Datornätverk A – lektion 5 Forts kapitel 5: Modem. Shannons regel. Kapitel 6: Transmission.

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1 Datornätverk A – lektion 5 Forts kapitel 5: Modem. Shannons regel. Kapitel 6: Transmission

2 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

3 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

4 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

5 5.2 Telephone Modems Modem Standards

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

7 Figure 5.18 Telephone line bandwidth

8 Modem stands for modulator/demodulator. Note:

9 Figure 5.19 Modulation/demodulation

10 Figure 5.20 The V.32 constellation and bandwidth

11 Figure 5.21 The V.32bis constellation and bandwidth

12 Figure 5.22 Traditional modems

13 Figure 5.23 56K modems

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

15 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

16 Digital modulation För att överföra N bit/symbol krävs M=2 N Vid M symboler överförs N=log 2 M bit/symbol. Baudrate f s = antal symboler per sekund. Enhet: baud eller symboler/sekund. Symbollängd T s = 1/f s. f s = 1/T s Bitrate R = datahastighet. Enhet: bps eller bit/s. R= f s log 2 M

17 Exempel: 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.

18 Forts exempel:

19 Table 5.1 Bit and baud rate comparison ModulationUnitsBits/Symbol 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

20 Figure 5.13 Relationship between baud rate and bandwidth in ASK, PSK, QAM (not FSK) without pulse shaping Vid många modulationsformer t.ex. s.k. ASK, PSK, och QAM är signalens bandbredd = symbolhastigheten. Vid FSK är bandbredden vanligen större. Bandbredden kan minskas genom s.k. pulsformning.

21 Maximal kanalkapacitet enligt Nyquist

22 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

23 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.

24 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

25 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

26 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.

27 Capacity Limits Maximum bit rate (capacity) depends on: ○The analog bandwidth available (in Hz) ○The quality of the channel The level of the noise is one of the characteristics of the channel. The ratio of the voltage of the signal sent and the noise present in the channel is important for the maximum data rate achieved. Shannon’s theorem determines the theoretical highest data rate of a noisy channel C = B log 2 (1 + S/N) S/N is the signal to noise ratio (often labeled as SNR)

28 Capacity Limits Maximum bit rate (capacity) depends on: ○The analog bandwidth available (in Hz) ○The quality of the channel The level of the noise is one of the characteristics of the channel. The ratio of the voltage of the signal sent and the noise present in the channel is important for the maximum data rate achieved. Shannon’s theorem determines the theoretical highest data rate of a noisy channel C = B log 2 (1 + S/N) S/N is the signal to noise ratio (often labeled as SNR)

29 Example: Problem: Given S/N ratio of 30.098756dB, bandwidth of 8Khz, compute maximum data rate. Answer: S/N = 30.098756dB = 10 ^ 3.0098756 = 1022.9999205  1023 C = 8 Khz * log 2 (1 + 1023 ) C = 8 Khz * log 2 (1024 ) C = 8 * 1000 cycles/second * 10 bits/cycle C = 80 Kbps

30 How to calculate log 2 x Calculators do not have a button for log 2 x calculation To calculate log 2 x use the following formula: log 2 x = log 10 x/log 10 2  log 10 (x)/0.3 Example: log 2 30 = log 30/log 2  1.477/0.3  4.9