UNIT I

BINARY SYSTEMS

1. DIGITAL SYSTEMS,BINARY NUMBERS & NUMBER BASE CONVERSION

A digital system can understand positional number system only where there are a few symbols called digits and these symbols represent different values depending on the position they occupy in the number.A value of each digit in a number can be determined using The digit The position of the digit in the number The base of the number system (where base is defined as the total number of digits available in the number system).The term digital refers to the fact that the signal is limited to only a few possible values. In general, digits signals are represented by only two possible voltages on a wire - 0 volts (which we called "binary 0", or just "0") and 5 volts (which we call "binary 1", or just "1"). We sometimes call these values "low" and "high", or "false" and "true".

Definition:

A digital system is an interconnection of digital modules and it is system that manipulates discrete elements of information that is represented internally in the binary form.

Characteristics of digital system:1. Digital systems manipulate elements of information2. Discrete elements are nothing but the digits such as 10 decimal digits ,26 letters of

alphbets 3. Digital systems use physical quantities called signals to represent discrete elements.4. In digital systems, the signals have two discrete values & therefore said to be binary.5. A signal in digital system represents one binary digit called a bit, the bit has value either

0 or 1.6. Discrete elements of information are represented with group of bits called binary codes.

Advantages of digital system:1. Dramatic cost reductions in digital devices have come about because of advances in digital integrated circuit technology.

2. Equipment built with digital integrated circuits can perform at a speed of hundreds of millions of operations per second.

3. Digital systems can be made to operate with extreme reliability by using error correcting codes.

Decimal Number System

The number system that we use in our day-to-day life is the decimal number system. Decimal number system has base 10 as it uses 10 digits from 0 to 9. In decimal number system, the successive positions to the left of the decimal point represent units, tens, hundreds, thousands and so on.Each position represents a specific power of the base (10). For example, the decimal number 1234 consists of the digit 4 in the units position, 3 in the tens position, 2 in the hundreds position, and 1 in the thousands position, and its value can be written as

(1×1000) + (2×100) + (3×10) + (4×l) (1×103) + (2×102) + (3×101) + (4×l00) = 1000 + 200 + 30 + 1 = 1234

Number System & Description

1 Binary Number System Base 2. Digits used: 0, 12 Octal Number System Base 8. Digits used: 0 to 73 Hexadecimal Number System Base 16. Digits used: 0 to 9, Letters used: A- F

Binary Number System

Characteristics Uses two digits, 0 and 1. Also called base 2 number system Each position in a binary number represents a 0 power of the base (2). Example: 20

Last position in a binary number represents an x power of the base (2). Example: 2 x where x represents the last position - 1.

ExampleBinary Number: 101012

Calculating Decimal Equivalent −Step Binary Number Decimal NumberStep 1 101012 ((1 × 24) + (0 × 23) + (1 × 22) + (0 × 21) + (1 × 20))10

Step 2 101012 (16 + 0 + 4 + 0 + 1)10

Step 3 101012 2110

Note: 101012 is normally written as 10101.

Octal Number SystemCharacteristics Uses eight digits, 0,1,2,3,4,5,6,7. Also called base 8 number system

Each position in an octal number represents a 0 power of the base (8). Example: 80

Last position in an octal number represents an x power of the base (8). Example: 8x where x represents the last position - 1.

ExampleOctal Number − 125708

Calculating Decimal Equivalent Step Octal Number Decimal NumberStep 1 125708 ((1 × 84) + (2 × 83) + (5 × 82) + (7 × 81) + (0 × 80))10

Step 2 125708 (4096 + 1024 + 320 + 56 + 0)10

Step 3 125708 549610

Note: 125708 is normally written as 12570.

Hexadecimal Number SystemCharacteristics Uses 10 digits and 6 letters, 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. Letters represents numbers starting from 10. A = 10, B = 11, C = 12, D = 13, E = 14, F = 15. Also called base 16 number system.

Each position in a hexadecimal number represents a 0 power of the base (16). Example 160. Last position in a hexadecimal number represents an x power of the base (16). Example 16x

where x represents the last position - 1.Example −Hexadecimal Number: 19FDE16

Calculating Decimal Equivalent −

StepBinary Number

Decimal Number

Step 1 19FDE16((1 × 164) + (9 × 163) + (F × 162) + (D × 161) + (E × 160))10

Step 2 19FDE16((1 × 164) + (9 × 163) + (15 × 162) + (13 × 161) + (14 × 160))10

Step 3 19FDE16 (65536 + 36864 + 3840 + 208 + 14)10

Step 4 19FDE16 10646210

Note − 19FDE16 is normally written as 19FDE.

Number Base Conversions

There are many methods or techniques which can be used to convert numbers from one base to another. We'll demonstrate here the following −

Decimal to Other Base System

Other Base System to Decimal

Other Base System to Non-Decimal

Shortcut method − Binary to Octal

Shortcut method − Octal to Binary

Shortcut method − Binary to Hexadecimal

Shortcut method − Hexadecimal to Binary

Decimal to Other Base SystemTwo methods

. 1) For integer part using successive division method. 2) For fractional part using successive multiplication method.

Decimal to binaryExample −Decimal Number: 2910

Calculating Binary Equivalent −Step Operation Result RemainderStep 1 29 / 2 14 1Step 2 14 / 2 7 0Step 3 7 / 2 3 1Step 4 3 / 2 1 1Step 5 1 / 2 0 1As mentioned in Steps 2 and 4, the remainders have to be arranged in the reverse order so that the first remainder becomes the Least Significant Digit (LSD) and the last remainder becomes the Most Significant Digit (MSD).

Decimal Number − 2910 = Binary Number − 111012.

Octal to decimal conversion:

Ex: Convert 4057.068 to octal

= 4x83+0x82+5x81+7x80+0x8-1+6x8-2

= 2048+0+40+7+0+0.093 = 2095.093710

Other Base System to Decimal System Steps

Step 1 − Determine the column (positional) value of each digit (this depends on the position of the digit and the base of the number system).

Step 2 − Multiply the obtained column values (in Step 1) by the digits in the corresponding columns.

Step 3 − Sum the products calculated in Step 2. The total is the equivalent value in decimal.

Example Binary Number − 111012

Calculating Decimal Equivalent −

Step Binary Number Decimal NumberStep 1 111012 ((1 × 24) + (1 × 23) + (1 × 22) + (0 × 21) + (1 × 20))10

Step 2 111012 (16 + 8 + 4 + 0 + 1)10

Step 3 111012 2910

Binary Number − 111012 = Decimal Number − 2910

Hexadecimal to Decimal conversion:

Ex: 5C716 to decimal

(5x162)+(C x161)+ (7 x160) = 1280+192+7.= 14710

Other Base System to Non-Decimal System Steps

Step 1 − Convert the original number to a decimal number (base 10). Step 2 − Convert the decimal number so obtained to the new base number.ExampleOctal Number − 258

Calculating Binary Equivalent −Step 1 − Convert to Decimal

Step Octal Number Decimal Number

Step 1 258 ((2 × 81) + (5 × 80))10

Step 2 258 (16 + 5 )10

Step 3 258 2110

Octal Number − 258 = Decimal Number − 2110

Step 2 − Convert Decimal to BinaryStep Operation Result RemainderStep 1 21 / 2 10 1Step 2 10 / 2 5 0Step 3 5 / 2 2 1Step 4 2 / 2 1 0Step 5 1 / 2 0 1

Decimal Number − 2110 = Binary Number − 101012

Octal Number − 258 = Binary Number – 1010

Binary to Octal conversion:

Starting from the binary pt. make groups of 3 bits each, on either side of the binary pt, & replace each 3 bit binary group by the equivalent octal digit.

Convert 110101.1010102 to octal Group of 3

110 101. 101 010

6 5 . 5 2

Hexadecimal to binary:Replace each hex digit by its 4-bit binary group.Ex: 4BAC10 to binary

=01001011101011002Shortcut method - Binary to Octal

Steps

Step 1 − Divide the binary digits into groups of three (starting from the right). Step 2 − Convert each group of three binary digits to one octal digit.Example

Binary Number − 101012

Calculating Octal Equivalent −Step Binary Number Octal NumberStep 1 101012 010 101

=65.528

4 B A C0100 1011 1010 1100

Step 2 101012 28 58

Step 3 101012 258

Binary Number − 101012 = Octal Number − 258

Shortcut method - Octal to Binary Steps

Step 1 − Convert each octal digit to a 3 digit binary number (the octal digits may be treated as decimal for this conversion).

Step 2 − Combine all the resulting binary groups (of 3 digits each) into a single binary number.

ExampleOctal Number − 258

Calculating Binary Equivalent –

Step Octal Number Binary NumberStep 1 258 210 510

Step 2 258 0102 1012

Step 3 258 0101012

Octal Number − 258 = Binary Number − 101012

Shortcut method - Binary to Hexadecimal Step 1 − Divide the binary digits into groups of four (starting from the right).

Step 2 − Convert each group of four binary digits to one hexadecimal symbol.

Example

Binary Number − 101012

Calculating hexadecimal Equivalent –

Step Binary Number Hexadecimal Number

Step 1 101012 0001 0101

Step 2 101012 110 510

Step 3 101012 1516

Binary Number − 101012 = Hexadecimal Number − 1516

Shortcut method - Hexadecimal to Binary Step 1 − Convert each hexadecimal digit to a 4 digit binary number (the hexadecimal

digits may be treated as decimal for this conversion).

Step 2 − Combine all the resulting binary groups (of 4 digits each) into a single binary

number.

Example: Hexadecimal Number − 1516

Calculating Binary Equivalent −

Step Hexadecimal Number Binary Number

Step 1 1516 110 510

Step 2 1516 00012 01012

Step 3 1516 000101012

Hexadecimal Number − 1516 = Binary Number − 101012

QUESTIONS:

1. Convert the decimal number 250.5 to base 2, base4, base7, base8 & base16.2. Convert the binary 1110001.110 to decimal, octal, & hexadecimal.3. Convert the hexadecimal 2AC5.D to decimal, octal & binary.4. Convert the octal 623.55 to decimal, binary & hexadecimal.5. List the first 16 numbers in base 12. Use the letters A and B to represent two

digits.6. Determine the base of the numbers in each case for the following operations

to be correct: (a) 14/2= 5, (b) 54/4 = 13, (c) 24 + 17 = 40, d) (211)x=(152)8.

7. What is the largest binary number that can be expressed with 14 bits? What are the equivalent decimal and hexadecimal numbers?

2. COMPLEMENTS & SIGNED BINARY NUMBERS

Complements

Complements are used in the digital computers in order to simplify the subtraction operation

and for the logical manipulations. For each radix-r system (radix r represents base of number

system) there are two types of complements.

S.N

.Complement Description

1 Radix ComplementThe radix complement is referred to as the r's

complement

2 Diminished Radix ComplementThe diminished radix complement is referred to as

the (r-1)'s complement

Binary system complementsAs the binary system has base r = 2. So the two types of complements for the binary system

are 2's complement and 1's complement.

1's complement

The 1's complement of a number is found by changing all 1's to 0's and all 0's to 1's. This is

called as taking complement or 1's complement. Example of 1's Complement is as follows.

2's complement

The 2's complement of binary number is obtained by adding 1 to the Least Significant Bit (LSB)

of 1's complement of the number.

2's complement = 1's complement + 1

Example of 2's Complement is as follows.

Binary arithmetic is essential part of all the digital computers and many other digital system. 9’s & 10’s Complements:

It is the Subtraction of decimal no.s can be accomplished by the 9‘s & 10‘s compliment methods similar to the 1‘s & 2‘s compliment methods of binary . the 9‘s compliment of a decimal no. is obtained by subtracting each digit of that decimal no. from 9. The 10‘s compliment of a decimal no is obtained by adding a 1 to its 9‘s compliment.

Example: 9‘s compliment of 3465 and 782.54 is 9999 999.99

-3465 -782.54---------- -----------6534 217.45

10‘s complement of 4069 is 9999

- 4069---------- 5930

+1---------- 5931-----------

SIGNED BINARY NUMBERS:

There are three ways of representing signed binary numbers.

1. signed-magnitude representation

2. signed 1’s complement representation

3. signed 2’s complement representation

1. signed-magnitude representation: In this representation the most significant bit indicates the sign, if it is 0 that represents +ve number & if it is 1 that represents –ve number and remaining bits represents magnitude.

2. signed 1’s complement representation: In this representation the most significant bit indicates the sign, if it is 0 that represents +ve number & if it is 1 that represents –ve number and the magnitude represents in1’s complement form.

3. signed 2’s complement representation: In this representation the most significant bit indicates the sign, if it is 0 that represents +ve number & if it is 1 that represents –ve number and the magnitude represents in2’s complement form.

UNSIGNED BINARY NUMBERS

Binary Addition and Binary SubtractionIt is a key for binary subtraction, multiplication, division. There are four rules of binary addition.

In fourth case, a binary addition is creating a sum of (1 + 1 = 10) i.e. 0 is written in the given column and a carry of 1 over to the next column.

Example − Addition

Subtraction and Borrow, these two words will be used very frequently for the binary subtraction. There are four rules of binary subtraction.

Example − Subtraction

Binary subtraction A − B can be executed by Adding A with two’s complement of B, provided we use two’s complementation for representation − ve numbers

Example: Find 129 − 128

0000 0000 1000 0001 [= +129 d ] 1111 1111 1000 0000 [= − 128 d ] -------------------------------------------0000 0000 0000 0001 [= + 1 d ]

Example Find + 16392 − (16392)

A= +16392 0100 0000 0000 1000 −B= −16392 1011 1111 1111 1000 --------------------------------------------------A + ( − B) 0000 0000 0000 0000

Example: Find +1020 − ( − 1017) 1111 1100 0000 0111 [ − 1017 d ] 0000 0011 1111 1001 [ − ( − 1017) d ] [+1017d] 0000 0011 1111 1100 [= +1020 d ] 0000 0111 1111 0101 [= +2037 d ]

Example: Find − 1017 − ( − 1020)1111 1100 0000 0111 [ − 1017 d ] 0000 0011 1111 1100 [ − ( − 1020) d ] --------------------------------------------------0000 0000 0000 0011 [= +3 d ]

Binary MultiplicationBinary multiplication is similar to decimal multiplication. It is simpler than decimal

multiplication because only 0s and 1s are involved. There are four rules of binary multiplication.

Example − Multiplication

Binary DivisionBinary division is similar to decimal division. It is called as the long division procedure.

Example − Division

QUESTIONS:

1. Obtain the 1’s complement and 2’s complement of the following binary numbersi) 101010001 ii) 00001001 iii) 1000111110

2. Obtain the 9’s complement and 10’s complement of the following numbersi) 13579 ii) 90090 iii) 00000

3. Perform the subtraction with the following decimal numbers using 10’s complementAnd 9’s complement. a) 5250-321 b) 753-864 c) 3570-2100

4. Find the 16's complement of B2FA.5. Convert decimal + 46 and + 29 to binary. using the signed-2' complement

representation and enough digits to accommodate the numbers. Then perform the binary equivalent of(+ 29) + (-49 ), ( - 29) + (+49) . and ( - 29) + ( - 49) . Convert the answers back In decimal and verify that they are correct.

3. BINARY CODES & BINARY STORAGE & REGISTERS

Binary codes

In the coding, when numbers, letters or words are represented by a specific group of symbols, it is said that the number, letter or word is being encoded. The group of symbols is called as a code. The digital data is represented, stored and transmitted as group of binary bits. This group is also called as binary code. The binary code is represented by the number as well as alphanumeric letter.

Advantages of Binary Code

Following is the list of advantages that binary code offers.

Binary codes are suitable for the computer applications.

Binary codes are suitable for the digital communications.

Binary codes make the analysis and designing of digital circuits if we use the binary codes.

Since only 0 & 1 are being used, implementation becomes easy.

Classification of binary codes

The codes are broadly categorized into following four categories.

Weighted Codes Non-Weighted Codes Binary Coded Decimal Code Alphanumeric Codes Error Detecting Codes Error Correcting Codes

Weighted CodesWeighted binary codes are those binary codes which obey the positional weight principle.

Each position of the number represents a specific weight. Several systems of the codes are used to express the decimal digits 0 through 9. In these codes each decimal digit is represented by a group of four bits.

Non-Weighted Codes

In this type of binary codes, the positional weights are not assigned. The examples of non-weighted codes are Excess-3 code and Gray code.

Excess-3 codeThe Excess-3 code is also called as XS-3 code. It is non-weighted code used to express decimal numbers. The Excess-3 code words are derived from the 8421 BCD code words adding (0011)2 or (3)10 to each code word in 8421. The excess-3 codes are obtained as follows −

Example

Excess-3 Addition:Add the xs-3 no.s by adding the 4 bit groups in each column starting from the

LSD. If there is no carry starting from the addition of any of the 4-bit groups , subtract 0011 from the sum term of those groups ( because when 2 decimal digits are added in xs-3 & there is no carry , result in xs-6). If there is a carry out, add 0011 to the sum term of those groups( because when there is a carry, the invalid states are skipped and the result is normal binary).

Excess -3 (XS-3) Subtraction:Subtract the xs-3 no.s by subtracting each 4 bit group of the subtrahend from the corresponding 4 bit group of the minuend starting form the LSD .if there is no borrow from the next 4-bit group add 0011 to the difference term of such groups (because when decimal digits are subtracted in xs-3 & there is no borrow , result is normal binary). I f there is a borrow , subtract 0011 from the differenceterm(b coz taking a borrow is equivalent to adding six invalid states , result is in xs-6)

Gray CodeIt is the non-weighted code and it is not arithmetic codes. That means there are no specific weights assigned to the bit position. It has a very special feature that, only one bit will change each time the decimal number is incremented as shown in fig. As only one bit changes at a time, the gray code is called as a unit distance code. The gray code is a cyclic code. Gray code cannot be used for arithmetic operation.

Application of Gray code Gray code is popularly used in the shaft position encoders. A shaft position encoder produces a code word which represents the angular position of

the shaft.

Binary to Gray conversion:

N bit binary no is represented by Bn Bn-1 ------- B1

Gray code equivalent is by Gn Gn-1 ------- G1

Bn,, Gn are the MSB‘s then the gray code bits are obtained from the binary code as

Gn=Bn Gn-1=Bn Bn-1 Gn-2=Bn- 1 Bn-2 ----------- G1=B2 B1

EX-OR symbol

Procedure: ex-or the bits of the binary no with those of the binary no shifted one position to the right . The LSB of the shifted no. is discarded & the MSB of the gray code no.is the same as the MSB of the original binary no.

Gray to Binary Conversion:If an n bit gray no. is represented by Gn Gn-1 ------- G1Its binary equivalent by Bn Bn-1 ------- B1 then the binary bits are obtained from gray bits as

Bn= Gn Bn-1=Bn Gn-1 Bn-2= Bn-1 Gn-2 ------- G1= B2 G1

To convert no. in any system into given no. first convert it into binary & then binary to gray. To convert gray no into binary no & convert binary no into require no system.

Ex: 10110010(gray) = 110111002= DC16=3348=22010

EX:1101 Gray: 1 1 0 1

Binary: 1 0 0 1

Binary Coded Decimal (BCD) codeIn this code each decimal digit is represented by a 4-bit binary number. BCD is a way to express each of the decimal digits with a binary code. In the BCD, with four bits we can represent sixteen numbers (0000 to 1111). But in BCD code only first ten of these are used (0000 to 1001). The remaining six code combinations i.e. 1010 to 1111 are invalid in BCD.

Advantages of BCD Codes It is very similar to decimal system. We need to remember binary equivalent of decimal numbers 0 to 9 only.

Disadvantages of BCD Codes

The addition and subtraction of BCD have different rules. The BCD arithmetic is little more complicated. BCD needs more number of bits than binary to represent the decimal number. So BCD is

less efficient than binary.

8421 BCD code (Natural BCD code):Each decimal digit 0 through 9 is coded by a 4 bit binary no. called natural binary codes. Because of the 8,4,2,1 weights attached to it. It is a weighted code & also sequential . it is useful for mathematical operations. The advantage of this code is its case of conversion to & from decimal. It is less efficient than the pure binary, it require more bits.Ex: 14 1110 in binary But as 0001 0100 in BCD code.The disadvantage of the BCD code is that, arithmetic operations are more complex than they are in pure binary. There are 6 illegal combinations 1010, 1011, 1100,1101,1110,1111 in these codes; they are not part of the 8421 BCD code system. The disadvantage of 8421 code is, the rules of binary addition 8421 no, but only to the individual 4 bit groups.BCD Addition:It is individually adding the corresponding digits of the decimal no,s expressed in 4 bit binary groups starting from the LSD . If there is no carry & the sum term is not an illegal code , no correction is needed .If there is a carry out of one group to the next group or if the sum term is an illegal code then 610(0100) is added to the sum term of that group & the resulting carry is added to the next group.

Ex: Perform decimal additions in 8421 code (a)25+13

In BCD 25= 0010 0101In BCD +13 =+ 0001 0011

38 0011 1000

No carry, no illegal code .This is the corrected sum (b). 679.6 + 536.8

679.6 = 0110 0111 1001 .0110 in BCD+536.8 = +0101 0011 0010 .1000 in BCD _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

1216.4 1011 1010 0110 . 1110 illegal codes+0110 + 0011 +0110 . + 0110 add 0110 to each

(1)0001 (1)0000 (1)0101 . (1)0100 propagate carry/

+1/

+1/

+1/

+1

0001 0010 0001 0110 . 0100 1 2 1 6 . 4

BCD Subtraction:Performed by subtracting the digits of each 4 bit group of the subtrahend the digits from the corresponding 4- bit group of the minuend in binary starting from the LSD . if there is no borrow from the next group , then 610(0110)is subtracted from the difference term of this group.

(a)38-15

In BCD 38= 0011 1000In BCD -15 = -0001 0101

23 0010 0011No borrow, so correct difference.

Alphanumeric codes

A binary digit or bit can represent only two symbols as it has only two states '0' or '1'. But this is not enough for communication between two computers because there we need many more symbols for communication. These symbols are required to represent 26 alphabets with capital and small letters, numbers from 0 to 9, punctuation marks and other symbols.

The alphanumeric codes are the codes that represent numbers and alphabetic characters. Mostly such codes also represent other characters such as symbol and various instructions necessary for conveying information. An alphanumeric code should at least represent 10 digits and 26 letters of alphabet i.e. total 36 items. The following three alphanumeric codes are very commonly used for the data representation.

American Standard Code for Information Interchange (ASCII). Extended Binary Coded Decimal Interchange Code (EBCDIC).

Five bit Bardot Code.

ASCII code table

ASCII code is a 7-bit code whereas EBCDIC is an 8-bit code. ASCII code is more commonly used worldwide while EBCDIC is used primarily in large IBM computers.

Error Codes

There are binary code techniques available to detect and correct data during data transmission.

ERROR DETECTING CODES:

When binary data is transmitted & processed, it is susceptible to noise that can alter or distort its contents. The 1‘s may get changed to 0‘s & 1‘s .because digital systems must be accurate to the digit, error can pose a problem. Several schemes have been devised to detect the occurrence of a single bit error in a binary word, so that whenever such an error occurs the concerned binary word can be corrected & retransmitted.

Parity: The simplest techniques for detecting errors is that of adding an extra bit known as parity bit to each word being transmitted. Two types of parity: Odd parity, even parity for odd parity, the parity bit is set to a ‗0‘or a ‗1‘a t the transmitter such that the total no. of 1 bit in the word including the parity bit is an odd no.For even parity, the parity bit is set to a ‗0‘or a ‗1‘at the transmitter such that the parity bit is an even no.

Decimal 8421 code Odd parity Even parity0 0000 1 01 0001 0 12 0010 0 13 0011 1 04 0100 0 15 0100 1 06 0110 1 07 0111 0 18 1000 0 19 1001 1 0

When the digit data is received. a parity checking circuit generates an error signal if the total no of 1‘s is even in an odd parity system or odd in an even parity system. This parity check can always detect a single bit error but cannot detect 2 or more errors with in the same word. Odd parity is used more often than even parity does not detect the situation. Where all 0‘s are created by a short ckt or some other fault condition.

Ex: Even parity scheme(a) 10101010 ( b) 11110110 (c) 10111001

Ans:(a) No. of 1‘s in the word is even is 4 so there is no error(b) No. of 1‘s in the word is even is 6 so there is no error(c) No. of 1‘s in the word is odd is 5 so there is error

Ex: odd parity(a)10110111 (b) 10011010 (c)11101010

Ans:(a) No. of 1‘s in the word is even is 6 so word has error(b) No. of 1‘s in the word is even is 4 so word has error(c) No. of 1‘s in the word is odd is 5 so there is no error

Error –Correcting Codes:A code is said to be an error –correcting code, if the code word can always be deduced from an erroneous word. For a code to be a single bit error correcting code, the minimum distance of that code must be three. The minimum distance of that code is the smallest no. of bits by which any two code words must differ. A code with minimum distance of 3 can‘t only correct single bit errors but also detect (can‘t correct) two bit errors, The key to error correction is that it must be possible to detect & locate erroneous that it must be possible to detect & locate erroneous digits. If the location of an error has been determined. Then by complementing the erroneous digit, the message can be corrected , error correcting , code is the Hamming code , In this , to each group of m information or message or data bits, K parity checking bits denoted by P1,P2,----------

pk located at positions 2 k-1 from left are added to form an (m+k) bit code word.To correct the error, k parity checks are performed on selected digits of each code

word, & the position of the error bit is located by forming an error word, & the error bit

is then complemented. The k bit error word is generated by putting a 0 or a 1 in the 2 k-1th position depending upon whether the check for parity involving the parity bit Pk is satisfied or not.Error positions & their corresponding values :

7-bit Hamming code:

To transmit four data bits, 3 parity bits located at positions 20 21&22 from left are added to make a 7 bit code word which is then transmitted.

The word format

P1 P2 D3 P4 D5 D6 D7

D—Data bits P-Parity bits

Ex: Encode the data bits 1101 into the 7 bit even parity Hamming Code The bit pattern is

P1 P2 D3 P4 D5 D6 D71 1 0 1

Bits 1,3,5,7 (P1 111) must have even parity, so P1 =1 Bits 2, 3, 6, 7 (P2 101) must have even parity, so P2=0 Bits 4,5,6,7 (P4 101)must have even parity, so P4=0

The final code is 1010101 15-bit Hamming Code: It transmit 11 data bits, 4 parity bits located 20 21 22 23

Word format is

P1 P2 D3 P4 D5 D6 D7 P8 D9 D10 D11 D12 D13 D14 D15

12-Bit Hamming Code: It transmit 8 data bits, 4 parity bits located at position 20 21 22 23 Word format is

P1 P2 D3 P4 D5 D6 D7 P8 D9 D10 D11 D12

Binary storage & Registers

A register is a memory device that can be used to store more than one bit of information.

A register is usually realized as several flip-flops with common control signals that control the movement of data to and from the register.

Common refers to the property that the control signals apply to all flip-flops in the same way A register is a generalization of a flip-flop. Where a flip- flop stores one bit, a register stores several bits The main operations on a register are the same as for any storage devices, namelyLoad or Store: Put new data into the register

Read: Retrieve the data stored in the register (usually without changing the stored data

QUESTIONS

1.Convert the following to binary and then to gray code. (a) (1111)12 (b) (BC54)16 (c) (237)8 (d) (164)10 2. The following is a string of ASCII characters whose bit patterns have been

converted into hexadecimal for compactness: 7F F4 E5 76 E5 FA EF 62 73. Of the eight bus in each pair of digits the leftmost is a parity bit. The remaining bits are the ASCII code.(a) Convert the string to bit form and decode the ASCII.

(b) Determine the parity used : odd or even ?3. The state of a 12·bit register is 1000 100 1011 1.What is its content if it

represents(a) Three decimal digits in BCD?(b) Three decimal digits in the excess-3 code?(c) Three decimal digit in the 84-2-1 code?(d) A binary number?

4. What bit must be complemented to change an ASCII letter from capital to lowercase and vice-versa?

4. BOOLEAN ALGEBRA & LOGIC GATESBASIC DEFINITIONS, AXIOMATIC DEFINITION OF BOOLEAN ALGEBRA, BASIC PROPERTIES OF BOOLEAN ALGEBRA

Boolean Algebra is used to analyze and simplify the digital (logic) circuits. It uses only the binary numbers i.e. 0 and 1. It is also called as Binary Algebra or logical Algebra. Boolean algebra was invented by George Boole in 1854.

Rule in Boolean Algebra Following are the important rules used in Boolean algebra.

Variable used can have only two values. Binary 1 for HIGH and Binary 0 for LOW.

Complement of a variable is represented by an overbar (-). Thus, complement of variable B is represented as . Thus if B = 0 then = 1 and B = 1 then = 0.

ORing of the variables is represented by a plus (+) sign between them. For example ORing of A, B, C is represented as A + B + C.

Logical ANDing of the two or more variable is represented by writing a dot between them such as A.B.C. Sometime the dot may be omitted like ABC.

Boolean Laws There are six types of Boolean Laws.

Commutative law Any binary operation which satisfies the following expression is referred to as commutative operation.

Commutative law states that changing the sequence of the variables does not have any effect on the output of a logic circuit.

Associative law This law states that the order in which the logic operations are performed is irrelevant as their effect is the same.

Distributive lawDistributive law states the following condition.

AND lawThese laws use the AND operation. Therefore they are called as AND laws.

OR law These laws use the OR operation. Therefore they are called as OR laws.

INVERSION lawThis law uses the NOT operation. The inversion law states that double inversion of variable results in the original variable itself.

AXIOMS & Laws of Boolean Algebra:Axioms or Postulates are a set of logical expressions i.e, without proof. & also we can build a set of useful theorems. Each axiom can be interpreted as the outcome of an operation performed by a logic gate.

AND OR NOT0.0=0 0+0=0 1=00.1=0 0+1=1 0=11.0=0 1+0=11.1=1 1+1=1

Simplification Using Algebraic FunctionsIn this approach, one Boolean expression is minimized into an equivalent expression by applying Boolean identities.Problem 1Minimize the following Boolean expression using Boolean identities −

F(A,B,C)=A′B+BC′+BC+AB′C′F(A,B,C)=A′B+BC′+BC+AB′C′SolutionGiven, F(A,B,C)=A′B+BC′+BC+AB′C′ F(A,B,C)=A′B+(BC′+BC′)+BC+AB′C′ [By idempotent law, BC’ = BC’ + BC’]F(A,B,C)=A′B+(BC′+BC)+(BC′+AB′C′) F(A,B,C)=A′B+B(C′+C)+C′(B+AB′) [By distributive laws]

F(A,B,C)=A′B+B.1+C′(B+A) [ (C' + C) = 1 and absorption law (B + AB')= (B + A)]F(A,B,C)=A′B+B+C′(B+A) [ B.1 = B ]F(A,B,C)=B(A′+1)+C′(B+A)F(A,B,C)=B.1+C′(B+A) [ (A' + 1) = 1 ]F(A,B,C)=B+C′(B+A) [ As, B.1 = B ]F(A,B,C)=B+BC′+AC′F(A,B,C)=B(1+C′)+AC′F(A,B,C)=B.1+AC′ [As, (1 + C') = 1]F(A,B,C)=B+AC′ [As, B.1 = B]

F(A,B,C)=B+AC′ is the minimized form.

Problem 2Minimize the following Boolean expression using Boolean identities −

F(A,B,C)=(A+B)(B+C)F(A,B,C)=(A+B)(B+C)SolutionGiven, F(A,B,C)=(A+B)(A+C)F(A,B,C)=A.A+A.C+B.A+B.C [Applying distributive Rule]F(A,B,C)=A+A.C+B.A+B.C  [Applying Idempotent Law]F(A,B,C)=A(1+C)+B.A+B.C  [Applying distributive Law]F(A,B,C)=A+B.A+B.C  [Applying dominance Law]F(A,B,C) = (A+1).A+B.C  [Applying distributive Law]F(A,B,C)=1.A+B.C  [Applying dominance Law]F(A,B,C)=A+B.CSo, F(A,B,C)=A+BC is the minimized form.

QUESTIONS

1. Simplify the following Bo Simplify the following Boolean expressions to minimum no. of literals.

i. ABC+A’B+ABC’ii. (BC’+A’D)(AB’+CD’)

2. Obtain the Dual of the following Boolean expressions.i. AB+A(B+C)+B’(B+D)ii. A’B+A’BC’+A’BCD+A’BC’D’Eiii. ABEF+ABE’F’+A’B’EF

3. Simplify the following Boolean expression: (i) F = (A+B)(A’+C)(B+C). (ii) F = A+B+C’+D (E+F)

4. Simplify the following expression Y = (A + B) (A + C’) (B' + C’)5. BOOLEAN FUNCTIONS, CANONICAL AND STANDARD

FORMS

Boolean functions & their representation:

A function of n Boolean variables denoted by f(x1,x2,x3------xn) is another variable denoted by & takes one of the two possible values 0 & 1.The various way of represent a given function is1. Sum of Product (SOP) form: It is called the Disjunctive Normal Form (DNF) Ex: f(A,B,C)=(A B+B C)2. Product of Sums (POS) form: It is called the Conjunctive Normal Form (CNF). Ex: f(A,B,C)=( A+ B)(B+C)3. Truth Table form:The function is specified by listing all possible combinations of values assumed by the variables & the corresponding values of the function.

Example:

# A B C F0 0 0 0 01 0 0 1 12 0 1 0 03 0 1 1 14 1 0 0 15 1 0 1 06 1 1 0 17 1 1 1 1

Table 1 – Arbitrary function defined via Truth Table

A "product" in Boolean algebra is a logical AND operation while a "sum" is a logical OR operation.

Minterm

A "minterm" is a Boolean expression that is true for the minimum number of combinations of inputs.Because we want to minimize the coverage, we want to use the Boolean operation that is the most restrictive, which is the AND operation. Since the AND is true only if ALL of the inputs are True, we can craft an expression that is True for only a single combination of inputs by including each input in the product. If the input is uncomplemented, then we

require that it be True, while if it is complemented, then we require it to be False. The minterm expression for each combination of inputs is therefore Keep in mind that each minterm is True only for the specific combination of inputs with which it is associated and is False for all others.

A B C minterm

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

Table 2 - Table of Minterms

In order to be a minterm, the product must cover exactly one of the possible combination of inputs. To do this, each input must be present in either uncomplemented or complemented form. Consider the system of three inputs in the table above. We can combine the last two minterms, namely ABC'+ABC, into a single product, namely AB. Using this "simplifies" the expression and it is still in SOP form, but it is not a minterm.

If we have an expression that is in SOP form but is not in standard form and we need it to be, then we can simply expand each product by ANDing it with products that are constructed by ORing one of the missing inputs and its complement. Thus, we can expand AB back to minterm form as follows:

Maxterm

A "maxterm" is a Boolean expression that is True for the maximum number of combinations of inputs. Because we want to maximize the coverage, we want to use the Boolean operation

that is the most permissive, which is the OR operation. While the OR is True if ANY of its inputs is True, for our purposes it is more convenient to recognize that the OR is False only if ALL of its inputs are False; thus, we can craft an expression that is False for only a single combination of inputs by including each input in the sum. If the input is uncomplemented, then we require that it be False, while if it is complemented, then we require it to be True. The maxterm expression for each combination of inputs is therefore Keep in mind that each maxterm is False only for the specific combination of inputs with which it is associated and is True for all others.

# A B C maxterm

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

Table 3 - Table of Maxterms

In order to be a maxterm, the sum must cover all but exactly one of the possible combinations of inputs. To do this, each input must be present in either uncomplemented or complemented form. Consider the system of three inputs in the table above. We can combine the last two maxterms, namely (A'+B'+C)(A'+B'+C'), into a single sum, namely (A'+B'). Using this "simplifies" the expression and it is still in POS form, but it is not a maxterm.

If we have an expression that is in POS form but is not in standard form and we need it to be, then we can simply expand each sum by OR ing it with terms that are constructed by AND ing one of the missing inputs and its complement. Thus, we can expand (A'+B') back to maxterm form as follows:

The above equation may look odd but that is because while we are used to multiplication being distributable over addition (and thus AND being distributable over OR looks natural), we are used to addition not being distributable over multiplication. However, OR is distributable over AND, hence

In the above relation,

Relationship between Labels, Minterms, and Maxterms

Consider the minterm and maxterm for a particular row. Since the minterm is True ONLY when that input combination is asserted while the maxterm is False ONLY when that combination is asserted, it is clear that each minterm is the logical inverse of the corresponding maxterm (and vice versa, of course).

As mentioned in the introduction, the SOP form takes on the appearance of being a sum of several terms, each of which is the product of several factors. To craft the SOP form of a Boolean logic function, we merely need to OR together the minterms associated with each combination of inputs for which the overall output should be True. By looking at Table 1 we see that we need to sum the minterms associated with rows {1,3,4,6,7}. This is often represented simply as

Standard POS (Product of Sums)

As mentioned in the introduction, the POS form takes on the appearance of being a product of several factors, each of which is the sum of several terms.To craft the POS form of a Boolean logic function, we merely need to AND together the maxterms associated with each combination of inputs for which the overall output should be False. By looking at Table 1 we see that we need to take the product of the maxterms associated with rows {0',2',5'}. This is often represented simply as

LOGIC GATES, OTHER LOGIC OPERATIONS, INTEGRATED

CIRCUITS

LOGIC GATES Logic gates are the basic building blocks of any digital system. It is an electronic

circuit having one or more than one input and only one output. The relationship between the input and the output is based on certain logic. Based on this, logic gates are named as AND gate, OR gate, NOT gate etc.

AND GateA circuit which performs an AND operation is shown in figure. It has n input (n >= 2) and one output.

Logic diagram

Truth Table

OR GateA circuit which performs an OR operation is shown in figure. It has n input (n >= 2) and one output.

Logic diagram

Truth Table

NOT GateNOT gate is also known as Inverter. It has one input A and one output Y.

Logic diagram

Truth Table

NAND GateA NOT-AND operation is known as NAND operation. It has n input (n >= 2) and one output.

Logic diagram

Truth Table

NOR GateA NOT-OR operation is known as NOR operation. It has n input (n >= 2) and one output.

Logic diagram

Truth Table

XOR GateXOR or Ex-OR gate is a special type of gate. It can be used in the half adder, full adder and subtractor. The exclusive-OR gate is abbreviated as EX-OR gate or sometime as X-OR gate. It has n input (n >= 2) and one output.

Logic diagram

Truth Table

XNOR GateXNOR gate is a special type of gate. It can be used in the half adder, full adder and subtractor. The exclusive-NOR gate is abbreviated as EX-NOR gate or sometime as X-NOR gate. It has n input (n >= 2) and one output.

Logic diagram

Truth Table

Boolean Expression & Logic Diagrams:Boolean expressions can be realized as hardware using logic gates. Conversely,

hardware can be translated into Boolean expressions for the analysis of existing circuits.

1. Converting Boolean Expressions to Logic:

To convert, start with the output & work towards the input.

Assume the expression (AB)’+A+(B +C)’ is to be realized using AOI logic. Start with this expression. (AB)‘ is the output of an inverter whose inputs is AB and (B+C)‘ must be the output of an inverter whose input is B+C. so, those two inverters are as Since it is three terms, it must be the output of a three-input OR gates. So, draw an OR gate with three inputs as

Now AB must be output of a two-input AND gate whose inputs are A and B . And B+C must be the output of a two-input OR gate whose inputs are B and C. so, an AND gate and an OR gate are as

2. Converting Logic to Boolean Expressions:

To convert logic to algebra, start with the input signals and develop the terms of the Boolean expression until the output is reached.

Integrated circuits

An integrated circuit is one in which circuit components such as transistors, diodes, resistors, capacitors etc. are automatically part of a small semiconductor chip.

An integrated circuit consists of a number of circuit components (e.g. transistors, diodes, resistors etc.) and their inter connections in a single small package to per- form a complete electronic function. These components are formed and connected within a small chip of semi- conductor material. The following points are about integrated circuits :

Advantages: Integrated circuits possess the following advantages over discrete circuits : Increased reliability due to lesser number of connections. Extremely small size due to the fabrication of various circuit elements Lesser weight and space requirement due to miniaturized circuit. Low power requirements. Greater ability to operate at extreme values of temperature. Low cost because of simultaneous production of hundreds of alike circuits on a small semi- conductor wafer.The circuit lay out is greatly simplified because integrated circuits are constrained to use minimum number of external connections.Disadvantages: The disadvantages of integrated circuits are :

If any component in an IC goes out of order, the whole IC has to be replaced by the new one.

In an IC, it is neither convenient nor economical to fabricate capacitances exceeding 30 pF.

It is not possible to fabricate inductors and transformers on the surface of semiconductor chip.

It is not possible to produce high power ICs (greater than 10 W). There is a lack of flexibility in an IC i.e., it is generally not possible to modify the

Parameters within which an integrated circuit will operate.

QUESTIONS:

1. Implement Boolean functions F=xy+xy’+y’z using basic gates2. Convert the following to the other canonical form F(x,y,z)=∑(1,3,7)3. Obtain the truth table of the function F= xy+xy’+y’z4. Convert the given expression in canonical SOP form Y = AC + AB + BC.5. Given two eight-bit strings A = 10110001 and B = 10101100, evaluate

the eight-bit result after the following logical operations: (a ) AND. (b) OR. (C)XOR. (d)NOT A, (e) NOT B .

6. Express the following function as a sum of minterms and as a product of maxterms:F(A ,B,C, D) = B'D + A'D + BD