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CS103 Guest Lecture

CS103 Guest Lecture. Number Systems & Conversion Bitwise Logic Operations. on. off. Why 1’s and 0’s. Transistors are electronic devices used to build computer hardware Like a switch (2 positions) Conducting / Non-conducting Output voltage of a transistor will either be high or low

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CS103 Guest Lecture

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  1. CS103 Guest Lecture Number Systems & Conversion Bitwise Logic Operations

  2. on off Why 1’s and 0’s • Transistors are electronic devices used to build computer hardware • Like a switch (2 positions) • Conducting / Non-conducting • Output voltage of a transistor will either be high or low • 1’s and 0’s are arbitrary symbols representing high and low voltage outputs. • 2 states of the transistor lead to only 2 values in computer hardware Circuit Diagram of a Switch Circuit Diagram of a Switch The voltage here circuit is open (off) – no current can flow determines if current 1 High Voltage +5V +12V can flow between Output drain and source ( Drain ) - - - - Controlling or Input - - - - ( Gate ) 0 circuit is closed (on) – current can flow Low Voltage 0V -12V Source Schematic Symbol of a Functional View of a Transistor Transistor as a Switch

  3. Positional Number Systems

  4. Interpreting Binary Strings • Given a string of 1’s and 0’s, you need to know the representation system being used, before you can understand the value of those 1’s and 0’s. • Information (value) = Bits + Context (System) 01000001 = ? Unsigned Binary system ASCII system 6510 ‘A’ASCII Signed System

  5. Binary Number System • Humans use the decimal number system • Based on number 10 • 10 digits: [0-9] • Because computer hardware uses digital signals with 2 states, computers use the binary number system • Based on number 2 • 2 binary digits (a.k.a bits): [0,1]

  6. Number Systems • Number systems consist of • A base (radix) r • r coefficients [0 to r-1] • Human System: Decimal (Base 10): 0,1,2,3,4,5,6,7,8,9 • Computer System: Binary (Base 2): 0,1 • Human systems for working with computer systems (shorthand for human to read/write binary) • Octal (Base 8): 0,1,2,3,4,5,6,7 • Hexadecimal (Base 16): 0-9,A,B,C,D,E,F (A thru F = 10 thru 15)

  7. Anatomy of a Decimal Number • A number consists of a string of explicit coefficients (digits). • Each coefficient has an implicit place value which is a power of the base. • The value of a decimal number (a string of decimal coefficients) is the sum of each coefficient times it place value radix (base) (934)10= (3.52)10=

  8. Anatomy of a Decimal Number • A number consists of a string of explicit coefficients (digits). • Each coefficient has an implicit place value which is a power of the base. • The value of a decimal number (a string of decimal coefficients) is the sum of each coefficient times it place value radix (base) (934)10= 9*102 + 3*101 + 4*100= 934 Implicit place values Explicit coefficients (3.52)10= 3*100 + 5*10-1 + 2*10-2= 3.52

  9. Positional Number Systems (Unsigned) • A number in base r has place values/weights that are the powers of the base • Denote the coefficients as: ai Left-most digit = Most Significant Digit (MSD) Right-most digit = Least Significant Digit (LSD) a 3 a 2 a 1 a 0 . a -1 a -2 ... ... r 3 r 2 r 1 r 0 r -1 r -2 Nr = Σi(ai*ri) = D10

  10. Examples (746)8= (1A5)16=

  11. Examples (746)8= 7*82 + 4*81 + 6*80= 448 + 32 + 16 = 48610 (1A5)16= 1*162 + 10*161 + 5*160= 256 + 160 + 5 = 42110

  12. Anatomy of a Binary Number • Same as decimal but now the coefficients are 1 and 0 and the place values are the powers of 2 Most Significant Digit (MSB) Least Significant Bit (LSB) (1011)2= 1*23 + 0*22 + 1*21 + 1*20 radix (base) place values = powers of 2 coefficients

  13. Binary Examples (1001.1)2= (10110001)2=

  14. Binary Examples (1001.1)2= 8 + 1 + 0.5 = 9.510 8 4 2 1 .5 (10110001)2= 128 + 32 + 16 + 1 = 17710 128 32 16 1

  15. Powers of 2 20 = 1 21 = 2 22 = 4 23 = 8 24 = 16 25 = 32 26 = 64 27 = 128 28 = 256 29 = 512 210 = 1024 1024 512 256 128 64 32 16 8 4 2 1

  16. Practice On Your Own • Decimal equivalent is… … the sum of each coefficient multiplied by its implicit place value (power of the base) = Σi(ai * ri) [ai = coefficient, r = base] (11010)2 = 1*24 + 1*23 + 1*21 = 16 + 8 + 2 = (26)10 (6523)8= 6*83 + 5*82 + 2*81 + 3*80 = 3072 + 320 + 16 + 3 = (3411)10 (AD2)16 = 10*162 + 13*161 + 2*160 = 2560 + 208 + 2 = (2770)10

  17. Shifting in Binary • Move bits to the left or right with 0's shoved in one side and dropping bits on the other • Useful for multiplying and dividing by power of 2. • Right shift by n-bits = Dividing by 2n • Left shift by n-bits = Multiplying by 2n 0 ... 0 1 1 0 0 = +12 Right Shift by 2 bits: Left Shift by 3 bits: 0’s shifted in… 0’s shifted in… 0 0 ... 0 0 1 1 = +3 ... 0 1 1 0 0 00 0 = +96

  18. Bottom-Up Conversion & Shifting • X = 0112 = • 01102 • 011012 • 0110102

  19. Unique Combinations • Given n digits of base r, how many unique numbers can be formed? rn • What is the range? [0 to rn-1] 100 combinations: 00-99 2-digit, decimal numbers (r=10, n=2) 0-9 0-9 1000 combinations: 000-999 3-digit, decimal numbers (r=10, n=3) 4-bit, binary numbers (r=2, n=4) 16 combinations: 0000-1111 0-1 0-1 0-1 0-1 64 combinations: 000000-111111 6-bit, binary numbers (r=2, n=6) Main Point: Given n digits of base r, rn unique numbers can be made with the range [0 - (rn-1)]

  20. Conversion from Decimal To Other Base

  21. Conversion: Base 10 to Base r • X10 = (?)r • General Method (base 10 to arbitrary base r) • Division Method for integer portion or number • Alternate Method • Left-to-right (greedy) approach (45)10= (?)r

  22. Division Method Explanation . a4 a3 a2 a1 a0 4510= 0 24 23 22 21 20 4510= a424 + a323 + a222 + a121 + a020 4510= a424 + a323 + a222 + a121 + a020 2 2 22.510= a423 + a322 + a221 + a120 + a02-1

  23. Binary Division Method Example 4510 = (??)2

  24. Binary Division Method Example 4510 = (??)2 2 45 LSB 2 22 rem. = 1 2 11 rem. = 0 2 5 rem. = 1 2 2 rem. = 1 2 1 rem. = 0 0 rem. = 1 MSB Keep dividing until you reach 0 Remainders form the number in base r (order from bottom up) 4510 = (101101)2

  25. Division Method • Converts integer portion of a decimal number to base r • Informal Algorithm • Repeatedly divide number by r until equal to 0 • Remainders form coefficients of the number base r • Remainder from last division = MSD (most significant digit) 19310 = (??)5 5 193 LSD rem. = 3 5 38 rem. = 3 5 7 rem. = 2 5 1 0 rem. = 1 MSD Keep dividing until you reach 0 Remainders form the number in base r (order from bottom up) 19310 = (1233)5

  26. How number conversion works 4510 = a4 a3 a2 a1 a0 24 23 22 21 20 • Each time we divide by r, another coefficient “falls out” and all the other place values are reduced by a factor of r. More bits may be required for this actual example, but we'll use 5 to illustrate… 4510 = a424 + a323 + a222 + a121 + a020 Rem. Quotient 4510 = a424 + a323 + a222 + a121 + a020 = a423 + a322 + a221 + a120 + a0 2 2 2 Quotient Rem. a423 + a322 + a221 + a120 = a422 + a321 + a220 + a1 2 2 This is just explanation for what you've learned…Focus on the conversion process outlined earlier

  27. How number conversion works • Each time we divide by r, another coefficient “falls out” and all the other place values are reduced by a factor of r. a0 is the remainder D10 written as coefficients * place values Factor r out of numerator terms with an-1 – a1 This is just explanation for what you've learned…Focus on the conversion process outlined earlier

  28. Left-To-Right Method • An alternative to the division method • To convert a decimal number, x, to binary: • Only coefficients of 1 or 0. So simply find place values that add up to the desired values, starting with larger place values and proceeding to smaller values and place a 1 in those place values and 0 in all others 0 1 1 0 0 1 2510= 32 16 8 4 2 1 For 2510 the place value 32 is too large to include so we include 16. Including 16 means we have to make 9 left over. Include 8 and 1.

  29. Left-To-Right Binary Examples 7310= 0 1 0 0 1 0 0 1 128 64 32 16 8 4 2 1 8710= 0 1 0 1 0 1 1 1 14510= 1 0 0 1 0 0 0 1 0.62510= 1 0 1 0 0 .5 .25 .125 .0625 .03125

  30. Left-To-Right In Other Bases • Can use the left-to-right method to convert a decimal number, x, to any base r: • Use the place values of base r (powers of r). Starting with largest place values, fill in coefficients that sum up to desired decimal value without going over. 7510 = 0 4 B hex 256 16 1

  31. Hexadecimal and Octal Shorthand for Binary

  32. Binary, Octal, and Hexadecimal • Octal (base 8 = 23) • 1 Octal digit ( _ )8 can represent: 0 – 7 • 3 bits of binary (_ _ _)2 can represent: 000-111 = 0 – 7 • Conclusion…1 Octal digit = 3 bits • Hex (base 16=24) • 1 Hex digit ( _ )16 can represent: 0-F (0-15) • 4 bits of binary (_ _ _ _)2 can represent: 0000-1111= 0-15 • Conclusion…1 Hex digit = 4 bits

  33. Binary to Octal or Hex • Make groups of 3 bits starting from radix point and working outward • Add 0’s where necessary • Convert each group of 3 to an octal digit • Make groups of 4 bits starting from radix point and working outward • Add 0’s where necessary • Convert each group of 4 to an octal digit 101001110.11 101001110.11 0 0 0 0 0 0 5 1 6 6 1 4 E C 516.68 14E.C16

  34. Octal or Hex to Binary • Expand each octal digit to a group of 3 bits • Expand each hex digit to a group of 4 bits 317.28 D93.816 011001111.0102 110110010011.10002 11001111.012 110110010011.12

  35. Logic Operations

  36. Bitwise Logical Operations F = X or ~X X B1 B1 B1 Z F XOR OR B2 AND Y B2 B2 Force '0' Pass Pass Force '1' Invert Pass 0 & x = 01 & x = xx & x = x 0 | x = x1 | x = 1x | x = x 0 ^ x = x1 ^ x = ~ xx ^ x = 0

  37. Logical Operations • Logic operations on numbers means performing the operation on each pair of bits 0xF0AND 0x3C0x30 1111 0000AND 0011 11000011 0000 0xF0OR 0x3C0xFC 1111 0000OR 0011 11001111 1100 0xF0XOR 0x3C0xCC 1111 0000XOR 0011 11001100 1100 NOT 0xAC0x53 NOT 1010 11000101 0011

  38. Logical Operations • The C language has two types of logic operations • Logical and Bitwise • Logical Operators (&&, ||, !) • Operate on the logical value of a FULL variable (char, int, etc.) interpreting that value as either True (non-zero) or False (zero) char x = 1, y = 2, z; z = x && y; • Result is z = 1; Why? • Bitwise Logical Operators (&, |, ^, ~) • Operate on the logical value of INDIVIDUAL bits in a variable char x = 1, y = 2, z; z = x & y; • Result is z = 0; Why?

  39. Logical Operations • Logic operations on numbers means performing the operation on each pair of bits 0x7AAND 0xEC0x30 0x36OR 0x910xFC 0x3C0x78XOR 0x3C0x78

  40. Look Toward LFSR PA • One of your next PA's will utilize the bitwise XOR operator and leverage the fact that: • a ^ b ^ a = b • Proof:

  41. Logical Operations Bitwise logic operations are often used for "bit fiddling" Change the value of a bit in a register w/o affecting other bits C operators: & = AND, | = OR, ^ = XOR, ~ = NOT Examples (Assume an 8-bit variable, v) Clear the LSB to '0' w/o affecting other bits v = v & 0xfe; Set the MSB to '1' w/o affecting other bits v = v | 0x80; Flip the LS 4-bits w/o affecting other bits v = v ^ 0x0f; Bit # 7 6 5 4 3 2 1 0 ? ? ? ? ? ? ? ? v _________________ & ? ? ? ? ? ? ? 0 v ? ? ? ? ? ? ? ? v _________________ | 1 ? ? ? ? ? ? ? v ? ? ? ? ? ? ? ? v 0 0 0 0 1 1 1 1 ^ ? ? ? ? ? ? ? ? v

  42. Exercises for Practice • Q1-Q15 on the posted worksheet • http://bits.usc.edu/files/cs103/coursework/NumberSys.pdf

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