1 digital convergence! smart phone: it + telecommunication + consumer electronics + entertainment...

1

Digital Convergence!• Smart Phone: IT + Telecommunication + Consumer Electronics + Entertainment

• Analog vs Digital system Ex: Watch, LP vs CD, Camera

- Why Digital?- Representation? Continuous vs Discrete Signals (value, time)

continuous: discrete-time, analog: digital- Noise?, Accuracy?Resolution: # of digits used to represent the signals

• Digital circuits: process digital input signals and output digital signalsref. Analog to Digital Converter, D/A Converter

• Digital logic: fundamental theories and practices for designing the digital circuits

• Course Objectives:- Learn to analyze and design digital circuits; namely, combinational circuits and sequential circuits.

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Digital Logic

• Professor:김 재희 교수B619, [email protected]

• Teaching Assistant:고준범 선생

2123-4537 B618

• Homepage:cherup.yonsei.ac.kr

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Digital Logic• Text: Fundamentals of Digital Logic, by Charles H. Roth, Jr., 6th Edition

Reference: Computer Engineering: Hardware Design, by Mano more..

• Course Objective: 컴퓨터를 포함한 여러 디지탈 시스템의 구성 요소가 되는 , 다양한 디지탈 회로를 이해하고 설계할 수 있는 기법을 익힘 .

• Summary of Course:Boolean Algebra, Logic Gates, Design of Combinational Circuits, Multiplexers, Programmable Logic Devices, Latches and Flip-Flops, Registers and Counters, Design of Sequential CircuitsChap 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16

• Related Courses (H/W): Digital Electronics, Computer Architectures, Operating System Digital Logic Experiments, Digital System Design

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Digital Logic

• Evaluation: 수업계획서의 일정을 원칙으로 실시 , 조정 가능

Exams (3 hrs.) : 300 (mid), 400(final)Homework, etc: 300Total: 1000

• Communications: One minute quiz, 수업 전 / 후 면담

• Note:Lecture-schedules may be adaptively changed.

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One Minute Quiz

• Date: 2012. Name: Student ID #:• Department:

• Important topic(s) learned today:

• Topic(s) hard to understand:

• Any comments to the professor:

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CHAPTER 1

INTRODUCTIONNUMBER SYSTEMS AND CONVERSION

This chapter in the book includes:ObjectivesStudy Guide

1.1 Digital Systems and Switching Circuits1.2 Number Systems and Conversion1.3 Binary Arithmetic1.4 Representation of Negative Numbers1.5 Binary Codes

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Objectives

Topics introduced in this chapter:

• Difference between Analog and Digital Systems

• Difference between Combinational and Sequential Circuits

• Binary number and digital systems

• Number systems and Conversion

• Add, Subtract, Multiply, Divide Positive Binary Numbers

• 1’s Complement, 2’s Complement for Negative binary number

• BCD code, 6-3-1-1 code, excess-3 code, ASCII code etc

8

1.1 Digital Systems and Switching Circuits

• Digital systems: used in computation, data processing, control, communication, measurement, etc

- Accurate, Reliable, Integration• Analog – Continuous

- Natural Phenomena

(Pressure, Temperature, Speed…)

- Difficulty in realizing, processing using electronics

• Digital – Discrete

- Binary Digit Signal Processing as ‘bit’ unit

- Easy in realizing, processing using electronics

- High performance due to Integrated Circuit Technology

김재희

accurate: digital calculator vs analog (1 % error, etc), more accuracy may be obtained by adding more digitsreliable: digital multimeter vs analog

9

Binary Digit?

• Binary:- Two values(0, 1)

- Each digit is called as a “bit”

- Number representation with only two values (0,1)

- Can be implemented with simple electronics devices

(ex: Voltage High(1), Low(0) ; positive logic <-> negative

Switch On (1) Off(0)…)

Good things in Binary Number

김재희

bit: unit in entrophy, uncertainty coin, head or tail?-> 1 bit rain or not after 10 days -> less than 1 bit

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Switching Circuit

• Combinational Circuit :

outputs depend on only present inputs, not on past inputs

Boolean algebra in Unit 2 & 3 used to describe I/O relations.

*takes on discrete values

11

Sequential Circuit

Memory

CombinationalCircuit

Sequential Circuit: - outputs depend on both present inputs and past inputs - have “memory” function

12

Design Levels of Digital Systems

• System Design:Memory unit, ALU, I/O devices, etc(Computer System Architectures)

• Logic Design:Logic gates, F/F, etc for a specific function(Digital Logic)

• Circuit Design:Resistors, diodes, transistors to form a gate, F/F, or other logic building block(Digital Electronics)Refer to Appendix A

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1.2 Number Systems and Conversion

33

22

11

00

11

22

33

44

32101234

).(

RaRaRa

RaRaRaRaRa

aaaaaaaaN R

10

10128

375.1038

373264838784813.147

2101210 10810710310510978.953

10

2101232

75.114

311

4

1

2

11208

21212121202111.1011

Decimal:

Binary:

Radix(Base),R

to Decimal:

Example:

Any number system to Decimal

*基數

14


10012

16 260715322560161516216102 FAHexa-Decimal:

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

15


01

12

21

10121 )( aRaRaRaRaaaaaaN nn

nnRnn

0111

22

11 remainder , aQaRaRaRa

R

N nn

nn

1221

33

121 remainder , aQaRaRaRa

R

Q nn

nn

2334

132 remainder , aQaRaRa

R

Q nn

nn

Decimal Number N to Base-R: integer part

Continue until to get an. Thus divide N by R until we have Q = 0, then

we will get (anan-1 … a2a1a0)R.

16


532

262

132

62

32

12

rem. = 1 = a0

rem. = 0 = a1

rem. = 1 = a2

rem. = 0 = a3

rem. = 1 = a4

0 rem. = 1 = a5

210 11010153

Example: Decimal to Binary Conversion: From 5310 to base 2?

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1 2 31 2 3 1 2 3(0. ) m

m R mF a a a a a R a R a R a R

1112

31

21 FaRaRaRaaFR mm

2221

321 FaRaRaaRF mm

333

32 FaRaaRF mm

)1(

250.1

2

625.

1

a

F

)0(

500.0

2

250.

2

1

a

F

)1(

000.1

2

500.

3

2

a

F

210 101.625.

Conversion of a decimal fraction F to Base-R

Example: From .625 to base 2

Thus multiplying F by R continuously, we will get (.a-1a-2…a-m)R.

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8).0(

2

4).0(

2

2).1(

2

6).1(

2

8).0(

2

4).1(

2

7.

Process starts repeating here because .4 was previouslyobtained

10 20.7 0.1 0110 0110 0110

Example: Convert 0.7 to binary

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24 10

3231.3 2 4 3 4 1 45.75

4

457

67

0 rem.6

rem.3

75).1(

7

25).5(

7

75).1(

7

25).5(

7

75.

710 5151.6375.45

Example: Convert (231.3)4 to

base-7

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Conversion of Binary to Octal, Hexa-decimal

• (101 011 010 111 )2

= ( 5 3 2 7 )8, octal

• (1010 1111 0010 0101)2

= ( A F 2 9 )16, Hexadecimal

{ { { {2 1654

1001101.010111 0100 1101 . 0101 1100 4 .5CD

D C

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1.3 Binary Arithmetic

01 1

10 1

11 0

00 0

and carry 1 to the next higher column

10

10

10

1 1 1 1

13 1 1 0 1

11 1 0 1 1

1 1 0 0 0 24

carries

Addition

Example:

22


01 1

10 1

11 0

00 0

and borrow 1 from the next higher column

1010

00111

11101

1

(indicatesa borrowFrom the3rd column)

1101

11

10000

11 1 1

borrows

101110

0111

111001

11 1

borrows

Subtraction

Example:

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*1.3 Binary Arithmetic

187

18

205

column 2 column 1

187]107 108 101[

]108 101) [

]101510)1010(10)12[(

]108 101 [

]10)510(10)10(102[

]108101 [

]105100102[18205

012

01

012

01

012

01

012

note borrow from column 1

note borrow from column 2

Subtraction Example with Decimal;

* borrow 1 from column n means subtract 1 from n and add 10 to column n-1

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111

001

010

000

1014310001111

1101

0000

1101

1101

1011

1101

1111

1101

1111

0000

(01111)

1111

(1001011)

1111

11000011

multiplicand

multiplier

first partial product

second partial product

sum of first two partial products

third partial product

sum after adding third partial product

fourth partial product

final product (sum after adding fourth partial

product)

Multiplication Multiply: (13 x11)10

*For easy hardwareimplementation,shifting andpartial productsare needed

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10

1011

1101

1011

1110

1011

10010001 1011

1101

The quotient is 1101 with a remainderof 10.

Division

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1.4 Representation of Negative Numbers: Integer

(a) Unsigned number: for positive only

(b) Signed number: for positive and negative both i) Sign and Magnitude

ii) 2’s Complementiii) 1’s Complement

* only negative numbers are represented differently

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1.4 Representation of Negative Numbers: Integerbn 1– b1 b0

Magnitude

MSB

(a) Unsigned number: for positive only

bn 1– b1 b0

MagnitudeSign

(b) Signed number: for positive and negative both i) Sign and Magnitude

bn 2–

0 denotes1 denotes

+

–MSB

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• for 4 bits,+0: 0000, +1: 0001, .. , +7: 0111

-0: 1000, -1: 1001, .. , -7: 1111• for n bits,

2n-1 positive numbers, 2n-1 negative numbers

2 zeros: 000..0 and 100..0

* Not good for arithmetic operations

1.4 Representation of Negative Numbers: Integer

i) Sign and Magnitude

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[1’s and 2’s complement are used for subtraction by adder]

NN n 2*

ii) 2’s Complement Representation of N for Negative Numbers

n: number of bits

Ex:

N=1100, n=6

Thus N* = 26 – 1100 = 1000000 -) 001100

-------------- 110100

as they arelast non zero, subtract from r (=2)subtract from r -1 (=1)

30

Signed Binary Integers (word size n=4)

1111

1110

1101

1100

1011

1010

1001

1000

-

- , 1)*

1111

1110

1101

1100

1011

1010

1001

1000

1000

1001

1010

1011

1100

1101

1110

1111

-

-0

-1

-2

-3

-4

-5

-6

-7

-8

0000

0001

0010

0011

0100

0101

0110

0111

+0

+1

+2

+3

+4

+5

+6

+7

1’s complement 2’s complement N*

Sign and

magnitude

Negative integers

-N

Positive

integers

(all systems)+N N

1)* There is no – 0, but + 0 is 000..0. So only 1 zero in 2’s complement.

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1.4 Representation of Negative Numbers

2 1 111111

001100

110011

n

N

N

11)12(2* NNNN nn

iii) 1’s complement representation for Negative Numbers

NN n )12(

Example:

== 2’s complement: 1’s complement + ‘1’

Thus 1’s complement of N isobtained by complementingeach bit of N.

So 2’s complement of N is obtained by adding1 to the 1’s complementof N.

32


2 * = 2 - (2 ) N

(2 1) (2 1) ((2 1) )

n n n

n n n

N N

N N N

Complement of N’s complement is N.In other words, magnitude of a negativenumber represented by a complement can be obtained by complementing it.

Ex: 0 0101 : +5, its 2’ complement is 100000 – 00101 = 11011 1 1011 : -5, its 2’ complement is 100000 – 11011 = 00101

Thus, Complement is not just to represent negative of positive numbers, but to represent negative of negative numbers.

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2’s complement of N

• Subtract from 2n

• Add 1 to 1’s complement of N• Starting from the right-most bit,

change the first 1 to 0 and complement all the left-side remaining.

for N=010011000, 1’s com.->101100111 2’s com.->101101000

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1.4 Representation of Negative Number

7

4

3

0111

0100

0011

(correct answer)

6

5

1011

0110

0101

wrong answer because of overflow (+11 requires5 bits including sign)

6

5

1111

1010

0101

(correct answer)

Case 1

Case 2

Case 3

Addition of 2’s complement Numbers

* Addition is carried out just as if all the numbers were positive, and any carry fromthe sign position is ignored. Subtraction is done by taking the complement of the Minuend. Overflow occurs when correct sum requires one more bit.

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7

4

3

1001)1(

1100

1101

correct answer when the last carry is ignored(this is not an overflow)

6

5

0101)1(

1010

1011

wrong answer because of overflow (-11 requires 5 bits including sign)

Case 5

Case 6


6

5

0001)1(

0110

1011

correct answer when the carry from the sign bitis ignored (this is not an overflow)

Case 4

Discard the carryCare about overflow

36

Detection of Overflow in 2’s

^ ^ : Overflow = carry Cn EOR Cn-1+5 0 1 0 1 -5 1 0 1 1+6 0 1 1 0 -6 1 0 1 0--------------------- -------------------+11 1 0 1 1 -11 (1) 0 1 0 1

1. Overflow occurs when the sign of sum is different from the sign of the two positive or negative numbers.

2. Overflow can be also detected by observing the carry bits:

=> Overflow = (Carry at Cn) EOR (Carry at Cn-1)

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1

6

5

1110

1001

0101

(correct answer)

(end-around carry)(correct answer, no overflow)

6

5

0001

1

0000 )1(

0110

1010

Case 3

Case 4


Instead of discarding the last carry, carry is added to the n-bit sum. => end-around carry

38


(end-around carry)(wrong answer because of overflow)

6

5

0100

1

0011 )1(

1001

1010Case 6


(end-around carry)(correct answer, no overflow)

3

4

1000

1

0111 )1(

1011

1100Case 5

39

Addition of end-carry

2’s com. : B – A => B + A* = B+ (2n – A) = 2n + (B – A) = 2n – (A - B) = (A – B)* Thus, when there is a carry 2n, just ignore the carry (2n) and the result becomes (B – A), correctly. Otherwise if there is no carry, result is 2’s complement of (A - B) which is - (A - B) = (B-A) It occurs when A > B.

1’s com. : B – A => B + A’ = B + (2n – 1 – A) = 2n + (B – A) – 1 = 2n – 1 – (A - B) = (A – B)’

Thus when there is a carry, we want to take (2n – 1) from the result so that it becomes (B – A). To take the carry means we take 2n.

So, we want to add 1 (end-carry) to the result.If there is no carry, result is (A – B) represented by

1’s complement form. It occurs when A >= B.

40


)20(

)11(

11110100

11101011

(1) 11011111

1

11100000 31 (end-around carry)

11

19

)8(

00010110)1(

00010011

11110001

(end-around carry)



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1.5 Binary Codes

5 2 . 7 3 9

0101 0010 . 0111 0011 1001

Decimal

Digit

8-4-2-1

Code

(BCD)

6-3-1-1

Code

Excees-3

Code

2-out-of-5

Code

Gray

Code

0

1

2

3

4

5

6

7

8

9

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

0000

0001

0011

0100

0101

0111

1000

1001

1011

1100

0011

0100

0101

0110

0111

1000

1001

1010

1011

1100

00011

00101

00110

01001

01010

01100

10001

10010

10100

11000

0000

0001

0011

0010

0110

1110

1010

1011

1001

1000

*To store and process decimal numbers in a computer, various types of coding arepossible.

42

Binary Codes for Decimal Digits

• 8-4-2-1 & 6-3-1-1 are weighted.• Excess-3 is self-complemented.• 2-out-of-5 is non-weighted and error

checked for only two bits are 1’s.• Gray code is for minimal change of

‘1’ bits to increase and also used for error correction.

43

1.5 Binary Codes

00112233 awawawawN

811110316 N

1010011 1110100 1100001 1110010 1110100

S t a r t

6-3-1-1 Code:

ASCII Code

Graphic Characters: 94 as in Table 1-3.Control Characters: 34 including ESC, DEL, SP, BS, etc.

Table 1-6 in Ref. Mano

44

Homework

• HW. Unit 1: 4, 7, 8. 13, 15, 25, 27

1 digital convergence! smart phone: it + telecommunication + consumer electronics + entertainment...

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