Datornätverk A – lektion 4 MKS B – lektion 4 Kapitel 5: Modulation.

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2 Datornätverk A – lektion 4 MKS B – lektion 4 Kapitel 5: Modulation.

3 4.3 Sampling Pulse Amplitude Modulation Pulse Code Modulation Sampling Rate: Nyquist Theorem How Many Bits per Sample? Bit Rate

4 Sampling och DA- omvandling Microfon- membranets läge Tid Viloläge 3 mm bakom 2 mm framför 2 mm bakom T = 0,4ms 0.1ms 0.2ms 0.3ms0.4ms 0.5ms 0.6ms T s = 0,1ms Nästan sinusformat ljud med periodtidT =0.4ms och frekvens f = 1/T = 2500 Hz = 2,5kHz. Samplingsperiod T s = 0.1ms,dvssamplingsfrekvensf s = 1/T =. Kvantisering (=avrundning) till 8 värden. Digitalisering ger 3 bit per värde: 101 000 010 110 000. 10000 sampels/sek = 10kHz

5 PCM = Pulse Code Modulation = Digitalisering av analoga signaler och seriell överföring Sampler AD-omvand- lare med seriell utsignal 011011010001... DA- omvandlare Antiviknings- filter Interpola- tionsfilter Sifferexempel från PSTN = publika telefonnätet: 3400- 4000Hz filter 8000 sampels per sek 8 bit per sampel dvs 64000 bps per tfnsamtal 2 8 = 256 spänningsnivåer 0 1 Mikrofon Högtalare

6 Exempel En 6 sekunder lång ljudinspelning digitaliseras. Hur stor är inspelningens informationsmängd? a) 22000 sampels/sekund, 256 kvantiseringsnivåer. b) 22000 sampels/sekund, 16 kvantiseringsnivåer. c) 5500 sampels/sekund, 256 kvantiseringsnivåer. 22000sampels * 6 s * 8 bit = 1056000bit. 22000sampels * 6 s * 4 bit = 528000bit. 5500sampels * 6 s * 8 bit = 264000bit.

7 Samplingsteoremet f < f s /2 Den högsta frekvens som kan samplas är halva samplingsfrekvensen. Om man samplar högre frekvens än f s /2 så byter signalen frekvens, dvs det uppstår vikningsdistorsion (aliasing). För att undvika vikningsdistorsion så har man ett anti-vikningsfilter innan samplingen, som tar bort frekvenser över halva samplingsfrekvensen. Interpolationsfiltret används vid rekonstruktion av den digitala signalen för att ”gissa” värden mellan samplen. Ett ideal interpolationsfilter skulle kunna återskapa den samplade signalen perfekt om den uppfyller samplingsteoremet. I verkligheten finns inga ideala filter. Följdregel: Nyqvist’s sats säger att max datahastighet = 2B 2 log M, där M är antal nivåer, och B är signalens bandbredd, oftast lika med signalens övre gränsfrekvens.

8 Figure 4.18 PAM

9 Pulse amplitude modulation has some applications, but it is not used by itself in data communication. However, it is the first step in another very popular conversion method called pulse code modulation. Note:

10 Figure 4.19 Quantized PAM signal

11 Figure 4.20 Quantizing by using sign and magnitude

12 According to the Nyquist theorem, the sampling rate must be at least 2 times the highest frequency. Note:

13 Example 4 What sampling rate is needed for a signal with a bandwidth of 10,000 Hz (1000 to 11,000 Hz)? Solution The sampling rate must be twice the highest frequency in the signal: Sampling rate = 2 x (11,000) = 22,000 samples/s

14 Example 5 A signal is sampled. Each sample requires at least 12 levels of precision (+0 to +5 and -0 to -5). How many bits should be sent for each sample? Solution We need 4 bits; 1 bit for the sign and 3 bits for the value. A 3-bit value can represent 2 3 = 8 levels (000 to 111), which is more than what we need. A 2-bit value is not enough since 2 2 = 4. A 4-bit value is too much because 2 4 = 16.

15 Example 6 We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? Solution The human voice normally contains frequencies from 0 to 4000 Hz. Sampling rate = 4000 x 2 = 8000 samples/s Bit rate = sampling rate x number of bits per sample = 8000 x 8 = 64,000 bps = 64 Kbps

16 Distorsion till följd av digitalisering Vikningsdistorsion ○Inträffar om man inte filtrerar bort frekvenser som är högre än halva samplingsfrekvensen. Kvantiseringsdistorsion (kvantiseringsbrus) ○Avrundningsfelet låter ofta som ett brus. ○Varje extra bit upplösning ger dubbelt så många spänningsnivåer, vilket ger en minskning av kvantiseringsdistorsionen med 6 dB. 16 bit upplösning ger ett signal-brus-förhållande på ca 16*6 = 96 dB (beroende på hur man mäter detta förhållande.) ○Svaga ljud avrundas bort, eller dränks i kvantiseringsbruset.

17 Informationsmängd N bit kan representera 2 N alternativa värden eller koder. Ex: ASCII-kodens 7 bitar kan representera 2 7 = 2·2 ·2 ·2 ·2 ·2 ·2 = 128 tecken. En kod som kan anta M alternativa värden har informationsmängden Ex: ISO-latinkodens 256 tecken kräver 2 log 256 = 8 bit per tecken.

18 Figure 5.25 Types of analog-to-analog modulation

19 Figure 5.26 Amplitude modulation

20 Figure 5.29 Frequency modulation

21 Modulation och demodulation Baudrate = antal symboler per sekund. Enhet: baud eller symboler/sekund. Bitrate = datahastighet. Enhet: bps eller bit/s. Vid många modulationsformer t.ex. s.k. ASK, PSK, och QAM är signalens bandbredd = symbolhastigheten. Vid FSK är bandbredden vanligen större.

22 Digitala modulationsmetoder Binär signal ASK = Amplitude Shift Keying (AM) FSK = Frequency Shift Keying (FM) PSK = Phase Shift Keying (PSK)

23 Exempel 1: Till höger visas fyra symboler som används av ett s.k. 4PSK- modem (PSK=Phase Shift Keying). De fyra symbolerna representerar bitföljderna 00, 01, 11 resp 10. a) Nedan visas utsignalen från det sändande modemet. Vilket meddelande, dvs vilken bitsekvens, överförs? b) Tidsaxeln är graderad i tusendels sekunder. Vad är symbolhastigheten i baud eller symboler/sekund? c) Vad är bithastigheten i bit per sekund (bps)? Svar: 1/1ms = 1000 symber per sekund = 1kbaud. Svar: 2000bps. Svar : 11 00 10 10.

24 Exempel 2: Nedan visas åtta symboler som används av ett s.k. 8QAM-modem (QAM=Quadrature Amplitude Modulation). Symbolerna i övre raden representerar bitföljderna 000, 001, 011 resp 010 (från vänster till höger). Undre raden representerar 100, 101, 111 resp 110.

25 Forts exempel 2:

26 Example 1 An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, find the baud rate and the bit rate Solution Baud rate = 1000 bauds per second (baud/s) Bit rate = 1000 x 4 = 4000 bps

27 Figure 5.4 Relationship between baud rate and bandwidth in ASK

28 Figure 5.3 ASK

29 Example 3 Find the minimum bandwidth for an ASK signal transmitting at 2000 bps. The transmission mode is half- duplex. Solution In ASK the baud rate and bit rate are the same. The baud rate is therefore 2000. An ASK signal requires a minimum bandwidth equal to its baud rate. Therefore, the minimum bandwidth is 2000 Hz.

30 Example 4 Given a bandwidth of 5000 Hz for an ASK signal, what are the baud rate and bit rate? Solution In ASK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But because the baud rate and the bit rate are also the same for ASK, the bit rate is 5000 bps.

31 Example 6 Find the minimum bandwidth for an FSK signal transmitting at 2000 bps. Transmission is in half-duplex mode, and the carriers are separated by 3000 Hz. Solution For FSK BW = baud rate + f c1  f c0 BW = bit rate + fc1  fc0 = 2000 + 3000 = 5000 Hz

32 Example 7 Find the maximum bit rates for an FSK signal if the bandwidth of the medium is 12,000 Hz and the difference between the two carriers is 2000 Hz. Transmission is in full-duplex mode. Solution Because the transmission is full duplex, only 6000 Hz is allocated for each direction. BW = baud rate + fc1  fc0 Baud rate = BW  (fc1  fc0 ) = 6000  2000 = 4000 But because the baud rate is the same as the bit rate, the bit rate is 4000 bps.

33 Figure 5.8 PSK

34 Figure 5.5 Solution to Example 5

35 Figure 5.6 FSK

36 Figure 5.9 PSK constellation

37 Figure 5.10 The 4-PSK method

38 Figure 5.11 The 4-PSK characteristics

39 Figure 5.12 The 8-PSK characteristics

40 Figure 5.13 Relationship between baud rate and bandwidth in PSK

41 Example 9 Given a bandwidth of 5000 Hz for an 8-PSK signal, what are the baud rate and bit rate? Solution For PSK the baud rate is the same as the bandwidth, which means the baud rate is 5000. But in 8-PSK the bit rate is 3 times the baud rate, so the bit rate is 15,000 bps.

42 Quadrature amplitude modulation is a combination of ASK and PSK so that a maximum contrast between each signal unit (bit, dibit, tribit, and so on) is achieved. Note:

43 Figure 5.14 The 4-QAM and 8-QAM constellations

44 Figure 5.15 Time domain for an 8-QAM signal

45 Figure 5.16 16-QAM constellations

46 Figure 5.17 Bit and baud

47 Table 5.1 Bit and baud rate comparison ModulationUnitsBits/Baud Baud rate Bit Rate ASK, FSK, 2-PSK Bit1NN 4-PSK, 4-QAM Dibit2N2N 8-PSK, 8-QAM Tribit3N3N 16-QAMQuadbit4N4N 32-QAMPentabit5N5N 64-QAMHexabit6N6N 128-QAMSeptabit7N7N 256-QAMOctabit8N8N

48 Example 10 A constellation diagram consists of eight equally spaced points on a circle. If the bit rate is 4800 bps, what is the baud rate? Solution The constellation indicates 8-PSK with the points 45 degrees apart. Since 2 3 = 8, 3 bits are transmitted with each signal unit. Therefore, the baud rate is 4800 / 3 = 1600 baud

49 Example 11 Compute the bit rate for a 1000-baud 16-QAM signal. Solution A 16-QAM signal has 4 bits per signal unit since log 2 16 = 4. Thus, 1000·4 = 4000 bps

50 Example 12 Compute the baud rate for a 72,000-bps 64-QAM signal. Solution A 64-QAM signal has 6 bits per signal unit since log 2 64 = 6. Thus, 72000 / 6 = 12,000 baud

51 5.2 Telephone Modems Modem Standards

52 A telephone line has a bandwidth of almost 2400 Hz for data transmission. Note:

53 Figure 5.18 Telephone line bandwidth

54 Modem stands for modulator/demodulator. Note:

55 Figure 5.19 Modulation/demodulation

56 Figure 5.20 The V.32 constellation and bandwidth

57 Figure 5.21 The V.32bis constellation and bandwidth

58 Figure 5.22 Traditional modems

59 Figure 5.23 56K modems

60 5.3 Modulation of Analog Signals Amplitude Modulation (AM) Frequency Modulation (FM) Phase Modulation (PM)

61 Figure 5.24 Analog-to-analog modulation

62 The total bandwidth required for AM can be determined from the bandwidth of the audio signal: BWt = 2 x BWm. Note:

63 Example 7 Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as Bit Rate = 2  3000  log 2 2 = 6000 bps

64 Shannons regel Kanalkapaciteten C är max antal bit per sekund vid bästa möjliga modulationsteknik och felrättande kodning: C = B log 2 (1+S/N), där B är ledningens bandbredd i Hertz (oftast ungefär lika med övre gränsfrekvensen), S är nyttosignalens medeleffekt i Watt och N (noice) är bruseffekten i Watt.

65 Example 8 Consider the same noiseless channel, transmitting a signal with four signal levels (for each level, we send two bits). The maximum bit rate can be calculated as: Bit Rate = 2 x 3000 x log 2 4 = 12,000 bps Bit Rate = 2 x 3000 x log 2 4 = 12,000 bps

66 Example 9 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

67 Example 10 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

68 Example 11 We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level? Solution C = B log 2 (1 + SNR) = 10 6 log 2 (1 + 63) = 10 6 log 2 (64) = 6 Mbps Then we use the Nyquist formula to find the number of signal levels. 4 Mbps = 2  1 MHz  log 2 L  L = 4 First, we use the Shannon formula to find our upper limit.