Datornätverk A – lektion 3 MKS B – lektion 3

1. Datornätverk A – lektion 3MKS B – lektion 3 Kapitel 3: Fysiska signaler. Kapitel 4: Digital transmission.

2. Repetition: The TCP/IP model H – header (pakethuvud): control data added at the front end of the data unit T – trailer (svans): control data added at the back end of the data unit Trailers are usually added only at layer 2. Computer Networks

3. PART II Physical Layer Computer Networks

4. Chapter 3 – Time and Frequency Domain Concept, Transmission Impairments

5. Figure 3.1Comparison of analog and digital signals Computer Networks

6. Note: 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. Computer Networks

7. Periodic vs. Non Periodic Signals • Periodicsignal • repeat over and over again, once per period • The period ( T ) is the time it takes to make one complete cycle • Non periodic signal • don’t repeat according to any particular pattern Computer Networks

8. Sinusvågor PeriodtidT = t2 - t1. Enhet: s. Frekvensf = 1/T. Enhet: 1/s=Hz. T=1/f. Amplitud eller toppvärdeÛ. Enhet: Volt. Fasläge: θ = 0 i ovanstående exempel. Enhet: Grader eller radianer. Momentan spänning: u(t)= Ûsin(2πft+θ) Computer Networks

9. Figure 3.4Period and frequency Computer Networks

10. Tabell 3.1 Enheter för periodtid och frekvens Exempel: En sinusvåg med periodtid 1 ns har frekvens 1 GHz. Computer Networks

11. Figure 3.6Sine wave examples Computer Networks

12. Figure 3.6Sine wave examples (continued) Computer Networks

13. Exempel Vilken frekvens i kHz har en sinusvåg med periodtid 100 ms? Lösning Alternativ 1: Gör om till grundenheten. 100 ms = 0.1 s f = 1/0.1 Hz = 10 Hz = 10/1000 kHz = 0.01 kHz Alternativ 2: Utnyttja att 1 ms motsvarar 1 kHz. f = 1/100ms = 0.01 kHz. Computer Networks

14. Figure 3.5Relationships between different phases Computer Networks

15. Measuring the Phase • The phase is measured in degrees or in radians. • One full cycle is 360o 360o (degrees) = 2p (radians) p  3.14 Example: A sine wave is offset one-sixth of a cycle with respect to time 0. What is the phase in radians? Solution:(1/6) 360 = 60 degrees = 60 x 2p /360 rad = 1.046 rad Computer Networks

16. Figure 3.6Sine wave examples (continued) Computer Networks

17. 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 2p /360 rad = 1.046 rad Computer Networks

18. Figure 3.7Time and frequency domains (continued) Computer Networks

19. Figure 3.7A DC signal (likspänning), i.e. a signal with frequency 0 Hz Computer Networks

20. Figure 3.8Square wave Computer Networks

21. Figure 3.9Three harmonics Computer Networks

22. Figure 3.10Adding first three harmonics Computer Networks

23. Figure 3.11Frequency spectrum comparison Computer Networks

24. Example: Square Wave Square wave with frequency fo Component 1: Component 3: Component 5: . . . . . . Computer Networks

25. Characteristic of the Component Signals in the Square Wave • Infinite number of components • Only the odd harmonic components are present • The amplitudes of the components diminish with increasing frequency Computer Networks

26. Figure 3.12Signal corruption Computer Networks

27. Figure 3.13Bandwidth Computer Networks

28. 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 = fh-fl = 900 - 100 = 800 Hz The spectrum has only five spikes, at 100, 300, 500, 700, and 900 (see Figure 13.4 ) Computer Networks

29. Figure 3.14Example 3 Computer Networks

30. 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. Computer Networks

31. Figure 3.16A digital signal Computer Networks

32. Note: A digital signal is a composite signal with an infinite bandwidth. Computer Networks

33. Figure 3.17Bit rate and bit interval Computer Networks

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 106ms = 500 ms Computer Networks

35. Media Filters the Signal Media INPUT OUTPUT Certain frequencies do not pass through What happens when you limit frequencies? Square waves (digital values) lose their edges -> Harder to read correctly. Computer Networks

36. Table 3.12 Bandwidth Requirement Computer Networks

37. Note: The bit rate and the bandwidth are proportional to each other. Computer Networks

38. Note: The analog bandwidth of a medium is expressed in hertz; the digital bandwidth, in bits per second. Computer Networks

39. Figure 3.19Low-pass and band-pass Computer Networks

40. Filtering the Signal • Filtering is equivalent to cutting all the frequiencies outside the band of the filter • Types of filters • Low pass • Band pass • High pass Low pass H(f) OUTPUT S2(f)= H(f)*S1(f) INPUT S1(f) H(f) f Band pass H(f) OUTPUT S2(f)= H(f)*S1(f) INPUT S1(f) H(f) f High pass H(f) INPUT S1(f) H(f) OUTPUT S2(f)= H(f)*S1(f) f Computer Networks

41. Note: Digital transmission (without modulation) needs a low-pass channel. Computer Networks

42. Note: Analog transmission (by means of modulation) can use a band-pass channel. Computer Networks

43. Figure 3.21Attenuation Computer Networks

44. Förstärkning mätt i decibel (dB) 1 gång effektförstärkning = 0 dB. 2 ggr effektförstärkning = 3 dB. 10 ggr effektförstärkning = 10 dB. 100 ggr effektförstärkning = 20 dB. 1000 ggr effektförstärkning = 30 dB. Osv. Computer Networks

45. Dämpning mätt i decibel • Dämpning 100 ggr = Dämpning 20 dB = förstärkning 0.01 ggr = förstärkning med – 20 dB. • Dämpning 1000 ggr = 30 dB dämpning = -30dB förstärkning. • En halvering av signalen = dämpning med 3dB = förstärkning med -3dB. Computer Networks

46. Measurement of Attenuation • Signal attenuation is measured in units called decibels (dB). • If over a transmission link the ratio of output power is Po/Pi, the attenuation is said to be –10log10(Po/Pi) = 10log10(Pi/Po) dB. • In cascaded links the attenuation in dB is simply a sum of the individual attenuations in dB. • dB is negative when the signal is attenuated and positive when the signal is amplified Computer Networks

47. What is dB? • A decibel is 1/10th of a Bel, abbreviated dB • Suppose a signal has a power of P1 watts, and a second signal has a power of P2 watts. Then the power amplitude difference in decibels, is: 10 log10 (P2 / P1) • As a rule of thumb: 10dB means power ratio 10/1 20dB means power ratio 100/1 • 30dB means power ratio1000/1 • 40dB means power ratio10000/1 Computer Networks

48. Example 12 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 log10 (P2/P1) = 10 log10 (0.5P1/P1) = 10 log10 (0.5) = 10(–0.3) = –3 dB Computer Networks

49. 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 log10 (P2/P1) = 10 log10 (10P1/P1) = 10 log10 (10) = 10 (1) = 10 dB Computer Networks

50. Figure 3.22Example 14: In cascaded links the amplification in dB is simply a sum of the individual amplifications in dB. Total amplification: –3dB + 7dB – 3dB = +1dB Computer Networks