Lecturer 吳安宇教授access.ee.ntu.edu.tw/course/logic_design_95first... · Graduate Institute...

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Graduate Institute of Electronics Engineering, NTU

CH1 Number Systems and ConversionCH1 Number Systems and Conversion

Lecturer：吳安宇教授

Date：2006/09/22


台灣大學吳安宇教授 pp.2

OutlineOutlineDigital Systems and Switching CircuitsNumber Systems and ConversionBinary ArithmeticRepresentation of Negative NumbersAddition of 2’s Complement NumbersAddition of 1’s Complement NumbersBinary Codes



PurposePurposeDesign a switching network (Logic Function)

：Binary number{ }1,0∈iX { }1,0∈iZ

Input OutputSwitching

Circuits

……

……

X1X2

Xm

Z1Z2

Zm



Number System and ConversionNumber System and ConversionDecimal (Base 10) Number (separated by decimal point):

953.7810(210.-1-2)

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

Binary (Base 2) Number (separated by binary point):

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 integerPositive integer NN with with Base (Radix) RBase (Radix) R：：

Name Decimal Binary Octal Hexadecimal

Radix 10 2 8 16

Digits 0,1,2,3,4,5,6,7,8,9 0,1 0,1,2,3,

4,5,6,7

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

C,D,E,F

Firstseventeenpositiveintegers

012345678910111213141516

0110111001011101111000100110101011110011011110111110000

01234567101112131415161720

0123456789ABCDEF10



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

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

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

EX：R=16, Digits = { 0,1,2,...,A,B,C,D,E,F }(A2F)16 = 10*162 + 2*161 + F*160

= (2607)10



Integer Conversion Using Integer Conversion Using ““Division MethodDivision Method”” (1/2)(1/2)



Integer Conversion Using Integer Conversion Using ““Division MethodDivision Method”” (2/2)(2/2)

EX：Convert (53)10 to Binary no.

MSB

(53)10 = (110101)2

MSB



Conversion of A Decimal Fraction Using Conversion of A Decimal Fraction Using ““Successive MultiplicationSuccessive Multiplication”” (1/3)(1/3)

Fi：Fraction Number




EX：Convert (0.625)10 to Binary number

(0.625)10 = (0.101)2

MSB MSB



EX：Convert (0.7)10 to Binary number

Repeated Process 0.8, 1.6, 1.2, 0.4, ….

(0.7)10 = 0.1 0110 0110 0110 …… (Base2)

No exact conversion !!!




Conversion (1/2)Conversion (1/2)EX：Convert (231.3)4 to Base 7 number(231.3)4 = 2*42 + 3*41 + 1*40 + 3*4-1 = (45.75)10

Integer Fraction

(45.75)10= (63.51 51 51 ……)2



Conversion (2/2)Conversion (2/2)Conversion from Binary (Base 2)

to Octal (Base 8)to Hexadecimal (Base 16)

(11010111110.0111 )2 = (3276.34)8

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

3 2 7 6 3 4(補0)

6 B E 3Binary point ( R=2 )

Hexadecimal point

decimal point (R=10)

00



Binary Arithmetic Binary Arithmetic ---- AdditionAdditionAddition

0 + 0 = 00 + 1 = 11 + 0 = 11 + 1 = 10 (sum 0 & carry 1)

EX：1111 (Carry)



Binary Arithmetic Binary Arithmetic ---- SubtractionSubtractionSubtraction

0 – 0 = 01 – 0 = 11 – 1 = 00 – 1 = 1 (with borrow 1 from next column)

EX：

1111 (Borrow)



Binary Arithmetic Binary Arithmetic ---- MultiplcationMultiplcationMultiplication

0 * 0 = 00 * 1 = 01 * 0 = 01 * 1 = 1

EX： (13)10

(11)10

copy of multiplicand if “1”

multiplicandmultiplier



Binary Arithmetic Binary Arithmetic ---- DivisionDivisionDivision



Signed Number RepresentationSigned Number Representation

S Magnitude

signed bit

bit bit bit bit bit bit bit bit

A computer “word”, wordlength n = 8 bits(other popular n = 16 bits, 32 bits)



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

N = ( s, an-2 ……a1 a0)2sm

EX：

N = -(13)10 = -(0,1101)2 = (1,1101)2sm

±

s = 0 if N 0

s = 1 if N 0≥≤

Signed magnitude

(2sm = Binary Singed Magnitude)



Radix ComplementRadix ComplementDefinition: The “radix complement [N]” of a number (N)r is defined as:

N* (notation in textbook) = [N]r = rn – (N)r

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

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



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

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

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

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

(carry)

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



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

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

[N]2 = 28 – (10110)2= (100000000)2 – (00010110)2= (11101010)2



22’’s Complements ComplementConvert (N)2 to [N]2：

Method 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 ComplementConvert (N)2 to [N]2：

Method 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 Flip then Add 1



22’’s Complements ComplementSigned Decimal Sign Magnitude

BinaryTwo’s Complement

SystemOne’s Complement

System

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

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

(1,0000)

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

(1,1111)

for n = 5



22’’s Complements ComplementSigned Decimal Sign Magnitude

BinaryTwo’s Complement

SystemOne’s Complement

System

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

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

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

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

for n = 5



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

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

EX：(n=8) Determine the decimal no. of N=(1,111,1010)2

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



Radix Complement ArithmeticRadix Complement ArithmeticEX：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)EX：compute (12)10 + (7)10

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

1 0 0 1 1 (-13)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 ArithmeticEX：(-9) – (5) = (-9) + (-5)

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

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

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

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



11’’s & 2s & 2’’s Complements Complement2’s complement is the main streamCheck SIGN for the overflow!

(+) + (+) (-)

(-) + (-) (+)overflow!!



Overflow ConditionOverflow Condition

A B A+B A-B

+ + - ˇ x

+ - x - ˇ

- + x +ˇ

- - + ˇ x

ˇ：overflowx ：no overflow



Diminished Radix ComplementDiminished Radix Complement1’s complement

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 Numbers“End-around carry” :

Instead 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 NumbersAddition of positive & negative numbers(a)

+5 0101-6 1001-1 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 NumbersAdding 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 NumbersEX：Addition for a word-length of 8(a) (-11) + (-20) in 1’s complement

+11 = 00001011 (-11) = 11110100+20 = 00010100 (-20) = 11101011

(1)11011111 1

(+31) 00011111 11100000 (-31)

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

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



Binary CodesBinary CodesBCD (Binary Coded Decimal) codesEX： 1 9 8 9

0001 1001 1000 1001

0： 00001： 00012： 00103： 00114： 0100

5： 01016： 01107： 01118： 10009： 1001



Binary CodesBinary CodesASCII codeskeyboard computer

Character Binary Code Hexadecimal Code

Digital

1000100110100111001111101001111010011000011101100

4469676974616C

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



ASCII Code (Table 1ASCII Code (Table 1--3 on p.22)3 on p.22)



Binary Codes for Decimal DigitsBinary Codes for Decimal Digits

Lecturer 吳安宇教授access.ee.ntu.edu.tw/course/logic_design_95first... · Graduate Institute...

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