MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 1 Overflow Signed binary is in fixed range -2 n-1 2 n-1 If the answer for addition/subtraction more than the

MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR

1

Overflow

• Signed binary is in fixed range

• -2n-1 2n-1

• If the answer for addition/subtraction more than the range, it is overflow

• Two situation where overflow can happen:– Positive + positive = negative (enough n-bit)– Negative + negative = positive(more than n-bit)


2

Overflow

• Example: Binary number 4-bit (second complement)

• Range : -2n-1 2n-1-1

• Range : (1000)2s(0111) 2s

• Range : (-8) 10 (+7) 10

• Two situation where overflow can happen:

– Positive + positive = negative (enough n-bit)

– Negative + negative = positive(more than n-bit)

0101 = 5 1001 = -7

0100 = 4 1010 = -6

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

1001 10011


3

Overflow

• Example: Binary number 4-bit (second complement)

• Range : -2n-1 2n-1-1

• Range : (1000)2s(0111) 2s

• Range : (-8) 10 (+7) 10

(Overflow exist)

(Overflow exist)

(ignore final carry)


4

Fixed Point Number• Signed number and unsigned number representation is given in fixed point number • Binary point is assumed to have fixed location, if it is located at the end of the number

• It can represent integer number between –128 to 127 (for 8-bit binary complement)

Binary point


5

Fixed Point Number• Generally, other locations in binary point position

• Example: If two fraction bit is used, we can represent:

Binary point

fraction


6

Floating Point Number• Fixed point number has limited range• To represent extremely large or extremely small number, we use floating point number (like

scientific number)• Example:

• 0.23X1023(really large number)• 0.1239X10-10(really small number)


7

Floating Point Number

• Floating point number is divided into three partsmantissa, base and exponent

• Base always fixed in number system• Therefore, only need mantissa and exponent

Mantissa exponent


8

Floating Point Number• Mantissa always in normalize form:

(base 10) 23X1021 is normalized to 0.23X1023

(base 10) –0.0017X1021 is normalized to -0.17X1019

(base 10) 0.01101X103 is normalized to 0.1101X102

• 16-bit floating point number might contain 10-bit mantissa and 6-bit exponent• More exponent, the greater its range• More mantissa, the greater its persistence


9

Arithmetic with Floating Point Number• Arithmetic with floating point number is much difficult• MULTIPLICATION

The steps:– multiply with the mantissa– Add its exponent– normalized


10

Arithmetic with Floating Point Number

• Example

(Normalization)


11


• ADDITION

Steps:– Equalize their exponent– Add their mantissa– Normalize them


12


• Example:

(0.12x102)10 + (0.0002x104 ) 10

= (0.12x102) 10 +(0.02x102) 10

= (0.12+0.02) 10 x 102

= (0.14x102 ) 10


13

Binary Coded Decimal (BCD)

• Decimal number is normally used by human. Binary number is normally used by computer. It is expensive to exchange between each other.

• If used only little calculation, we can use coding scheme for decimal number.

• One of the scheme is BCD, or also called 8421 code.

• Which represent every decimal digit with 4-bit binary code.


14


• There are code which is not used, e.g. (1010)BCD,(1011)BCD,….,(1111)BCD. This code is said to be an error.

• Easy to convert but the arithmetic is hard• Suitable as interface such as keyboard input

and digital reading

Decimal Digit

Decimal Digit


15


• Example:

Notes: BCD is not similar to binary

Example: (243)10=(11101010)2

Decimal Digit

Decimal Digit


16

Gray Code

• No weight• Only one bit change from one code number to the

others• Suitable for error detection

Decimal Binary Gray Code Decimal Binary Gray Code


17

Gray Code


18

Gray Code


19

Convert Binary Code to Gray Code

• Fixed MSB• From left to right, add each coupled binary code

bit next to each other to get Gray code bit, ignore carry

• Example: convert binary 10110 to Gray code


20

Convert Gray Code to Binary Code

• Fixed MSB• From left to right, add each coupled binary code

executed with Gray code bit at the next position, ignore carry

• Example: convert Gray 10110 to binary code


21

Other Decimal Code

• Self compliment code: excess-3 code, 84-2-1, 2*421• Error detection code: Biquinary code (bi=two, quinary=five)


22

Self Compliment Code

• Example: Excess-3, 84-2-1, 2*421• Code represented by coupled compliment-digit

which compliment each other


23

Alphanumeric Code

• Part of numbers, computer also handle textual data• Set which always used includes:

Letters : ‘A’,…..,‘Z’ and ‘a’,…..,‘z’

Digits : ‘0’,…..,‘9’

Special Characters: ‘$’, ‘’, ‘!’, ‘,’, ‘.’,….

Not Printable: SOH, NULL, BELL,….• Most of the time, it is represented by 7 or 8-bit


24

Alphanumeric Code

• Two standard that are frequently used

ASCII (American Standard Code for Information Interchange)

EBCDIC (Extended BCD Interchange Code)• ASCII: 7-bit, add with parity bit for error

detection (odd,even parity)• EBCDIC: 8-bit


25

Alphanumeric Code

• ASCII Table


26

Error Detection Code

• Error can exist in transmission. It must be detected so that retransmission can be requested

• With binary number, mostly exist 1-bit error. Example: 0010 is transmitted incorrectly as 0011, or 0000, or 0110, or 1010

• Biquinary using additional 3-bit to detect error. For one error detection, only one extra bit is needed


27


• Parity Bit– Even parity: number of bit 1 is even

– Odd parity: number of bit 1 is odd

• Example: Odd parity


28


• Parity Bit can detect odd error and not even error (if odd is set)

Example: For odd parity number

10011=>10001 (detected)

10011=>10101 (not detected)• Parity bit can also be used on data block


29


• Sometimes, it is not enough to detect code, we need to correct it

• Error correction is expensive in practical, we only need to use one bit error correction

• Popular technique: Hamming Code– Add k-bit to n-bit number to produce n+k bit

– Number the bit 1 on bit n+k

– Every parity bit is on the number range


30


• E.g: For 8-bit number, we need 4 parity bit

12 bit number are 0001,0011,…,1100. Every 4 bit parity is used to detect group of bit. Every parity bit is for themselves and has bit ‘1’ on certain position bit


31


• Therefore:

P1= parity for bit {3,5,7,9,11}

P2= parity for bit {3,6,7,10}




32


• Given 8-bit number: 1100 0100

• Assume even parity is

P1= parity for bit {3,5,7,9,11} = 0

P2= parity for bit {3,6,7,10} = 0

P4= parity for bit {5,6,7,12} = 1

P8= parity for bit {9,10,11,12} = 1


33

Error Detection Code• To check error, execute checking code

C1= XOR {1,3,5,7,9,11}

C2= XOR {2,3,6,7,10}

C4= XOR {4,5,6,7,12}

C8= XOR {8,9,10,11,12}

If C8 C4 C2 C1=0000 therefore no error, if otherwise C8 C4 C2 C1 show position, there is an error for only one bit

• Example

Documents

MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 1 Overflow Signed binary is in fixed range -2 n-1 2 n-1 If the answer for addition/subtraction more than the