1. Topic 14 Digital Technology

2. Analogue and Digital Signals

3. Base 10 5037

4. 10 3 10 2 10 1 10 0 5037

5. Base 10 5037 = 5 x 10 3 + 0 x 10 2 + 3 x 10 1 + 7 x 10 0 10 3 10 2 10 1 10 0 5037

6. Base 2 (binary numbers) 1 x 2 4 + 0 x 2 3 + 1 x 2 2 + 0 x 2 1 + 1 x 2 0 = 21 2626 2525 2424 23232 2121 2020 10101

7. Base 2 (binary numbers) ? 2626 2525 2424 23232 2121 2020 100001

8. 1 x 2 5 + 1 x 2 0 = 33 2626 2525 2424 23232 2121 2020 100001

9. Four-bit words 1 x 2 3 + 1 x 2 0 = 9 2626 2525 2424 23232 2121 2020 1001

10. Four bit words 1001, 1110, etc. A three bit word can be made into a four bit word by adding a zero to the front 4 = 100 (3 bit word) = 0100 (4 bit word)

11. Four bit words With 4 bit words we can represent a total of 16 numbers in binary numbers (0 to 15) Total number of numbers = 2 4 Total number of numbers = 2 (# of bits)

12. Example question How many numbers can be represented with 6 bit words?

13. Example question How many numbers can be represented with 6 bit words? # numbers = 2 6 = 64

14. Another example question Write the decimal number 65 using eight bits.

15. Another example question Write the decimal number 65 using eight bits. 65 = 1x2 6 + 0x2 5 + 0x2 4 + 0x2 3 + 0x2 2 + 0x2 1 + 1x2 0 65 = 1000001 = 01000001 (eight bits)

16. Another example question Write the decimal number 65 using eight bits. 65 = 1x2 6 + 0x2 5 + 0x2 4 + 0x2 3 + 0x2 2 + 0x2 1 + 1x2 0 65 = 1000001 = 01000001 (eight bits) Least significent bit (LSB) Most significent bit (MSB)

17. Analogue signals Analogue signals are continuous signals that vary in proportion to the physical mechanism that created the signal.

18. Digital signal is a coded form of signal that takes the discrete values of 0 or 1 only

19. Converting from analogue to digital

20. Another example t/ms V/V 2 4 6 8 1234

21. Another example t/ms V/V 2 4 6 8 1234 Take a sample of the signal every 1 ms (sampling frequency or rate of 1000 Hz)

22. Another example t/ms V/V 2 4 6 8 1234 This is called a pulse amplitude modulated signal (PAM) Time/msPAM signal /V 00 12 24 36 48

23. Another example t/ms V/V 2 4 6 8 1234 Note the duration of each sample is is very short (1 μs or less) Time/msPAM signal /V 00 12 24 36 48

24. Converting into 2 bit words t/ms V/V 2 4 6 8 1234 Time/msPAM signal /V 00 12 24 36 48

25. Converting into 2 bit words t/ms V/V 2 4 6 8 1234 Binary code PAM signal /V 000 ≤ V < 2 012 ≤ V < 4 104 ≤ V < 6 116 ≤ V < 8 The maximum number of words is 2 2 = 4, so we have to split the voltages into 4 levels (0- 2,2-4,4-6,6-8)

26. Converting into 2 bit words t/ms V/V 2 4 6 8 1234 So in 2-bit words we can transmit the signal as 00011011 Binary code PAM signal /V 000 ≤ V < 2 012 ≤ V < 4 104 ≤ V < 6 116 ≤ V < 8

27. Converting into 2 bit words t/ms V/V 2 4 6 8 1234 Binary code PAM signal /V 0000 ≤ V < 1 0011 ≤ V < 2 0102 ≤ V < 3 0113 ≤ V < 4 1004 ≤ V < 5 1015 ≤ V < 6 1106 ≤ V < 7 1117 ≤ V < 8 We can reduce the loss of information by sampling every 0.5ms and using 3 bit words (maximum of 2 3 = 8)

28. Quantization Binary code PAM signal /V 0000 ≤ V < 1 0011 ≤ V < 2 0102 ≤ V < 3 0113 ≤ V < 4 1004 ≤ V < 5 1015 ≤ V < 6 1106 ≤ V < 7 1117 ≤ V < 8 The process of dividing the range of the analogue system into levels is called quantization. The number of quantization levels used depends on the number of bits used. Number of levels = 2 n

29. Quantization error Binary code PAM signal /V 0000 ≤ V < 1 0011 ≤ V < 2 0102 ≤ V < 3 0113 ≤ V < 4 1004 ≤ V < 5 1015 ≤ V < 6 1106 ≤ V < 7 1117 ≤ V < 8 In this case the quantization error is 1 V. Two analogue signals that differ than less than the quantization error will be assigned the same binary number.

31. Compact disks Paths (1.6μm apart) pits land

32. A spiral of unconnected pits are sensed with a laser beam. The pattern of pits on a CD store information as a series of binary '1's and '0's. As the disc rotates a laser beam is used to produce a '1' every time it finds a pit edge.

33. Destructive interference when path difference is λ/2 (depth is λ/4) This corresponds to a binary ‘1’

35. Example question Calculate the pit depth of a CD being read by a laser of λ = 600 nm.

36. Example question Calculate the pit depth of a CD being read by a laser of λ = 600 nm. Pit depth = λ/4 = 150 nm

37. Another example Information is imprinted on a CD at a rate of 44 100 words per second. The information is in 32-bit words. A CD lasts for 74 minutes. Calculate the storage capacity of the CD

38. Another example Information is imprinted on a CD at a rate of 44 100 words per second. The information is in 32-bit words. A CD lasts for 74 minutes. Calculate the storage capacity of the CD # of bits = 44100 x 32 x 60 x 74 = 6.27 x 10 9 bits 1 byte = 8 bits 6.27 x 10 9 bits = (6.27 x 10 9 )/8 bytes = 780 Mb

39. Questions!