1. Part II. Physical Layer and Media Chapter 3. Data and Signals COMP 3270 Computer Networks Computing Science Thompson Rivers University

2. Learning Objectives Interpret three characteristics of a sine wave. Explain the basic idea of encoding/decoding. Define the time domain plot and the frequency domain plot. Interpret the concept of bandwidth. Interpret a composite signal. Relate the bandwidth of a signal and the bandwidth of a transmission medium when data is transmitted. List the types of noise. Use decibel to calculate the attenuation of a signal. Use Nyquist bit rate and Shannon capacity formulas to determine the appropriate bit rate and signal level when a channel is given. Interpret the concept of bandwidth-delay product.

3. Position of the Physical Layer

4. Main responsibility/role: node-to-node delivery of bits

5. To be transmitted, data (i.e., bits) must be transformed to electric or electromagnetic signals. Note:

6. 1. Analog and Digital Analog and Digital Data Analog and Digital Signals Periodic and Aperiodic Signals

7. Analog and Digital Data

8. Signals can be analog or digital. Analog signals can have an infinite number of values (energy strength) in a range; digital signals can have only a limited number of values. Analog and Digital Signals

9. AC electric signals, Electromagnetic signals DC electric signals

10. In data communication, we commonly use periodic analog signals and aperiodic digital signals.

11. 2. Periodic Analog Signals Sine Wave Phase Examples of Sine Waves Time and Frequency Domains Composite Signals Bandwidth

12. Three characteristics: Peak Amplitude Frequency Phase Sine Wave A most fundamental form of a periodic analog signal

13. Amplitude: energy level of a signal

14. 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. The number of sine waves in a second

15. Frequency and period are inverses of each other.

16. Period and frequency f = 1/T, T = 1/f

18. Units of periods and frequencies UnitEquivalentUnitEquivalent Seconds (s)1 shertz (Hz)1 Hz Milliseconds (ms)10 –3 skilohertz (KHz)10 3 Hz Microseconds (μs)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

19. Example Express a period of 100 ms in microseconds, and express the corresponding frequency in kilohertz. Solution From the previous table 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

20. If a signal does not change at all, its frequency is zero. If a signal changes instantaneously, its frequency is infinite.

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

22. Relationships between different phases

23. Example 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 = 1.046 rad

24. Wavelength

25. Sine wave examples s(t) = A sin (2π f t + φ)

28. An analog signal is best represented in the frequency domain plot. Time and Frequency Domain Plots

31. A single-frequency sine wave is not useful in data communications; we need to change one or more of its characteristics to make it useful. ☺ What characteristics? ☺ Basic idea of encoding/decoding? The change of a characteristic Composite Signals

32. When we change one or more characteristics of a single-frequency signal, it becomes a composite signal made of many frequencies. Any non-single-frequency sine wave is a composite signal.

33. According to Fourier analysis, any composite signal (even digital signal) can be represented as a combination of simple sine waves with different frequencies, phases, and amplitudes.

34. Example: Square wave (electric signal with two different voltage levels)

35. The first three sine waves in the Fourier transform of the previous square wave The sine waves in the Fourier transform of the previous square wave

37. Adding first three harmonics Not the exact original square wave! It is still possible to decode the information.

38. The time and frequency domains of a nonperiodic signal

39. Signal corruption A transmission medium can pass only some frequencies, not all of them. ☺ ☺ What does this mean?

40. Intermediate summary Two types of signals o Analog and digital signal Three characteristics of analog signals o Peak amplitude, frequency, phase The basic idea of encoding/decoding Composite signals A transmission medium can pass only some frequencies, not all of them.

41. 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. Frequencies of at least one half of the peak amplitude. Bandwidth

42. The bandwidth is also a property of a composite signal: It is the difference between the highest and the lowest frequencies.

43. We use the term bandwidth to refer to the property of a medium, or the property of a medium, or the width of a single spectrum. the width of a single spectrum.

44. Composite signal at the destination Composite signal from the source

45. Example 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 (see Figure 13.4 )

47. Example 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

49. Example 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.

50. Bandwidth of transmission media Bandwidth of a signal

51. Intermediate summary Two types of signals o Analog and digital signal Three characteristics of analog signals o Peak amplitude, frequency, phase The basic idea of encoding/decoding -> Composite signals -> A transmission medium can pass only some frequencies, not all of them. -> Bandwidths of transmission media

52. 3. Digital Signals Bit Interval and Bit Rate As a Composite Analog Signal Through Wide-Bandwidth Medium Through Band-Limited Medium

53. Examples: Send bits using different amplitude levels.

54. Bit Rate and Bit Interval Bit rate = 1 / bit interval

55. Example A digital signal has a bit rate of 2000 bps (bits per second). 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 = 0.5 ms = 0.000500 x 10 6  s = 500  s

56. A digital signal is a composite signal with an infinite bandwidth. Digital Signal as a Composite Analog Signal

57. Low-pass and band-pass channels (links) More popular

58. Digital transmission needs a low-pass channel. ☺ Why? ☺ How to send digital information on a band-pass channel? We will discuss this issue later.

59. 4. Transmission Impairment Transmission impairment: Signals are changed during transmission. Attenuation Distortion Noise

60. Impairment types

61. Attenuation Loss of energy: Converted to heat

62. Decibel The relative strengths of two signals or a signal at two different points dB = 10 × log 10 (P 2 /P 1 )

63. Example Imagine a signal travels through a transmission medium and its power is reduced to half. This means that P2 = 1/2 P1. dB = 10 x log 10 (P2/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

64. Example Imagine a signal travels through an amplifier and its power is increased ten times. This means that P2 = 10 × P1. dB = 10 x log 10 (P2/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

65. Example 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 the next figure, 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.

66. dB = –3 + 7 – 3 = +1

67. Example The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with -0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km? The loss is the cable is -0.3 dB/km x 5 km = -1.5 dB -1.5 dB = 10 log 10 (P 2 /2mW) P 2 = 2mW x 10 -0.15 ≈ 1.4 mW

68. Distortion Composite signals Different frequencies have different propagation speeds. Some frequencies would be blocked by the transmission medium.

69. Noise Thermal noise: the random motion of electrons Induced noise: from outside such as motors Crosstalk: from the adjacent wire Impulse noise: high spikes such as lightning Noise changes amplitude of signals.

70. Signal-to-Noise Ratio (SNR) The power of a signal is 10 mW and the power of the noise is 1 μW; what are the values of SNR and SNR dB ? Solution: The values of SNR and SNR dB can be calculated as follows: µ

71. The effects of different SNRs

72. 5. Data Rate Limit Noiseless Channel: Nyquist Bit Rate Noisy Channel: Shannon Capacity Using Both Limits

73.  We need to modulate signals to send 0s and 1s. The modulated signal becomes a composite signal.  Any transmission medium has a bandwidth, i.e., low and high frequencies cannot faithfully pass through the medium.  There are always noise and signal distortion. That can be expressed as SNR.  When a transmission medium is given, its bandwidth and SNR can be obtained from experiments.  What is the capacity of a given medium?  What is the proper number of signal levels in digital transmission to use the capacity? Intermediate review

74. Noisy channel: Shannon capacity formula Capacity = Bandwidth x log 2 (1 + SNR), where SNR is the signal-to-noise ratio. Each transmission link has its own SNR and bandwidth, i.e. fixed capacity.

75. Example Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity is calculated as C = B log 2 (1 + SNR) ≈ B log 2 (1 + 0) = B log 2 (1) = B  0 = 0

76. Example We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 Hz to 3300 Hz). The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as C = B log 2 (1 + SNR) = 3000 log 2 (1 + 3162) = 3000 log 2 (3163) C = 3000  11.62 = 34,860 bps ☺ ☺ Then, how can ADSL, such as Telus Internet service, send data faster than the above?

77. Noiseless channel: Nyquist bit rate formula BitRate = 2 x Bandwidth x log 2 L, where L is the number of signal levels to represent bits and must be a power of 2. L=2 => 1 bit per a signal level L=4 => 2 bits per a signal level L=8 => 3 bits per a signal level L=16 => 4 bits per a signal level

78. Example Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a digital signal with two signal levels. The maximum bit rate can be calculated as (Note that a voice channel uses a 3KHz bandwidth.) Bit Rate = 2  3,000  log 2 2 = 6,000 bps

79. Example We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many different signal levels do we need?  Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate.  If we use 128 (=2 ??? ) levels, the bit rate is 280 kbps.  If we use 64 (=2 ??? ) levels, the bit rate is 240 kbps.  Then how many signal levels do we need to use? How many bits will be encoded on each signal level?

80. Example We have a channel with a 1 MHz bandwidth. The SNR for this channel is 61; what is the appropriate maximum bit rate and signal levels? BitRate = 2 x Bandwidth x log 2 L; Capacity = Bandwidth x log 2 (1 + SNR) Solution C = B log 2 (1 + SNR) = 10 6 log 2 (1 + 61) = 10 6 log 2 (62) ≈  6 Mbps For the better performance we choose something lower. Then we use the Nyquist bit rate formula to find the number of signal levels. L = 2 -> 2 Mbps = 2  1 MHz  log 2 L L = 4 -> 4 Mbps = 2  1 MHz  log 2 L L = 8 -> 6 Mbps = 2  1 MHz  log 2 L First, we use the Shannon capacity formula to find our upper limit.

81. 6. Performance Throughput Transmission speed, and transmission time Propagation Time Bandwidth-Delay Product

82. Bandwidth Shannon capacity formula Bandwidth in Hz is the range of frequencies contained in a composite signal or the range of frequencies a channel can pass. Bandwidth in bps is the bits per second that a channel or a network can transmit.

83. Throughput Throughput is the actual data transmission rate, not the data transmission speed that the network can achieve.

84. Example The throughput is almost one-fifth of the bandwidth in this case. The utilization rate is ??? A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network? Solution: We can calculate the throughput as

85. Propagation time depends on the medium and the frequency. Network power = Throughput / propagation time Propagation Delay (or Time)

86. Example What are the propagation time and the transmission time for a 2.5-Kbyte message (an e-mail) if the bandwidth of the network is 1 Gbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 10 8 m/s. The total time to send the message to the destination is 50ms + 0.02ms = 50.02ms. Note that in this case, because the message is short and the bandwidth is high, the dominant factor is the propagation time, not the transmission time. The transmission time can be ignored.

87. Bandwidth Delay Product

88. 5 5 5 5 5 5 25

89. The bandwidth-delay product defines the number of bits that can fill the link. This measurement is important when we need to send data in bursts and wait for the acknowledgement of each burst before sending the next one – Chapter 11.