6.2
Bandwidth utilization is the wise use of
available bandwidth to achieve
specific goals.
Efficiency can be achieved by
multiplexing; privacy and anti-jamming
can be achieved by spreading.
Note
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Multiplexing I-Basic Data Communication Systems-Lecture Slides, Slides of Digital Systems Design
This lecture is part of lecture series on Data Communication Systems. It was delivered by Prof. Prajin Ahuja at Birla Institute of Technology and Science. Its main points are: Anti-Jamming, Telecommunication, Frequency, Wavelength, Synchronous, Statistical, Mux, Demux, Slots,Multilevel
Typology: Slides
2011/2012
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Bandwidth utilization is the wise use of6.
available bandwidth to achieve
specific goals.
Note Efficiency can be achieved bymultiplexing; privacy and anti-jammingcan be achieved by spreading.
6-1^ MULTIPLEXING^ Whenever the bandwidth of a medium linking twodevices is greater than the bandwidth needs of thedevices, the link can be shared. Multiplexing is the setof^ techniques 6.
that^
allows^
the^ simultaneous
transmission of multiple signals across a single datalink. As data and telecommunications use increases, sodoes traffic. Topics discussed in this section: Frequency-Division MultiplexingWavelength-Division MultiplexingSynchronous Time-Division MultiplexingStatistical Time-Division Multiplexing
Figure 6.2 6.
Categories of multiplexing
Figure 6.3 6.
Frequency-division multiplexing
Figure 6.4 6.
FDM process
Figure 6.5 6.
FDM demultiplexing example
Five channels, each with a 100-kHz bandwidth, are to bemultiplexed together. What is the minimum bandwidth ofthe link if there is a need for a guard band of 10 kHzbetween the channels to prevent interference?SolutionFor five channels, we need at least four guard bands.This means that the required bandwidth is at least 6.
5 × 100 + 4 × 10 = 540 kHz,
Example 6.2 as shown in Figure 6.7.
Four data channels (digital), each transmitting at 1Mbps, use a satellite channel of 1 MHz. Design anappropriate configuration, using FDM.SolutionThe satellite channel is analog. We divide it into fourchannels, each channel having a 6.
250-kHz bandwidth.
Example 6.3 Each digital channel of 1 Mbps is modulated such thateach 4 bits is modulated to 1 Hz. One solution is 16-QAMmodulation. Figure 6.8 shows one possible configuration.
- Example 6.1 Assume that a voice channel occupies a bandwidth of 4kHz. We need to combine three voice channels into a linkwith a bandwidth of 12 kHz, from 20 to 32 kHz. Show theconfiguration, using the frequency domain. Assume thereare no guard bands.SolutionWe shift (modulate) each of the three voice channels to adifferent bandwidth, as shown in Figure 6.6. We use the20- to 24-kHz bandwidth for the first channel, the 24- to28-kHz bandwidth for the second channel, and the 28- to32-kHz bandwidth for the third one. Then we combinethem as shown in Figure 6.6. 6.
- Figure 6.6 6.
- Example 6.
- Figure 6.7 6.
- Example 6.
- Figure 6.8 6.
- Example 6.
The Advanced Mobile Phone System (AMPS) uses twobands. The first band of 824 to 849 MHz is used forsending, and 869 to 894 MHz is used for receiving.Each user has a bandwidth of 30 kHz in each direction.How^ 6. many^ people
can^ use
their^
cellular
phones
simultaneously?SolutionEach band is 25 MHz. If we divide 25 MHz by 30 kHz, weget^ 833.33.
In^ reality,
the^ band
is^ divided
into^832
Example 6.4 channels. Of these, 42 channels are used for control,which means only 790 channels are available for cellularphone users.
Figure 6.10 6.
Wavelength-division multiplexing
Figure 6.11 6.
Prisms in wavelength-division multiplexing and demultiplexing
Figure 6.12 6.
TDM