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:
Modem stands for
modulator/demodulator.
Figure 5.19
Modulation/demodulation
Figure 5.20
The V.32 constellation and bandwidth
Figure 5.21
The V.32bis constellation and bandwidth
Figure 5.22
Traditional modems
Figure 5.23
56K modems
Note:
The total bandwidth required for AM
can be determined from the bandwidth
of the audio signal:
BWt = 2 x BWm.
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
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
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.
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
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 signalto-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
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
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)
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)
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
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