Boolean Algebra and Logic Series - WordPress.com • Able to convert number among number system (Binary, Octal, Decimal, and Hexadecimal) • Able to present negative number in 1’s

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Boolean Algebra and Logic Series

CLO2-Week 6-Number System

Outline

• Able to convert number among number system (Binary, Octal, Decimal, and Hexadecimal)

• Able to present negative number in 1’s complement and 2’s complement form

• Able to solve summation and substraction based on complement number

• Able to present BCD number and floating point number

• Able to solve summation and substraction in various number system (Binary, Octal, Decimal, Hexadecimal, and BCD)

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Binary, Octal & Hexadecimal Number (1)

• Binary Number: binary system is based on powers of 2, constituent figures numbers : 0 & 1

• Converstion Example Decimal to binary: 1410 = X2 , X= ?

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Hence 1410= 11102


• Decimal to binary conversion can also be done by

considering the weight of each decimal number '1' in every

position binary numbers.

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• The number '1' inserted into the largest weighing

boxes still smaller than the remaining decimals


• Octal Numbers: Based on number eight, the numbers

making up numbers: 0,1,2,3,4,5,6 & 7

• Hexadecimal Numbers: Numbers base 16, the numbers

making up numbers : 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E & F

• How converting decimal number to octal and hex basically

the same as converting into binary, but it needs to be

converted into a divider 8 and 16.

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• Examples of conversion binary, octal and hex to decimal

1. 1012 = 1*22 + 0*21 + 1*20 = 510

2. 288 = 2*81 + 8*80 = 1810

3. 1716 = 1*161 + 7*160 = 2310

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• Converting a binary number to octal (hexa)

1. For to octal (hexa), segment the binary number threes

(four-four), starting from the right. If a lot of bits is not a

multiple of 3 (4), plug in some numbers '0' on the left of the

number to be a multiple of 3 (4).

2. Converting each group to octal (hexa)

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• Example

1. Binary to Octal:

101102 = 0101102 = 010 | 110 = 2 | 6 = 268

2. Binary to Hexa:

101102 = 000101102 = 0001 | 0110 = 1 | 6 = 1616

• For the opposite conversion, the steps above to stay behind.

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Some terms in binary numbers:

• Bit: binary digit

• Nibble: 4 bits

• Byte: 8 bits

• MSB: most significant bit (leftmost bit, the greatest weight)

• LSB: least significant bit (rightmost bit, the smallest weight)

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Binary Coded Decimal (BCD) (1)

• On the basis of the BCD system, one decimal point (0-9)

equivalent with the binary system of weighted 8-4-2-1

Used for the conversion Binary Decimal numbers can

still be easily understood by humans

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• Other bit

scheme(1010,

1011, 1100, 1101,

1110, 1111) are

not used

Binary Coded Decimal (BCD) (2)

Example: 863.9810 = 1000 0110 0011.1001 1000BCD

(8) (6) (3) (9) (8)

Weigth: (800, 400, 200, 100)(80, 40, 20, 10)(8, 4, 2, 1)(0.8, 0.4,

0.2, 0.1)(0.08, 0.04, 0.02, 0.01)

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XS3 BCD (3)

• Eccess 3 BCD (XS3 BCD/ XS3)

Binary numbers for the system is obtained by adding to the

number 00112 BCD base

XS3 = BCD + 00112

0

Ekivalen Desimal

Pola BitBCD

00110100 1

234

0101

5

Ekivalen Desimal

Pola BitBCD

01100111

10001001 6

789

101010111100

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• Other bits

schemes (0000,

0001, 0010, 1101,

1110, 1111) are

not used

XS3 BCD (3)

Example: 863.9810 = 1011100101110.11001011XS3

(8) (6) (3) (9) (8)

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Negative Binary Number (1)

• Negative numbers required for arithmetic operations on

digital machines based on the sum. For reduction, a concept

that is used : A – B = A + (-B)

• Negative numbers are mathematically characterized by

administering a '-' in front of the number in question. In the

digital machine, generally negative numbers expressed in

three systems / ways

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1. Sign Bit System: MSB = 0 for positive numbers and MSB = 1 for

negative numbers. Example (4-bit system):

+410 = 01002SM +0 = 00002SM

-410 = 11002SM -0 = 10002SM

Weakness

No number 'zero' is unique and requires a long time in the

operations of addition and subtraction. System Bit Markers

typically used in system Floating Point numbers

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2. Radix Complement System / Base. For a number X AND

numbers, Complement Radix :

LSDX

XRX

R

RN

RC

1

X

01234567

BinerX

10

DesimalX

98765

HexaX

FEDCBA98

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LSD: Least Significant DigitXR: Complement Number


2. Radix/Basis Complement System (continued)

Example:

1. 10's complement from 47,83 is equal to N10 + 1LSD = 52,17.

2. 2's complement from 0101101,101 is equal to N2 + 1LSD = 1010010,011.

3. 16's complement from A3D is equal to N16 + 1LSD = 5C2 + 1 = 5C3.

Note:

For the two bases, the above system is called 2’s Complement

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• Basic reduction in negative numbers by means of 2's

Complement analogous to Odometer (estimator Mileage)

on motor vehicles. Suppose an Odometer have three-digit

numbers then Odometer they measure up to 1000km.

After reaching the number '999' next Odometer will be

back to 000.

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0 0 0

0 0 1

0 0 2

9 9 9

9 9 8

0

+1

+2

-1

-2

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3. Radix Complement System / Small Base

System are Radix Complement System / Base but without reduction 1LSD. The tables shown earlier is of this system

BinerX DesimalX HexaX

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Deduced how to obtain small radix complement and complement regular radix


3. Radix Complement System / Small Base (continued)

For base 2: This system is called the 1's Complement (pair of 2's

complement)

For base 10: This system is called the 9's Complement (pair of

10's complement)

For base 16: This system is called 15's Complement (pair of 16's

complement)

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3. Radix Complement System / Small Base (continued)

Base 2 example:

+710= 0000 01112 (8-bit system, focus on MSB)

-710= 1111 10002 (8-bit system, focus on MSB)


• When the numbers marked on the system width will be

treated with a certain bit width bit larger system then steps

are performed:

A. For positive numbers: fill in the blank (left) with '0' to all

unused bits.

B. For negative numbers: fill in the blank (left) with '1' to all

unused bits.

00012 (+110, 4 bit) = 0000 00012 (+110, 8bit)

10112 (-510, 4 bit) = 1111 10112 (-510, 8bit)

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Fraction Number (1)

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A. Converting fractions binary / hex to decimal

Every binary / hex on the right commas mean weight

(rank) of its negative

0,0112 = 0*20 +0*2-1 +1*2-2 +1*2-3 = 0 +0,2510 +0,12510

= 0,37510

0,1016 = 0*160 + 1*16-1 + 0*16-2 = 0 +0,062510 +0

= 0,062510

Fraction Number (2)

B. Converting decimal to binary fractions

0,X10 = 0,Y2

Y10 = 0,X10*2n; n: many binary digits to the right of the comma which are desired

Example

0,7510 = 0,Y2 ; wants four digits

Y10 = 0,7510* 24 = 0,7510*1610 = 1210 = 11002

0,7510 = 0,11002

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Fraction Number (3)

C. Conversion of hexadecimal to decimal fractions

0,X10 = 0,Y16

Y10 = 0,X10*16n; n: many binary digits to the right of the comma which are desired

Example

0,7510 = 0,Y16 ; Y desired two-digit

Y10 = 0,7510* 162 = 0,7510*25610 = 19210 = C016

0,7510 = 0,C016

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Fraction Number (5)

D. Converting binary fractions hexadecimal

Same way a conversion of binary integer hexadecimal

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The IEEE standard for floating point arithmetic

• Standard for floating point representation in:single precision (32-bit)double precision (64-bit)

• Floating point format differences affect the floating point

computation .

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Single Precision

• Single precision floating point standard representation 32 bit word, represented from 0 s / d 31, from left to right.

• The first bit is the sign bit (S), then the next 8 bits are the bits exponent (E), and the 23 remaining bits are fraction(F):

S EEEEEEEE FFFFFFFFFFFFFFFFFFFFFFF

0 1 8 9 31

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Single Precision

• V-value is represented by the following word :

– If E = 255 and F is not 0 (nonzero), then V = NaN ("Not a number")

– If E = 255 and F is 0, and S = 1, then V = -infinity

– If E = 255 and F are 0 and S = 0, then V = Infinity

– If 0 <E <255 then V = (- 1) ^ S * 2 ^ (E-127) * (1.F) where "1.F" is intended to represent the binary number created by prefixing F with an implicit leading 1 and a binary point.

– if E = 0 and F is not 0 (nonzero), then V = (- 1) ^ S * 2 ^ (-126) * (0.F). This includes the value "un-normalized".

– if E = 0 and F = zero and S = 1, then V = -0

– if E = 0 and F = zero and S = 0, then V = 0

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Single Precision

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Double Precision

• The IEEE standard for double precision floating point representation requires a 64 bit word, 0 to 63, from left to right.

• The first bit sign bit, S, 11 bits subsequent exponent bits, 'E', and the remaining 52 bits is a fraction 'F'

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Double Precision

• Value V is represented by a group of word that can be shown as follows:

– if E = 2047 and F = nonzero, then V = NaN ("Not a number")

– if E = 2047 and F = zero and S = 1, then V = -infinity

– if E = 2047 and F = zero and S = 0, then V = Infinity=

– If 0 <E <2047 then V = (- 1) ** S * 2 ** (E-1023) * (1.F) where "1.F" is intended to represent the binary number created by prefixing F with an implicit leading one and a binary point.

– If E = 0 and F = nonzero, then V = (- 1) ** S * 2 ** (-1022) * (0.F) This includes the "unnormalized" values.

– If E = 0 and F = zero and S = 1, then V = -0

– if E = 0 and F = zero and S = 0, then V = 0

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Example

• The binary representation of - 0.75 in IEEE single precision format is as follows:

(-1) s x (1+M) x 2E

• Answer:

• decimal representation : - 0.75 = -3/4 = -3/22

• binary representation : - 0.75 = - 0.11 = -1.1 x 2-1

– (-1)s x (1 + M) x 2E

– Sign bit = 1

– (1+M) = (1+ .1000….)

– Exponent = (-1 + 127) = 126

Note: -1 obtained from the rank of rank 2-1 exponent is -1127 obtained from the bit length eksp 28-1-1 = 127

– So, : 10111111010000000000000000000000

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Example 2

• -118.625 Binary representation of the IEEE single precision format is as follows :

(-1) s x (1+M) x 2E

• Answer :

• binary representation : - 118.625 = - 1110110.101

= - 1.110110101 x 26

– (-1)s x (1 + M) x 2E

– Sign bit = 1– (1+M) = (1+ .11011010100000000000000)

– Exponent = (6 + 127) = 133Note : 6 obtained from the rank of powered exponent 26 is 6

127 obtained from the bit length exponent 28-1-1 = 127– So : 11000010111011010100000000000000

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Binary Summation & Substraction (1)

1. Direct Summation

• Add two binary numbers + can be done as a decimal (base

10). When the sum of two bits more than 012, a carry bit is

added to the next MSB; This process continues until all bits

are summed

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1. Directly Summation (continued)

Example: Addition of 8-bit binary number

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• Reduction of the 2's Complement: The most widely used in computing.

1310

710-0 11012

0 01112-0 11012

1 10012+== 710- dalam 2's Complement

0 011021

Bit Penanda: PositifOverflow dibuang

1710

310-0

11012001112

-0 01112

1 00112+== 1310- dalam 2's Complement

1 101020

Bit Penanda: NegatifOverflow dibuang

=

610+=

610-=

=

dalam 2's Complement

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• 1’s Complement Substraction

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• 1’s Complement Substraction (continued)

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How to read the numbers with negative marker bit ('1') in the negative decimal numbers?


• Summation BCD: Summation ranging from LSD and ending at MSD. If the sum exceeds 10012 (910) (including Overflow per number) be corrected by adding 01 102 (610). Carry (00012) is added to the next MSD.

05610

06910+0000 0101 0110BCD

0000 0110 1001BCD+

0000 1011 1111 0110 0110

Hasil penjumlahan

Koreksi

Carry

0001 0010 0101BCD

1 1 01 1111

Hasil: 12510

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• Reduction BCD: Done by making unfounded negative BCD 10's complement system (10C). If negative, the results should be negated back.

08,2510

13,5210-

0000 1000 , 0010 0101BCD

1000 0110 , 0100 1000BCD+

1000 1110 , 0110 1101 0110 , 0110

Hasil penjumlahan

Koreksi

Carry

1001 0100 , 0111 0011BCD10C

11

Hasil: 94,7310C

08,2510

86,4810C+

05,2710- 94,7310C+

11 1

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• Substraction BCD (continued)

Results +94,7310C to be negated by the 10's Complement

method to obtain correct results.

+94,7310C = - 05,2710

Conclusion

For binary addition and subtraction, 2's complement shows

the simplest steps.

2's complement is not necessarily the simplest to other

arithmetic operations (multiplication and division)

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Documents

Boolean Algebra and Logic Series - WordPress.com • Able to convert number among number system (Binary, Octal, Decimal, and Hexadecimal) • Able to present negative number in 1’s