BANDWIDTH UTILIZATION
Multiplexing and Spectrum Spreading
• Frequency-divisionmultiplexing(FDM),
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• Wavelength-divisionmultiplexing(WDM),
• Time-division multiplexing (TDM),
• Frequencyhoppingspreadspectrum(FHSS), • Directsequencespreadspectrum(DSSS),
Multiplexing
• Wheneverthebandwidthofamediumlinkingtwo devices is greater than the bandwidth needs of the devices, the link can be shared.
• Multiplexing is the set of techniques that allow the simultaneous transmission of multiple signals across a single data link
• Inamultiplexedsystem,nlinessharethebandwidthof one link
Categories of Multiplexing
Frequency-Division Multiplexing
• Frequency-divisionmultiplexing(FDM)isananalog technique that can be applied when the bandwidth of a link (in hertz) is greater than the combined bandwidths of the signals to be transmitted
• signalsgeneratedbyeachsendingdevicemodulate different carrier frequencies.
Frequency-Division Multiplexing
• Modulatedsignalsarethencombinedintoasingle composite signal that can be transported by the link.
– Carrier frequencies are separated by sufficient bandwidth to accommodate the modulated signal separated by strips of unused bandwidth—guard bands—to prevent signals from overlapping
Multiplexing Process
Demultiplexing Process
• Thedemultiplexerusesaseriesoffiltersto decompose the multiplexed signal into its constituent component signals.
• Theindividualsignalsarethenpassedtoademodulator that separates them from their carriers and passes them to the output lines
Demultiplexing Process
• Assumethatavoicechanneloccupiesabandwidthof4 kHz. We need to combine three voice channels into a link with a bandwidth of 12 kHz, from 20 to 32 kHz. Show the configuration, using the frequency domain. Assume there are no guard bands.
• Solution: We shift (modulate) each of the three voice channels to a different bandwidth. We use the [24,20] kHz bandwidth for the first channel, the [28,24] kHz bandwidth for the second channel, and the [32,28] kHz bandwidth for the third one. Then we combine them as shown in Figure
• Fivechannels,eachwitha100-kHzbandwidth,areto be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 kHz between the channels to prevent interference?
• Solution: For five channels, we need at least four guard bands. This means that the required bandwidth is at least
– 5×100+4×10=540kHz
5×100+4×10=540kHz
• Fourdatachannels(digital),eachtransmittingat1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration, using FDM.
• Solution: The satellite channel is analog. We divide it into four channels, each channel having a 250-kHz bandwidth. Each digital channel of 1 Mbps is modulated such that each 4 bits is modulated to 1 Hz. One solution is 16-QAM modulation. Next Figure shows one possible configuration.
The Analog Carrier System
• Tomaximizetheefficiencyoftheirinfrastructure, telephone companies have traditionally multiplexed signals from lower-bandwidth lines onto higher- bandwidth lines.
• Inthisway,manyswitchedorleasedlinescanbe combined into fewer but bigger channels
• Oneofthesehierarchicalsystemsusedbytelephone companies is made up of groups, supergroups, master groups, and jumbo groups
Analog Hierarchy
Applications of FDM
• AverycommonapplicationofFDMisAMandFMradio broadcasting
• Aspecialbandfrom530to1700kHzisassignedtoAM radio. All radio stations need to share this band. Each AM station needs 10 kHz of bandwidth
• ThesituationissimilarinFMbroadcasting.However,FM has a wider band of 88 to 108 MHz because each station needs a bandwidth of 200 kHz
• AnothercommonuseofFDMisintelevision broadcasting. Each TV channel has its own bandwidth of 6 MHz
Applications of FDM
• ThefirstgenerationofcellulartelephonesusedFDM.
• Eachuserisassignedtwo30-kHzchannels,onefor sending voice and the other for receiving. The voice signal, which has a bandwidth of 3 kHz (from 300 to 3300 Hz), is modulated by using FM.
• RememberthatanFMsignalhasabandwidth10times that of the modulating signal, which means each channel has 30 kHz (10 × 3) of bandwidth
• Therefore,eachuserisgiven,bythebasestation,a60- kHz bandwidth in a range available at the time of the call
• TheAdvancedMobilePhoneSystem(AMPS)usestwo bands. The first band of 824 to 849 MHz is used for sending, and 869 to 894 MHz is used for receiving. Each user has a bandwidth of 30 kHz in each direction. How many people can use their cellular phones simultaneously?
• Solution: Each band is 25 MHz. If we divide 25 MHz by 30 kHz, we get 833.33. In reality, the band is divided into 832 channels. Of these, 42 channels are used for control, which means only 790 channels are available for cellular phone users.
Wavelength-Division Multiplexing
• Wavelength-divisionmultiplexing(WDM)isdesignedto use the high-data-rate capability of fiber-optic cable.
• Theopticalfiberdatarateishigherthanthedatarateof metallic transmission cable, but using a fiber-optic cable for a single line wastes the available bandwidth.
• Multiplexingallowsustocombineseverallinesintoone
• WDMisconceptuallythesameasFDM,exceptthatthe multiplexing and demultiplexing involve optical signals transmitted through fiber-optic channels
Wavelength-Division Multiplexing
Wavelength MUX/DEMUX
Prisms in wavelength-division multiplexing and demultiplexing
Time-Division Multiplexing (TDM)
• Time-divisionmultiplexing(TDM)isadigitalprocess that allows several connections to share the high bandwidth of a link. Instead of sharing a portion of the bandwidth as in FDM, time is shared
• Eachconnectionoccupiesaportionoftimeinthelink
• NotethatthesamelinkisusedasinFDM
• WecandivideTDMintotwodifferentschemes: synchronous and statistical
Synchronous TDM
• Eachinputconnectionhasanallotmentintheoutput even if it is not sending data
• Thedataflowofeachinputconnectionisdividedinto units, where each input occupies one input time slot
• Eachinputunitbecomesoneoutputunitandoccupies one output time slot
• The duration of an output time slot is n times shorter than the duration of an input time slot, n is number of channels sharing the time
Synchronous TDM
• InpreviousFigure,thedatarateforeachinput connection is 1 kbps. If 1 bit at a time is multiplexed (a unit is 1 bit), what is the duration of (a) each input slot, (b) each output slot, and (c) each frame?
• Solution:
a. The data rate of each input connection is 1 kbps. This means that the bit duration is 1/1000 s or 1 ms. The duration of the input time slot is 1 ms (same as bit duration).
• InpreviousFigure,thedatarateforeachinput connection is 1 kbps. If 1 bit at a time is multiplexed (a unit is 1 bit), what is the duration of (a) each input slot, (b) each output slot, and (c) each frame?
• Solution:
b. The duration of each output time slot is one-third of the input time slot. This means that the duration of the output time slot is 1/3 ms.
• InpreviousFigure,thedatarateforeachinput connection is 1 kbps. If 1 bit at a time is multiplexed (a unit is 1 bit), what is the duration of (a) each input slot, (b) each output slot, and (c) each frame?
• Solution:
c. Each frame carries three output time slots. So the
duration of a frame is 3 × 1/3 ms, or 1 ms.
The duration of a frame is the same as the duration of an input unit.
• SynchronousTDMof4channelswithadatastreamfor each input and one data stream for the output. The unit of data is 1 bit. Find (a) the input bit duration, (b) the output bit duration, (c) the output bit rate, and (d) the output frame rate.
Example2 • Solution:
a. The input bit duration is the inverse of the bit rate: 1/1 Mbps = 1 μs.
Example2 • Solution:
b. The output bit duration is one-fourth of the input bit duration, or 1⁄4 μs.
Example2 • Solution:
c. The output bit rate is the inverse of the output bit duration or 1/(4μs) or 4 Mbps.
Example2 • Solution:
d. The frame rate is always the same as any input rate. So the frame rate is 1,000,000 frames per second.
Data Rate Management • Multilevel Multiplexing
Data Rate Management • Multiple-Slot Allocation
Data Rate Management • Pulse Stuffing
Digital Signal Service
• TelephonecompaniesimplementTDMthrougha hierarchy of digital signals, called digital signal (DS) service or digital hierarchy.
Digital Signal Service
• TelephonecompaniesimplementTDMthrougha hierarchy of digital signals, called digital signal (DS) service or digital hierarchy.
• Toimplementthoseservices,thetelephonecompanies use T lines (T-1 to T-4).
T-1 line for multiplexing telephone lines
T-1 frame structure
Statistical Time-Division Multiplexing
• insynchronousTDM,eachinputhasareservedslotin the output frame.
• Thiscanbeinefficientifsomeinputlineshavenodata to send.
• Instatisticaltime-divisionmultiplexing,slotsare dynamically allocated to improve bandwidth efficiency.
• Onlywhenaninputlinehasaslot’sworthofdatato send is it given a slot in the output frame.
Statistical Time-Division Multiplexing
• Instatisticalmultiplexing,thenumberofslotsineach frame is less than the number of input lines.
• Themultiplexercheckseachinputlineinround-robin fashion; it allocates a slot for an input line if the line has data to send; otherwise, it skips the line and checks the next line
TDM slot comparison
Addressing
A major difference between slots in synchronous TDM and statistical TDM is that
– an output slot in synchronous TDM is totally occupied by data;
– in statistical TDM, a slot needs to carry data as well as the address of the destination
Addressing
• Instatisticalmultiplexing,thereisnofixedrelationship between the inputs and outputs because there are no preassigned or reserved slots.
• Weneedtoincludetheaddressofthereceiverinside each slot to show where it is to be delivered.
• Theaddressinginitssimplestformcanbenbitsto define N different output lines with
– n = log2N
Spread Spectrum
• Multiplexingcombinessignalsfromseveralsourcesto achieve bandwidth efficiency
• In spread spectrum (SS), we combine signals from different sources to fit into a larger bandwidth, but our goals are to prevent eavesdropping and jamming. To achieve these goals, spread spectrum techniques add redundancy.
• IftherequiredbandwidthforeachstationisB,spread spectrum expands it to Bss, such that Bss >> B
Spread Spectrum
The expanding of the original bandwidth B to the bandwidth Bss must be done by a process that is independent of the original signal. In other words, the spreading process occurs after the signal is created by the source
Frequency Hopping Spread Spectrum
• Thefrequencyhoppingspreadspectrum(FHSS) technique uses M different carrier frequencies that are modulated by the source signal
• A pseudorandom code generator, called pseudorandom noise (PN), creates a k-bit pattern for every hopping period Th
• Thefrequencytableusesthepatterntofindthe frequency to be used for this hopping period and passes it to the frequency synthesizer.
• Thefrequencysynthesizercreatesacarriersignalof that frequency, and the source signal modulates the carrier signal
FHSS Cycles
Bandwidth sharing
Direct Sequence Spread Spectrum
• Thedirectsequencespreadspectrum(DSSS)technique also expands the bandwidth of the original signal
• InDSSS,wereplaceeachdatabitwithnbitsusinga spreading code. In other words, each bit is assigned a code of n bits, called chips, where the chip rate is n times that of the data bit
• thespreadingcodeis11chipshavingthepattern 10110111000 (in this case)
• Thismeansthattherequiredbandwidthforthespread signal is 11 times larger than the bandwidth of the original signal
• Thespreadsignalcanprovideprivacyiftheintruder does not know the code. It can also provide immunity against interference if each station uses a different code
References
• DataCommunicationsandNetworking5thedition– 2013, Behrouz A. Forouzan; Chapter 6
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