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Introduction to Electrical Engineering, II

Instructor: Andrew B. Kahng (lecture)Email: abk@ece.ucsd.eduTelephone: 858-822-4884 office, 858-353-0550 cellOffice: 3802 AP&MLecture: TuThu 3:30pm – 4:50pm, HSS, Room 2250Discussion: Wed 6:00pm-6:50pm, Peterson Hall, 108

Class Website: http://vlsicad.ucsd.edu/courses/ece20b/wi03

Login: ece20bPassword: b02ece

ECE 20B, Winter 2003

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Purpose of CourseIntroduction to design of digital systems and computer hardwareBasic to CS, EE, CEMajor topics– Information representation and manipulation– Logic elements and Boolean algebra– Combinational Logic– Arithmetic Logic– Sequential Logic– Registers, Counters, Memories– Control

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AdministrationLab instructor: Prof. Mohan TrivediTextbook: Mano and Kime, 2nd edition (updated)– Goal: cover MK Chapters 1-5, (6), 8– Rizzoni (Sections 12.1-2) used only for op amps

Labs– This week: (1) show up and verify partner (2) if you need

a partner, talk to Prof. Trivedi– If you need to switch lab sections, go to undergrad office

Adding ECE 20B– Must get stamp from undergrad office in EBU I– Prerequisites rigidly enforced

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Course StructureHomework: assigned but not collected– All problems and solutions posted on web– Exams are based on homework problems– Do problems before looking at solutions!

Discussion: Wed 6:00-6:50pm, Peterson 108– Will typically go over the previous lectures and problems

Grading– 40%: 2 in-class midterms (Jan 30, Feb 25), – 40%: 1 final– 20%: un-announced in-class quizzes– Exams cover both lecture (~3/4) and lab (~1/4)

• For each lab, a set of prelab questions will be assigned. These must be included in your lab notebook.

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Course ConductResources– Email– Discussion session (TA)– Lab sessions (readers)– http://www.prenhall.com/mano/

Questions about course – see me– Broader consultation – see academic advisor

Academic misconduct: do not let this happen

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IntroductionAssigned reading: Chapters 1, 2 of MK (see website for specific sections)Homework: Check website for problems/solutionsToday– Concept of “digital”– Number systems

Next lecture– Binary logic– Boolean algebra

We will spend ~3 weeks going through the first 3 chapters of MK.

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Digital SystemTakes a set of discrete information inputs and discrete internal information (system state) and generates a set of discrete information outputs.

DiscreteInformationProcessingSystem

DiscreteInputs Discrete

Outputs

System State

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Types of SystemsWith no state present– Combinational logic system– Output = Function (Input)

With state present– State updated at discrete times (e.g., once per clock tick)

Synchronous sequential system– State updated at any time

Asynchronous sequential system

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Example: Digital Counter (e.g., Odometer)

1 30 0 5 6 4

Inputs: Count Up, ResetOutputs: Visual DisplayState: “Value” of stored digitsIs this system synchronous or asynchronous?

UPRESET

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Example: Digital Computer

Inputs: keyboard, mouse, modem, microphoneOutputs: CRT, LCD, modem, speakersIs this system synchronous or asynchronous?

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SignalsInformation variables mapped to physical quantitiesIn digital systems, the quantities take on discretevalues – Two-level, or binary, values are the most prevalent values

in digital systems– Binary values are represented abstractly by digits 0 and 1

Signal examples over time:Analog

Asynchronous

Synchronous

(Time)

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Physical Signal Example - Voltage

Threshold Region

Other physical signals representing 1 and 0– CPU Voltage– Disk Magnetic field direction– CD Surface pits / light– Dynamic RAM Charge

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Threshold in the News

Punched = 1Not punched = 0What about the rest?

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Number SystemsDecimal Numbers– What does 5,634 represent?– Expanding 5,634:

5 x 103 = 5,000+ 6 x 102 = 600+ 3 x 101 = 30+ 4 x 100 = 4 5,634

– What is “10” called in the above expansion?The radix.

– What is this type of number system called? Decimal.

– What are the digits for decimal numbers?0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

– What are the digits for radix-r numbers?0, 1, …, r-1.

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Powers of 2Noteworthy powers of 2:• 210 = kilo- = K • 220 = mega- = M• 230 = giga- = G• 240 = tera- = T• 250 = peta- = P

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General Base Conversion

“n” integer digits an-1,…,a0

(Number)r

= ( ) ( )ΣΣ +j = -1

j = -m

jj

ii = n-1

i = 0i rara

(Integer Portion) + (Fraction Portion)

Number Representation

an-1 an-2 an-3 … a2 a1 a0 . a-1 a-2 … a-m

Given a number of radix r of

and “m” fractional digits a-1,…,a-m

written as :

has value:

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Commonly Occurring BasesRadix DigitsName

Binary 2 0,1

Octal 8 0,1,2,3,4,5,6,7

Decimal 10 0,1,2,3,4,5,6,7,8,9

Hexadecimal 16 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

Decimal (Base 10)

Binary (Base 2)

Octal (Base 8)

Hexadecimal (Base 16)

00 00000 00 0001 00001 01 01

13 01101 15 0D14 01110 16 0E15 01111 17 0F16 10000 20 10

.

...

.

...

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Converting Binary to DecimalTo convert to decimal, use decimal arithmetic to sum the weighted powers of two

Converting 110102 to N10N10 = 1 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20

= 26

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Converting Decimal to BinaryMethod 1 (Method 2 – repeated division – next slide)– Subtract the largest power of 2 that gives a positive result and

record the power.– Repeat, subtracting from the prior result, until the remainder is

zero.– Place 1’s in the positions in the binary result corresponding to

the powers recorded; in all other positions place 0’s.Example: 62510 10011100012

• 625 – 512 = 113 9• 113 – 64 = 49 6• 49 – 32 = 17 5• 17 – 16 = 1 4• 1 – 1 = 0 1

– Place 1’s in the the positions recorded and 0’s elsewhereConverting binary to decimal: sum weighted powers of 2 using decimal arithmetic, e.g., 512 + 64 + 32 + 16 + 1 = 625

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Conversion Between BasesConvert the Integral Part– Repeatedly divide the number by the radix you want to convert to and

save the remainders. The new radix digits are the remainders in reverse order of computation.

Why does this work? This works because, the remainder left in the division is alwaysthe coefficient of the radix’s exponent.

• If the new radix is > 10, then convert all remainders > 10 to digits A, B, … Convert the Fractional Part– Repeatedly multiply the fraction by the radix and save the integer digits

that result. The new radix fraction digits are the integer numbers in computed order. Why does this work?To convert fractional part, it should be divided by reciprocal of radix, which is same as multiplying with radix.

• If the new radix is > 10, then convert all integer numbers > 10 to digits A, B, …

Join together with the radix point

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Example: Convert 46.687510 To Base 2Convert 46 to Base 2

46/2 = 23 remainder = 0 23/2 = 11 remainder = 1 11/2 = 5 remainder = 1 5/2 = 2 remainder = 1 2/2 = 1 remainder = 0 1/2 = 0 remainder = 1Read off in reverse order: 1011102

Convert 0.6875 to Base 2:0.6875 * 2 = 1.3750 int = 10.3750 * 2 = 0.7500 int = 00.7500 * 2 = 1.5000 int = 1 0.5000 * 2 = 1.0000 int = 10.0000Read off in forward order: 0.10112

Join together with the radix point: 1011110.10112

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Converting Among Octal, Hexadecimal, BinaryOctal (Hexadecimal) to Binary:– Restate the octal (hexadecimal) as three (four) binary digits,

starting at radix point and going both waysBinary to Octal (Hexadecimal):– Group the binary digits into three (four) bit groups starting at

the radix point and going both ways, padding with zeros as needed in the fractional part

– Convert each group of three (four) bits to an octal (hexadecimal) digit

Example: Octal to Binary to Hexadecimal 6 3 5 . 1 7 7 8

= 110|011|101 . 001|111|111 2= 1|1001|1101 . 0011|1111|1(000)2 (regrouping)= 1 9 D . 3 F 816 (converting)

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Non-numeric Binary CodesGiven n binary digits (called bits), a binary codeis a mapping from a subset of the 2n binary numbers to some set of represented elements.

Example: Abinary codefor the sevencolors of therainbow

Binary Number Color000 Red001 Orange010 Yellow011 Green100 (Not mapped)101 Blue110 Indigo111 Violet

Flexibility of representation: can assign binary code word to any numerical or non- numerical data as long as data uniquely encoded.

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Number of Bits RequiredGiven M elements to be represented by a binary code, the minimum number of bits, n, needed satisfies the following relationships:– 2n ≥ M > 2n – 1

– n = ceil(log2 M) where ceil(x) is the smallest integer greater than or equal to x

Example: How many bits are required to represent decimal digits with a binary code?– M = 10 n = 4

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Number of Elements RepresentedGiven n digits in radix r, there are rn distinct elements that can be represented.But, can represent m elements, m < rn

Examples:– Can represent 4 elements in radix r = 2 with n = 2 digits: (00, 01,

10, 11)– Can represent 4 elements in radix r = 2 with n = 4 digits: (0001,

0010, 0100, 1000)• This code is called a "one hot" code

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Binary Codes for Decimal DigitsThere are over 8,000 ways that you can chose 10 elements from the 16 binary numbers of 4 bits. A few are useful:

Decimal 8,4,2,1 Excess3 8,4,-2,-1 Gray0 0000 0011 0000 00001 0001 0100 0111 01002 0010 0101 0110 01013 0011 0110 0101 01114 0100 0111 0100 01105 0101 1000 1011 00106 0110 1001 1010 00117 0111 1010 1001 00018 1000 1011 1000 10019 1001 1100 1111 1000

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Binary Coded Decimal (BCD)The BCD code is the 8,4,2,1 code.This code is the simplest, most intuitive binary code for decimal digits and uses the same weights as a binary number, but only encodes the first ten values from 0 to 9.Example: 1001 (9) = 1000 (8) + 0001 (1)How many “invalid” code words are there?What are the “invalid” code words?

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Excess-3 Code and 8, 4, –2, –1 Code

111111009100010118100110107101010016101110005010001114010101103011001012011101001000000110

8, 4, –2, –1Excess-3Decimal

What property is common to these codes?– These are reflected codes; complementing is performed simply by

replacing 0’s by 1’s and vice-versa

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Gray Code

What property does this Gray code have?– Counting up or down changes only one bit at a time

(including counting between 9 and 0)

Decimal 8,4,2,1 Gray0 0000 00001 0001 01002 0010 01013 0011 01114 0100 01105 0101 00106 0110 00117 0111 00018 1000 10019 1001 1000

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Gray Code: Optical Shaft Encoder

000

001010

011

100

101 110

111

(a) Binary Code for Positions 0 through 7

000

001011

010

110

111 101

100

(b) Gray Code for Positions 0 through 7

B0B1

B2

G0

G1

G2

Shaft encoder: Capture angular position (e.g., compass)For binary code, what values can be read if the shaft position is at boundary of “3” and “4” (011 and 100) ?For Gray code, what values can be read ?

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Warning: Conversion or Coding?Do NOT mix up conversion of a decimal number to a binary number with coding a decimal number with a BINARY CODE.

1310 = 11012 (This is conversion)13 ⇔ 0001|0011 (This is coding)

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Binary ArithmeticSingle Bit Addition with CarryMultiple Bit AdditionSingle Bit Subtraction with BorrowMultiple Bit SubtractionMultiplicationBCD Addition

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Single Bit Binary Addition with CarryGiven two binary digits (X,Y), a carry in (Z) we get the following sum (S) and carry (C): Carry in (Z) of 0: Carry in (Z) of 1: Z 1 1 1 1

X 0 0 1 1+ Y + 0 + 1 + 0 + 1C S 0 1 1 0 1 0 1 1

Z 0 0 0 0X 0 0 1 1

+ Y + 0 + 1 + 0 + 1C S 0 0 0 1 0 1 1 0

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Multiple Binary AdditionExtending this to two multiple bit examples:Carries 00000 01100Augend 01100 10110 Addend +10001 +10111Sum 11101 101101Note: The 0 is the default Carry-In to the least significant bit.

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Single Bit Binary Subtraction with BorrowGiven two binary digits (X,Y), a borrow in (Z) we get the following difference (S) and borrow (B):Borrow in (Z) of 0:

Borrow in (Z) of 1:

Z 1 1 1 1X 0 0 1 1

- Y -0 -1 -0 -1BS 11 1 0 0 0 1 1

Z 0 0 0 0X 0 0 1 1

- Y -0 -1 -0 -1BS 0 0 1 1 0 1 0 0

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Multiple Bit Binary SubtractionExtending this to two multiple bit examples:Borrows 00000 00110Minuend 10110 10110 Subtrahend - 10010 - 10011Difference 00100 00011Notes: The 0 is a Borrow-In to the least significant bit. If the Subtrahend > the Minuend, interchange and append a – to the result.

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Binary MultiplicationThe binary multiplication table is simple:

0 ∗ 0 = 0 | 1 ∗ 0 = 0 | 0 ∗ 1 = 0 | 1 ∗ 1 = 1 Extending multiplication to multiple digits:

Multiplicand 1011Multiplier x 101Partial Products 1011 0000 - 1011 - -Product 110111

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Error-Detection CodesRedundancy (e.g. extra information), in the form of extra bits, can be incorporated into binary code words to detect and correct errors. A simple form of redundancy is parity, an extra bit appended onto the code word to make the number of 1’s odd or even. Parity can detect all single-bit errors and some multiple-bit errors.A code word has even parity if the number of 1’s in the code word is even.A code word has odd parity if the number of 1’s in the code word is odd.

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3-Bit Parity Code ExampleFill in the even and odd parity bits:

The binary codeword "1111" has even parity and the binary code "1110" has odd parity. Both could be used to represent data.

Even Parity Odd ParityMessage -Parity Message -Parity

000 - 000 -001 - 001 -010 - 010 -011 - 011 -100 - 100 -101 - 101 -110 - 110 -111 - 111 -

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ASCII Character CodesAmerican Standard Code for Information InterchangeThis code is the most popular code used to represent information sent as character-based data. It uses 7-bits to represent:– 94 Graphic printing characters.– 34 Non-printing characters

Some non-printing characters are used for text format (e.g. BS = Backspace, CR = carriage return)Other non-printing characters are used for record marking and flow control (e.g. STX and ETX start and end text areas).ASCII is a 7-bit code, but most computers manipulate 8-bit quantity called byte. To detect errors, the 1st bit is used as a parity bit.E.g., ASCII ‘A’ = 1000001 (7 bits)

ASCII ‘A’ with parity bit = 01000001 (8 bits) (Note: even parity is used)