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Datornätverk A – lektion 5

Datornätverk A – lektion 5 . Forts kapitel 5: Modem. Shannons regel. Kapitel 6: Transmission. 5.2 Telephone Modems. Modem Standards. Note:. A telephone line has a bandwidth of almost 2400 Hz for data transmission. Figure 5.18 Telephone line bandwidth. Note:.

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Datornätverk A – lektion 5

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

  2. 5.2 Telephone Modems Modem Standards

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

  4. Figure 5.18Telephone line bandwidth

  5. Note: Modem stands for modulator/demodulator.

  6. Figure 5.19Modulation/demodulation

  7. Figure 5.20The V.32 constellation and bandwidth

  8. Figure 5.21The V.32bis constellation and bandwidth

  9. Figure 5.22Traditional modems

  10. Figure 5.2356K modems

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

  12. 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 BitRate = 2  3000  log2 2 = 6000 bps

  13. 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 log2 4 = 12,000 bps

  14. Shannons regel Kanalkapaciteten C är max antal bit per sekund vid bästa möjliga modulationsteknik och felrättande kodning: C = B log2 (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.

  15. 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 log2 (1 + SNR) = B log2 (1 + 0)= B log2 (1) = B  0 = 0

  16. 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 log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163) C = 3000  11.62 = 34,860 bps

  17. 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 First, we use the Shannon formula to find our upper limit. C = B log2 (1 + SNR) = 106 log2 (1 + 63) = 106 log2 (64) = 6 Mbps Then we use the Nyquist formula to find the number of signal levels. 4 Mbps = 2  1 MHz  log2L L = 4

  18. 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 log2 (1 + S/N) S/N is the signal to noise ratio (often labeled as SNR)

  19. 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 log2 (1 + S/N) S/N is the signal to noise ratio (often labeled as SNR)

  20. 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 * log2 (1 + 1023 ) C = 8 Khz * log2 (1024 ) C = 8 * 1000 cycles/second * 10 bits/cycle C = 80 Kbps

  21. How to calculate log2 x • Calculators do not have a button for log2 x calculation • To calculate log2 x use the following formula: log2 x = log10 x/log102  log10 (x)/0.3 Example: log230 = log 30/log 2 1.477/0.3 4.9

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