CH1 Number Systems and Conversionaccess.ee.ntu.edu.tw/course/logic_design_94first...Graduate Institute of Electronics Engineering, NTU 台灣大學吳安宇教授 pp.5 Number System

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Graduate Institute of Electronics Engineering, NTU

CH1 Number Systems and ConversionCH1 Number Systems and Conversion

Lecturer：吳安宇Date：2005/9/23


台灣大學吳安宇教授 pp.2

OutlineOutlinevDigital Systems and Switching CircuitsvNumber Systems and ConversionvBinary ArithmeticvRepresentation of Negative Numbers

Addition of 2’s Complement NumbersAddition of 1’s Complement NumbersvBinary Codes



PurposePurposev Design a switching network (Logic Function)

：Binary number{ }1,0∈iX { }1,0∈iZ







Number System and ConversionNumber System and Conversionv Decimal (Base 10) Number：

953.7810(210.-1-2)

= 9*102 + 5*101 + 3*100 + 7*10-1 +8*10-2

v Binary (Base 2) Number：1011.112(3210.-1-2)

= 1*23 + 0*22 + 1*21 + 1*20 + 1*2-1 + 1*2-2

= 8 + 0 +2 +1+1/2 +1/4= (11.75)10



Generalized Representation of an Generalized Representation of an Positive integer N with Base (Radix) RPositive integer N with Base (Radix) R：：

0123456789ABCDEF10

01234567

101112131415161720

011011

10010111011110001001101010111100110111101111

10000

0123456789

10111213141516

Firstseventeenpositiveintegers

0,1,2,3,4,5,6,7,8,9,A,B,C,D,

E,F

0,1,2,3,4,5,6,7,0,10,1,2,3,4,5,6,

7,8,9Digits

168210Radix

HexadecimalOctalBinaryDecimalName



Generalized Representation of an Positive Generalized Representation of an Positive integer N with Base (Radix) Rinteger N with Base (Radix) R：：

v N= (a4 a3 a2 a1 a0 . a-1 a-2 a-3)R= a4*R4 + a3*R3 + a2*R2 + a1*R1 + a0*R0

+ a-1*R-1 + a-2*R-2 + a-3*R-3

v EX：R=8 , Digits = { 0,1,2,3,4,5,6,7 }(147.3)8 = 1*82 + 4*81 + 7*80 + 3*8-1

= (103.375)10

v EX：R=16, Digits = { 0,1,2,...,A,B,C,D,E,F }(A2F)16 = 10*162 + 2*101 + F*160

= (2607)10



Integer Conversion using Integer Conversion using ““Division MethodDivision Method””



Integer Conversion using Integer Conversion using ““Division MethodDivision Method””

v EX：Convert (53)10 to Binary no.

MSB

(53)10 = (110101)2

MSB



Fi：Fraction Number

Conversion of A Decimal Fraction Conversion of A Decimal Fraction Using Using ““Successive MultiplicationSuccessive Multiplication””



v EX：Convert (0.625)10 to Binary no.

(0.625)10 = (0.101)2





v EX：Convert (0.7)10 to Binary no.

Repeated Process 1.6, 1.2, 0.4, 0.8

(0.7)10 = 0.0 0110 0110 0110 …… (Base2)

No exact conversion !!!



ConversionConversionv EX：Convert (231.3)4 to Base 7 number

(231.3)4 = 2*42 + 3*41 + 1*40 + 3*4-1 = (45.75)10

ØInteger ØFraction

(45.75)10= (63.51 51 51 ……)2



ConversionConversionvConversion from Binary (Base 2)vto Octal (Base 8)vto Hexadecimal (Base 16)

(11010111110.0111)2 = (3276.14)8

(11010111110.0011)2 = (6BE.3)16

3 2 7 6 1 4(補0)

6 B E 3 Binary point ( R=2 )

Hexadecimal point

decimal point (R=10)







Binary ArithmeticBinary Arithmeticv Additionv 0 + 0 = 0v 0 + 1 = 1v 1 + 0 = 1v 1 + 1 = 10 (sum 0 & carry 1)

v EX：



Binary ArithmeticBinary Arithmeticv Subtractionv 0 – 0 = 0v 1 – 0 = 1v 1 – 1 = 0v 0 – 1 = 1 (with borrow 1 from next column)

v EX：



Binary ArithmeticBinary ArithmeticvMultiplicationv 0 * 0 = 0v 0 * 1 = 0v 1 * 0 = 0v 1 * 1 = 1

v EX：(13)10

(11)10

copy of multiplicand



Binary ArithmeticBinary Arithmeticv Division







Signed Number RepresentationSigned Number Representation

S Magnitude

signed bit



Singed Magnitude NumbersSinged Magnitude Numbersv N = (an-1 an-2 ……a1 a0)r

N = ( s an-1 an-2 ……a1 a0)rsm

v EX：

N = -(13)10 = -(1101)2 = (1,1101)2sm

±

s = 0 if N 0

s = 0 if N 0≥≤

sign magnitude



Radix ComplementRadix ComplementvDef. The “radix complement [N]” of a number

(N)r is defined as

[N]r = rn – (N)r

where n is the number of digits in (N)r

vThe largest positive number (positive full scale) = rn-1 – 1vThe most negative number (negative full

scale) = - rn-1



22’’s Complements Complement[N]2 = 2n – (N)2

v EX：2’s complement of (N)2 = (01100101)2

[N]2 = [01100101]2= 28 – (01100101)2= (100000000)2 – (01100101)2= (10011011)2

v EX：show that (N)2 + [N]2 = 0011001011001101100000000

(carry)

+1 [N]2 = - (N)2



22’’s Complements ComplementvEX：check (N)2 = [ [N]2 ]2 (by yourself)

vEX：2’s complement of (N)2 = (10110)2 for n=8

[N]2 = 28 – (01100101)2= (100000000)2 – (10110)2= (11101010)2



22’’s Complements Complementv Convert (N)2 to [N]2：way 1

N = 0 1 1 0 0 1 0 1

[N]2 = 1 0 0 1 1 0 1 1

N = 1 1 0 1 0 1 0 0

[N]2 = 0 0 1 0 1 1 0 0

First nonzero digit

First nonzero digit



22’’s Complements Complementv Convert (N)2 to [N]2：way 2

N = 0 1 1 0 0 1 0 1 1 0 0 1 1 0 1 0 complement the bits

1 add 11 0 0 1 1 0 1 1

→→

→1001

,kk aa then add 1



22’’s Complements Complement

0,11110,11100,11010,11000,10110,10100,10010,10000,01110,01100,01010,01000,00110,00100,00010,0000

(1,1111)

0,11110,11100,11010,11000,10110,10100,10010,10000,01110,01100,01010,01000,00110,00100,00010,0000

0,11110,11100,11010,11000,10110,10100,10010,10000,01110,01100,01010,01000,00110,00100,00010,0000

(1,0000)

+15+14+13+12+11+10+9+8+7+6+5+4+3+2+10

One’s Complement System

Two’s Complement System

Sign Magnitude BinarySigned Decimal

for n = 5



22’’s Complements Complement

1,11101,11011,11001,10111,10101,10011,10001,01111,01101,01011,01001,00111,00101,00011,0000--------

1,11111,11101,11011,11001,10111,10101,10011,10001,01111,01101,01011,01001,00111,00101,00011,0000

1,00011,00101,00111,01001,01011,01101,01111,10001,10011,10101,10111,11001,11011,11101,1111--------

-1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16

One’s Complement System

Two’s Complement System

Sign Magnitude BinarySigned Decimal

for n = 5



22’’s Complements Complementv EX：2’s complement of –(13)10 for n = 8

(13)10 = (1011)2 = (00001101)2-(00001101)2 = [00001101]2 = (11110011)2

v EX：Determine the decimal no. of N=(1,1111,1010)2

1111010 (?) (-6)0000110 (6)



Radix Complement ArithmeticRadix Complement Arithmeticv EX：compute (9)10 + (5)10 for 5-bit 2’s complement

0 1 0 0 1 (+9)+ 0 0 1 0 1 (+5)

0 1 1 1 0 (+14)v EX：compute (12)10 + (7)10

0 1 1 0 0 (+12)+ 0 0 1 1 1 (+7)

1 0 0 1 1 (-13)v EX：compute (12)10 – (5)10 = (12) + (-5)

0 1 1 0 0 (+12)+ 1 1 0 1 1 (2’s complement of (5)2)1 0 0 1 1 1 (+7)

discard the carry

Add two positive no. and obtain a negative no.(overflow occurs!)



Radix Complement ArithmeticRadix Complement Arithmeticv EX：(-9) – (5) = (-9) + (-5)

9 = 0 1 0 0 1 à -9 = 1 0 1 1 15 = 0 0 1 0 1 à -5 = 1 1 0 1 1

1 0 1 1 1 à (-9)+ 1 1 0 1 1 à (-5)1 1 0 0 1 0 à (-14)

discard (why?)v EX：(-12) – (5) = (-12) + (-5)

1 0 1 0 0 à (-12)+ 1 1 0 1 1 à (-5)1 0 1 1 1 1 à (+15) (overflow occurs)



Radix Complement ArithmeticRadix Complement Arithmetic

x+ ˇ--

+ˇx+-

-ˇx-+

x-ˇ++

A-BA+BBA

ˇ：overflowx ：no overflow



Diminished Radix ComplementDiminished Radix Complementv 1’s complement

v EX： 1 0 1 1 0 1 0 0 (N)2

0 1 0 0 1 0 1 1 1’s complement of (N)2

→→

→1001

,kk aa



Addition of 1Addition of 1’’s Complement Numberss Complement Numbersv“End-around carry” : vInstead of discarding the last carry (as in 2’s

complement), it is added to the n-bit sum in the position furthest to the right.



Addition of 1Addition of 1’’s Complement Numberss Complement NumbersvAddition of positive & negative numbers

(a)+5 0101-6 1011-1 1110 (correct)

(b)-5 1010

+6 0110+1 1 0000

1 (end-around carry)0001 (correct, no overflow)



Addition of 1Addition of 1’’s Complement Numberss Complement Numbersv Adding two negative numbers

(a) -3 1100-4 1011-7 1 0111

1 (end-around carry)1000 (-7) (correct, no overflow)

(b) -5 1010-6 1001

-11 1 00111 (end-around carry)

0100 (wrong, overflow!!)



Addition of 1Addition of 1’’s Complement Numberss Complement Numbersv EX：Addition for a word-length of 8

(a) (-11) + (-20) in 1’s complement+11 = 00001011 ßà (-11) = 11110100+20 = 00010100 ßà (-20) = 11101011

(1)110111111

(+31) 00011111 11100000 (-31)

(b) (-8) + (+19) in 2’s complement11111000 (-8)00010011 (+19)

(1)00001011 (+11)discard the last carry



11’’s & 2s & 2’’s Complements Complementv2’s complement is the main streamvcheck SIGN for the overflow!

(+) + (+) à (-)

(-) + (-) à (+)overflow!!







Binary CodesBinary Codesv BCD (Binary Coded Decimal) codesv EX： 1 9 8 9

0001 1001 1000 1001

5： 01016： 01107： 01118： 10009： 1001

0： 00001： 00012： 00103： 00114： 0100



Binary CodesBinary Codesv ASCII codes

keyboard à computer

4469676974616C

1000100110100111001111101001111010011000011101100

Digital

Hexadecimal CodeBinary Code Character

Encode the word Digital in ASCII code, representing each character by two hexadecimal digits



Binary CodesBinary Codesv ASCII character code

000~111 (c6c5c4)

0000~1111(c3c2c1c0)



Binary Codes for Decimal DigitsBinary Codes for Decimal Digits



ASCII CodeASCII Code

Documents

CH1 Number Systems and Conversionaccess.ee.ntu.edu.tw/course/logic_design_94first...Graduate Institute of Electronics Engineering, NTU 台灣大學吳安宇教授 pp.5 Number System