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

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Presentation on theme: "Datornätverk A – lektion 3 MKS B – lektion 3  Kapitel 3: Fysiska signaler.  Kapitel 4: Digital transmission."— Presentation transcript:

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

2 Summer 2006Computer Networks2 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.

3 Summer 2006Computer Networks3 Physical Layer PART II

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

5 Summer 2006Computer Networks5 Figure 3.1 Comparison of analog and digital signals

6 Summer 2006Computer Networks6 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. Note:

7 Summer 2006Computer Networks7  Periodic signal  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 Periodic vs. Non Periodic Signals

8 Summer 2006Computer Networks8 Sinusvågor Periodtid T = t 2 - t 1. Enhet: s. Frekvens f = 1/T. Enhet: 1/s=Hz. T=1/f. Amplitud eller toppvärde Û. Enhet: Volt. Fasläge: θ = 0 i ovanstående exempel. Enhet: Grader eller radianer. Momentan spänning: u(t)= Ûsin(2πft+θ)

9 Summer 2006Computer Networks9 Figure 3.4 Period and frequency

10 Summer 2006Computer Networks10 Tabell 3.1 Enheter för periodtid och frekvens EnhetEkvivalentEnhetEkvivalent Sekunder (s)1 sHertz (Hz)1 Hz Millisekunder (ms)10 –3 sKilohertz (kHz)10 3 Hz Mikrosekunder (μs)10 –6 sMegahertz (MHz)10 6 Hz Nanosekunder (ns)10 –9 sGigahertz (GHz)10 9 Hz Pikosekunder (ps)10 –12 sTerahertz (THz)10 12 Hz Exempel: En sinusvåg med periodtid 1 ns har frekvens 1 GHz.

11 Summer 2006Computer Networks11 Figure 3.6 Sine wave examples

12 Summer 2006Computer Networks12 Figure 3.6 Sine wave examples (continued)

13 Summer 2006Computer Networks13 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.

14 Summer 2006Computer Networks14 Figure 3.5 Relationships between different phases

15 Summer 2006Computer Networks15 Measuring the Phase  The phase is measured in degrees or in radians.  One full cycle is 360 o 360 o (degrees) = 2  (radians)  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 = rad

16 Summer 2006Computer Networks16 Figure 3.6 Sine wave examples (continued)

17 Summer 2006Computer Networks17 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 = rad

18 Summer 2006Computer Networks18 Figure 3.7 Time and frequency domains (continued)

19 Summer 2006Computer Networks19 Figure 3.7 A DC signal (likspänning), i.e. a signal with frequency 0 Hz

20 Summer 2006Computer Networks20 Figure 3.8 Square wave

21 Summer 2006Computer Networks21 Figure 3.9 Three harmonics

22 Summer 2006Computer Networks22 Figure 3.10 Adding first three harmonics

23 Summer 2006Computer Networks23 Figure 3.11 Frequency spectrum comparison

24 Summer 2006Computer Networks24 Example: Square Wave Square wave with frequency f o Component 1: Component 5: Component 3:

25 Summer 2006Computer Networks25 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

26 Summer 2006Computer Networks26 Figure 3.12 Signal corruption

27 Summer 2006Computer Networks27 Figure 3.13 Bandwidth

28 Summer 2006Computer Networks28 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 (see Figure 13.4 )

29 Summer 2006Computer Networks29 Figure 3.14 Example 3

30 Summer 2006Computer Networks30 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.

31 Summer 2006Computer Networks31 Figure 3.16 A digital signal

32 Summer 2006Computer Networks32 A digital signal is a composite signal with an infinite bandwidth. Note:

33 Summer 2006Computer Networks33 Figure 3.17 Bit rate and bit interval

34 Summer 2006Computer Networks34 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 = s = x 10 6  s = 500  s

35 Summer 2006Computer Networks35 Media Filters the Signal Media INPUTOUTPUT Certain frequencies do not pass through What happens when you limit frequencies? Square waves (digital values) lose their edges -> Harder to read correctly.

36 Summer 2006Computer Networks36 Table 3.12 Bandwidth Requirement Bit Rate Harmonic 1 Harmonics 1, 3 Harmonics 1, 3, 5 Harmonics 1, 3, 5, 7 1 Kbps500 Hz1,5 KHz3 KHz4,5 KHz 10 Kbps5 KHz15 KHz30 KHz45 KHz 100 Kbps50 KHz150 KHz300 KHz450 KHz

37 Summer 2006Computer Networks37 The bit rate and the bandwidth are proportional to each other. Note:

38 Summer 2006Computer Networks38 The analog bandwidth of a medium is expressed in hertz; the digital bandwidth, in bits per second. Note:

39 Summer 2006Computer Networks39 Figure 3.19 Low-pass and band-pass

40 Summer 2006Computer Networks40 Filtering the Signal  Filtering is equivalent to cutting all the frequiencies outside the band of the filter High pass INPUT S 1 (f) H(f) OUTPUT S 2 (f)= H(f)*S 1 (f) Low pass INPUT S 1 (f) H(f) f OUTPUT S 2 (f)= H(f)*S 1 (f) Band pass INPUT S 1 (f) H(f) OUTPUT S 2 (f)= H(f)*S 1 (f)  Types of filters  Low pass  Band pass  High pass f f

41 Summer 2006Computer Networks41 Digital transmission (without modulation) needs a low-pass channel. Note:

42 Summer 2006Computer Networks42 Analog transmission (by means of modulation) can use a band-pass channel. Note:

43 Summer 2006Computer Networks43 Figure 3.21 Attenuation

44 Summer 2006Computer Networks44 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 ggr effektförstärkning = 30 dB. Osv.

45 Summer 2006Computer Networks45 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.

46 Summer 2006Computer Networks46 Measurement of Attenuation  Signal attenuation is measured in units called decibels (dB).  If over a transmission link the ratio of output power is P o /P i, the attenuation is said to be –10log 10 (P o /P i ) = 10log 10 (P i /P o ) 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

47 Summer 2006Computer Networks47 What is dB?  A decibel is 1/10th of a Bel, abbreviated dB  Suppose a signal has a power of P 1 watts, and a second signal has a power of P 2 watts. Then the power amplitude difference in decibels, is: 10 log 10 (P 2 / P 1 )  As a rule of thumb: 10dB means power ratio 10/1 20dB means power ratio 100/1 30dB means power ratio 1000/1 40dB means power ratio 10000/1

48 Summer 2006Computer Networks48 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 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

49 Summer 2006Computer Networks49 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

50 Summer 2006Computer Networks50 Figure 3.22 Example 14: In cascaded links the amplification in dB is simply a sum of the individual amplifications in dB. –3dB + 7dB – 3dB = +1dB Total amplification:

51 Summer 2006Computer Networks51 Figure 3.23 Symbol distortion

52 Summer 2006Computer Networks52 Figure 3.24 Noise

53 Summer 2006Computer Networks53 Noise and Interference  Noise is present in the form of random motion of electrons in conductors, devices and electronic systems (due to thermal energy) and can be also picked up from external sources (atmospheric disturbances, ignition noise etc.)  Interference (cross-talk) generally refers to the unwanted signals, picked up by communication link due to other transmissions taking place in adjacent frequency bands or in physically adjacent transmission lines

54 Summer 2006Computer Networks54 Signal-brus-förhållande  Ett signal-brus-förhållande på 100 dB innebär att den starkaste signalen är 100 dB starkare än bruset.  Ljud som är svagare än bruset hörs inte utan dränks i bruset.  Ljudets dynamik skillnaden mellan den starkaste ljudet och det svagaste ljudet som man kan höra, och är vanligen ungefär detsamma som signal- brus-förhållandet.

55 Summer 2006Computer Networks55

56 Summer 2006Computer Networks56 Throughput (Genomströmningshastighet)

57 Summer 2006Computer Networks57 Delay (Time, Latency)  When data are sent from one node to next node (without intermediate points), two types of delays are experienced:  transmission time (Paketsändningstid)  propagation delay (Utbredningsfördröjning)  When data pass through intermediate nodes four types of delay (latency) are experienced:  transmission time  propagation delay  queue time  processing time

58 Summer 2006Computer Networks58 Figure 3.26 Propagation time

59 Summer 2006Computer Networks59 Transmission Time (Paketsändningstid)  The transmission time is the time necessary to put the complete message on the link (channel).  The transmission time depends on the length of the message and the bit rate of the link and is expressed as: length of packet (bits) bit rate (bits/sec)

60 Summer 2006Computer Networks60 Propagation Delay (Time)  The propagation delay is the time needed for the signal to propagate (travel) from one end of a channel to the other.  The transmition time depends on the distance between the two ends and the speed of the signal and is expressed as distance (m) / speed of propagation (m/s)  Through free space signals propagate at the speed of light which is 3 * 10 8 m/s  Through wires signals propagate at the speed of 2 * 10 8 m/s


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