U of I CS 231 - Intro & Base Conversions

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CS 231 Intro & Base ConversionsAnnouncementsTips for CS 231Slide 4Digits vs. bitsBinary to decimalDecimal to binarySlide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Hexadecimal baseSlide 17Binary to HexOctal baseSlide 20Binary to OctalOther conversionsSummaryBasic Boolean OperationsCircuit AnalysisBoolean expressionsTruth TableSimplification with axiomsSlide 29CS 231Intro & Base ConversionsSangkyum KimSep 1, 2006AnnouncementsMallard is up. If you already registered but cannot log on, let me know it. Sometimes course webpage is downMallard link (https://mallard2.cites.uiuc.edu/CS231/)First homework assignment is released Due: 9/11(Mon) 5:00pm inside SC 0212Tips for CS 231Don’t hesitate to:Ask questionsInterrupt when you don’t understandBe comfortable with binary and hexLearn device inputs & outputsAllows abstractionEasier to understandTips for CS 231Efficiency is important!Neatness is important!Unreadable solutions are considered wrongReadable ones are neat, short (big favor for both you and the graders)Digits = powers of 10… 100, 10, 1,1/10,1/100,1/1000 …… 102, 101, 100, 10-1, 10-2, 10-3 …Ex: (36.25)10 = 3*10 + 6*1 + 2*1/10 + 5* 1/100Bits = powers of 2… 8, 4, 2, 1,1/2,1/4,1/8 …… 23, 22, 21, 20, 2-1, 2-2, 2-3 …Ex: (100100.01)2 = 1*32 + 1*4 + 1* 1/4Digits vs. bitsBinary to decimaladd powers that have a 1(101001)2 = 32 + 8 + 1(0.1001)2 = 1/2 + 1/16(10110.1011)2 = ?16 + 4 + 2 + 1/2 + 1/8 + 1/16= 22.6875Decimal to binaryLeft of decimal pointRepeatedly divide integer part by 2 until you get 0Read remainders bottom to up22 = (?)22211 R 05 R 12 R 11 R 00 R 1Decimal to binaryLeft of decimal pointRepeatedly divide integer part by 2 until you get 0Read remainders bottom to up22 = (10110)22211 R 05 R 12 R 11 R 00 R 1Decimal to binaryRight of decimal pointRepeatedly multiply fractional part by 2 until you get 1Read integer portion top to bottom0.8125 = (?)20.81251.62501.250.51.0Decimal to binaryRight of decimal pointRepeatedly multiply fractional part by 2 until you get 1Read integer portion top to bottom0.8125 = (0.1101)20.81251.62501.250.51.0Decimal to binaryWhat if there are both left and right of the decimal point?Do them separately and combine22.8125 = (?)2Decimal to binaryWhat if there are both left and right of the decimal point?Do them separately and combine22.8125 = (?)22211 R 05 R 12 R 11 R 00 R 10.81251.62501.250.51.0Decimal to binaryWhat if there are both left and right of the decimal point?Do them separately and combine22.8125 = (10110.1101)22211 R 05 R 12 R 11 R 00 R 10.81251.62501.250.51.0up downDecimal to binaryToy example 3.25 = (?)231 R 10 R 10. 250.501.0Decimal to binaryToy example 3.25 = (11.01)231 R 10 R 10. 250.501.0Hexadecimal baseHex digits = powers of 16… 256, 16, 1,1/16,1/256 …… 162, 161, 160, 16-1, 16-2 …use digits 0-9, A-FA=10, B=11, C=12, D=13, E=14, F=15often preceded by 0xbook subscript notation (24.4)16Ex: (24.4)16 = 2*16 + 4*1 + 4* 1/16Hexadecimal baseHex (hexadecimal)Hex digit is a group of 4 bitsMemorize this table!!dec. hex binary0 0 00001 1 00012 2 00103 3 00114 4 01005 5 01016 6 01107 7 01118 8 10009 9 100110 A 101011 B 101112 C 110013 D 110114 E 111015 F 1111Hex (hexadecimal)Group from decimal point outwardPad with zeros to get groups of 4(1101101001010.101001)2(0001 1011 0100 1010 . 1010 0100)2 1 B 4 A . A 4(1101101001010.101001)2 = (1B4A.A4)16Binary to HexOctal digits = powers of 8… 64, 8, 1,1/8,1/64 …… 82, 81, 80, 8-1, 8-2 …use digits 0-7sometimes preceded by 0book subscript notation (24.4)8Ex: (44.2)8 = 4*8 + 4*1 + 2* 1/8Octal baseOctalOctal digits are groups of 3 bitsPad with zerosdec. octal binary0 0 0001 1 0012 2 0103 3 0114 4 1005 5 1016 6 1107 7 111Octal baseOctalGroup from decimal point outwardPad with zeros to get groups of 3(1101101001010.101001)2(001 101 101 001 010 . 101 001)2 1 5 5 1 2 . 5 1(1101101001010.101001)2 = (15512.51) 8Binary to OctalWhat about other conversions such as:Octal  HexDecimal  Hex…Use other conversions you already knowOctal  Binary  HexDecimal  Binary  HexOther conversionsI. Decimal  BinaryBinary  DecimalII. Binary  HexBinary  OctalIII. Other conversionsUse the conversions you already knowSummaryBasic Boolean Operationsx y xy0 0 00 1 01 0 01 1 1x y x+y0 0 00 1 11 0 11 1 1x x’0 11 0AND (product)of two inputsOR (sum) of two inputsNOT(complement)on one inputxy, or x-y x + y x’Operation:Expression:Truth table:Logic gate:Circuit AnalysisCircuit analysis involves figuring out what some circuit does.Every circuit computes some function, which can be described with Boolean expressions or truth tables.The first thing to do is figure out what the inputs and outputs of the overall circuit are.Boolean expressionsWe did a little simplification for the top AND gate.You can see the circuit computes f(x,y,z) = xz + y’z + x’yz’Truth Tablex y z f0 0 0 00 0 1 10 1 0 10 1 1 01 0 0 01 0 1 11 1 0 01 1 1 1Simplification with axiomsx’y’ + xyz + x’y= x’(y’ + y) + xyz [ Distributive; x’y’ + x’y = x’(y’ + y) ]= x’-1 + xyz [ Axiom 7; y’ + y = 1 ]= x’ + xyz [ Axiom 2; x’-1 = x’ ]= (x’ + x)(x’ + yz) [ Distributive ]= 1 - (x’ + yz) [ Axiom 7; x’ + x = 1 ]= x’ + yz [ Axiom 2 ]1. x + 0 = x 2. x - 1 = x3. x + 1 = 1 4. x - 0 = 05. x + x = x 6. x - x = x7. x + x’ = 1 8. x - x’ = 09. (x’)’ = x10. x + y = y + x 11. xy = yx Commutative12. x + (y + z) = (x + y) + z 13. x(yz) = (xy)z Associative14. x(y + z) = xy + xz 15. x + yz = (x + y)(x + z) Distributive16. (x + y)’ = x’y’ 17. (xy)’ = x’ + y’ DeMorgan’sSimplification with axiomsyz + y(x+z)’ + x(x+y)= y + x1. x + 0 = x 2. x - 1 = x3. x + 1 = 1 4. x - 0 = 05. x + x = x 6. x - x = x7. x + x’ = 1 8. x - x’ = 09. (x’)’ = x10. x + y = y + x 11. xy = yx Commutative12. x + (y + z) = (x + y) + z 13. x(yz) = (xy)z Associative14. x(y + z) = xy + xz 15. x + yz = (x + y)(x + z) Distributive16. (x + y)’ = x’y’ 17. (xy)’ = x’ + y’


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U of I CS 231 - Intro & Base Conversions

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