1. 1 Outline Analog and Digital Data Analog and Digital Signals Amplitude Modulation (AM) Frequency Modulation (FM)

2. 2 What is a Signal ? Signals can be analog or digital. Analog signals can have an infinite number of values in a range; digital signals can have only a limited number of values.

3. 3 WCB/McGraw-Hill  The McGraw-Hill Companies, Inc., 1998 Periodic Signals

4. 4 WCB/McGraw-Hill  The McGraw-Hill Companies, Inc., 1998 Aperiodic Signals

5. 5 Analog and Digital Signals In data communication, we commonly use periodic analog signals and aperiodic digital signals.

6. 6 Periodic signal: Sine Wave – Period(T) & Frequency(F) T = 1 / F Peak Amplitude

7. 7 Frequency is the rate of change with respect to time. Change in a short span of time means high frequency. Change over a long span of time means low frequency. Frequency

8. 8 Frequencies UnitEquivalentUnitEquivalent Seconds (s)1 shertz (Hz)1 Hz Milliseconds (ms)10 –3 skilohertz (KHz)10 3 Hz Microseconds (ms)10 –6 smegahertz (MHz)10 6 Hz Nanoseconds (ns)10 –9 sgigahertz (GHz)10 9 Hz Picoseconds (ps)10 –12 sterahertz (THz)10 12 Hz

9. 9 Example 1 Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz. Solution From Table 3.1 we find the equivalent of 1 ms.We make the following substitutions: 100 ms = 100  10 -3 s = 100  10 -3  10   s = 10 5  s Now we use the inverse relationship to find the frequency, changing hertz to kilohertz 100 ms = 100  10 -3 s = 10 -1 s f = 1/10 -1 Hz = 10  10 -3 KHz = 10 -2 KHz

10. 10 Phase Phase describes the position of the waveform relative to time zero.

11. 11 Example 2 A sine wave is offset one-sixth of a cycle with respect to time zero. What is its phase in degrees and radians? Solution We know that one complete cycle is 360 degrees. Therefore, 1/6 cycle is (1/6) 360 = 60 degrees = 60 x 2  /360 rad =  /3 rad or 1.046 rad

12. 12 where A = Amplitude f = frequency  = phase Time-Domain Signal Representation

13. 13

14. 14 An analog signal is best represented in the frequency domain.

15. 15 Time-Frequency Domain

16. 16 Time-Frequency Domain

17. 17 Time-Frequency Domain

18. 18 WCB/McGraw-Hill  The McGraw-Hill Companies, Inc., 1998 Examples

19. 19 Fourier Decomposition for Periodic Signals

20. 20 Example(1)

21. 21 Example(2)

22. 22 Example(3)

23. 23 Frequency Content of a Square Wave

24. 24 Figure 4-19 WCB/McGraw-Hill  The McGraw-Hill Companies, Inc., 1998 Harmonics of a Digital Signal

25. 25 Transmission Medium Imperfection The bandwidth is a property of a medium: It is the difference between the highest and the lowest frequencies that the medium can satisfactorily pass.

26. 26 Frequency Response of a Medium Signal frequency content Frequency response of the medium

27. 27 Example 3 If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is the bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V. Solution B = f h  f l = 900  100 = 800 Hz The spectrum has only five spikes, at 100, 300, 500, 700, and 900

28. 28

29. 29 Example 4 A signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all integral frequencies of the same amplitude. Solution B = f h  f l 20 = 60  f l f l = 60  20 = 40 Hz

30. 30

31. 31 Example 5 A signal has a spectrum with frequencies between 1000 and 2000 Hz (bandwidth of 1000 Hz). A medium can pass frequencies from 3000 to 4000 Hz (a bandwidth of 1000 Hz). Can this signal faithfully pass through this medium? Solution The answer is definitely no. Although the signal can have the same bandwidth (1000 Hz), the range does not overlap. The medium can only pass the frequencies between 3000 and 4000 Hz; the signal is totally lost.

32. 32 Digital Signals

33. 33 Bit rate (R) & period of a bit (T) The capacity of a data communication system can be expressed in terms of the number of data bits sent per second in time. – bits per second or (bits/sec) This is also refers to the data rate (R) of the system – speed of transmission. If the period of a data bit is T then R is the inverse of T. The period of data bit is called bit interval.

34. 34 Example 6 A digital signal has a bit rate of 2000 bps. What is the duration of each bit (bit interval) Solution The bit interval is the inverse of the bit rate. Bit interval = 1/ 2000 s = 0.000500 s = 0.000500 x 10 6  s = 500  s

35. 35

36. 36 Transmission Impairment Attenuation Distortion Noise

37. 37 Attenuation

38. 38 Imagine a signal travels through a transmission medium and its power is reduced to half. This means that P2 = 1/2 P1. In this case, the attenuation (loss of power) can be calculated as Solution 10 log 10 (P2/P1) = 10 log 10 (0.5P1/P1) = 10 log 10 (0.5) = 10(–0.3) = –3 dB 10 log 10 (P2/P1) = 10 log 10 (0.5P1/P1) = 10 log 10 (0.5) = 10(–0.3) = –3 dB Example 12

39. 39 Example 13 Imagine a signal travels through an amplifier and its power is increased ten times. This means that P2 = 10 ¥ P1. In this case, the amplification (gain of power) can be calculated as 10 log 10 (P2/P1) = 10 log 10 (10P1/P1) 10 log 10 (P2/P1) = 10 log 10 (10P1/P1) = 10 log 10 (10) = 10 (1) = 10 dB = 10 log 10 (10) = 10 (1) = 10 dB

40. 40 Example 14 One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are talking about several points instead of just two (cascading). In Figure 3.22 a signal travels a long distance from point 1 to point 4. The signal is attenuated by the time it reaches point 2. Between points 2 and 3, the signal is amplified. Again, between points 3 and 4, the signal is attenuated. We can find the resultant decibel for the signal just by adding the decibel measurements between each set of points.

41. 41 dB = –3 + 7 – 3 = +1

42. 42 Distortion

43. 43 Noise

44. 44 Analog Modulation

45. 45 Amplitude Modulation

46. 46 Amplitude Modulation

47. 47 Amplitude Modulation The total bandwidth required for AM can be determined from the bandwidth of the audio signal: rule-of thumb: BWt = 2 x BWm.

48. 48 AM band allocation

49. 49 Example 13 We have an audio signal with a bandwidth of 4 KHz. What is the bandwidth needed if we modulate the signal using AM? Solution An AM signal requires twice the bandwidth of the original signal: BW = 2 x 4 KHz = 8 KHz

50. 50 Frequency Modulation

51. 51 Frequency Modulation

52. 52 Frequency Modulation The total bandwidth required for FM can be determined from the bandwidth of the audio signal: rule-of thumb: BWt = 10 x BWm.

53. 53 Frequency Modulation The bandwidth of a stereo audio signal is usually 15 KHz. Therefore, an FM station needs at least a bandwidth of 150 KHz. The FCC requires the minimum bandwidth to be at least 200 KHz (0.2 MHz).

54. 54 FM Band Allocation

55. 55 Example 14 We have an audio signal with a bandwidth of 4 MHz. What is the bandwidth needed if we modulate the signal using FM? Solution An FM signal requires 10 times the bandwidth of the original signal: BW = 10 x 4 MHz = 40 MHz

56. 56 Tutorial and Examples 1- Describe the three characteristics of a sine wave. a. amplitude, b. frequency/period, c. phase

57. 57 Tutorial and Examples 2- Describe a sine signal using the unit circle. where x is in radians.

58. 58 Tutorial and Examples 2- Describe a cosine signal using the unit circle. where x is in radians.

59. 59 Tutorial and Examples 3-What are the amplitude, phase, and frequency characteristics of the cosine signal that is shown below. Amplitude= 10 Frequency=100 Hz Phase= Π/3 radians=60 degrees

60. 60 Tutorial and Examples 4-What is the Fourier transform of the signal that is shown below. Dirac-Delta function Dirac-Delta function is very useful as an approximation for tall narrow spike functions.

61. 61 Properties of Dirac-Delta functions: 1- 2- 0 As red lines approaches to zero, the green line approaches to infinity.

62. 62 t x(t) AcAc 0

63. 63 Tutorial and Examples 5-What is the Fourier transform of the cosine signal shown below? Euler’s Equations

64. 64 To prove Euler’s equations

65. 65 Hence, the Fourier transform of :