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Digital Systems

Digital Systems. Discrete quantities: weekly salaries, income taxes, letters or alphabet, digits Digital systems: digital telephones, digital cameras, digital computers and etc. Digital systems have the ability to manipulate discrete quantities.

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Digital Systems

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  1. Digital Systems Discrete quantities: weekly salaries, income taxes, letters or alphabet, digits Digital systems: digital telephones, digital cameras, digital computers and etc. Digital systems have the ability to manipulate discrete quantities. Some digital systems can operate with extreme reliability by using error-correcting code. For example: DVD

  2. Numbering Systems • Decimal numbering systems: 102 is the combination of 2 digits from the values of 0,1,2,3,4,5,6,7,8 and 9 . In general for any numbering system the number of combination of n digits can be determined by: Bn where B is the base of numbering system and n is the number of digits to be combined. • For example for two decimal digits the largest number is 99 so the two decimal digits end at 99. Using the formula of Bn the total number of combinations for two decimal digits would be 102 = 100

  3. Numbering Systems • A decimal number such as 7,392.42 is equal to: 7 * 103 + 3 * 102 + 9 * 101 + 2 * 100 + 4 * 10-1 + 2 * 10-2 • In general for any numbering system an an-1 an-2…… a1 a0. a-1 a-2……. a-m the equal decimal number is: an* rn + an-1* rn-1 + ….a1* r + a0 + a-1* r-1+… am* rm where aj is the coefficient and the number expressed in a base-r system

  4. Octal Numbering System • The Octal numbering system is a base-eight numbering system with eight digits of 0,1,2,3,4,5,6,7 Decimal Octal ------------ --------- 0 0 1 1 7 7 8 10 9 11

  5. Octal Numbering System • For example to count in octal the digits combine after reaching a count of 7 1,..7,10,11,12,…,15,16,17,20,21,…,75,76,77,100 for two octal digits the largest number is 77 so the two octal digits end after at 77. Using the formula of Bn the total number of combinations for two octal digits would be 82 = 64 • To find the decimal number equal to an octal number: (127.4)8= 1* 82 +2 * 81 + 7 * 80 + 4 * 8-1 = (87.5)10

  6. The Hexadecimal Numbering System • It is a base-sixteen numbering system. That is there are 16 digits in this system: Hexadecimal Decimal ------------------ ----------- 0 0 1 1 2 2 …….. …… 9 9 A 10 B 11 C 12 D 13 E 14 F 15

  7. The Hexadecimal Numbering System • For example to count in hexadecimal …F,10,11,12,…19,1A,1B,1C,1D,1E,1F,20, 21,….99,9A,…,9F,A0,A1…..,FE,FF,100 for two hexadecimal digits the largest number is FF so the two hexadecimal digits end after at FF. Using the formula of Bn the total number of combinations for two hexadecimal digits would be 162 = 256. • To find the decimal number equal to a hexadecimal number: (B65F)16= 11 * 163 + 6 * 162 + 5 * 161 + 5 * 160 = (46,687)10

  8. The Binary Numbering System Only two digits required {1,0}

  9. The Binary Numbering System • For four binary digits the largest number is 1111 so the four binary digits end at 1111. Using the formula of Bn the total number of combinations for four binary digits would be 24 = 16. • To find the decimal number equal to a binary number: (110101)2 = 32 + 16 + 4 + 1 = (53)10 (0.1101)2=(1*.5 +1*.25 +0*.125 +1*.0625 )10 =(0.8125) 10

  10. Adding Binary Numbers Same rule as decimal Example: carry

  11. Multiplication of Binary numbers Multiplication Table Example:

  12. Number Base Conversions • For converting decimal number to binary number REPETETIVE DIVISION is used as follows: Algorithm that generates binary digits from 0 to n: Q = decimal number While Q is not equal to 0 do the following Binary digit = Q mod 2 Q = Q / 2 (quotient) End While

  13. Conversion Example Convert (58)10 to (?)2 58 mod 2 = 0 29 29 mod 2 = 1 14 14 mod 2 = 0 7 7 mod 2 = 1 3 3 mod 2 = 1 1 1 mod 2 = 1 1 Ans: (111010)2

  14. Conversion Example • To convert a fraction, it multiples by 2 to give an integer and a fraction and only the fraction again multiplies by 2. This process continues until the faction becomes zero or until the number of digits have sufficient accuracy: (0.6875) 10 = ( ? ) 2 0.6875 * 2 = 1 + 0.3750 0.3750 * 2 = 0 + 0.7500 0.7500 *2 = 1 + 0.5000 0.5000 * 2 = 1 + 0.0000 == result is (0.1011)2

  15. Hex to Binary Converting from Hex to Binary is easy: Every hex digit becomes 4 binary digits (1AF5) 16 =(0001 1010 1111 0101) 2 =(1101011110101)2 (306.D)16 = ( 0011 0000 0110. 1101 )2

  16. Binary to Hex Just as simple, reverse to process (111001010101011101) 2 =(0011 1001 0101 0101 1101) 2 =(3955D) 16

  17. Octal to Binary Converting from Octal to Binary is trivial: Every octal digit becomes 3 binary digits (175 ) 8 =(001 111 101) 2 =(1111101) 2

  18. Binary to Octal Just as simple, reverse to process (11001010101011101) 2 =(011 001 010 101 011 101) 2 =(312535) 8 • Using the hex and octal equivalent instead of binary numbers are more convenient and less prone to errors.

  19. Complements Complements are using for two purposes: • For simplifying subtraction operation • For logical operations There are two types of complements for Base-r: • r’s complement ( radix complement ) • (r-1)’s complement ( diminished radix complement)

  20. (r-1)’s Complements (r-1)’s complement for a number N in base-r with n digits is (rn -1 ) – N For example for decimal numbers r =10 ( 10n – 1) – N 9’s complement of 12389 is (105 -1 ) - 12389 = 99999 -12389= 87610 9’s complement of 2389 is (104 -1) – 2389 = 9999 -2389 = 7610

  21. (r-1)’s Complements 1’s complement is defined as (2n – 1 ) – N For example for 1011001 it is equal to : ( 27 – 1 ) – 1011001 = 0 100110 Therefore, 1’s complement of binary numbers is formed by changing 1’s to 0’s and 0’s to 1’s For example 0001111 becomes 1110000

  22. (r’s complement) (r’s) complement for number N in base r with n digits is (rn – N) also r’s complement= ( r-1)’s complement +1 For example: 10’s complement 2389= 7610 + 1 = 7611 2’s complement 101100= 010011 1’s complement + 1 -------------- 010100 Or 2n – N = 1000000 - 0101100 ---------------------- 0010100

  23. Subtraction with complements (Unsigned Numbers) For doing the subtraction on two n digits unsigned minuend M - subtrahend N in base r 1- add M to r’s complement of N 2- If M>=N subtract 10n form the result 3- if M<N put “-” and r’s complement of result

  24. Subtraction with complements (Unsigned Numbers) • For example 3250 -72532 M = 03250 10’s complement of N = + 27468 ------------ 30718 - (10’s complement 30718) =-69282

  25. Subtraction with complements (Unsigned Numbers) • For Binary numbers the algorithm is same • Subtraction also can be done by (r-1)’s complement. In that case if M>=N, after discarding the end carry one should be added to result (end-around carry): X = 1010100 1010100 2’scopml ( Y= 1000011)+ Y = 0111101 1’s +0111100 ----------------- --------------- 10010001 10010000 -10000000 - 10000000 1 -------------------- ---------------- 0010001 0010001

  26. Signed Binary Numbers A negative binary number assumed to be signed or unsigned. For example: 01001 unsigned binary = 9 signed binary = +9 11001 unsigned binary = 25 signed binary = -9 Signed-magnitude convention ( 0 for + and 1 for -)

  27. Signed Binary Numbers • Another system to represent negative numbers is signed complement system For example +9 is 00001001 in two systems but -9 is 10001001 in signed-magnitude 11110110 in signed-1’s-complement (complementing all bits include sign bit or subtracting from 2n -1) 11110111 in signed-2’s-complement ( 2’complement of all bits or subtracting from 2n )

  28. Decimal signed-2’s signed-1’s signed complement complement magnitude ---------- ----------- ------------- ----------- + 7 0111 0111 0111 +2 0010 0010 0010 +1 0001 0001 0001 +0 0000 0000 0000 -0 --- 1111 1000 -1 1111 1110 1001 -2 1110 1101 1010 -7 1001 1000 1111 -8 1000 ------- ------

  29. Signed Binary Numbers • Representation method only matters when we are talking about negative numbers • All negative numbers have 1 in leftmost bit • The signed magnitude is mostly used in ordinary arithmetic • The 1’s complement is mostly used in logical operations • The 2’s complement is mostly used in computer arithmetic

  30. Arithmetic Addition (Signed Numbers) • The addition of binary numbers in signed-magnitude system follows the same rules as ordinary arithmetic. (sign of larger number) • For adding the numbers in 2’s complement form can be obtained by adding the two numbers including their sign bits. The carry of sign bit is discarded and negative results are automatically in 2’s complement form. • In order to get the negative result we must find the 2’s complement form • When adding 2 numbers of same sign (pos or neg) in 2’s complement form if result cannot be shown in bits available, then overflow occurs that indicates result is wrong.

  31. Arithmetic Addition (2’s Comp) • Two positive numbers 0 1001 0 1001 + 0 0100 0 1000 ---------------- ----------- 0 1101 1 0001 Neg result wrong, overflow

  32. Arithmetic Addition (2’s Comp.) • For example: +9 0 1001 + -4 1 1100 -------------------------------- 1 0 0101 (5) Discarded final carry. Note that sign bit also participates in the process

  33. Arithmetic Addition (2’s Comp.) • For example: +6 0 0000110 + -13 1 1110011 -------------------------------- 1 1111001 => 0000111 (2’s comp.) it is equal to 7 so the result is -7

  34. Arithmetic Addition (2’s Comp.) • For example: -9 1 0111 + -4 1 1100 -------------------------------- 11 0011 => 1101 (2’s comp.) it is equal to 13 so the result is -13

  35. Arithmetic Subtraction • In 2’s complement format it is very simple : Take the 2’s complement of sabtrahend (the second number) including the sign bit and add it to minuend (the first number) including the sign bit and discard a carry out of sign bit • By taking 2’s complement of the subtrahend its sign can be changed. Thus, we can change subtraction to addition operation.

  36. Arithmetic Subtraction (2’s Comp) • For example: (+9) 0 1001 -(+4) 0 0100 is changed to (+9) 0 1001 +(-4) 1 1100 ---------------- 1 0 0101 (+5) discard carry

  37. Arithmetic Subtraction (2’s Comp) • For example: (-4) 1 1100 -(+9) 0 1001 is changed to (-4) 1 1100 + 1 0111 ---------------- 1 1 0 011 (2’s comp 01101 means magnitude is 13) discard carry => (-13)

  38. Arithmetic Subtraction • For example: -6 11111010 • -13 11110011 => is changed to -6 11111010 + +13 00001101 ----------------- 100000111 => discard carry  00000111 is +7

  39. Binary Coded Decimal (BCD) • 4 bits used to encode one decimal digit • For example (4321)10 = 0100 0011 0010 0001 • The problem is BCD can not be used for conversion from decimal to binary and from binary to decimal. Because there is no decimal digits for 1010, 1011,1100,1101,1110,1111 For example : (0111 0010 1100)2= (72?) there is no decimal digits (0..9) for 1100

  40. BCD Arithmetic • Add 2 BCD numbers using regular binary addition • Check each nibble ( 4-bit) if result is greater than 9 add 6 to it If there is carry between 2 nibble or coming from 2th nibble add 6

  41. BCD Arithmetic Example: 27 0010 0111 + 34 0011 0100 ----------------- 0101 1011 > 9 + 0110 ---------------- 01100001

  42. BCD Arithmetic Example: 1carry 59 0101 1001 + 39 0011 1001 ---------------- 1001 0010 0110 ---------------- 1001 1000 (98)

  43. BCD Arithmetic 1 1 carry 98 1001 1000 + 89 1000 1001 --------------- 1 0010 0001 + 01100110 --------------- 11000 0111 (187)

  44. Character Representation ASCII – American Standard Code for Information Interchange 128 characters (7 bits required) Contains: • Control characters (non-printing) • Printing characters (letters, digits, punctuation)

  45. ASCII – Characters

  46. ASCII – Characters

  47. Error-Detecting Code • To detect the error in data communication, an eighth bit is added to ASCII character to indicate its parity. • A parity bit is an extra bit included with a message to make the total number of 1’s either even or odd • The 8-bit characters included parity bits (with even parity) are transmitted to their destination. If the parity of received character is not even it means at least one bits has been changed.

  48. Error-Detecting Code Example: even parity odd parity ASCII A = 1000001 01000001 11000001 ASCII T = 1010100 11010100 01010100 • The method of error checking with even parity detects any odd combinations of errors in each character that is transmitted. An even combination of errors is undetected with this method.

  49. Binary Storage and Registers • Register is a group if binary cells that are responsible for storing and holding the binary information. • Register transfer operation is transferring binary operation from one set of registers to another set of registers. • Digital logic circuits process the binary information stored in the registers.

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