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Transcript

  • 1. Higher Computing Mr Arthur
  • 2. Course Outline
    • 3 Main Units
      • Computer Systems = 40 hours
      • Software Development = 40 hours
      • Artificial Intelligence = 40 hours
    • Assessment
      • 3 End of Unit Assessments (NABS)
      • Practical Coursework Tasks (/60 or 30%)
      • Written Exam (/140 or 70%)
  • 3. Computer Systems
    • 5 units in the Computer Systems Section
      • Data Representation = 6 hours
      • Computer Structure = 7 hours
      • Peripherals = 5 hours
      • Networking = 9 hours
      • Computer Software = 9 hours
  • 4. Aims of Lesson 1
    • How are numbers, text and images represented inside the computer system?
    • Discussing the 2 state computer system
    • Converting positive whole numbers to binary and vice versa
    • Playing Binary Bingo
  • 5. Data Representation 100 billion switches per sq. cm
  • 6. Data Storage
    • Numbers, Text, and Images are all stored as a series of 1s and 0s inside the computer system.
    • These series of 1s and 0s are made up of pulses of electricity from 1 volt to 5 volts
  • 7. Decimal Counting System
    • When we represent numbers we use the decimal counting system, for example
    • 123,000
    • 100,000 10,000 1,000 100 10 1
    • 1 2 3 0 0 0
    • Since the computer is 2 state, the binary counting system goes up by the power 2, rather than 10 i.e
    • 256 128 64 32 16 8 4 2 1
  • 8. How Positive Whole Numbers are Stored
    • 34
    • 128 64 32 16 8 4 2 1
    • 0 0 1 0 0 0 1 0
    • = 32 + 2
    • 134
    • 128 64 32 16 8 4 2 1
    • 1 0 0 0 0 1 1 0
    • = 128 + 4 + 2
  • 9. Binary back to Decimal
    • 1011 0011
    • 128 64 32 16 8 4 2 1
    • 1 0 1 1 0 0 1 1
    • = 128 + 32 + 16 + 2 + 1
    • = 179
  • 10. Binary to Decimal
    • What is the decimal representation of the following 8 bits using 2s complement
    • (a) 0001 0110
    • (b) 1000 1100
    • (c) 0111 0011
    • What is the 8 bit representation of the following decimal numbers
    • (a) 174
    • (b) 121
    • (c) 71
  • 11. Binary Bingo
    • 42
    • 81
    • 21
    • 16
    • 121
    • 73
    • 101
    • 75
    • 127
    • 13
    • 209
    • 32
    • 56
    • 175
    • 192
    • 186
    • 176
    • 121
  • 12. Data Storage
    • 1 or 0 = 1 bit
    • 8 bits = 1 byte
    • 1024 bytes = 1 kilobyte
    • 1024 kilobytes = 1 megabyte
    • 1024 megabytes = 1 gigabyte
  • 13. Aims of Lesson 2
    • Representation of negative whole numbers
    • The 2s complement system
  • 14. Representing Negative Numbers
    • The signed bit method
    • 0000 0001 = 1
    • 0000 0000 = 0
    • 1000 0001 = -1
    • 1000 0010 = -2
    • 1000 0011 = -3
    • 1000 0100 = -4
  • 15. Representing Negative Numbers
    • There is a problem with this method??
    • Using 8 bits you can only store the decimal numbers from
    • 128 64 32 16 8 4 2 1
    • 1 1 1 1 1 1 1 1
    • = 64 +32+16+8+4+2+1 = -127
    • 128 64 32 16 8 4 2 1
    • 0 1 1 1 1 1 1 1
    • =64+32+16+8+4+2+1=127
    • Rather than -255 to 255
  • 16. 2s Complement
    • What is the 8 bit two’s complement representation of the decimal number -101
    • 101
    • 128 64 32 16 8 4 2 1
    • 0 1 1 0 0 1 0 1
    • Invert numbers
    • 1 0 0 1 1 0 1 0
    • +1
    • -101
    • 1 0 0 1 1 0 1 1
  • 17. Negative Whole Numbers
    • What is the decimal representation of the following 8 bits using 2s complement
    • 1 0 1 0 1 1 1 1
    • You invert every number
    • 0 1 0 1 0 0 0 0
    • Then add 1
    • 0 1 0 1 0 0 0 1
    • 128 64 32 16 8 4 2 1
    • 64+16+1
    • -81
  • 18. 2s Complement Questions
    • What is the decimal representation of the following 8 bits using 2s complement
    • (a) 1000 1011
    • (b) 1100 1100
    • (c) 1001 0111
    • (d) 1110 1100
    • What is the 8 bit two’s complement representation of the following decimal numbers
    • (a) -45
    • (b) -121
    • (c) -176
    • (d) -71
  • 19. Aims of Lesson 3
    • So far we have looked at representing positive and negative whole numbers using binary
    • We are now going to look at the representation of non whole numbers using the floating point system
  • 20. Representing Non Whole Numbers
    • How do we represent the number 128.75 in binary?
    • 128 + 0.5 + 0.25
    • = 128.75
    128 64 32 16 8 4 2 1 0.5 0.25 0.125 0.0625 1 0 0 0 0 0 0 0 1 1 0 0
  • 21. Mantissa and Exponent
    • Mantissa
    • Exponent
    • 8
    • 8 4 2 1
    • 1 0 0 0
    128 64 32 16 8 4 2 1 0.5 0.25 0.125 0.0625 1 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0
  • 22.
    • Mantissa
    • Exponent
    • 6
    • 8 4 2 1
    • 0 1 1 0
    1 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 0 1 0 How do we represent the number 38.125 using floating point 32 16 8 4 2 1 0.5 0.25 0.125 0.0625
  • 23. Representing Non Whole Numbers
    • Mantissa relates to the precision of the number you can represent i.e 34.44454321
    • Exponent relates to the range of the number
    • 1111 = 15
    • 1111 1111 = 255
    8 4 2 1 0.5 0.25 0.125 0.075 0.0375 0.01875 0.009375
  • 24.
    • What is the decimal number if the Mantissa is
    • 10010011 and the exponent is 0101
    • Exponent
    • 8 4 2 1
    • 0 1 0 1
    • = 5
    • Mantissa
    • 1 0 0 1 0 0 1 1
    Mantissa and Exponent 16 8 4 2 1 0.5 0.25 0.125 16 + 2 + 0.25 + 0.125 = 18.375
  • 25. Aims of Lesson 4
    • So far we have looked at representing positive and negative whole numbers using binary
    • We have also looked at representing non whole numbers using floating point.
    • Today we are going to practice converting storage capacities from bit, byte, kilobyte, megabyte, gigabyte, terabyte
    • Discuss how text is represented in a computer system
  • 26. Storage Capacities 0 or 1 = 1 bit 8 bits = 1 byte 1024 bytes = 1 Kilobyte 1024 Kilobytes = 1 Megabyte 1024 Megabytes = 1 Gigabyte 1024 Gigabytes = 1 Terabyte
  • 27. Storage Conversions
    • I have a 2 Gigabyte IPOD Classic. How many 512Kb songs can I store on the IPOD?
    • Convert 2Gb to Kb
    • 2 X 1024 = 2048Mb
    • 2048 X 1024 = 2,097,152Kb
    • 512Kb
    • 4096 Songs
  • 28. Storage Conversion Questions
    • I have a memory card for a Digital Camera with a capacity of 4Gb. How many 460Kb images can I store on the memory card?
    • Mr Haggarty has recently been working as a DJ at weekends. He has bought an external hard disk to back up songs. How many 4Mb songs would he be able to fit on the 80Gb hard disk?
  • 29. Solutions
    • 4Gb X 1024 = 4096Mb
    • 4096 X 1024 = 4,194,304Kb
    • 460Kb
    • = 9118 images
    • 80Gb X 1024 = 81920Mb
    • 4Mb
    • = 20,480 songs
  • 30. How is Text Represented
    • ASCII
      • Each key on the keyboard is converted into a binary code using 7 bits
      • Using 7 bits i.e 2 = 128 characters can be represented
    • Character Set
      • A list of all the characters which the computer can process
    • Control Characters
      • Codes 0 to 31 are non printable characters
    7 97 110 0001 a 65 100 0001 A 49 011 0001 1 34 010 0010 ‘ 33 010 0001 ! 32 010 0000 space 13 000 1101 return 9 000 1001 tab Decimal Binary Character
  • 31. How is Text Represented
    • Unicode (Universal Code)
      • Each key on the keyboard is converted into a binary code using 16 bits
      • Using 16 bits i.e 2 = 65,536 characters can be represented
      • Can represent Latin, Roman, Japanese characters
    • Advantages
      • More characters can be represented
    • Disadvantages
      • Takes up more than twice as much space for each character
    16
  • 32. Aims of Lesson 5
    • Last Lessons
    • Representing positive whole numbers as binary
    • Representing negative whole numbers using 2s complement
    • Non whole numbers using mantissa and exponent
    • Storage calculations
    • Looked at how text is represented using ASCII and Unicode
    • Today’s Lesson
    • Discuss graphic representation
    • Calculate storage capacities of colour Bit Map graphics
    • Bit Map v Vector
  • 33. BIT Map Graphics SCREEN MEMORY PIXEL MEMORY REQUIRED 8 BITS X 8 BITS = 64 BITS = 8 BYTES Bit Map = the graphic is made up from a series of pixels
  • 34. Graphics Resolution
    • The smaller the size of the pixels, the finer the detail of the image
    • 800 x 600 pixels lower quality than 1024 x 768
    • As the number of pixels increases so does the storage space required
    Pixel Pattern using 8x8 grid Pixel Pattern using 16x16 grid
  • 35. Calculating Storage Capacities of Bit Mapped Images
    • Storage Requirements = total number of pixels * number of bits used for each pixel
    • This picture of Mr Haggarty has a resolution of 300dpi. The image is 2 inches by 4 inches in 128 colours
    • 300 X 2 = width 600 pixels
    • 300 X 4 = height 1200 pixels
    • Total pixels = 600 X 1200 = 720,000 pixels
    • Each pixel = 7 bits i.e. 2 = 128 colours
    • 720,000 X 7 = 5,040,000 bits / 8 = 630,000 bytes
    • 630,000 / 1024 = 615Kb
    7
  • 36. Bit Map V Vector Graphics
    • Bit Map Graphic
      • Bit map packages paint pictures by changing the colour of the pixels
      • Known as “Paint Packages”
      • When shapes overlap, the one on top rubs out the other
      • When you save a file the whole screen is saved
      • The resolution of the image is fixed when you create the image
    • Vector Graphic
      • Work by drawing objects on the screen
      • Known as “Draw Packages”
      • When shapes overlap they remain as separate objects
      • Only the object attributes are stored taking up much less space
      • Resolution Independent
  • 37. Aims of Lesson 6
    • Last Lessons
    • Representing positive whole numbers as binary
    • Representing negative whole numbers using 2s complement
    • Non whole numbers using mantissa and exponent
    • Storage calculations
    • Looked at how text is represented using ASCII and Unicode
    • Discuss graphic representation
    • Calculate storage capacities of colour Bit Map graphics
    • Bit Map v Vector
    • Today’s Lesson
    • Discuss true colour
    • Today’s Tasks
    • Complete Data Representation Questions
    • Read chapter in the book
  • 38. True Colour
    • Bit Depth (Colour Depth)
      • The number of bits used to represent colours in the graphic
        • 1 bit = black or white
        • 2 bits = 4 colours
        • 3 bits = 8 colours
        • 8 bits = 256 colours
        • 24 bits = 16,777,216 colours this is true colour
    • True Colour
      • 24 bits
        • 8 bits for red
        • 8 bits for blue
        • 8 bits for green
    Bit Depth = 1 bit Human eye cannot distinguish between adjacent shades of grey when looking at more than 200 shades between black and white Bit Depth = 2 bit
  • 39. Bit Depths Bit Depth = 2 bits 01 10 11 00
  • 40. Solutions
    • Question 1
    • 2 inches X 90 = 180 pixels
    • 2 inches X 90 = 180 pixels
    • 180 X 180 = 32,400 pixels in total
    • 256 colours = 2 power 8
    • 32,400 X 8 = 259,200 bits
    • 259,200/8 = 32,400 bytes
    • 32,400 / 1024 = 31.6Kb
    • Question 2
    • 5 inches X 200 = 1000 pixels
    • 3 inches X 200 = 600 pixels
    • 1000 X 600 = 600,000 pixels in total
    • 128 colours = 2 power 7
    • 600,000 X 7 = 4,200,000 bits
    • 4,200,000/8 = 525,000 bytes
    • 525,000 / 1024 = 512.7Kb
  • 41. Aims of Lesson 7
    • Last Lessons
    • Representing positive whole numbers as binary
    • Representing negative whole numbers using 2s complement
    • Non whole numbers using mantissa and exponent
    • Storage calculations
    • Looked at how text is represented using ASCII and Unicode
    • Discuss graphic representation
    • Calculate storage capacities of colour Bit Map graphics
    • Bit Map v Vector
    • True Colour
    • Today’s Lesson
    • Data Compression
    • Today’s Tasks
    • Complete Compression task
    • Issue Scholar logins
    • Complete Data Representation Questions Sheet
    • Read chapter in the book
  • 42. Compression
    • Data compression means reducing the size of a file in order to save backing storage space.
    • 2 types of compression
      • Lossless compression
      • Lossy compression
  • 43. Lossless Compression
    • Lossless means that none of the original data is lost
    • One method of lossless compression involves counting repeating pixels
    COLOUR = 10011000 11100000 e.g. 16 bits NUMBER OF THE SAME PIXELS = 32 100000 STORAGE REQUIRED = 16 BITS + 6 BITS = 22 BITS
  • 44. Lossy Compression
    • Lossy compression involves sacrificing some of the data in order to reduce the file size
    • Deliberately losing some types of information that our eyes and brains usually ignore
    • Lossy is only suitable if the loss of data will not cause the file to become useless
    • JPEG is a file format that uses lossy compression to reduce file sizes
  • 45. Data Representation – Learning Aims
    • Representation of positive numbers in binary up to 32 bits
    • Conversion from binary to decimal and vice versa
    • Representation of negative numbers using 2s complement
    • Representation of non whole numbers using floating point with mantissa and exponent
    • Conversion to and from bit, byte, kilobyte, megabyte, gigabyte, terabyte
  • 46. Data Representation – Learning Aims
    • Unicode and its advantages over ASCII
    • Description of the bit map method of graphics representation
    • Description of the relationship between bit depth and the number of colours represented up to 24 bit depth
    • Vector graphics
    • Relationship between bit depth and file size