16 x 2 = 0 .32

2 12 - 0 0.32 x 2 = 0 .64

2 6 - 0

2 3 - 1

1

12310= 1 1 0 0 0 0 1.12

1.2.2

Decimal to Octal ConversionsThe decimal number will divide by 8 until minimum values. If the decimal number

have the decimal point, so each the decimal point value must be multiply by 8.

Example:

Convert 39610to it octal equivalent.


8 49 remainder of 1

6

39610= 6 1 48MSD LSD

Example:

Convert 172.3110to it octal equivalent.

Ans: 8 172 - 4 0.31 x 8 = 2 .48

8 21 - 5 0.48 x 8 = 3 .84

2 0.84 x 8 = 6 .72

0.72 x 8 = 5 . 76

172.3110= 254.23658


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1.2.3 Decimal to Hexadecimal ConversionsThe decimal number will divide by 16 until minimum values. If the decimal number

have the decimal point, so each the decimal point value must be multiply by 16.

Example:

Convert 698310to it hexadecimal equivalent.


16 436 remainder of 4

16 27 remainder of 11

1

698310= 1 B 4 716MSD LSD

Example:

Convert 5932.7610to it binary equivalent.

Ans: 16 5932 - 12 0.76 x 16 = 12 .16

16 370 - 2 0.16 x 16 = 2 .56

16 23 - 14 0.56 x 16 = 8 .96

1 0.96 x 16 = 15 .36

5932.7610= 1 E 2 C . C 2 8 F16

1.2.4 Binary to Decimal Conversions

Example :

Convert 1012to it decimal equivalent.

Ans: 1012 = (1 x 22) + (0 x 21) + (1 x 20)

= (1 x 4) + (0 x 2) + (1 x 1)

= 4 + 0 + 1

= 510

Binary System

.25 24 23 22 21 20 2-1 2-2 2-3..

.32 16 8 4 2 1 0.5 0.25 0.125.


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Example :

Convert 1011.012to it decimal equivalent.

Ans: 1011.0112 = (1 x 23) + (0 x 22) + (1 x 21) + (1 x 20) + (0 x 2-1) + (1 x 2-2) + (1 x 2-3)

= (1 x 8) + (0 x 4) + (1 x 2) + (1 x 1) + (0 x 0.5) + (1 x 0.25) + (1 x 0.125)

= 8 + 0 + 2 + 1 + 0 + 0.25 + 0.125

= 11.37510

1.2.5 Octal to Decimal Conversions

Example :

Convert 6258to it decimal equivalent.

Ans: 6258 = (6 x 82) + (2 x 81) + (5 x 80)

= (6 x 64) + (2 x 8) + (5 x 1)

= 384 + 16 + 5

= 40510

Example :

Convert 31.78to it decimal equivalent.

Ans: 31.78 = (3 x 81) + (1 x 80) + (7 x 8-1)

= (3 x 8) + (1 x 1) + (7 x 0.125)

= 24 + 1 + 0.875

= 25.87510

1.2.6 Hexadecimal to Decimal Conversions

Example :

Convert 1D216to it decimal equivalent.

Ans: 1D216 = (1 x 162) + (D x 161) + (2 x 160)

= (1 x 256) + (13 x 16) + (2 x 1)

= 256 + 208 + 2

= 46610

Octal System

.....83 82 81 80 8-1 8-2..

512 64 8 1 0.125 0.015625.

Hexadecimal System

.....163 162 161 160 16-1 16-2..

4096 256 16 1 0.0625 0.0039.


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Example :

Convert A69.5C16to it decimal equivalent.

ANS: A69.5C16 = (A x 162) + (6 x 161) + (9 x 160) + (5 x 16-1) + (C x 16-2)

= (10 x 256) + (6 x 16) + (9 x 1) + (5 x 0.0625) + (12 x 0.0039)

= 2560 + 96 + 9 + 0.3125 + 0.0468

= 2665.359310

1.2.7 Binary to Octal ConversionsAssign the each bit binary numbers into 3 bit from LSB to MSB. It is because octal

value need 3 bit of binary value to complete the systems.

Example :

Convert 110101002to it octal equivalent.

Ans:

110101002=3248

Example :

Convert 10110.01012to it octal equivalent.

Ans:

10110.01012=26.248

Binary

System

Octal

System

000 0001 1

010 2

011 3

100 4

101 5

110 6

111 7

MSB LSB

0 1 1 0 1 0 1 0 0

3 2 4

0 1 0 1 1 0 0 1 0 1 0 0

2 6 2 4


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1.2.8 Binary to Hexadecimal ConversionsAssign the each bit binary numbers into 4 bit from LSB to MSB. It is because

hexadecimal value need 4 bit of binary value to complete the systems.

Example :


Ans:

10110110102=2DA16

Example :

Convert 11110001.0101112to it hexadecimal equivalent.

Ans:

11110001.0101112 =F1.5C16

Binary

System

Hexadecimal

System0000 0

0001 1

0010 2

0011 3

0100 4

0101 5

0110 6

0111 7

1000 8

1001 9

1010 A

1011 B

1100 C

1101 D

1110 E

1111 F

MSB LSB

0 0 1 0 1 1 0 1 1 0 1 0

2 D A

1 1 1 1 0 0 0 1 0 1 0 1 1 1 0 0

F 1 5 C


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1.2.9 Hexadecimal to Binary ConversionsExample:

Convert 195D16to it binary equivalent.

Ans:

195D16 = 11001010111012

Example :

Convert 8E2.A616to it binary equivalent.

Ans:

8E2.A616 =100011100010.10100112

1.2.10Hexadecimal to Octal ConversionsThere are two methods to convert the hexadecimal to octal systems. The methods

are:

1. HexadecimalBinaryOctal2. HexadecimalDecimalOctal

Example:

Convert 62E316to it octal equivalent.

ANS:

Method 1: Hexadecimal Binary Octal

62E316 = 613438

Method 2: Hexadecimal Decimal Octal

Hexadecimal Decimal :62E316= (6 x 163) + (2 x 162) + (14 x 161) + (3 x 160)

= (6 x 4096) + (2 x 256) + (14 x 16) + (3 x 1)

= 24576 + 512 + 224 + 3

= 2531510

1 9 5 D0 0 0 1 1 0 0 1 0 1 0 1 1 1 0 1

MSB LSB

8 E 2 A 6

1 0 0 0 1 1 1 0 0 0 1 0 1 0 1 0 0 1 1 0

Hexa 6 2 E 3

Binary 0 0 0 1 1 0 0 0 1 0 1 1 1 0 0 0 1 1

Octal 0 6 1 3 4 3

This method is easier and it can reduce calculation careless.


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Decimal Octal : 2531510=613438

62E316 = 613438

Example :

Convert 2C.1916to it octal equivalent.

Ans:

Method 1: Hexadecimal Binary Octal

2C.1916 = 54.0628

Method 2: Hexadecimal Decimal Octal

Hexadecimal Decimal : 2C.1916= (2 x 161) + (12 x 160) + (1 x 16-1) + (9 x 16-2)

= (2 x 16) + (12 x 1) + (1 x 0.0625) + (9 x 0.00391)

= 32 + 12 + 0.0625 + 0.03519

= 44.0976910

Decimal Octal : 0.09769 x 8 = 0 .78152

0.78152 x 8 = 6 .25216

0.25216 x 8 = 2 .01728

So, it will be: 44.0976910=54.0628

2C.1916 = 54.0628

8 25315 - 3

8 3164 - 4

8 395 - 3

8 49 - 1

6

Hexa 2 C 1 9

Binary 0 0 0 1 0 1 1 0 0 0 0 0 1 1 0 0 1 0

Octal 0 5 4 0 6 2

8 44 - 4

5


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1.2.11Octal to Hexadecimal ConversionsThere are two methods to convert the hexadecimal to octal systems. The methods

are:

1. OctalBinaryHexadecimal2. OctalDecimalHexadecimal

Example :


Ans:

Method 1: Octal Binary Hexadecimal

42578 = 8AF16

Method 2: Octal Decimal Hexadecimal

Octal Decimal : 42578 = (4 x 83) + (2 x 82) + (5 x 81) + (7 x 80)

= (4 x 512) + (2 x 64) + (5 x 8) + (7 x 1)

= 2048 + 128 + 40 + 7

= 222310

Decimal Hexadecimal:

222310 =8AF16

42578 = 8AF16

Example :

Convert 741.258to it hexadecimal equivalent.

Ans:

Method 1: Octal Binary Hexadecimal

741.258 = 1E1.5416

Octal 4 2 5 7

Binary 1 0 0 0 1 0 1 0 1 1 1 1

Hexa 8 A F

16 2223 - F

16 138 - A

8

Octal 7 4 1 2 5

Binary 0 0 0 1 1 1 1 0 0 0 0 1 0 1 0 1 0 1 0 0

Hexa 1 E 1 5 4

This method is easier and it can reduce calculation careless.


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Decimal

Division

Binary

Octal

Hexadecimal

Multiplication

Decimal

REMEMBERCONVERSION NUMBER SYSTEMS

Number system positions

.N5 N4 N3 N2 N1 N0 N-1 N-2 N-3..

Method 2: Octal Decimal Hexadecimal

Octal Decimal : 741.258 = (7 x 82) + (4 x 81) + (1 x 80) + (2 x 8-1) + (5 x 8-2)

= (7 x 64) + (4 x 8) + (1 x 1) + (2 x 0.125) + (5 x 0.015625)

= 448 + 32 + 1 + 0.25 + 0.078125

= 481.3281210

Decimal Hexadecimal:

0.32812 x 16 = 5 .24992

0.24992 x 16 = 3 .99872

0.99872 x 16 = F .97952

481.3281210=1E1.53F16

741.258 = 1E1.53F16

16 481 - 1

16 30 - E

1

Binary

Octal

Hexadecimal

Multiply

Decimal

Division Binary

Octal

Hexadecimal

Note: For decimalvalue must be multiply

by their base


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1.3 Arithmetic of Number Systems1.3.1 Binary Arithmetic1.3.1.1 Addition

Note: The rules of binary addition (without carries) are the same as the truths of the

XORgate.

Example :

1. 10102+ 1112= 100012

2. 111102+ 100112= 1100012

1.3.1.2 Subtraction

Example :

1. 1010121102= 11112

1 0 1 02 = 1010

+ 1 1 12 = 710

1 0 0 012 = 1710

1 1 1 1 02 = 3010

+ 1 0 0 1 12 = 19101 1 0 0 0 12 = 4910

1 0 1 0 12 = 2110

- 1 1 02 = 610

1 1 1 12 = 1510

0

Rules of Binary Addition:0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 0 and carry 1 to the next more significant bit

Rules of Binary Subtraction:

0 - 0 = 0

0 - 1 = 1 and borrow 1 from the next more significant

1 - 0 = 1

1 - 1 = 0

11

12+ 12= 102

110+ 110= 210

111

100101 10
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1.3.1.3 Multiplication

Note: The rules of binary multiplication are the same as the truths of theANDgate.

Example :

1. 1012 112= 11112

1.3.1.4 DivisionBinary division is the repeated process of subtraction, just as in decimal division.

Example:

1. 1010102 1102= 1112 1 1 11 1 01 0 1 0 1 0

- 1 1 0

1 0 0 1

- 1 1 01 1 0

1 1 0

1 0 1 = 510

1 1 = 310

1 0 1 1510

+ 1 0 1

1 1 1 1

Rules of Binary Multiplication:

0 0 = 0

0 1 = 0

1 0 = 01 1 = 1 and no carry or borrow bits

. . .
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2. 100001112 1012= 1101121 1 0 1 1

1 0 11 0 0 0 0 1 1 1- 1 0 1

1 1 0

- 1 0 11 1

- 01 1 1

- 1 0 11 0 1

- 1 0 1

1.3.2 Octal Arithmetic1.3.2.1 Addition

Example:

1. 6538+ 1228=7758

2. 7168+ 1548=7758

1.3.2.2 Subtraction

Example:

1. 13648- 1238=12418

6 5 38

+ 1 2 28

7 7 58

7 1 68

+ 1 5 48

1 0 7 28

1 3 6 48

- 1 2 38

1 2 4 18

. . .

6 + 4 = 10 8= 2

7 + 1 = 8 8= 0

11


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2. 36458- 7518=26748

1.3.3 Hexadecimal Arithmetic1.3.3.1 Addition

Example:

1. 952116+ 1A616=96C716

2. 3F216+ 6E16=46016

1.3.3.2 Subtraction

Example:

1. 2B991639816=280116

2. 6EF316FF16=6DF48

3 6 4 58

- 7 5 18

2 6 7 48

9 5 2 116

+ 1 A 616

9 6 C 716

3 F 216

+ 6 E16

4 6 016

2 B 9 916

- 3 9 816

2 8 0 116

6 E F 316

- F F16

6 D F 416

4 +8= 12 5 = 7

5 + 8= 13 7 = 6

58

2 8

2 + 14 = 16 16= 0

15 + 1 + 6 = 22 16= 6

11

16 +3 = 19 15 = 4

16 + 14 = 30 - 15 = F

D E 1616


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1.4 Sign NumbersIn digital systems, the binary numbers are represented by a set of binary such as a

microprocessor register 8 bits. The most significant bit (MSB) is the sign bit. If this

bit is 0, then the number is (+) positive. However, if the sign bit is 1, then the

number is (-) negative. The other 7 bits in this 8-bit register represent the

magnitude of the number.

Sign bit 0 = (+)

1 = ()

MSB LSB

Magnitude

Figure below :Representation of signed numbers in sign-magnitude form.

A7 A6 A5 A4 A3 A2 A1 A0

0 0 1 0 1 1 0 1 = +4510

Sign

bit

(+)

True binary

A7 A6 A5 A4 A3 A2 A1 A0

1 1 1 0 1 0 1 1 = -4510

Sign

bit

(-)

2s complement

1.5 2s Complement NumbersThe 2s complement method of representing numbers is widely used inmicroprocessor based equipment. Until now, we have assumed that all numbers are

positive. However, microprocessors must process both positive and negative

numbers. By using 2s complement representation, the sign as well as the magnitudeof a number can be determined.

1s Complement

The 1s complement of a binary number is obtained by changing each 0 to a 1 andeach 1 to a 0. In other words, change each bit in the number to its complement.

Example:

Original binary number 0 0 1 0 1 1 0 1

1s complement 1 1 0 1 0 0 1 0


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2s Complement

The 2s complement of a binary number is formed by taking the 1s complement ofthe number and adding 1 to the least-significant-bit position.

Example:

Original binary number 0 0 1 0 1 1 0 1

1s complement 1 1 0 1 0 0 1 0

Add 1 to form 2s complement + 12s complement of original number 1 1 0 1 0 0 1 1

Example:

Represent the signed decimal numbers below as a signed binary number in 2scomplement system. Use a total of five bits including the sign bit.

a) + 13b) 9

Ans:

a) + 13 = 01101

sign bit

b) + 9change to 9 by using 2s complement

9 = 10111

1.5.1 Arithmetic of 2s ComplementCase I: Two Positive Numbers

The addition of two positive numbers is straight forward.

Example: + 9 + 4 = + 13

+ 9 0 1001+ 4 + 0 0100

0 1101

sign bits

+ 9 010011s complement 10110

Add 1 + 1

2s complement 10111


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Case II: Positive Number and Smaller Negative Number

Consider the addition of +9 and 4. Remember that the 4 will be in its 2scomplement form. Thus, +4 (00100) must be converted to 4 (11100).

Example: + 9 4 = + 5sign bits

+ 9 0 1001- 4 + 1 1100

1 0 0101

This carry is disregarded; the result is 00101 (sum = +5)

Case III: Positive Number and Larger Negative Number

Consider the addition of 9 and + 4.

Example: - 9 + 4 = - 5

- 9 1 0111+ 4 + 0 0100

1 1011 (sum = - 5)

negative sign bits

Case IV: Two Negative Numbers

Consider the addition of 9 and 4.

Example: - 9 4 = - 13sign bits

- 9 1 0111- 4 + 1 1100

1 1 0011

This carry is disregarded; the result is 10011 (sum = - 13)

Case V: Equal and Opposite Numbers

Example: - 9 + 9 = + 0

sign bits

- 9 1 0111+ 9 + 0 1001

1 0 0000

This carry is disregarded; the result is 00000 (sum = + 0)


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1.5.2 Arithmetic OverflowExample:

incorrect sign bits

+ 9 0 1001+ 8 + 0 1000

1 0001

The answer has a negative sign bit, which is obviously incorrect since we are adding

two positive numbers. The answer should be +17, but the magnitude 17 requires

more than four bits and therefore overflows into the sign-bit position. This overflow

condition can occur only when two positive or two negative numbers are being

added, and it always produces an incorrect result. Overflow can be detected by

checking to see that the sign bit of the result is the same as the sign bits of the

numbers being added.

1.6 Code System1.6.1 BCD 8421

Decimal BCD 8421

8 4 2 1

0 0 0 0 0

1 0 0 0 1

2 0 0 1 0

3 0 0 1 14 0 1 0 0

5 0 1 0 1

6 0 1 1 0

7 0 1 1 1

8 1 0 0 0

9 1 0 0 1

Example 1:

Convert the following decimal value to BCD code.

a) 593110b) 28.9410

Ans:

a) 593110= 101100100110001BCDDecimal 5 9 3 1

BCD code 0 1 0 1 1 0 0 1 0 0 1 1 0 0 0 1


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b) 28.9410= 101000.100101BCD

Example 2:

Convert the following BCD code to it decimal value equivalent.

a) 10101010101000101BCDb) 101001100.1001011BCD

Ans:

a) 10101010101000101BCD= 1554510

b) 101000011.1001011BCD= 143.9610

1.6.1.1 Arithmetic of BCD code

BCD addition follows the same rules asbinary addition.However, if the addition

produces a carry and/or creates an invalid BCD number, an adjustment is required

to correct the sum. The correction method is to add 6 to the sum in any digit

position that has caused an error.

Case I: Sum Equals 9 or Less

Example:

1.

2.

Decimal 2 8 9 4

BCD code 0 0 1 0 1 0 0 0 1 0 0 1 0 1 0 0

Decimal 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 1

BCD code 1 5 5 4 5

Decimal 0 0 0 1 0 1 0 0 0 0 1 1 1 0 0 1 0 1 1 0

BCD code 1 4 3 9 6

5 0101BCD

+ 4 + 0100BCD

9 1001BCD

45 0100 0101BCD

+ 33 + 0011 0011BCD

78 0111 1000BCD
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3.

Case II: Sum Greater Than 9

Example:

1.

2.

1

3.

BCD subtraction follows the same rules asbinary subtraction.However, if the

subtraction causes a borrow and/or creates an invalid BCD number, an adjustment

is required to correct the answer. The correction method is to subtract 6from the

difference in any digit position that has caused an error.

25 0010 0101BCD

+ 12 + 0001 0010BCD

37 0011 0111BCD

6 0110BCD

+ 7 + 0111BCD

13 1101BCD invalid code group for BCD+ 0110BCD add 6 for correction

0001 0011BCD

1 3

47 0100 0111BCD

+ 35 + 0011 0101BCD

82 0111 1100BCD invalid sum in first digit+ 1 0110BCD add 6 for correction

1000 0010BCD correct BCD sum8 2

59 0101 1001BCD

+ 38 + 0011 1000BCD

97 1001 0001BCD invalid sum in first digit+ 0110BCD add 6 for correction

1001 0111BCD correct BCD sum

9 7

1
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1.6.2 ASCII CodeASCII code is used to transmit the digital data through telephone line. Table 1 shown

the sequential of ASCII code table that have alphabets, numbers and some symbols.

ASCII code has 7 bit code. For example, if the data was transmitted is 100 0001, that

means the data is A. So, 100 0001 = A.

Table 1: ASCII code

How to determine the ASCII code:

Example 1:

Y = 59 (hexadecimal)

Y = 101 1001 (binary)


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Example 2:

The following is a message encoded in ASCII code. What is the message?

1001000 10001010 1001100 1010000

Ans:

4 8 4 5 4 C 5 0

1 0 0 1 0 0 0 1 0 0 0 1 0 1 1 0 0 1 1 0 0 1 0 1 0 0 0 0

H E L P

chapter 1 number and code systems

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