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1. Binary System

1. Binary System. Binary Number. a 5 a 4 a 3 a 2 a 1 a 0 .a -1 a -2 a -3 = a n r n +a n-1 r n-1 +...+a 2 r 2 +a 1 r+a 0 +a -1 r -1 +a -2 r -2 +...+a -m r -m 7392 = 7 × 10 3 + 3 × 10 2 + 9 × 10 1 +2 × 10 0 (11010.11) 2 = 1×2 4 +1×2 3 +0×2 2 +1×2 1 +0×2 0 +1×2 -1 +1×2 -2

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1. Binary System

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  1. 1. Binary System

  2. Binary Number • a5a4a3a2a1a0.a-1a-2a-3 = anrn+an-1rn-1+...+a2r2+a1r+a0+a-1r-1+a-2r-2+...+a-mr-m 7392= 7 × 103 + 3 × 102 + 9 × 101 +2 × 100 (11010.11)2 =1×24+1×23+0×22+1×21+0×20+1×2-1+1×2-2 = (26.75)10 210 220 230 = 1Kilo = 1Mega = 1Giga

  3. Binary Number

  4. Number Base Conversions • Ex 1-1) Convert decimal 41 to binary. answer : (41)10 = (a5a4a3a2a1a0)2 = (101001)2 Answer =101001

  5. Number Base Conversions • Ex 1-2) Convert decimal 153 to octal. • Ex 1-3) Convert (0.6875)10 to binary. Answer:(0.6875)10 = (0.a-1a-2a-3a-4)2 = (0.1011)2  Ex 1-4) 해보세요 .

  6. Octal and Hexadecimal Numbers ( 10110001101011 .111100000110 )2 = (26153.7460)8 2 6 1 5 3 7 4 0 6 ( 10110001101011 .11110010 )2 = (2C6B.F2)16 2 C 6 B F 2 * PC의 계산기 기능

  7. Complements –Diminished Radix Complement • (r-1)’s complements of an n-digit N is (rn-1)-N • r=10, r-1=9, 9’complements of N is (10n-1)-N Ex) the 9’s complements of 546700 is 999999-546700 = 453299 the 9’s complements of 012398 is 999999-012398 = 987601 • For binary number, r=2, r-1=1 1’complements of N is (2n-1)-N Ex) the 1’s complements of 1011000 is 0100111 the 1’s complements of 0101101 is 1010010

  8. Complements –Radix Complement • The r’s complements of an n-digit number N is rn-N, N≠0 0, N=0 • rn-N=[(rn-1)-N]+1 => The r’s complements is obtained by adding 1 to the (r-1)’s complements • Ex) The 10’s complements of 012398 is 987602 The 10’s complements of 246700 is 753300 The 2’s complements of 1101100 is 0010100 The 2’s complements of 0110111 is 1001001 • Ex) The 10’s complements and 9’s complements of 086632 ? The 2’s complements and 1’s complements of 0110101 ? • 1’s complement 와 2’s complement 중 어느쪽이 계산 편리? • CPU는 (-) 연산 기능 없슴  (+) 과 complement 기능으로 (-) 연산 수행

  9. Complements –Subtraction with Complements • Ex1-5) using 10’s complement, subtract 72532-3250. • 72532 + (100000-3250) : carry 발생 • Ex1-6) Using 10’s complement, subtract 3250-72532. • 3250 -72532+100000 = 3250 + (100000-72532) : no carry There is no end carry Therefore, the answer is –(10’s complement of 30718)=-69282

  10. Complements –Subtraction with Complements • Ex1-7) X=1010100, Y=1000011, (a) X-Y, (b) Y-X (a) X-Y (b) Y-X There is no carry. The answer is Y-X = -(2’s complement of 1101111)=-0010001

  11. Complements –Subtraction with Complements (a) X-Y = 1010100-1000011 • Ex1-8) Repeat Example 1-7 using 1’complement. (b) Y-X = 1000011-1010100 There is no carry. The answer is Y-X = -(1’s complement of 1101110)=-0010001

  12. Signed Binary Numbers • Ex) 8bit로서 나타낼 수 있는 숫자 : 0-255 (모두 양수인경우). 만약 반은 음수, 반은 양수로 나타내고 싶을 때의 방법은? (즉 -127 ~ +127(8) 로 나태 내고 싶을 때의 방법은?) • Ex) The number 9 represented in binary with eight bit +9 : 00001001 -9 : 10001001 (signed-magnitude representation) 11110110 (signed-1’s-complement representation) 11110111 (signed-2’s-complement representation) • Complement 의 complement 는 원래의 수가 됨 • 덧셈, 뺄셈의 결과가 모순이 없고, 편리하게 되는지의 여부

  13. Signed Binary Numbers

  14. Signed Binary Numbers • Arithmetic Addition - signed-magnitude system follows the rules of ordinary arithmetic. -같은 부호의 경우 : 덧셈후 해당 부호 부여 –예 : … -다른 부호의 경우 : 큰 수에서 작은 수를 뺀 후 큰 수의 부호를 부여-예:.. - signed-complement system requires only addition (no deed of subtraction) – 2’s complement의 경우(carry 는 discard). • Arithmetic Subtraction (±A)-(+B) = (±A)+(-B) (±A)-(-B) = (±A)+(+B) • Overflow 문제 : paper에서는 자리수를 증가하여 해결가능하나, 컴퓨터 시스템에서는 제한된 bit 길이 고려하여야 함

  15. Arithmetic Addition의 예(2’s complement) 000 0 0 001 1 1 010 2 2 011 3 3 100 4 -4 101 5 -3 110 6 -2 111 7 -1 -4 ~ 3 안에서의 결과가 나오는 연산은 O.K.

  16. Binary Code-BCD code - the 4-bit code for one decimal (185)10 = (0001 1000 0101)BCD = (10111001)2 • BCD Addition - if the binary sum is greater or equal to 1010, we add 0110 to obtain the correct BCD

  17. Binary Code-Other Decimal Codes 2421 : 같은 숫자 2개의 코드 가능, ex) 4의 경우 0100, 1010 모두 가능 Excess-3 : BCD+3임, 9’complement를 구할 때는 bit의 1,0을 바꾸어주면 됨

  18. Binary Code-Other Decimal Codes 예 ) 0100을 다음의 형식대로 해석을 하면 얼마를 나타내는 수인가? - BCD 8421 : 0X8 + 1X4 + 0X2 + 0X1 =4 - 2421 : 0X2 + 1X4 + 0X2 + 0X1 = 4 - Excess-3 : 4-3 = 1 - 8 4 -2 -1 : 0X8 + 1X4 + 0X(-2) + 0X(-1) = 4 예 ) 1001을 BCD 8421, Excess-3, (8 4 -2 -1)의 형식으로 해석하면 각각 얼마를 나타내는 수인가? 예 ) 2421 코드에서 7을 나타낼 수 있는 코드 2개를 찾으시오.

  19. Binary Code-Gray Code 연속적인 값에 1bit 만이 변함,아날로그 값 입력시의 오류나 애매함을 방지. 즉 연속적인 값 입력시에 2bit 이상이 차이가 나면 오류임예 ) 3 bit의 gray code를 design 하시오.

  20. Binary Code-Gray Code 예 ) 3 bit의 gray code를 design 하시오. 000 0 001 1 011 2 010 3 110 4 111 5 101 6 100 7

  21. Binary Code- ASCII Character Code A : 100 0001 = 0x41a : 110 0001 = 0x61DEL : 111 1111 = 0x7FSP : 010 0000 = 0x20

  22. Binary Code • Error-Detecting Code

  23. Binary Storage and Registers • Registers – A register with n cells can store any discrete quantity of information that contains n bits. • Register Transfer

  24. Binary Logic • Definition of Binary Logic • Logic Gates

  25. Binary Logic

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