DKT 122 / 3DKT 122 / 3DIGITAL SYSTEMS 1DIGITAL SYSTEMS 1

CHAPTER 2 (b) : NUMBER SYSTEMS OPERATION NUMBER SYSTEMS OPERATION

AND CODES m.rizal@unimap.edu.my

sitizarina@unimap.edu.my@ p y

DECIMAL VALUE OF SIGNED NUMBERSDECIMAL VALUE OF SIGNED NUMBERS

SIGN-MAGNITUDE:

Decimal values of +ve & -ve no are Decimal values of +ve & ve no are determined summing the weights in all themag. bit positions where there are 1’s and i i th iti h th ignoring those positions where there are zeros.

The sign is determined by examinationOf the sign bit

DECIMAL VALUE OF SIGNED NUMBERSDECIMAL VALUE OF SIGNED NUMBERS

Sign Magnitudeg g- Determine the decimal value of this signed binary

number expressed in sign-magnitude1001010110010101

26 25 24 23 22 21 20 summing the weight where there is 1s

0 0 1 0 1 0 1 16 + 4 + 1 = 21

The sign bit is 1; dec no is 21The sign bit is 1; dec no is -21

DECIMAL VALUE OF SIGNED NUMBERSDECIMAL VALUE OF SIGNED NUMBERS

1’s COMPLEMENT:

Decimal values of +ve no are determined byysumming the weights in all bit positions wherethere are 1s and ignoring those positions wherethere are zerosthere are zeros.

Decimal number of –ve no are determined byl h h f bassigning a –ve value to the weight of sign bit,

summing all the weights where there are 1s and adding 1 to the resultsg

DECIMAL VALUE OF SIGNED NUMBERSDECIMAL VALUE OF SIGNED NUMBERS

1’s Complement form (example: +ve value)1 s Complement form (example: +ve value)- Determine the decimal value of this signed binary

number expressed in 1’s compliment00010111

-27 26 25 24 23 22 21 20 summing the weight where there is 1s-2 2 2 2 2 2 2 2 g g

0 0 0 1 0 1 1 1 16 + 4 + 2 + 1 = +23

DECIMAL VALUE OF SIGNED NUMBERSDECIMAL VALUE OF SIGNED NUMBERS

1’s Complement form (example: ve value)1 s Complement form (example: -ve value)- Determine the decimal value of this signed binary

number expressed in 1’s complimentp p11101000

-27 26 25 24 23 22 21 20 summing the weight where there is 1s

1 1 1 0 1 0 0 0 -128 + 64 + 32 + 8 = -2424

Adding 1 to the result, the final decimal no isAdding 1 to the result, the final decimal no is-24 + 1 = -23

DECIMAL VALUE OF SIGNED NUMBERSDECIMAL VALUE OF SIGNED NUMBERS

2’S COMPLEMENT:

Decimal values of +ve and ve no are determinedDecimal values of +ve and –ve no are determinedby summing the weights in all by summing theweight in all positions where there are 1s and

h h hignoring those positions where there are zeros

The weight of the sign bit in a –ve no is giveng g ga –ve value

DECIMAL VALUE OF SIGNED NUMBERSDECIMAL VALUE OF SIGNED NUMBERSDECIMAL VALUE OF SIGNED NUMBERSDECIMAL VALUE OF SIGNED NUMBERS

2’s Complement form (example: +ve value)p ( p )- Determine the decimal value of this signed binary

number expressed in 1’s compliment01010110

27 26 25 24 23 22 21 20 summing the weight where there is 1s-27 26 25 24 23 22 21 20 summing the weight where there is 1s

0 1 0 1 0 1 1 0 64 + 16 + 4 + 2 = +86

DECIMAL VALUE OF SIGNED NUMBERSDECIMAL VALUE OF SIGNED NUMBERS

2’s Complement form (example: -ve value)p ( p )- Determine the decimal value of this signed binary

number expressed in 1’s compliment10101010

27 26 25 24 23 22 21 20 summing the weight where there is 1s-27 26 25 24 23 22 21 20 summing the weight where there is 1s

1 0 1 0 1 0 1 0 -128 + 32 + 8 + 2 = -86

THIS WEEK:THIS WEEK:

• Arithmetic operation with signed numbers• Arithmetic operation with signed numbers• Hexadecimal Numbers• Octal Numbers• Binary Coded Decimal (BCD)• Digital Codes

ARITHMETIC OPERATION WITH SIGNED NOs

Addition:

Conditions:

1.Both number +ve

2. +ve number with magnitude larger than –ve no

b h d l h3.-ve number with magnitude larger than +ve no

4.Both no –ve

ARITHMETIC OPERATION WITH SIGNED NOs

Substraction:

Remember: The sign of a +ve or –ve binary g ynumber is changed by taking its2’s complement

To substract two signed numbers, take the 2’sl f h b h d d ddcomplement of the subtrahend and add.

Discard any final carry bity y

ARITHMETIC OPERATION WITH SIGNED NOsMultiplication:p

2 methods:

• Direct addition – lengthy

2 P i l d 2. Partial product – most common

Partial Product:

Step 1 – determine if the sign of the multiplicandare the same or different That’s willare the same or different. That s willdetermines the end result

If th i th th d t - If the sign are the same, the product = +ve- If the sign are different, the product = -ve

Step 2 – Change any number to uncomplemented form. Usually from2’s complemented form to true numbernumber

Step 3 – Do a partial product multiplication. Useonly the mag bits Ignore sign bitonly the mag. bits. Ignore sign bit

Step 4 – Add each successive partial product to get the final product.

Step 5 – If sign bit is –ve, take the 2’s complementStep 5 If sign bit is ve, take the 2 s complementof the product. Else just leave it as the final result. Don’t forget to add the sign bit

ARITHMETIC OPERATION WITH SIGNED NOsDivision:

Step 1 – determine if the sign of the divident anddivisor are the same or different That’s willdivisor are the same or different. That s willdetermines the sign of the quotient

If th i th th ti t - If the sign are the same, the quotient = +ve- If the sign are different, the quotient = -ve

Step 2 – Substract the divisor from the dividend using 2’s compliment addition to get the 1’s partial remainder and add 1 to quotientpartial remainder and add 1 to quotient

- If the partial remainder is +ve, go to step 3. If –ve,di i i i l t ddivision is completed.

Step 3 – Substract the divisor from the partial remainder and add 1 to quotient. Ifresult = +ve, repeat for the next partialremainder. remainder.

If result = 0 or –ve, division is complete.

0000000000000BinaryOctalHexDec

000000000000000100000010

000001002

012

012

000000110000010000000101

003004005

345

345 00000101

000001100000011100001000

005006007010

5678

5678 00001000

0000100100001010

010011012

89A

8910

000010110000110000001101

013014015

BCD

111213 00001101

0000111000001111

015016017

DEF

131415

Significant DigitsSignificant DigitsSignificant DigitsSignificant DigitsBinary: 11101101Binary: 11101101

Most significant digit Least significant digitMost significant digit Least significant digit

Hexadecimal: 1D63A7AHexadecimal: 1D63A7A

Most significant digit Least significant digitMost significant digit Least significant digit

Hexadecimal Number Hexadecimal Number SystemSystemSystemSystem

• Base 16 system• Base 16 system• Uses digits 0-9 &l tt A B C D E Fletters A,B,C,D,E,F• Groups of four bits

represent eachbase 16 digit

Hexadecimal to Decimal Hexadecimal to Decimal ConversionConversionConversionConversion

Convert 3B4F16 to its decimal equivalent:Convert 3B4F16 to its decimal equivalent:

Hex Digits 3 B 4 Fxxx

163 162 161 160Positional Valuesx

163 162 161 160

12288 +2816 + 64 +15

Positional Values

Products

15,18310

Decimal to Hexadecimal Decimal to Hexadecimal ConversionConversionConversionConversion

Convert 83010 to its hexadecimal equivalent:Convert 83010 to its hexadecimal equivalent:

830 / 16 = 51 R14 = E in Hex830 / 16 = 51 R1451 / 16 = 3 R33 / 16 = 0 R33 / 16 = 0 R3

33E16

Number ConversionNumber Conversion

• Binary to Hexadecimal Conversion (vice versa))

1. Grouping the binary position in 4-bit groups, starting from the least groups, starting from the least significant position.

Binary to Hexadecimal Binary to Hexadecimal C iC iConversionConversion

• The easiest method for converting binary The easiest method for converting binary to hexadecimal is to use a substitution code

• Each hex number converts to 4 binary digits

Number ConversionNumber Conversion

Example:Example:– Convert the following binary numbers to

their hexadecimal equivalent (vice versa)their hexadecimal equivalent (vice versa).a) 10000.12

b) 1F.C16b) 1F.C16

– Answer:a) 10.816) 16

b) 00011111.11002

Substitution CodeSubstitution CodeConvert 0101011010101110011010102 to

Substitution CodeSubstitution CodeConvert 0101011010101110011010102 to

hex using the 4-bit substitution code :

5 6 A E 6 A

0101 0110 1010 1110 0110 1010

56AE6A16

0000000000000BinaryOctalHexDecN

000000000000000100000010

000001002

012

012

UMB 00000011

0000010000000101

003004005

345

345

BER

000001100000011100001000

006007010

678

678

R

S000010000000100100001010

010011012

89A

8910

YST 00001011

0000110000001101

013014015

BCD

111213

TEM

0000111000001111

016017

EF

1415

MS

Octal Number SystemOctal Number SystemOctal Number SystemOctal Number System• Also known as the Base 8 SystemAlso known as the Base 8 System• Uses digits 0 - 7• Readily converts to binary • Readily converts to binary • Groups of three (binary) digits can

be used to represent each octal digitbe used to represent each octal digit• Also uses multiplication and division

algorithms for conversion to and algorithms for conversion to and from base 10

Octal to Decimal Octal to Decimal ConversionConversionConversionConversion

Convert 6538 to its decimal equivalent:

6 5 3xxx

82 81 80Positional Values

Octal Digits

82 81 80

384 + 40 + 3

Positional Values

Products

42710

Decimal to Octal ConversionDecimal to Octal ConversionDecimal to Octal ConversionDecimal to Octal ConversionConvert 42710 to its octal equivalent:Convert 42710 to its octal equivalent:

427 / 8 = 53 R3 Divide by 8; R is LSD427 / 8 53 R3 Divide by 8; R is LSD53 / 8 = 6 R5 Divide Q by 8; R is next digit6 / 8 = 0 R6 Repeat until Q = 0

6538

Number ConversionNumber Conversion

• Binary to Octal Conversion (vice versa)y ( )1. Grouping the binary position in groups

of three starting at the least significant of three starting at the least significant position.

Octal to Binary ConversionOctal to Binary ConversionOctal to Binary ConversionOctal to Binary ConversionEach octal number converts to 3 binary Each octal number converts to 3 binary

digits

To convert 6538 to binary, just substitute code:

6 5 3

110 101 011

Number ConversionNumber Conversion

E lExample:– Convert the following binary numbers to their

octal equivalent (vice versa)octal equivalent (vice versa).a) 1001.11112 b) 47.38

) 1010011 11011c) 1010011.110112

– Answer:) 11 74a) 11.748

b) 100111.0112

c) 123.668

Substitution CodeSubstitution CodeSubstitution CodeSubstitution Code

Substitution code can also be used to convert binary to octal by using 3-bit groupings:

2 5 5 2 7 1 5 2

010 101 101 010 111 001 101 010

255271528

Digital CodesDigital Codes• BCD (Binary Coded Decimal) Code

1. Represent each of the 10 decimal 1. Represent each of the 10 decimal digits (0~9) as a 4-bit binary code.

Example:Example:– Convert 15 to BCD.

1 5

0001 0101 C0001 0101BCD– Convert 10 to binary and BCD.

Digital CodesDigital CodesDigital CodesDigital Codes• ASCII (American Standard Code for

Information Interchange) Codeg )1. Used to translate from the keyboard

characters to computer languagep g g

Digital CodesDigital CodesDigital CodesDigital Codes• The Gray Code Decim Binary Gray The Gray Code

– Only 1 bit changes – Can’t be used in

aly y

Code0 0000 0000

arithmetic circuits• Binary to Gray Code

d i

1 0001 0001

2 0010 0011and vice versa. 2 0010 0011

3 0011 0010

4 0100 01104 0100 0110

5 0101 0111

6 0110 0101