digital design, number system and data...

Digital Design,Number System

and Data Representation

Jin-Tai Yan Introduction to Digital Logic

Digital DesignDigital& Analog

Number System & OperationsBase 2, 10 and 16

Data RepresentationCharacter, Integer, Floating‐Point


OutlinesOutlines

Digital Devices


2000: Pentium® 4 ProcessorUsers of Pentium® 4 processor-based PCs can create professional-quality movies; deliver TV-like video via the Internet; communicate with real-time video and voice; render 3D graphics in real time; quickly encode music for MP3 players; and simultaneously run several multimedia applications while connected to the Internet. The processor debuted with 42 million transistors and circuit lines of 0.18 microns. Intel's first microprocessor, the 4004, ran at 108 kilohertz (108,000 hertz), compared to the Pentium® 4 processor's initial speed of 1.5 gigahertz (1.5 billion hertz). If automobile speed had increased similarly over the same period, you could now drive from San Francisco to New York in about 13 seconds.

Integrated Circuits that operate on Digital Data are in 95% of every electrical powered device

Analog & Digital


103.5

Analog Voltage meter

Digital Voltage meter

About 100

Continuous & Discrete


Ambiguity & Interpretation


Analog Clock

1:56 pmDigital Clock

About 2:00

1:50

1:56

1:56

Advantage of Digital Data

• Digital data can have additional data added to it to allow for detection and correction of errors– Scratch a CDROM ‐ will still play fine– Scratch, stretch an analog tape ‐ throw it away

• Digital data can be transmitted over a medium that introduces errors that are corrected at receiving end– Satellite transmission of Direct TV ‐ each ‘screen’ image is digitally encoded; errors corrected when it reaches your digital Set Top receiver, shows up as a ‘Perfect’ Picture.


Binary Representation

• The basis of all digital data is binary representation.

• Binary ‐ means ‘two’– 1, 0

– True, False

– Hot, Cold

– On, Off

• More than just values are handled for real worlds– 1, 0, 56

– True, False, Maybe

– Hot, Cold, LikeWarm, Cool

– On, Off, Leaky


Binary Codes


One binary digit (one bit) can take on values 0, 1.

(0 = hot, 1 = cold), (1 = True, 0 = False), (1 = on, 0 = off).Two binary digits (two bits) can take on values of 00, 01, 10, 11.

(00 = hot, 01 = warm, 10 = cool, 11 = cold).

Three binary digits (three bits) can take on values of 000, 001, 010, 011, 100, 101, 110, 111.

000 = Black, 001 = Red, 010 = Pink, 011 = Yellow,

100 = Brown, 101 = Blue, 110 = Green , 111 = White.

N bits (or N binary digits) can represent 2N different values.

Base 10 and Base 2

• A digit in base 10 ranges from 0 to 9.

• A digit in base 2 ranges from 0 to 1 (binary number system). A digit in base 2 is also called a ‘bit’.

• ‘%’ will be used for base 2 (%10101111)


Base 2, 8 and 16

• A digit in Base 16 can range from 0 to 16‐1 (0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F).

• Use letters A‐F to represent values 10 to 15.

• Base 16 is also called Hexadecimal or just ‘Hex’.

• ‘$’ will be used for base 16 ( $A2F)


81 = 8 = 23

161 = 16 = 24

Number Positional Notation(Polynomial Notation)


Number value is determined by multiplying each digit by a weight and then summing. The weight of each digit is a POWER of the BASE and is determined by position.

953.78 = 9 x 102 + 5 x 101 + 3 x 100 + 7 x 10-1 + 8 x 10-2

= 900 + 50 + 3 + 0.7 + 0.08

= 953.78

Powers of 2


2-3 = 0.1252-2 = 0.252-1 = 0.520 = 121 = 222 = 423 = 824 = 16

160 = 1 = 20

161 = 16 = 24

162 = 256 = 28

163 = 4096 = 212

210 = 1024 = 1 K220 = 1048576 = 1 M (1 Megabits) = 1024 K

230 = 1073741824 = 1 G (1 Gigabits)

25 = 3226 = 6427 = 12828 = 25629 = 512210 = 1024211 = 2048212 = 4096

Any Base to Base 10(Series Substitution Conversion)


Base 2 to Base 10

Base 16 to Base 10

% 1011.11 = 1x23 + 0x22 + 1x21 + 1x20 + 1x2-1 + 1x2-2

= 8 + 0 + 2 + 1 + 0.5 + 0.25 = 11.75

$ A2F = 10x162 + 2x161 + 15x160

= 10 x 256 + 2 x 16 + 15 x 1 = 2560 + 32 + 15

= 2607

Base 10 to Base 2


Integer Conversion:Radix Divide Method

NI = bn-1xBn-1 + … + b1xB1 + b0xB0

NI/B = ( bn-1xBn-1 + … + b1xB1 + b0xB0 )/B= bn-1xBn-2 + … + b1xB0 + b0

Fraction Conversion:Radix Multiply Method

Number, N = Integer, NI + Fraction, NF

NF = b-1xB-1 + b-2xB-2 + … + b-mxB-m

NF x B = ( b-1xB-1 + b-2xB-2 + … + b-mxB-m )xB= b-1 + b-2xB-1 + … + b-mxB-(m-1)

Base 10 to Base 2


Integer Conversion:53/2 = 26, remainder = 126/2 = 13, remainder = 013/2 = 6, remainder = 16 /2 = 3, remainder = 03/2 = 1, remainder = 11/2 = 0, remainder = 1

0.625x2 = 1.25, carry = 10.25x2 = 0.5 , carry = 00.5x2 = 1.0, carry = 1

53.625 = % 110101.101= 1x25 + 1x24 + 0x23 + 1x22 + 0x21 + 1x20 + 1x2-1 + 0x2-2 + 1x2-3

= 32 + 16 + 0 + 4 + 0 + 1 + 0.5 + 0 + 0.125

Fraction Conversion:

Number = Integer + Fraction

53.625 = 53 + 0.625

Base 16 to Base 2


Each Hex digit represents 4 bits. To convert a Hex number to Binary, simply convert each Hex digit to four binary bits.

Single Hex digit to four binary bits:$ 0 = % 0000$ 1 = % 0001$2 = % 0010$3 = % 0011$4 = % 0100$5 = % 0101$6 = % 0110$7 = % 0111

$ 8 = % 1000$ 9 = % 1001$ A = % 1010$ B = % 1011$ C = % 1100$ D = % 1101$ E = % 1110$ F = % 1111

161 = 16 = 24

Conversion of Base 16 and 2


$ A2F.B2 = % 1010 0010 1111.1011 0010

$ 345.C79 = % 0011 0100 0101.1100 0111 1001

= % 11 0100 0101.1100 0111 1001

% 1010001.0011011 = % 0101 0001. 0011 0110

= $ 51.36

Padded with a zero

Base 16 to Base 2

Base 2 to Base 16

Binary Addition & Subtraction


Binary Addition 0 + 0 = 0, carry = 0

1 + 0 = 1, carry = 0

0 + 1 = 1, carry = 0

1 + 1 = 0, carry = 1

Binary Subtraction 0 - 0 = 0, borrow = 0

1 - 0 = 1, borrow = 0

0 - 1 = 1, borrow = 1

1 - 1 = 0, borrow = 0

% 101011

+ % 000001---------------

101100

% 100

- % 001-------

011

Data Representation


Character & String– ASCII Code, Unicode

Integer Number– Unsigned

• Binary Codes, BCD Codes, Gray Code – Signed

• Signed-Magnitude,• 1’s Complement, 2’s Complement

Floating Point Number– IEEE 754 Standard Code

ASCII Codes


ASCII code (American Standard Code for Information Interchange) is a 7-bit code for Character data.

7 bits can only represent 27 different values (128). This enough to represent the Latin alphabet (A-Z, a-z, 0-9, punctuation marks, some symbols like $)

8 bits are actually used with the 8th bit being zero or used for error detection (parity checking).

1 Byte = 8 bits ‘A’ = % 01000001 = $41‘&’ = % 00100110 = $26

ASCII Codes


Unicode


UNICODE is a 16-bit code for representing alphanumeric data. With 16 bits, can represent 216 or 65536 different symbols.

16 bits = 2 Bytes per character.

$0041-005A A-Z$0061-4007A a-z

Some other alphabet/symbol ranges

$3400-3d2d Korean Hangul Symbols$3040-318F Hiranga, Katakana, Bopomofo, Hangul$4E00-9FFF Han (Chinese, Japenese, Korean)

UNICODE used by Web browsers, Java.

Binary Code


N binary digits (N bits) can represent unsigned integers from 0 to 2N-1.

4 bits = 0 to 158 bits = 0 to 25516 bits = 0 to 65535

Binary Code0 = % 00001 = % 00012 = % 00103 = % 00114 = % 01005 = % 01016 = % 01107 = % 0111

8 = % 10009 = % 100110 = % 101011 = % 101112 = % 110013 = % 110114 = % 111015 = % 1111

BCD Codes


BCD Code0 = % 00001 = % 00012 = % 00103 = % 00114 = % 01005 = % 01016 = % 01107 = % 01118 = % 10009 = % 1001

Each decimal digit simply represented by 4 bits.

Advantage: Easy to convertDisadvantage: Takes more bits to store a number

255 = % 1111 1111 = $ FF (binary code)255 = % 0010 0101 0101 = $ 255 (BCD code)

Gray Code


Gray Code0 = % 00001 = % 00012 = % 00113 = % 00104 = % 01105 = % 01116 = % 01017 = % 01008 = % 1100

9 = % 1101

A Gray code changes by only 1 bit for adjacent values. This can help reduce circuit complexity and also reduce signal noise.

Unsigned Overflow


Overflow occurs when two unsigned numbers added or subtracted, and the correct result is a number that is outside of the range of allowable numbers for that precision.

8 bits -- unsigned integers 0 to 28 -1 or 0 to 255.

16 bits -- unsigned integers 0 to 216-1 or 0 to 65535

Signed Magnitude Number


The Most Significant Bit(MSB) is used to represent the sign. ‘1’ is used for a ‘-’ (negative sign), a ‘0’ for a ‘+’ (positive sign).

The format of a SM number in 8 bits is:

% smmmmmmm

where ‘s’ is the sign bit and the other 7 bits represent the magnitude.

-5 = % 1 0000101 = $ 85+5 = % 0 0000101 = $ 05+127 = % 0 1111111 = $ 7F-127 = % 1 1111111 = $ FF+ 0 = % 0 0000000 = $ 00- 0 = % 1 0000000 = % 80

SM is Correct ?


8-bit number can represent the signed integers -127 to +127.

N-bit number can represent the signed integers

-(2(N-1) – 1) to + 2(N-1) - 1

Two problems for SM representation:

(1) Two ways of representing 0 (-0, and +0) .

(2) Addition of K + (-K) does not give Zero

-5 + 5 = $ 85 + $ 05 = $8A = -10

Complementary Number


Radix Complement(r’s Complement):

The radix complement [N]r of a number (N)r is defined as

[N]r = rn - (N)r ,

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

(r-1)’s Complement:

The (r-1)’s complement [N]r of a number (N)r is defined as

[N]r = rn – 1 - (N)r ,

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

1’s Complement Number


To encode a negative number, get the binary representation of its magnitude, then COMPLEMENT each bit.

Complementing each bit mean that 1s are replaced with 0s, 0s are replaced with 1s.

+5 = % 00000101 = $ 05-5 = % 11111010 = $ FA+127 = % 01111111 = $ 7F-127 = % 10000000 = $ 80+ 0 = % 00000000 = $ 00- 0 = % 11111111 = $ FF

1’s Complement is Correct ?




-(2(N-1) – 1) to + 2(N-1) - 1

One Problem for 1’s complement representation

(1) two ways of representing 0 (-0, and +0),

K + 0 = K only works if we use +0, does not work if we use -0.

5 + (+0) = $05 + $00 = $05 = 5 (ok)

5 + (-0) = $05 + $FF = $04 = 4 !!! (wrong)

2’s Complement Number


To encode a negative number, get the binary representation of its magnitude, COMPLEMENT each bit, then ADD 1.

(get Ones complement, then add 1).

+5 = % 00000101 = $ 05-5 = % 11111011 = $ FB+127 = % 01111111 = $ 7F-127 = % 10000001 = $ 81-128 = % 10000000 = $80

+ 0 = % 00000000 = $ 00- 0 = % 00000000 = $ 00

2’s Complement is Correct ?




-2(N-1) to + 2(N-1) - 1

There is only one zero, and K + (-K) = 0.

-5 + 5 = $ FB + $ 05 = $00 = 0

2’s complement is a correct representation of signed numbers.

Representation & Meaning


$FE as an 8 bit unsigned integer = 254$FE as an 8 bit signed magnitude integer = -126$FE as an 8 bit 1’s complement integer = -1$FE as an 8 bit 2’s complement integer = -2

$7F as an 8 bit unsigned integer = 127$7F as an 8 bit signed magnitude integer = +127$7F as an 8 bit 1’s complement integer = +127$7F as an 8 bit 2’s complement integer = +127

Integer & Representation


+34 as an 8 bit SM number is % 00100010 +34 as an 8 bit 1’s complement number is % 00100010 +34 as an 8 bit 2’s complement number is % 00100010

-20 as an 8 bit SM number is % 10010100 -20 as an 8 bit 1’s complement number is % 11101011 -20 as an 8 bit 2’s complement number is % 11101100

Integer Representation


Binary Number Sign-Magnitude 1’s Complement 2’s Complement% 0000 +0 +0 +0 +0% 0001 +1 +1 +1 +1% 0010 +2 +2 +2 +2% 0011 +3 +3 +3 +3% 0100 +4 +4 +4 +4% 0101 +5 +5 +5 +5% 0110 +6 +6 +6 +6% 0111 +7 +7 +7 +7

% 1000 +8 -0 -7 -8% 1001 +9 -1 -6 -7% 1010 +10 -2 -5 -6% 1011 +11 -3 -4 -5% 1100 +12 -4 -3 -4% 1101 +13 -5 -2 -3% 1110 +14 -6 -1 -2% 1111 +15 -7 -0 -1

Signed Overflow


Signed overflow occurs in

Add two POSITIVE numbers and get a NEGATIVE resultAdd two NEGATIVE numbers and get a POSITIVE result

$ FF = -1

+ $ 80 = -128--------

$ 7F = +127 (incorrect!!)

Excess(Biased) Representation


2’s Complement Excess-8+7 0111 1111+6 0110 1110+5 0101 1101+4 0100 1100+3 0011 1011+2 0010 1010+1 0001 1001+0 0000 1000

-1 1111 0111-2 1110 0110-3 1101 0101-4 1100 0100-5 1011 0011-6 1010 0010-7 1001 0001-8 1000 0000

Single Precision Floating Point


Numbers are represented by X = (-1)S x 2E-127 x 1.M

S = Sign bit

E = Biased integer, 0 ≤ E ≤ 255

M = Normalized fraction with hidden 1

Sign

32 bits

23 bits of mantissaexcess-127exponent

8-bit

S M E

+0 denotes–1 denotes

Underflow & Overflow


Positive Minimal Number : s = 0, E = 0, M = 0,Npmin = (-1)0 x 2-127 x 1.0 = 2-127

Positive Maximal Number : s = 0, E = 255, M = 11...11, Npmax = (-1)0 x 2128 x (2 - 2-23) = (1 - 2-24) x2129

Negative Minimal Number : s = 1, E = 0, M = 0,Npmin = (-1)1 x 2-127 x 1.0 = -2-127

Negative Maximal Number : s = 1, E = 255, M = 11...11, Npmax = (-1)1 x 2128 x (2 - 2-23) = -(1 - 2-24) x2129

Underflow = -2-127 ~ 2-127

Overflow = -∞ ~ -(1 - 2-24) x2129 and +(1 - 2-24) x2129 ~ +∞

Double Precision Floating Point


52 bits of mantissa11-bit excess-1023exponent

64 bits

Sign

S M E

Numbers are represented by X = (-1)S x 2E-1023 x 1.M

S = Sign bit

E = Biased integer, 0 ≤ E ≤ 2047

M = Normalized fraction with hidden 1

Underflow & Overflow


Positive Minimal Number : s = 0, E = 0, M = 0,Npmin = (-1)0 x 2-1023 x 1.0 = 2-1023

Positive Maximal Number : s = 0, E = 255, M = 11...11, Npmax = (-1)0 x 21024 x (2 - 2-52) = (1 - 2-53) x21025

Negative Minimal Number : s = 1, E = 0, M = 0,Npmin = (-1)1 x 2-1023 x 1.0 = -2-1023

Negative Maximal Number : s = 1, E = 255, M = 11...11, Npmax = (-1)1 x 21024 x (2 - 2-52) = -(1 - 2-53) x21025

Underflow = -2-1023 ~ 2-1023

Overflow = -∞ ~ -(1 - 2-53) x21025 and +(1 - 2-53) x21025 ~ +∞

Data Representation in C Language


char ---> 8 bit ASCII code

unsigned int ----> 16 bit unsigned integer

(signed) int ---> 16 bit 2’s complement integer

unsigned long int ---> 32 bit unsigned integer

(signed) long int ---> 32 bit 2’s complement integer

float ---> 32 bit single precision floating point number

double ---> 64 bit double precision floating point number

Error Detection Codes


Simple Parity Code

Even-Parity Code

Odd-Parity Code

2-out-of-5 Code

0 = % 000111 = % 001012 = % 010013 = % 100014 = % 001105 = % 010106 = % 100107 = % 011008 = % 101009 = % 11000

Parity bit

Information bits

Homework

• 1.1, 1.3, 1.4, 1.5, 1.10, 1.12, 1.14, 1.16, 1.18


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