ECEN 248: INTRODUCTION TO DIGITAL DESIGN Lecture Set A Dr. S.G.Choi Dept. of Electrical and Computer Engineering.

ECEN 248: INTRODUCTION TO DIGITAL DESIGN

Lecture Set A

Dr. S.G.Choi

Dept. of Electrical and Computer Engineering

Instructor:

Office 333G WERC Office Hours MWF 10-11 AM Email: gchoi@ece.tamu.edu Lab Page: http://

people.tamu.edu/~rajballavdash/ecen_labpage.html

Required textbook:

Required Textbook: "Fundamentals of Digital Logic Design" by Brown and Vranesic. VERILOG VERSION

Supplemental texts: "Digital Design: Principles and Practices" by John Wakerly. "Contemporary Logic Design" by Randy Katz.

Course info

Mailing list: Emails will be sent periodically to neo accounts Announcements:

Lecture cancellations Deadline extension Updates, etc.

Course website http://ece.tamu.edu/~gchoi/248.htm All slides, labs, assignments, etc.

Grading Policy:

Assignments and Labs 25% 3 Hour Exams

20% for 1st exam 25% for 2nd

30% for 3rd No Final Exam Quizzes:

Each quiz 1% of the final Taken into account only if it improves the final grade Can improve the grade, but no extra points

Final grading may or may not be curved up

Quizzes:

3-6 quizzes Multiple choice questions 10-15 minutes Each quiz 1% of the final No quiz makeups

Grading scale

A standard grading scale will be utilized. The tentative grading scale for the course is: 

A       90-100% B         80-89% C        70-79% D         60-69% F        Below 59%

Course Goals

Study methods for Representation, manipulation, and optimization for

both combinatorial and sequential logic Solving digital design problems Study HDL description language (verilog)

Topics

Number bases Logic gates and Boolean Algebra Gate –label minimization Combinational Logic Sequential logic (Latches, Flip-flops,

Registers, and Counters) Memory and Programmable Logic HDL language (Verilog)

Attention over time!t

Attention over time!

~5min

t

Peer Instruction

Increase real-time learning in lecture, test understanding of concepts vs. details

Complete a “segment” with multiple choice questions 1-2 minutes to decide yourself 3 minutes in pairs/triples to reach consensus.

Teach others! 5-7 minute discussion of answers, questions,

clarifications

The Evolution of Computer Hardware

When was the first transistor invented?

Modern-day electronics began with the invention in 1947 of the transfer resistor - the bi-polar transistor - by Bardeen et.al at Bell Laboratories

The Evolution of Computer Hardware

When was the first IC (integrated circuit) invented? In 1958 the IC was born when Jack Kilby at Texas

Instruments successfully interconnected, by hand, several transistors, resistors and capacitors on a single substrate

The Underlying Technologies

Year Technology Relative Perf./Unit Cost

1951 Vacuum Tube 1

1965 Transistor 35

1975 Integrated Circuit (IC) 900

1995 Very Large Scale IC (VLSI) 2,400,000

2005 Ultra VLSI 6,200,000,000

The PowerPC 750

Introduced in 1999

3.65M transistors 366 MHz clock

rate 40 mm2 die size 250nm

technology

ECEN 248

Layers of abstraction

I/O systemProcessor

CompilerOperatingSystem(Mac OSX)

Application (ex: browser)

Digital DesignCircuit Design

Instruction Set Architecture

Datapath & Control

transistors

MemoryHardware

Software Assembler

Year

Tra

nsis

tors

1000

10000

100000

1000000

10000000

100000000

1970 1975 1980 1985 1990 1995 2000

i80386

i4004

i8080

Pentium

i80486

i80286

i8086

Technology Trends: Microprocessor Complexity

2X transistors/ChipEvery 1.5 years

Called “Moore’s Law”

Alpha 21264: 15 millionPentium Pro: 5.5 millionPowerPC 620: 6.9 millionAlpha 21164: 9.3 millionSparc Ultra: 5.2 million

Moore’s Law

Athlon (K7): 22 Million

Itanium 2: 41 Million

DIGITAL SYSTEMS

Overview

Logic functions with 1’s and 0’s. Building digital circuitry.

Truth tables. Logic symbols and waveforms. Boolean algebra. Properties of Boolean Algebra

Reducing functions. Transforming functions.

Digital Systems

Analysis problem:

Determine binary outputs for each combination of inputs

Design problem: given a task, develop a circuit that accomplishes the task

Many possible implementations. Try to develop “best” circuit based on some criterion (size,

power, performance, etc.)

··

··

LogicCircuitInputs Outputs

Toll Booth Controller

Consider the design of a toll booth controller. Inputs: quarter, car sensor. Outputs: gate lift signal, gate close signal

If driver pitches in quarter, raise gate. When car has cleared gate, close gate.

LogicCircuit

$·25

Car?

Raise gate

Close gate

Describing Circuit Functionality: Inverter

Basic logic functions have symbols. The same functionality can be represented with truth

tables· Truth table completely specifies outputs for all input

combinations. The above circuit is an inverter.

An input of 0 is inverted to a 1. An input of 1 is inverted to a 0.

A Y

0 1

1 0

Input Output

A Y

Symbol

Truth Table

The AND Gate

This is an AND gate. So, if the two inputs signals

are asserted (high) the output will also be asserted.Otherwise, the output willbe deasserted (low).

A B Y

0 0 0

0 1 0

1 0 0

1 1 1

A

BY

Truth Table

The OR Gate

This is an OR gate. So, if either of the two

input signals are asserted, or both of them are, the output will be asserted.

A B Y

0 0 0

0 1 1

1 0 1

1 1 1

AB

Y

Describing Circuit Functionality: Waveforms

Waveforms provide another approach for representing functionality.

Values are either high (logic 1) or low (logic 0). Can you create a truth table from the waveforms?

A B Y

0 0 0

0 1 0

1 0 0

1 1 1

AND Gate

Consider three-input gates

3 Input OR Gate

Ordering Boolean Functions

How to interpret A · B+C? Is it A · B ORed with C ? Is it A ANDed with B+C ?

Order of precedence for Boolean algebra: AND before OR.

Note that parentheses are needed here :

Boolean Algebra

A Boolean algebra is defined as a closed algebraic system containing a set K or two or more elements and the two operators, · and +·

Useful for identifying and minimizing circuit functionality

Identity elements a + 0 = a a · 1 = a

0 is the identity element for the + operation· 1 is the identity element for the · operation·

Commutativity and Associativity of the Operators

The Commutative Property:

For every a and b in K, a + b = b + a a · b = b · a

The Associative Property:

For every a, b, and c in K, a + (b + c) = (a + b) + c a · (b · c) = (a · b) · c

The Distributive Property

The Distributive Property:

For every a, b, and c in K, a + ( b · c ) = ( a + b ) · ( a + c ) a · ( b + c ) = ( a · b ) + ( a · c )

Distributivity of the Operators and Complements

The Existence of the Complement:

For every a in K there exists a unique element called a’ (complement of a) such that, a + a’ = 1 a · a’ = 0

To simplify notation, the · operator is frequently omitted. When two elements are written next to each other, the AND (·) operator is implied… a + b · c = ( a + b ) · ( a + c ) a + bc = ( a + b )( a + c )

Duality

The principle of duality is an important concept. This says that if an expression is valid in Boolean algebra, the dual of that expression is also valid.

To form the dual of an expression, replace all + operators with · operators, all · operators with + operators, all ones with zeros, and all zeros with ones.

Duality

The principle of duality is an important concept· This says that if an expression is valid in Boolean algebra, the dual of that expression is also valid·

To form the dual of an expression, replace all + operators with · operators, all · operators with + operators, all ones with zeros, and all zeros with ones·

Form the dual of the expressiona + (bc) = (a + b)(a + c)

Following the replacement rules…a(b + c) = ab + ac

Take care not to alter the location of the parentheses if they are present·

Duality

The principle of duality is an important concept· This says that if an expression is valid in Boolean algebra, the dual of that expression is also valid·

To form the dual of an expression, replace all + operators with · operators, all · operators with + operators, all ones with zeros, and all zeros with ones·

Form the dual of the expressiona + (bc) = (a + b)(a + c)

Following the replacement rules…a(b + c) = ab + ac

Take care not to alter the location of the parentheses if they are present·

Involution

This theorem states:

a’’ = a Remember that aa’ = 0 and a+a’=1.

Therefore, a’ is the complement of a and a is also the complement of a’.

As the complement of a’ is unique, it follows that a’’=a·

Taking the double inverse of a value will give the initial value.

Absorption

This theorem states:

a + ab = a a(a+b) = a To prove the first half of this theorem:

a + ab = a · 1 + ab

= a (1 + b)

= a (b + 1)

= a (1)

a + ab = a

DeMorgan’s Theorem

A key theorem in simplifying Boolean algebra expression is DeMorgan’s Theorem. It states:

(a + b)’ = a’b’ (ab)’ = a’ + b’

Complement the expression

a(b + z(x + a’)) and simplify.

(a(b+z(x + a’)))’ = a’ + (b + z(x + a’))’= a’ + b’(z(x + a’))’= a’ + b’(z’ + (x + a’)’)= a’ + b’(z’ + x’a’’)= a’ + b’(z’ + x’a)

Postulates and Basic Theorems

Summary

Basic logic functions can be made from AND, OR, and NOT (invert) functions.

The behavior of digital circuits can be represented with waveforms, truth tables, or symbols.

Primitive gates can be combined to form larger circuits Boolean algebra defines how binary variables can be

combined. Rules for associativity, commutativity, and distribution are

similar to algebra. DeMorgan’s rules are important.

Will allow us to reduce circuit sizes.

NUMBER SYSTEMS

Overview

Understanding decimal numbers Binary and octal numbers

The basis of computers! Conversion between different number systems

Digital Computer Systems

Digital systems consider discrete amounts of data. Examples

26 letters in the alphabet 10 decimal digits

Larger quantities can be built from discrete values: Words made of letters Numbers made of decimal digits (e.g. 239875.32)

Computers operate on binary values (0 and 1) Easy to represent binary values electrically

Voltages and currents. Can be implemented using circuits Create the building blocks of modern computers

Understanding Decimal Numbers

Decimal numbers are made of decimal digits: (0,1,2,3,4,5,6,7,8,9)

Number representation: 8653 = 8x103 + 6x102 + 5x101 + 3x100

What about fractions? 97654.35 = 9x104 + 7x103 + 6x102 + 5x101 + 4x100 + 3x10-

1 + 5x10-2

Informal notation -> (97654.35)10

Understanding Octal Numbers

Octal numbers are made of octal digits: (0,1,2,3,4,5,6,7)

Number representation: (4536)8 = 4x83 + 5x82 + 3x81 + 6x80 = (1362)10

What about fractions? (465.27)8 = 4x82 + 6x81 + 5x80 + 2x8-1 + 7x8-2

Octal numbers don’t use digits 8 or 9

Understanding Binary Numbers

Binary numbers are made of binary digits (bits): 0 and 1

Number representation: (1011)2 = 1x23 + 0x22 + 1x21 + 1x20 = (11)10

What about fractions? (110.10)2 = 1x22 + 1x21 + 0x20 + 1x2-1 + 0x2-2

Groups of eight bits are called a byte (11001001) 2

Groups of four bits are called a nibble. (1101) 2

Why Use Binary Numbers?

° Easy to represent 0 and 1 using electrical values.

° Possible to tolerate noise.

° Easy to transmit data

° Easy to build binary circuits.

AND Gate

1

00

Conversion Between Number Bases

Decimal(base 10)

Octal(base 8)

Binary(base 2)

Hexadecimal

(base16)

Convert an Integer from Decimal to Another Base

1. Divide decimal number by the base (e.g. 2)

2. The remainder is the lowest-order digit

3. Repeat first two steps until no divisor remains.

For each digit position:

Example for (13)10:

IntegerQuotient

13/2 = 6 1 a0 = 1 6/2 = 3 0 a1 = 0 3/2 = 1 1 a2 = 1 1/2 = 0 1 a3 = 1

Remainder Coefficient

Answer (13)10 = (a3 a2 a1 a0)2 = (1101)2

Convert an Fraction from Decimal to Another Base

1. Multiply decimal number by the base (e.g. 2)

2. The integer is the highest-order digit

3. Repeat first two steps until fraction becomes zero.

For each digit position:

Example for (0.625)10:

Integer

0.625 x 2 = 1 + 0.25 a-1 = 10.250 x 2 = 0 + 0.50 a-2 = 00.500 x 2 = 1 + 0 a-3 = 1

Fraction Coefficient

Answer (0.625)10 = (0.a-1 a-2 a-3 )2 = (0.101)2

The Growth of Binary Numbers

n 2n

0 20=1

1 21=2

2 22=4

3 23=8

4 24=16

5 25=32

6 26=64

7 27=128

n 2n

8 28=256

9 29=512

10 210=1024

11 211=2048

12 212=4096

20 220=1M

30 230=1G

40 240=1T

Mega

Giga

Tera

Kilo

Binary Addition

Binary addition is very simple. This is best shown in an example of adding two

binary numbers…

1 1 1 1 0 1+ 1 0 1 1 1---------------------

0

1

0

1

1

1111

1 1 00

carries

Binary Subtraction

° We can also perform subtraction (with borrows in place of carries).

° Let’s subtract (10111)2 from (1001101)2…

1 100 10 10 0 0 10

1 0 0 1 1 0 1- 1 0 1 1 1------------------------ 1 1 0 1 1 0

borrows

Binary Multiplication

Binary multiplication is much the same as decimal multiplication, except that the multiplication operations are much simpler…

1 0 1 1 1X 1 0 1 0----------------------- 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 1 1 1----------------------- 1 1 1 0 0 1 1 0

Convert an Integer from Decimal to Octal

1. Divide decimal number by the base (8)

2. The remainder is the lowest-order digit

3. Repeat first two steps until no divisor remains.

For each digit position:

Example for (175)10:

IntegerQuotient

175/8 = 21 7 a0 = 7 21/8 = 2 5 a1 = 5 2/8 = 0 2 a2 = 2

Remainder Coefficient

Answer (175)10 = (a2 a1 a0)2 = (257)8

Convert an Fraction from Decimal to Octal

1. Multiply decimal number by the base (e.g. 8)

2. The integer is the highest-order digit

3. Repeat first two steps until fraction becomes zero.

For each digit position:

Example for (0.3125)10:

Integer

0.3125 x 8 = 2 + 5 a-1 = 20.5000 x 8 = 4 + 0 a-2 = 4

Fraction Coefficient

Answer (0.3125)10 = (0.24)8

Summary

Binary numbers are made of binary digits (bits) Binary and octal number systems Conversion between number systems Addition, subtraction, and multiplication in binary

HEXADECIMAL NUMBERS

Overview

Hexadecimal numbers Related to binary and octal numbers

Conversion between hexadecimal, octal and binary Value ranges of numbers Representing positive and negative numbers Creating the complement of a number

Make a positive number negative (and vice versa) Why binary?

Understanding Hexadecimal Numbers

Hexadecimal numbers are made of 16 digits: (0,1,2,3,4,5,6,7,8,9,A, B, C, D, E, F)

Number representation: (3A9F)16 = 3x163 + 10x162 + 9x161 + 15x160 = 1499910

What about fractions? (2D3.5)16 = 2x162 + 13x161 + 3x160 + 5x16-1 = 723.312510

Note that each hexadecimal digit can be represented with four bits. (1110) 2 = (E)16

Groups of four bits are called a nibble. (1110) 2

Putting It All Together

° Binary, octal, and hexadecimal similar

° Easy to build circuits to operate on these representations

° Possible to convert between the three formats

Converting Between Base 16 and Base 2

° Conversion is easy!

Determine 4-bit value for each hex digit

° Note that there are 24 = 16 different values of four bits

° Easier to read and write in hexadecimal.

° Representations are equivalent!

3A9F16 = 0011 1010 1001 11112

3 A 9 F

Converting Between Base 16 and Base 8

1. Convert from Base 16 to Base 2

2. Regroup bits into groups of three starting from right

3. Ignore leading zeros

4. Each group of three bits forms an octal digit.

352378 = 011 101 010 011 1112

5 2 3 73

3A9F16 = 0011 1010 1001 11112

3 A 9 F

How To Represent Signed Numbers

• Plus and minus sign used for decimal numbers: 25 (or +25), -16, etc.

• For computers, desirable to represent everything as bits..

• Three types of signed binary number representations: signed magnitude, 1’s complement, 2’s complement.

• In each case: left-most bit indicates sign: positive (0) or negative (1).Consider signed magnitude:

000011002 = 1210

Sign bit Magnitude

100011002 = -1210

Sign bit Magnitude

One’s Complement Representation

• The one’s complement of a binary number involves inverting all bits.

• 1’s comp of 00110011 is 11001100

• 1’s comp of 10101010 is 01010101

• For an n bit number N the 1’s complement is (2n-1) – N.

• Called diminished radix complement by Mano since 1’s complement for base (radix 2).

• To find negative of 1’s complement number take the 1’s complement.

000011002 = 1210

Sign bit Magnitude

111100112 = -1210

Sign bit Magnitude

Two’s Complement Representation

• The two’s complement of a binary number involves inverting all bits and adding 1.

• 2’s comp of 00110011 is 11001101

• 2’s comp of 10101010 is 01010110

• For an n bit number N the 2’s complement is (2n-1) – N + 1.

• Called radix complement by Mano since 2’s complement for base (radix 2).

• To find negative of 2’s complement number take the 2’s complement.

000011002 = 1210

Sign bit Magnitude

111101002 = -1210

Sign bit Magnitude

Two’s Complement Shortcuts

Algorithm 1 – Simply complement each bit and then add 1 to the result. Finding the 2’s complement of (01100101)2 and of its 2’s

complement… N = 01100101 [N] = 10011011

10011010 01100100 + 1 + 1

--------------- --------------- 10011011 01100101

Algorithm 2 – Starting with the least significant bit, copy all of the bits up to and including the first 1 bit and then complementing the remaining bits. N = 0 1 1 0 0 1 0 1

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

Finite Number Representation

Machines that use 2’s complement arithmetic can represent integers in the range

-2n-1 <= N <= 2n-1-1where n is the number of bits available for representing N. Note that 2n-1-1 = (011..11)2 and –2n-1 = (100..00)2

o For 2’s complement more negative numbers than positive.

o For 1’s complement two representations for zero.o For an n bit number in base (radix) z there are zn

different unsigned values.(0, 1, …zn-1)

1’s Complement Addition

Using 1’s complement numbers, adding numbers is easy.

For example, suppose we wish to add +(1100)2 and +(0001)2.

Let’s compute (12)10 + (1)10. (12)10 = +(1100)2 = 011002 in 1’s comp. (1)10 = +(0001)2 = 000012 in 1’s comp.

0 1 1 0 0 + 0 0 0 0 1-------------- 0 0 1 1 0 1 0-------------- 0 1 1 0 1

Add carry

Final Result

Step 1: Add binary numbersStep 2: Add carry to low-order bit

Add

1’s Complement Subtraction

Using 1’s complement numbers, subtracting numbers is also easy.

For example, suppose we wish to subtract +(0001)2 from +(1100)2.

Let’s compute (12)10 - (1)10. (12)10 = +(1100)2 = 011002 in 1’s comp. (-1)10 = -(0001)2 = 111102 in 1’s comp.

1’s Complement Subtraction

0 1 1 0 0 - 0 0 0 0 1--------------

0 1 1 0 0 + 1 1 1 1 0-------------- 1 0 1 0 1 0 1-------------- 0 1 0 1 1

Add carry

Final Result

Step 1: Take 1’s complement of 2nd operandStep 2: Add binary numbersStep 3: Add carry to low order bit

1’s comp

Add

2’s Complement Addition

Using 2’s complement numbers, adding numbers is easy.

For example, suppose we wish to add +(1100)2 and +(0001)2.

Let’s compute (12)10 + (1)10. (12)10 = +(1100)2 = 011002 in 2’s comp. (1)10 = +(0001)2 = 000012 in 2’s comp.

2’s Complement Addition

0 1 1 0 0 + 0 0 0 0 1-------------- 0 0 1 1 0 1

FinalResult

Step 1: Add binary numbersStep 2: Ignore carry bit

Add

Ignore

2’s Complement Subtraction

Using 2’s complement numbers, follow steps for subtraction

For example, suppose we wish to subtract +(0001)2 from +(1100)2.

Let’s compute (12)10 - (1)10. (12)10 = +(1100)2 = 011002 in 2’s comp. (-1)10 = -(0001)2 = 111112 in 2’s comp.

2’s Complement Subtraction

0 1 1 0 0 - 0 0 0 0 1--------------

0 1 1 0 0 + 1 1 1 1 1-------------- 1 0 1 0 1 1

Final Result

Step 1: Take 2’s complement of 2nd operandStep 2: Add binary numbersStep 3: Ignore carry bit

2’s comp

Add

IgnoreCarry

2’s Complement Subtraction: Example #2

Let’s compute (13)10 – (5)10. (13)10 = +(1101)2 = (01101)2

(-5)10 = -(0101)2 = (11011)2

Adding these two 5-bit codes…

Discarding the carry bit, the sign bit is seen to be zero, indicating a correct result. Indeed,

(01000)2 = +(1000)2 = +(8)10.

0 1 1 0 1 + 1 1 0 1 1-------------- 1 0 1 0 0 0

carry

2’s Complement Subtraction: Example #3

Let’s compute (5)10 – (12)10. (-12)10 = -(1100)2 = (10100)2

(5)10 = +(0101)2 = (00101)2

Adding these two 5-bit codes…

Here, there is no carry bit and the sign bit is 1. This indicates a negative result, which is what we expect. (11001)2 = -(7)10.

0 0 1 0 1 + 1 0 1 0 0-------------- 1 1 0 0 1

Binary numbers

Binary numbers can also be represented in octal and hexadecimal

Easy to convert between binary, octal, and hexadecimal

Signed numbers represented in signed magnitude, 1’s complement, and 2’s complement

2’s complement most important (only 1 representation for zero).

Important to understand treatment of sign bit for 1’s and 2’s complement.

Binary Coded Decimal

Binary coded decimal (BCD) represents each decimal digit with four bits Ex· 0011 0010 1001 = 32910

This is NOT the same as 0011001010012

Why do this? Because people think in decimal.

Digit BCD Code

Digit BCD Code

0 0000 5 0101

1 0001 6 0110

2 0010 7 0111

3 0011 8 1000

4 0100 9 1001

3 2 9

Putting It All Together

° BCD not very efficient.

° Used in early computers (40s, 50s).

° Used to encode numbers for seven-segment displays.

° Easier to read?

Gray Code

Gray code is not a number system. It is an alternate way to represent

four bit data

Only one bit changes from one decimal digit to the next

Useful for reducing errors in communication.

Can be scaled to larger numbers.

Digit Binary Gray Code

0 0000 0000

1 0001 0001

2 0010 0011

3 0011 0010

4 0100 0110

5 0101 0111

6 0110 0101

7 0111 0100

8 1000 1100

9 1001 1101

10 1010 1111

11 1011 1110

12 1100 1010

13 1101 1011

14 1110 1001

15 1111 1000

ASCII Code

American Standard Code for Information Interchange

ASCII is a 7-bit code, frequently used with an 8th bit for error detection (more about that in a bit).Character ASCII (bin) ASCII (hex) Decimal Octal

A 1000001 41 65 101

B 1000010 42 66 102

C 1000011 43 67 103

…

Z

a

…

1

‘

ASCII Codes and Data Transmission

° ASCII Codes

° A – Z (26 codes), a – z (26 codes)

° 0-9 (10 codes), others (@#$%^&*…·)

° Complete listing in Mano text

° Transmission susceptible to noise.

° Typical transmission rates (1500 Kbps, 56.6 Kbps)

° How to keep data transmission accurate?

Parity Codes

Parity codes are formed by concatenating a parity bit, P to each code word of C.

In an odd-parity code, the parity bit is specified so that the total number of ones is odd.

In an even-parity code, the parity bit is specified so that the total number of ones is even.

Information BitsP

1 1 0 0 0 0 1 1

Added even parity bit

0 1 0 0 0 0 1 1

Added odd parity bit

Parity Code Example

Concatenate a parity bit to the ASCII code for the characters 0, X, and = to produce both odd-parity and even-parity codes.

Character ASCII Odd-Parity ASCII Even-Parity ASCII

0 0110000 10110000 00110000

X 1011000 01011000 11011000

= 0111100 10111100 00111100

Binary Data Storage

• Binary cells store individual bits of data

• Multiple cells form a register.

• Data in registers can indicate different values

• Hex (decimal)

• BCD

• ASCII

Binary Cell

0 0 1 0 1 0 1 1

Register Transfer

Data can move from register to register. Digital logic used to process data. We will learn to design this logic.

Register A Register B

Register C

Digital Logic Circuits

Transfer of Information

Data input at keyboard Shifted into place Stored in memory

NOTE: Data input in ASCII

Building a Computer

We need processing We need storage We need communication

You will learn to use and design these components.

Summary

Although 2’s complement most important, other number codes exist.

ASCII code used to represent characters (including those on the keyboard).

Registers store binary data. Next time: Building logic circuits!