Time and Frequency Domain in Computer Networks

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

2. Computer Networks Chapter 3 – Time and Frequency Domain Concept, Transmission Impairments

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

4. 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 • signals don’t repeat according to any particular pattern Computer Networks

5. 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

6. Figure 3.6Sine wave examples Computer Networks

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

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

9. 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

10. Figure 3.5Relationships between different phases Computer Networks

11. 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

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

13. 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

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

15. Figure 3.7Time and frequency domains Computer Networks

16. Example: Sine waves Frequency domain Time domain f t f1 T1=1/f1 t f 3f1 T2=1/3f1 f t 5f1 T5=1/f5 Computer Networks

17. Example: A signal with frequency 0 Frequency domain Time domain . . . f t 0 Computer Networks

18. Figure 3.8Square wave Computer Networks

19. Figure 3.9Three harmonics Computer Networks

20. Figure 3.10Adding first three harmonics Computer Networks

21. Figure 3.11Frequency spectrum comparison Computer Networks

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

23. 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

24. Examples • If digital signal has bit rate of 2000 bps, what is the duration of each bit? bit interval = 1/2000 = 0.0005 = 500ms • 2.If a digital signal has a bit interval of 400 ns, what is the bit rate? • bit rate = 1/(400 ·10-9) = 25 ·106 = 25 Mbps Computer Networks

25. Bandwidth Requirements for a Digital Signal 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 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 = fh- fl 20 = 60 - fl fl = 60 - 20 = 40 Hz Computer Networks

31. Figure 3.15Example 4 Computer Networks

32. 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

33. Figure 3.16A digital signal Computer Networks

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

35. Figure 3.17Bit rate and bit interval Computer Networks

36. 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

37. Filtering the Signal • Filtering is equivalent to cutting all the frequiencies outside the band of the filter • Types of filters • Low pas • 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

38. 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

39. Figure 3.18Digital versus analog Computer Networks

40. Table 3.12 Bandwidth Requirement Computer Networks

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

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

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

44. Note: Digital transmission needs a low-pass channel. Computer Networks

45. Note: Analog transmission can use a band-pass channel. Computer Networks

46. Figure 3.20Impairment types Computer Networks

47. Figure 3.21Attenuation Computer Networks

48. 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

49. 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

50. 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