Lect - 3 & 4.ppt

7/27/2019 Lect - 3 & 4.ppt

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1. Add the minuend M to the r’s complement of the subtrahend N.

2. Inspect the result obtained in step1 for an end carry

3. If an end carry occurs, discard it or

4. If end carry does not occur ,take the r’s complement of thenumber obtained in step 1

Case 1 M> = N

M = 72532 and N= 03250

72532

+ 96750 ( 10’s complement of N)

1 69282

end carry discard it

Answer : 69282

Subtraction using r’s complement

7/27/2019 Lect - 3 & 4.ppt


Case 2:M<N

• Subtract 3250 – 72532M=03250 N=72532

10’s complement for N is 27468

So 03250+ 27468

_____________

30718 (no carry)So answer will be –(10’s complement of 30718)= -69282

7/27/2019 Lect - 3 & 4.ppt


Given X= 1010100 and Y= 1000011, perform the

subtraction (a) X-Y (b) Y-X using 2’s complement

(a) M=1010100 N=1000011

2’s complement for N is 0111101

So 1010100

+ 0111101

_____________

10010001

Discard end carry, So answer will be 0010001.

7/27/2019 Lect - 3 & 4.ppt


(b) Y= 1000011 X= 1010100

• So 1000011

+ 0101100

_____________

1101111 No end carry, Take 2’s complement of answer and put –

ve sign.

• So answer will be -0010001.

7/27/2019 Lect - 3 & 4.ppt


Subtraction using (r-1)’s complement

1. Add the minuend M to the (r-1)’s complement of the subtrahend N. 2. Inspect the result obtained in step1 for an end carry if an end carry

occurs, add 1 to the least significant digit (end-around carry).

3. If end carry does not occur, take the (r-1)’s complement of the number obtained in step 1.

(a) M = 72532 and N= 03250

72532

+ 96749 ( 9’s complement of N)

1 69281

+1 end-around carry69282 answer

7/27/2019 Lect - 3 & 4.ppt


(b)1’s Complement Subtraction

• Using 1’s complement numbers, subtracting numbers is also easy.

• For example, suppose we wish to subtract +(0001)2 from +(1100)2.

• Let’s compute (12)10 - (1)10.

– (12)10 = +(1100)2 = 011002 in 1’s comp.

– (-1)10 = -(0001)2 = 111102 in 1’s comp.0 1 1 0 0

- 0 0 0 0 1

--------------

0 1 1 0 0

+ 1 1 1 1 0

--------------

1 0 1 0 1 0

1

--------------

0 1 0 1 1

Add carry

Final

Result

Step 1: Take 1’s complement of 2nd operandStep 2: Add binary numbers

Step 3: Add carry to low order bit

1’s comp

Add

7/27/2019 Lect - 3 & 4.ppt


Binary Addition (Direct)

• Binary addition is very simple.• This is best shown in an example of adding

two binary numbers…

1 1 1 1 0 1

+ 1 0 1 1 1

---------------------

0

1

0

1

1

1111

1 1 00

carries

7/27/2019 Lect - 3 & 4.ppt


Binary Subtraction (Direct)° We can also perform subtraction (with borrows in place of

carries).° Let’s subtract (10111)2 from (1001101)2…

1 10

0 10 10 0 0 10

1 0 0 1 1 0 1

- 1 0 1 1 1

------------------------

1 1 0 1 1 0

borrows

7/27/2019 Lect - 3 & 4.ppt


Binary Multiplication

• Binary Multiplication is similar to decimal

multiplication

• In binary each partial product is zero when

multiplication of (1 or 0 ) by 0.

• In binary each partial product is one when

multiplication of 1 by 1

7/27/2019 Lect - 3 & 4.ppt


Example:

• Multiply 1 0 0 1 by 1 1 0 1

1 0 0 1

* 1 1 0 1

1 0 0 1

0 0 0 0 *

1 0 0 1 * *1 0 0 1 * * *

1 1 1 0 1 0 1

7/27/2019 Lect - 3 & 4.ppt


Codes

• Computer and other digital circuits processdata in the binary formats.

• Various binary codes are used to representdata which may be numeric, alphabets or special characters.

• A user must be very careful about the code being used while interpreting informationavailable in the binary format.

7/27/2019 Lect - 3 & 4.ppt


Example:

• 1000001 represents (65)10 in straight binary

• 01000001 represents (41)10 in BCD.

• 1000001 represents A in ASCII Code.

7/27/2019 Lect - 3 & 4.ppt


Some commonly used Codes

1. Straight Binary Codes

2. Natural BCD Codes

3. Excess-3 Codes

4. Gray – Codes

5. Alphanumeric Codes

7/27/2019 Lect - 3 & 4.ppt


Straight Binary Codes

• One Binary Digit (one bit) can take on values

0, 1.

• This is used to represent numbers usingnatural( Straight) Binary form as discussed

earlier

• (65)10

in straight binary represented by

1000001

7/27/2019 Lect - 3 & 4.ppt


More Examples:-

• We can represent TWO values:

True = 1, False = 0

On = 1 , off = 0

• Two Binary digits (two bits) can take on

values of 00, 01, 10, 11.

7/27/2019 Lect - 3 & 4.ppt


• Three Binary digits (three bits) can take on

values of 000, 001, 010, 011, 100, 101, 110,

111. We can represent 8 values

• N bits (or N binary Digits) can represent 2 N

different values.(for example, 4 bits can

represent 24 or 16 different values)N bits

can take on unsigned decimal values from 0to 2 N-1.

7/27/2019 Lect - 3 & 4.ppt


Natural BCD Codes

• In this Codes, decimal digit 0 through 9 are represented by their natural binary equivalents using four bits andeach decimal digit of a decimal number is represented

by this four bits code individually

• It is also known as 8,4,2,1 code.

• 8,4,2,1 are the weights of the four bit of the decimaldigit similar to straight binary number system

7/27/2019 Lect - 3 & 4.ppt


Examples:-

• (23)10 is represented by 0010 0011

• (08)10 is represented by 0000 1000

• (921)10 is represented by 1001 0010 0001

7/27/2019 Lect - 3 & 4.ppt


Putting It All Together

7/27/2019 Lect - 3 & 4.ppt


BCD Addition

Represent the unsigned decimal numbers 965 and 672 in BCD

and then show the steps necessary to form their sum .

965= 1001 0110 0101672= 0110 0111 0010

+___ ___ ____

1 0000 1101 0111

+0110+0110

+_________________ 0001 0110 0011 0111 (1637)10

7/27/2019 Lect - 3 & 4.ppt


Excess-3 Codes

• This is another from of BCD Code , in which each

decimal digit is coded into a 4-bit binary code

• The code for each decimal digit is obtained by adding

decimal 3 to the BCD code of the digit.

• Decimal 2 is coded as 0010 + 0011= 0101

• It is not a weighted code

• It is a self complementing code, means 1’s complement

of the coded number yields 9’s complement of the

number it self

7/27/2019 Lect - 3 & 4.ppt


Example:-

• Excess – 3 code of decimal 2 is 0101 its 1’s

complement is 1010 which is excess-3 code

for decimal 7 , which is 9’s complement of 2.

• This property helps in performing subtraction

operation in digital systems.

7/27/2019 Lect - 3 & 4.ppt


Other Decimal Codes

• Many ways to assign 4-bit code to 10 decimal digits

• Each code uses only 10 combinations out of 16

• BCD and 8, 4, -2, -1 are

weighted codes

• Excess-3 and 8,4,-2,-1 are

self-complementing codes

• Note that BCD is NOTself-complementing

DecimalBCD

8,4,2,1Excess-3 8,4,-2,-1

0 0000 0011 0000

1 0001 0100 0111

2 0010 0101 0110

3 0011 0110 0101

4 0100 0111 01005 0101 1000 1011

6 0110 1001 1010

7 0111 1010 1001

8 1000 1011 1000

9 1001 1100 1111

7/27/2019 Lect - 3 & 4.ppt


Gray Code

• It is a very useful code in which a decimal

number is represented in binary form in such a

way so that each gray code number differs fromthe preceding and the succeeding numbers by a

single bit.

• It is not a weighted code.• It is a reflected code.

7/27/2019 Lect - 3 & 4.ppt


Construction of Gray Code

• A 1-bit gray code has two code words 0 and1 representing decimal numbers 0 and 1.

• An n-bit (n >=2) Gray Code will have first2n-1 Gray Codes with n-1 bits (LSB) writtenin order with a leading 0 appended.

• The last 2n-1 Gray Codes with n-1 bits (LSB)written in reverse order( Mirror image) witha leading 1 appended.

7/27/2019 Lect - 3 & 4.ppt


Examples:-

• 1 Bit Gray Code

Decimal Number Gray Code

0 01 1

• 2 Bit Gray Code

0 0 0

1 0 1

2 1 1

3 1 0

7/27/2019 Lect - 3 & 4.ppt


More Examples:-

• 3- bit Gray Code

Decimal Number Gary Code

0 0 0 0

1 0 0 12 0 1 1

3 0 1 0

4 1 1 0

5 1 1 1

6 1 0 1

7 1 0 0

7/27/2019 Lect - 3 & 4.ppt


Conversion From Binary to Gray

• Start with the most significant bit of the binarynumber.

• Copy this bit as the MSB of the gray code

number.• Add the MSB of the binary to the next bit of

the binary number.• The sum (ignoring carry) is the next bit of the

gray code number.• Continue adding each bit of the binary to the

next bit to its right to get the gray code for that

position as shown in the next slide

7/27/2019 Lect - 3 & 4.ppt


Conversion From Binary to Gray Cont…

d

7/27/2019 Lect - 3 & 4.ppt


Gray Code

• Gray code is not a number

system. – It is an alternate way to represent four

bit data

• Only one bit changes fromone decimal digit to the

next

• Useful for reducing errorsin communication.

• Can be scaled to larger

numbers.

Digit Binary GrayCode

0 0000 0000

1 0001 0001

2 0010 0011

3 0011 0010

4 0100 0110

5 0101 0111

6 0110 0101

7 0111 0100

8 1000 1100

9 1001 1101

10 1010 1111

11 1011 1110

12 1100 1010

13 1101 1011

14 1110 1001

15 1111 1000

7/27/2019 Lect - 3 & 4.ppt


Direct Conversion From Gray to Binary

Example:

Gray 1 1 0 0

1

Binary

1 0 0 0 1

7/27/2019 Lect - 3 & 4.ppt


Alphanumeric Code

• In Many situations, digital systems are

required to handle data that may consist of

numerals, letters and special symbols• If we use an n-bit binary code, we can

represent 2n elements using this code

therefore to represent 10 digits 0 through 9and 26 alphabets A through Z , we need a

minimum of 6-bits ( 26 = 64 ) not 5 bits

7/27/2019 Lect - 3 & 4.ppt


• Frequently there is a need to represent more

than 64 characters including the lower caseletters and the special control characters for

the transmission of the digital information

• For this reason the following two codes are

normally used

– Extended BCD Interchange ( EBCDIC )

– American Standard Code for Info. Interchange (

ASCII )

7/27/2019 Lect - 3 & 4.ppt


American Standard Code for

Info. Interchange ( ASCII )

• Required to represent more than 64 characters

• 7- bit code word.

• Maximum128 characters different characters

can be represented by this code.

• 100 0001 is ASCII representation of alphabet A

ASCII C d

7/27/2019 Lect - 3 & 4.ppt


ASCII Code• American Standard Code for Information Interchange

• ASCII is a 7-bit code, frequently used with an 8th bit for error

detection (more about that in a bit).

Character ASCII (bin) ASCII (hex) Decimal Octal

A 1000001 41 65 101

B 1000010 42 66 102

C 1000011 43 67 103

…

Z

a

…

1

‘

ASCII C d

7/27/2019 Lect - 3 & 4.ppt


ASCII Codes

° ASCII Codes

° A – Z (26 codes), a – z (26 codes)

° 0-9 (10 codes), others (@#$%^&*….)

° Complete listing in Mano text

° Transmission susceptible to noise

° Typical transmission rates (1500 Kbps, 56.6 Kbps)

° How to keep data transmission accurate?

7/27/2019 Lect - 3 & 4.ppt


Parity Codes• Parity codes are formed by concatenating a parity bit, P to

each code word of C.• In an odd-parity code, the parity bit is specified so that the

total number of ones is odd.

• In an even-parity code, the parity bit is specified so that the

total number of ones is even.

Information BitsP

1 1 0 0 0 0 1 1

Added even parity bit

0 1 0 0 0 0 1 1

Added odd parity bit

7/27/2019 Lect - 3 & 4.ppt


Extended BCD Interchange ( EBCDIC )

• Also used to represent more than 64 characters

• 8- bit code word.

• 8th bit (MSB) is invariably added for parity.• Maximum128 characters different characters can

be represented by this code.

• A Parity bit is an extra bit included with themessage to make the total number of 1’s either odd or even.

7/27/2019 Lect - 3 & 4.ppt


• Odd parity to make total number of 1’s odd

( including parity bit)

• Even parity to make total number of 1’s even

( including parity bit)

• 1100 0001 is EBCDIC representation of

alphabet A.

7/27/2019 Lect - 3 & 4.ppt


Parity Bit & Error Detection• Information in form digital signals are transmitted

from one location (transmitter) to another location(receiver), transmission errors may occur because of electrical noise in the channel.

• Due to such error a signal transmitted as 0 may bereceived as 1 or vice-versa.

• It is desired to detect the error in the received data

word, locate its bit position and correct it.• Usually a probability of occurrence of error in at least

1 bit position is always there, so we restrict our discussion to the detection of single error

7/27/2019 Lect - 3 & 4.ppt


• To ensure that the occurrence of single error does not lead

to incorrect interpretation at the receiver, the code used to

transmit the information should possess the property that

the occurrence of any single error is easily detected.

• This property could be achieved if an extra bit is attached

to the n- bit code word to make the number of bit as “n+1”in such a way so as to make a number of ones in the

resulting (n+1)-bit code even or odd, it will certainly a

error – detecting code.

7/27/2019 Lect - 3 & 4.ppt


• For detection of error an extra bit known as

parity bit is attached to each code word to

make the number of ones in the code even(even parity) or odd ( odd parity).

7/27/2019 Lect - 3 & 4.ppt


Parity

• Even parity: set parity bit to 0 if there are an

even number of 1’s in the code.

• Odd parity: set parity bit to 1 if there are aneven number of 1’s in the code.

Even-parity Odd-parity

1 1 1 0 1 0

7/27/2019 Lect - 3 & 4.ppt


• Parity will not detect 2 bit errors because

changing 2 bits does not affect the parity.• Parity will detect ODD numbers of bit errors,

e.g, 1,3 etc.

• Note that 1 bit error are more likely than 2 bit

etc.

7/27/2019 Lect - 3 & 4.ppt


Parity Generation for 4 bit data

Documents

Lect - 3 & 4.ppt