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

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

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 Summer 2006Computer Networks3 Figure 3.1 Comparison of analog and digital signals

4 Summer 2006Computer Networks4  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  signals don’t repeat according to any particular pattern Periodic vs. Non Periodic Signals

5 Summer 2006Computer Networks5 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)

6 Summer 2006Computer Networks6 Figure 3.6 Sine wave examples

7 Summer 2006Computer Networks7 Figure 3.6 Sine wave examples (continued)

8 Summer 2006Computer Networks8 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.

9 Summer 2006Computer Networks9 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.

10 Summer 2006Computer Networks10 Figure 3.5 Relationships between different phases

11 Summer 2006Computer Networks11 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

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

13 Summer 2006Computer Networks13 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

14 Summer 2006Computer Networks14 Figure 3.7 Time and frequency domains (continued)

15 Summer 2006Computer Networks15 Figure 3.7 Time and frequency domains

16 Summer 2006Computer Networks16 Example: Sine waves Time domain t t t f f T 1 =1/f 1 T 5 =1/f 5 T 2 =1/3f 1 3f 1 5f 1 Frequency domain f f1f1

17 Summer 2006Computer Networks17 Example: A signal with frequency 0 Time domain t Frequency domain f 0...

18 Summer 2006Computer Networks18 Figure 3.8 Square wave

19 Summer 2006Computer Networks19 Figure 3.9 Three harmonics

20 Summer 2006Computer Networks20 Figure 3.10 Adding first three harmonics

21 Summer 2006Computer Networks21 Figure 3.11 Frequency spectrum comparison

22 Summer 2006Computer Networks22 Example: Square Wave Square wave with frequency f o Component 1: Component 5: Component 3:

23 Summer 2006Computer Networks23 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

24 Summer 2006Computer Networks24 Examples 1. If digital signal has bit rate of 2000 bps, what is the duration of each bit? bit interval = 1/2000 = = 500  s 2. If a digital signal has a bit interval of 400 ns, what is the bit rate? bit rate = 1/(400 ·10 -9 ) = 25 ·10 6 = 25 Mbps

25 Summer 2006Computer Networks25 Bandwidth Requirements for a Digital Signal Bit Rate Harmonic 1 Harmonics 1, 3 Harmonics 1, 3, 5 Harmonics 1, 3, 5, 7 1 Kbps500 Hz1.5 KHz2.5 KHz3.5 KHz 10 Kbps5 KHz15 KHz25 KHz35 KHz 100 Kbps50 KHz150 KHz250 KHz350 KHz

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 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

31 Summer 2006Computer Networks31 Figure 3.15 Example 4

32 Summer 2006Computer Networks32 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.

33 Summer 2006Computer Networks33 Figure 3.16 A digital signal

34 Summer 2006Computer Networks34 A digital signal is a composite signal with an infinite bandwidth. Note:

35 Summer 2006Computer Networks35 Figure 3.17 Bit rate and bit interval

36 Summer 2006Computer Networks36 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

37 Summer 2006Computer Networks37 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 pas  Band pass  High pass f f

38 Summer 2006Computer Networks38 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.

39 Summer 2006Computer Networks39 Figure 3.18 Digital versus analog

40 Summer 2006Computer Networks40 Table 3.12 Bandwidth Requirement Bit Rate Harmonic 1 Harmonics 1, 3 Harmonics 1, 3, 5 Harmonics 1, 3, 5, 7 1 Kbps500 Hz2 KHz4.5 KHz8 KHz 10 Kbps5 KHz20 KHz45 KHz80 KHz 100 Kbps50 KHz200 KHz450 KHz800 KHz

41 Summer 2006Computer Networks41 The bit rate and the bandwidth are proportional to each other. Note:

42 Summer 2006Computer Networks42 The analog bandwidth of a medium is expressed in hertz; the digital bandwidth, in bits per second. Note:

43 Summer 2006Computer Networks43 Figure 3.19 Low-pass and band-pass

44 Summer 2006Computer Networks44 Digital transmission needs a low-pass channel. Note:

45 Summer 2006Computer Networks45 Analog transmission can use a band- pass channel. Note:

46 Summer 2006Computer Networks46 Figure 3.20 Impairment types

47 Summer 2006Computer Networks47 Figure 3.21 Attenuation

48 Summer 2006Computer Networks48 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.

49 Summer 2006Computer Networks49 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.

50 Summer 2006Computer Networks50 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

51 Summer 2006Computer Networks51 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, symbolized S dBP, is: S dBP = 10 log 10 (P 2 / P 1 )  As a rule of thumb: S/N ratio of 10dB means 10/1 S/N ratio of 20dB means 100/1 S/N ratio of 30dB means 1000/1 S/N ratio of 40dB means 10000/1

52 Summer 2006Computer Networks52 Examples: 1. A signal that travels through a transmission medium is reduced to half. This means that P 2 = (1/2)P 1 The attenuation can be calculated as follows: 10log 10 (P 2 /P 1 )=10 log 10 (0.5 P 1 /P 1 )=10log 10 (0.5)=  3 dB 2. Imagine a signal goes through an amplifier and its power is increased 10 times. This means that P 2 = 10P 1 The amplification is: 10 log 10 (10 P 1 /P 1 )= =10log 10 10=10dB

53 Summer 2006Computer Networks53 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

54 Summer 2006Computer Networks54 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

55 Summer 2006Computer Networks55 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.

56 Summer 2006Computer Networks56 Figure 3.22 Example 14 dB = –3 + 7 – 3 = +1

57 Summer 2006Computer Networks57 Figure 3.23 Distortion

58 Summer 2006Computer Networks58 Figure 3.24 Noise

59 Summer 2006Computer Networks59 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

60 Summer 2006Computer Networks60 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.

61 Summer 2006Computer Networks61

62 Summer 2006Computer Networks62 Genomströmningshastighet (throughput)

63 Summer 2006Computer Networks63 Figure 3.26 Propagation time

64 Summer 2006Computer Networks64 Delay (Time, Latency)  When data are sent from one point to the other point (without intermediate points), two types of delays are experienced:  transmission delay (time)  propagation delay (time)  When data pass through intermediate points four types of delay (latency) are experienced:  transmission delay (time)  propagation delay (time)  queue time  processing time

65 Summer 2006Computer Networks65 Transmission Delay (Time)  The transmission time is the time necessary to put the message on the link (chanel).  The transmission time depends on the length of the message and the throughput (bit rate) of the link and is expressed as: length of message (bits) bit rate (bits/sec)

66 Summer 2006Computer Networks66 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

67 Summer 2006Computer Networks67 Queue and Processing time  Queue time  When the intermediate nodes are busy processing other data, the data arrived at the node are queued. Queue time is the time spent waiting in the queue.  Processing time  This is the time needed for the data to be processed at the intermediate nodes.


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