Chap. 2 Number Systems and Codes

1. Chap. 2 Number Systems and Codes • Chapter Outcomes (Objectives) • Convert a number from one number system (decimal, binary, hexadecimal) to its equivalent in one of the other number systems. • Cite the advantages of the hexadecimal number system. • Count in hexadecimal. • Represent decimal numbers using the BCD code; cite the pros and cons of using BCD. • Explain the difference between BCD and straight binary. • Explain the purpose of alphanumeric codes such as the ASCII code. • Explain the parity method for error detection. • Determine the parity bit to be attached to a digital data string.

2. Chap. 2 Number Systems and Codes • Data Representation in Digital System (Data Types) • Numbers used in arithmetic computations • Letters of the alphabet used in data processing (ASCII Code) • Other discrete symbols used for specific purpose • 위의 Number 와 Letter 이외 모두 예) control word, gray code, error detection code, … 01 0 0 00 0 1 65? A? control word? BCD code?

3. 2-1 Binary to Decimal Conversions • Binary-to-Decimal Conversions • 1. Weighted Sum method 1011.1012 = (1 x 23) + (0 x 22)+ (1 x 21) + (1 x 2o) + (1 x 2-1) + (0 x 2-2) + (1 x 2-3) = 810+ 0 + 210 + 110 + 0.510 + 0 + 0.12510 = 11.62510 • 2. Double Dabble method

4. 2-2 Decimal to Binary Conversions • Decimal-to-Binary Conversions • Method 1 • Decimal number is simply expressed as a sum of powers of 2 4510 = 3210 + 810 + 410 + 110 = 25 + 0 + 23 + 22 + 0 + 2o = 1 0 1 1 0 12 • Method 2 : Exam. 2-1 • Repeated division(See Fig. 2-1) 37 / 2 = 18 remainder 1 (binary number will end with 1) : LSB 18 / 2 = 9 remainder 0 9 / 2 = 4 remainder 1 4 / 2 = 2 remainder 0 2 / 2 = 1 remainder 0 1 / 2 = 0 remainder 1 (binary number will start with 1) : MSB Read the result upward to give an answer of 3710 = 1001012 • Counting Range(Using N bits) • 0 to 2N-1 value range, 2N different values • Exam. 2-2 • 소수점 변환 • 0.375x 2 = 0.750 integer 0 MSB • 0.750 x 2 = 1.500 integer 1 . • 0.500 x 2 = 1.000 integer 1 LSB • Read the result downward .37510 = .0112

5. 2-3Hexadecimal Number System • Hex-to-Decimal Conversion 2AF16 = (2 x 162) + (10 x 161) + (15 x 16o) = 51210 + 16010 + 1510 = 68710 • Decimal-to-Hex Conversion : Exam. 2-3 42310 / 16 = 26 remainder 7 (Hex number will end with 7) : LSB 2610 / 16 = 1 remainder 10 110 / 16 = 0 remainder 1 (Hex number will start with 1) : MSB Read the result upward to give an answer of 42310 = 1A716 • Hex-to-Binary Conversion 9F216 = 9 F 2 = 1001 1111 0010 = 1001111100102 • Counting in Hexadecimal • Once a digit position reaches the value F, it is reset to 0, and the next digit position is incremented • With N hex digit : decimal 0 to 16N -1, 16N different values Table 2-1 Hex Binary Decimal 0 0000 0 1 0001 1 2 0010 2 3 0011 3 4 0100 4 50101 5 6 0110 6 7 0111 7 8 1000 8 9 1001 9 A 1010 10 B 1011 11 C 1100 12 D 1101 13 E 1110 14 F 1111 15 • Binary-to-Hex Conversion 1 1 1 0 1 0 0 1 1 02 = 0 0 1 1 1 0 1 0 0 1 1 0 3 A 6 = 3A616

6. Usefulness of Hex and Octal • To illustrate the advantage of hex/octal representation of a binary string • Printout the contents of 50 memory locations, and check it against a list • Each memory location is 16-bit number • Which one would you be more apt to read incorrectly? • Check 50 numbers like 010111001100111 or 6E67 • Digital circuits all work in binary. Hex and Octal are simply used as a convenience for the humans involved • Exam. 2-4 : Dec 378 → Hex 17A→ Bin 0000 0001 0111 1010 (16bit), and Exam. 2-5 • Summary of Conversions • Decimal  Binary, Octal, Hex • 1. Bin, Oct, Hex  Dec : Weighted sum or Double dabble • 2. Dec Bin, Oct, Hex : Repeated divide by 2, 8, 16 • Binary  Octal, Hex • 3. Bin  Oct, Hex : Group the bits in 3(Oct) or 4(Hex), and convert each group into the correct Octal or Hex digit. • 4. Oct, Hex Bin : Convert each digit into its 3 or 4 bit equivalent • Oct  Hex :  Oct  Bin, Bin  Hex

7. Octal Number System • Octal Number System • Composed of 8 symbols or numerals : 0, 1, 2, 3, 4, 5, 6, 7, • Octal to Decimal Conversion • ( 83) + ( 82)+ ( 81) + (8o) • (8-1) + ( 8-2) + ( 8-3) • Weighted system : octal point • Decimal to Octal Conversion : repeated division by 8 • Octal to Binary Conversion • Binary to Octal Conversion • Counting in Octal : 000 ~ 777 • Decimal to Binary Conversion • 1. By first converting to octal • 2. Decimal to Octal : 17710→ 2618 • 3. Octal to Binary : 2618→ 010 110 0012

8. 2-4BCD Code • Code • Special group of symbols(Number, letter, words) are being encoded • Straight binary coding • A decimal number is represented by its equivalent binary number • Binary-Coded-Decimal Code • Each digit of a decimal number is represented by its binary equivalent 8 7 4 (Decimal) 1000 0111 0100 (BCD) • only the four bit binary numbers from 0000 through 1001 are used • Comparison of BCD and Binary • 13710 = 100010012 (Binary) - require only 8 bits : straight binary coding • 13710 = 0001 0011 0111BCD(BCD) - require 12 bits : BCD coding • Exam. 2-6 : BCD to Dec. & Exam. 2-7 : forbidden code 1100 (12) • Exam. 2-8 : ATM ( the amount of cash you wish to withdraw in decimal ) • Exam. 2-9 : Cell phone ( 10 decimal-digit phone number input )

9. 2-5The Gray Code • Gray Code : to represent a sequence of numbers • when the three-bit binary number for 3 changes to 4, all three bits must change state : 011 → 100 • reduce the likelihood of a digital circuit misinterpreting a changing input • used in applications where numbers change rapidly • only one bit changes between two successive numbers : Tab. 2-2 Tab. 2-2 three-bit binary and Gray code equivalents BinaryGray Code B2 B1 B0 G2 G1 G0 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 1 1 1 1 1 0 1 0 1 1 1 1 1 0 0

10. Conversion from Binary to Gray code : Fig. 2-2(a) • 1. Binary MSB (B2) is used as Gray MSB (G2) : G2 • 2. Compare B2 (MSB) with B1 : G1 • Same: G1= 0 • Different: G1= 1 • 3. Compare B1 with B0 : G0 • Same: G0= 0 • Different: G0= 1 • Conversion from Gray code to Binary : Fig. 2-2(b) • 1. Gray MSB (G2) is used as Binary MSB (B2) : B2 • 2. Compare B2 (MSB) with G1 : B1 • Same: B1= 0 • Different: B1= 1 • 3. Compare B1 with G0 : B0 • Same: B0= 0 • Different: B0= 1 • An eight-position, three-bit shaft encoder : Fig. 2-3

11. Fig. 2-2 Converting (a)binary to Gray code and (b) Gray to binary Fig. 2-3 An eight-position, three-bit shaft encoder

12. Quadrature Encoders • The most common application of the Gray code • To determine which direction the shaft is rotating : speed or position • As the shaft rotates, this device produces a two-bit Gray code sequence • The two-bit Gray code from a quadrature shaft encoder : Tab. 2-3 • A mechanical contact quadrature encoder : Fig. 2-4 • Operation of a quadrature encoder : Fig. 2-5 Fig. 2-5 Operation of a quadrature encoder

13. 2-6 Putting it al together • Putting it all together : Table 2-4

14. 2.7 Byte, Nibble, and Word • 1 byte • A string of eight bits • 1 Byte = 8 Bits • 1 nibble = 4 bits • 1 word = size depends on data pathway size. • Word size in a simple system may be one byte (8 bits) • Word size in a PC is eight bytes (64 bits) • Exam. 2-10 ~ Exam. 2-14

15. 2-8 Alphanumeric Codes2-9 Parity Method for Error Detection Transmitter Receiver  ~ ~ • Alphanumeric Codes • In addition to numerical data, computer should recognize alphabet letters, punctuation marks, and other special characters • ASCII Code : Table 2-5 • American Standard Code for Information Interchange(ASCII) • 7 bit code, 27 = 128 possible codes • Exam. 2-15 ~ Exam. 2-16 • Parity Method for Error Detection • Binary Data Communication Examples • The transmission of digitized voice over a microwave link • The data retrieval from external memory devices(magnetic tape/disk) • The information transmission from a computer to a remote terminal by modem • Electrical noise : spurious fluctuations in voltage/current Fig. 2-6 Noise causing an error in the transmission of digital data

16. Parity method • One of the simplest and most widely used schemes for error detection • Checksum, CRC(cyclic redundant check), ECC(error correct code) : 데이터 통신 • Parity Bit • An extra bit attached to a code group • Even-parity method • The value of the parity bit is chosen so that the total number of 1s (including the parity bit) is aneven number 1 1 0 0 0 0 1 1 • Odd-parity method • Exactly the same way except that the total number of 1s is an oddnumber 1 1 0 0 0 0 0 1 • Error Detection • Fig. 4-25(p.189) • Can not tell which bit in error • Can detect only single bit error • Exam. 2-17 2-10. Applications : App). 2-1, 2-2, 2-3, 2-4, 2-5, 2-6 Added parity bit Added parity bit 1 1 0 0 0 0 1 1 “C”1 1 0 0 0 0 1 0 ”B” (Even-parity Generator) (Even-parity Checker)

17. Application 2-5 • ASCII Code : #→ 23 →1010 0011 • MSB D7 D6 D5 D4 D3 D2 D1 D0 LSB 1 0 1 0 0 0 1 1 • LSB D0 D1 D2 D3 D4 D5 D6 D7 MSB 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1