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Example 338
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We can calculate the theoretical highest bit rate of a regular telephone line A telephone line normally has a bandwidth of 3000 Hz (300 to 3300 Hz) assigned for data communications The signal-to-noise ratio is usually 3162 For this channel the capacity is calculated as C = B log2 (1
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+ SNR) =3000 log2 (l + 3162) =3000 log2 3163
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:::::: 3000 x 1162 = 34,860 bps
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This means that the highest bit rate for a telephone line is 34860 kbps If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-tonoise ratio
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DATA AND SIGNALS
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Example 339
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The signal-to-noise ratio is often given in decibels Assume that SN~B = 36 and the channel bandwidth is 2 MHz The theoretical channel capacity can be calculated as
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SNRdB = 10 loglO SNR SNR = lOSNRoB/10
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SNR::; 10 36
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= 3981
C =B log2 (1+ SNR) = 2 X 10 X log2 3982 = 24 Mbps
Example 340
For practical purposes, when the SNR is very high, we can assume that SNR + I is almost the same as SNR In these cases, the theoretical channel capacity can be simplified to SNR dB
C=BX - - =
For example, we can calculate the theoretical capacity of the previous example as 36 C= 2 MHz X - =24 Mbps 3
Using Both Limits
In practice, we need to use both methods to find the limits and signal levels Let us show this with an example
Example 341
We have a channel with a I-MHz bandwidth The SNR for this channel is 63 What are the appropriate bit rate and signal level
Solution
First, we use the Shannon formula to find the upper limit C =B log2 (l + SNR) = 106 log2 (1 + 63) = 106 10g2 64 = 6 Mbps The Shannon formula gives us 6 Mbps, the upper limit For better performance we choose something lower, 4 Mbps, for example Then we use the Nyquist formula to find the number of signal levels 4Mbps=2x 1 MHz x log2 L
L=4
The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need
SECTION 36
PERFORMANCE
PERFORMANCE
Up to now, we have discussed the tools of transmitting data (signals) over a network and how the data behave One important issue in networking is the performance of the network-how good is it We discuss quality of service, an overall measurement of network performance, in greater detail in 24 In this section, we introduce terms that we need for future chapters
Bandwidth
One characteristic that measures network performance is bandwidth However, the term can be used in two different contexts with two different measuring values: bandwidth in hertz and bandwidth in bits per second
Bandwidth in Hertz
We have discussed this concept Bandwidth in hertz is the range of frequencies contained in a composite signal or the range of frequencies a channel can pass For example, we can say the bandwidth of a subscriber telephone line is 4 kHz
Bandwidth in Bits per Seconds
The term bandwidth can also refer to the number of bits per second that a channel, a link, or even a network can transmit For example, one can say the bandwidth of a Fast Ethernet network (or the links in this network) is a maximum of 100 Mbps This means that this network can send 100 Mbps
Relationship
There is an explicit relationship between the bandwidth in hertz and bandwidth in bits per seconds Basically, an increase in bandwidth in hertz means an increase in bandwidth in bits per second The relationship depends on whether we have baseband transmission or transmission with modulation We discuss this relationship in s 4 and 5
In networking, we use the term bandwidth in two contexts
The first, bandwidth in hertz, refers to the range of frequencies in a composite signal or the range of frequencies that a channel can pass The second, bandwidth in bits per second, refers to the speed of bit transmission in a channel or link
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