1. Repetition kapitel 5: Modem. Shannons regel.
Lektion 5
Repetition kapitel 5: Modem. Shannons regel.

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 23 = 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
log216 = 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
log2 64 = 6.
Thus,
72000 / 6 = 12,000 baud

5. 5.2 Telephone Modems
Modem Standards

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

7. Figure 5.18 Telephone line bandwidth

8. Modem stands for modulator/demodulator.
Note:
Modem stands for modulator/demodulator.

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 K modems

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

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  log2 2 = 6000 bps

16. Digital modulation För att överföra N bit/symbol krävs M=2N
Vid M symboler överförs N=log2 M bit/symbol.
Baudrate fs= antal symboler per sekund. Enhet: baud eller symboler/sekund.
Symbollängd Ts= 1/fs. fs= 1/Ts
Bitrate R = datahastighet. Enhet: bps eller bit/s.
R= fslog2 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
Modulation
Units
Bits/Symbol
Baud rate
Bit Rate
ASK, FSK, 2-PSK
Bit 1 N
4-PSK, 4-QAM
Dibit 2 2N
8-PSK, 8-QAM
Tribit 3 3N
16-QAM
Quadbit 4 4N
32-QAM
Pentabit 5 5N
64-QAM
Hexabit 6 6N
128-QAM
Septabit 7 7N
256-QAM
Octabit 8 8N

20. Vid FSK är bandbredden vanligen större.
Figure 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 log2 4 = 12,000 bps

23. Shannons regel C = B log2 (1+S/N),
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.

24. C = B log2 (1 + SNR) = B log2 (1 + 0) = B log2 (1) = B  0 = 0
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

25. C = B log2 (1 + SNR) = 3000 log2 (1 + 3162) = 3000 log2 (3163)
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 For this channel the capacity is calculated as
C = B log2 (1 + SNR) = 3000 log2 ( ) = 3000 log2 (3163)
C = 3000  = 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
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  log2 L  L = 4

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

29. Example:
Problem: Given S/N ratio of dB, bandwidth of 8Khz, compute maximum data rate.
Answer: S/N = dB = 10 ^ =  1023
C = 8 Khz * log2 ( )
C = 8 Khz * log2 (1024 )
C = 8 * 1000 cycles/second * 10 bits/cycle
C = 80 Kbps

30. 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