Multiple access 281

(a)

f*grl.v-r -El -

F_ :

- -

ffiNI-

t

l frequency

'l ffi_.TDMA

cDMA trme

Figure 5.3 The principle of multiple access. (a) Frequency division multiple access (FDMA). (b)Time division multiple access ODMA). (c) Code division multiple access (CDMA). (B: channel(transponder) bandwidth.)

The use of such codes has the effect of broadening the carrier spectrum incomparison with that which it would have if modulated only by the usefulinformation. This is why CDMA is also sometimes called Spread SpectrumMultiple Access (SSMA).

Several types of multiple access as defined above can be combined; Figure 6.4illustrates the range of combinations.

(b)

Code

282 Multiple access

BASIC TECHNIOUES

FRECIJENCYDIVISION (FDMA)

FREOUEI.ICY /T]ME0tvtstoN (FoITDMAI

FREOUENCY/ TIME OIVISION(ToMA)(FolcoMA) (FDIToICoMA )SHOWING SIGNALOCCUPANCY IN TIMEFREOUENCY PLANE

t-T-t

cooE olvtsloN(CDMA)

Figure 6.4 Combination of the three fundamental types of multiple access into hybrid accesstypes.

5.3.2 Multiple access to the satellite repeater

Multiple access to a particular repeater channel (transponder) implies prior multipleaccess to the satellite repeater. Access to a satellite repeater is achieved as a functionof frequency and polarisation of the carrier. For every carrier with givenpolarisation and frequency there is an obligatory FDMA access to the repeatertogether with FDMA, TDMA or CDMA access to each channel. The correspondingcombinations of Figure 6.4 can thus be considered as representative of multipleaccess to a satellite repeater. In all cases, the spectral occupation of a carrier mustnot exceed the channel bandwidth.

6.4 FREQUENCY DIVISION MULTIPLE ACCESS (FDMA)The bandwidth of a repeater channel is divided into sub-bands; each sub-band isassigned to one of the carriers transmitted by the earth stations. With this type ofaccess, the earth stations transmit continuously and the channel conveys severalcarriers simultaneously at different frequencies. It is necessary to provide guardinten'als betrt'een each band occupied by a carrier to avoid interference as a result ofimperfections of oscillators and filters. The receiver selects the required carrier inaccordance with the appropriate frequency. The Intermediate Frequency 0F)amplifier provides the filtering.

(Jztr, ,AAH l|!

riit :

ir

|r

FRAME PERIOD

I-lTI I(SYSTEMBANDW|DTH)I t_l+ TIME

iiFIFt:

290 Multiple access

Figure 6.10 Variation of (C/Ng)rM as a function of back-off and number of carriers.

carriers increases, the bandwidth allocated to each carrier must decrease and thisleads to a reduction of the capacity of the modulating multiplexed signal. As thetotal capacity is the product of the capacity of each carrier and the number ofcarriers, it could be imagined that the total capacity would remain constant. But it isnot; the total capacity decreases as the number of carriers increases. This results

THROUGHPUTf/4

NUMBER OF ACCESSES

Figure 6.11 Throughput of an FDMA transmission; the curve indicates the relative variation ofthe total capacity of a transponder with a bandwidth of 36 MHz as a function of the number ofaccesses/ that is tl're number of carriers of FDM /FM/FDMA type. The value indicated as 700Vorepresents the total capacity of the multiplex which modulates the carrier for the case of singleaccess to the repeater channel, operated at saturation.

INTERMODULATION PRODUCTSARE GENERATED

2?2 (24 cHAFTER 6 MULn?tE AccEss

ffi---'+

TDMA streamfrom satellite

One frame

FIGURE 6.7 l l lustrat ion of TDMA with three earth stat ions. Transmitt ing earth stat ionsmust t ime their burst transmissions so that they arr ive at the satel l i te in the correct se-quence. The signal transrnit ted by the satel l i te is a continuous sequence of bursts separaby short guard t imes.

TDMA is an RF multiple access technique that allows a single transponder trshared in time between RF carriers from different earth stations. In a TDMA systemRF carrier from each earth station sharing a transponder is sent as a burst at a spetime. At the satellite, bursts from different earth stations arrive sequentiall% so the transder carries a near continuous signal made up of a sequence of short bursts corningdifferent earth stations. The principle of TDMA is illustrated in Figure 6.7.

The burst transmission is assembled at a transmitting earth station so that it wilrectly fit into the TDMA frame at the satellite. The frame has a length from 125 ps tomilliseconds, and the burst from the earttr station must be transmined at the correct tiarrive at the satellite in the correct position within the TDMA frame. This requireqhronization g_f all the earth stations in a TDMA network, adding considerable comrto the equipment at the transmining station. Each station mu.st know exactly when tcmit, typically within a microsecond, so that the RF bunts arriving at the satellite frrferent earth stations do not ovedap. (A time overlap of nvo RF signals is called a cand results in data in both signals being lost. Collisions must not be allowed to cc,TDMA system.)

A receiving earth station must synchronize its receiver to each of the scbursts in the TDMA signal and recover the transmission from each uplink earthThe uplink transmjssions are then broken down to extract the data bits, which aland reassembled into theii original bit streams for onward transmission. The irtransmissions from different uplink earth stations are usually sent using npS[.and will inevitably have small differences in carrier and clcck frequencies, andcarrier phases. The receiving earth station must synchronize its PSK demodulatrburst of signal within a few microseconds, and then synchronize its bit clock itfew microseconds so that a bit stream can be recovered. In high-speed TDMAoperating at 120 Mbps, for example, these are demanding requirements.

Bits, Symbols, and ChannelsA potential source of confusion in the discussion of TDMA systems is that QPisibly QAM) modulation is typically used by transmitting earth stations, and da

Satellite

Incoming bit

Time diaisian multiple nccess (TDMA) 293

user bit streams to ABa

bit rate : R3

Figure 5.13 Burst generation with the 'one carrier per transmitting station' technique. R i : Userrate (bit/s), Ru: information rate of the multiplex (bit/s): IRi, R: rate in each burst (bit/s),Is : burst duration (s), Tr :frame duration (s).

The value of R is high when the burst duration is short and consequently thetransmission duty cycle (TBITF) of the station is low. Hence, for example, ifRu:2Mbit /s and ( fF/Ta) :10, modulat ion occurs at 2 jMbit ls. Not ice that Rrepresents the total capacity of the network; that is the sum of the station capacitiesin bitls. If all stations have the same capacity, the duty cycle QFITil represents thenumber of stations on the network.

It can now be seen why this type of access is always associated with digitaltransmission; it is easy to store bits for a frame period and to empty a digitalmemory in the shorter period of one burst. Performing this type of processing onanalogue information is not easy.

The structure of a burst can be seen in Figure 6.13 and is further detailed in Figure6.14. This consists of a header, or preamble, and a traffic field. The header has severalfunctions:

-To permit the demodulator of the receiving earth station, in the case of coherentdemodulation, to synchronise its local oscillator to the received carrier. For this

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232 cH236 cHAPTER 6 MULTTPLE AccEss

without disrupting its operation. They must also be able to track changes in the timirthe frame caused by motion of the satellite toward or away from the earth station. learth station must also be able to extract the data bits and other information from Itransmissions of other earth stations in the TDMA network. The transmitted-bunQe'ontaig ;ynchronizatioa aryl-ldeqgf-q4-tlgl in{oryation--that-help-reeelving-Ea4LStar

Synchronization of the TDMA network is achieved with the portion of the preamble trmitted by each earth station that contains carrier and bit clock synchronization wavefoIn some systems, a separate reference burst may be transmitted by one of the statjdesignated as the master station. A reference burst is a preamble followed by no trbits" Traffrc bits are the revenue producing portion of each frame, and the preamblereference bursts represent overhead. The smaller the-_overhead, the more efficientTDMA.sys-t9m, buf the sleqte{ {r9 ditrig[lty!ffiqufilg *a maftiai!fit"{1ql11@chronization" The preamble of each station's burst transmission requires a fixed uniiSsion time. A longer frame contains proportionally less preamble time than a sframe, so more revenue producing data bits can be carried in a long frame. Early TDsystems were designed around 125 p,s frames, to match the sample rate of digital sprin telephone systems, in exactly the same way that Tl 24-channel systems operater Aital telephone channel generates one 8-bit digital word every 125 ps (8 kllz sampling r,so a I 25-p,s frame transmits one word from each speech channel. However, it is morrficient to lengthen the frame to 2 ms or longer so that the proportion of overhead to rsage transmission tirne is reduced. It must be remembered that a longer frame reqrmultiple 8-bit words when transmitting digital speech. For example, in a time perio2 ms, a digital terrestrial channel will deliver sixteen 8-bit words, so a 2-ms TDMA frrequires sixteen 8-bit words from each terrestrial channel in each transmitted burst.

Figure 6.9 shows a typical TDMA framc with 2.0 ms duration used by some estations operating in TDMA through Intelsat satellites. All of the blocks at the start ol

CBTR UW fiY SC vow I vow Digital speech channels

M Satellite channels

1 2 3 4 5 6 7 M

Sixteen 8-bit samples form each satellite channelFIGURE 6.9 Structure of an Intelsat traffic data burst. A satellite channel is a block of siteen 6-bit samples from one terrestrial speech channel" Other blocks in the traffic burst arused to synchronize the PSK demodulator, the bit clock, and the frame clock in the receivr{CBTR, UW) and to provide communication links between earth stations (TTY SC, and VCCBTR, carr ier and bit t iming recovery; UW, unique word; TTY teletype; SC, satel l i te chanrVOW, voice order wire.

1 2 3 4 5 6 7 8 I 1 0 1 1 12 1 3 1 41 5 16

zln cHAprER 6 MuLnpLE AccEss

EXAMPLE 6.3.2 TDMA in a VSAT NetworkAs an example, consider a typical VSAT earth station in the United States which is part of a TDt,l,t'network using a 54 MHz bandwidth transponder on a domestic Ku-band GEO sateuite. The VSATearth station has a i m antenna that transmits a single 64 kbps signal at 14 GHz. Let's assume dthe TDMA network uses QPSK modulation and that all transmitters have a symbol rate of 30 MbauC,We will set (C/N),,' at 20 dB, and then calculate the required uplink transmit power. The followingsystem pafameters will be used: .1

Earth station antenna gain : 41.5 dB, satellite antenna gain (on axis) : 32.0 dB, edgebeam loss = 3 dB, path loss at 14 GHz : 2O7 .ldB, receiver noise bandwidth : 30 MHz. transpon,noise temperature : 500 K, atmospheric and other losses = 1.0 dB.

The uplink power and noise budgets are

Earth station transmit power : p, dBWEarth station antenna gain at ll GHz = 41.5 dBSatellite antennaEdge of beam lossOther lossesPath loss at l I GHzPower at transponder inputBoltzmann's constant

: 32.0 dB: -3.0 dB: - 1 . 0 d B= -207.1 dB- Pt - 137.6 dBw= -228.6 dBWK/Hz

Transponder noise bandwidth : 74.g dBHzTransponder noise temperature - 2?.0 dBKTransponder inpur noise power : -126.9 dBW

we require (c/N)"p : P,f kT,Bn: 20 dB: hence p, - I 37.6 + 126.g = 20 dBw and p, =dBW or P, : 1200 W

Now consider the same earth station transmitting the same 64-kbps signal in a SCpC-FDM{VSAT network using QPSK with a symbol rate of 32 kbautl and a receiver noise bandwidth oi32 kllz. The uplink power budget is unchanged, but the noise powcr in the transponder.in a bandwidth of 32 kHz is - 156.5 dBW.

To achieve (C1w)"o : 2o clB in the transponder now requires an uplink transmitter powerP , : 2 0 + 1 3 7 . 6 - 1 5 6 . 5 : 1 . 1 d B W : 1 . 3 W .

The above example illustrates a key problem with TDMA for any small earth sHtion: uplink transrnit power. No one is going to equip a l-m VSAT station with a 1200-[transtnitter. Apart frorn the excessive cost, FCC regulation in the United States do notsmall VSAT statiotts to transmit more than 2 W to limit interference to adjacent satelli

If we change the multiple access technique for just two earth stations, so that etranstnits a burst of QPSK signal at 64 kbaud for half the tirne, the uplink transmipower requirement is doubled to 4.1 dBW or 2.6 W. This makes wideband TDMA anlikely choice in VSAT networks, and limits the number of stations that can share aframe in a low earth orbit satellite telephone system. The lridium LEO system was d,signed to use a hybrid TDMA-FDMA multiple access scheme at L band to combinesmall number of digital telephone transmissions inro a 50-kbps QPSI< signal. Similarniques are used in some VSAT networks.

EXAilPLE o.3.3 TDMA in a Fixcd Earth stttion NotworkIn Example 6"2.1, three identical large ea:th stations shared a single 36-M*zbandwidthder using FDMA. The three earth stations transmitted signals with powers and bandwidths

rich is Part of a TDMAiO satellite. The VSAT3Hz. Let's assume thatmbol rate of 30 Mbaud'it power. The following

is) : 32.0 dB, edge of= 30 MHz, transPonder

!wKlHz

- 20 dBW and P, = 30'8

, signal in a SCPC-FDMA,.iJ", noise bandwidth of

the transPonder, measured

plink transmitter Power ofI

for anY small td {- ,i-;;;" with a lzoGw'rj

gtwork

i6-MHz bandwidth-tntJl".r, *o bandwidths

6.3 NME DIVISION MULNPLE ACCESS (TDMA) 245

by

Station A:Station B:Station C:

15 MHzl0 MHz5 MHz

B _f r , -fi, =

P, : 125 W : 21 .0 dBWP , : 8 3 W : l 9 . 2 d B WP, = 42W : 16.2 dBW

16 dBW with 3-dB output backoff and 105-dBThe transponder total power output wastransponder gain.

The three earth station accesses to the transponder are changed to TDMA, with a frame lengthof 1.0 ms, a preamble time of l0 p.s, and a guard time of 2 pr.s. There is no reference burst in theTDMA frame. The signals are transmined using QPSK, and within the earth stations the bit ratesof the signals are

Ru = 15.0 MbpsRu = 10.0 Mbps i fi .':\Rr = 5.0 Mbps

Calculate the burst duration and symbol rate f*o-r e:c_h-qanlr slation, and the_earth llaggll trans-mltterp-uEurpo**jequired if the qansponQgr outp11,q U-1ci9f is set at 1.0 On ana-tnJ gain of thetransponder with this output backoff is 104 dB. Compare the uplink (C/N) ratios in the transponderfor FDMA and TDMA operation given that station A's transmission has a (C/N)"p ratio of 34 dBwhen the transponder is operated in FDMA.

The transponder must carry a total bit rate of 15 't l0 + 5 = 30 Mbps within the 1"0-msframes. Thus each frame canies 30 Mbps x 0.001 s : 30 kb" The three preamble and guard timestake up 3 X (10 + 2) : 36 p,s in each frame, leaving 1000 * 36 = 964 ps for transmission ofdata' Hence the burst bit rate is

r, rrl-. '

'

Rbburr, : 30kb/9& p,s : 31.12 Mbps.

Since we are using QPSK for the transmissions, the burst symbol rate on the link is

R,bursr : 31.12 Mbps/2 : 15.56 Msps

Each of the stations must transmit at the same burst rate of 15.56 Msps. The burst lengthscan be calculated from the time available in each frame for data transmission and the number ofbits each station must send in a I ms TDMA frame" The time available for data transmission is 964ps, which must be shared in proportion to the number of bits each station sends in a frame. Thenumber orbits in a rram;;,::T:' '"

ff$il:;;; T:: ;:'f a rrame is given berow

Station B: Ru = 10,@0 bits Ts = 321.3 psStz"tion C: Ru - 5,000 bits Tc = 160"7 ps

We'can easily check to see if these results are correct. Each earth station rnust have the statedaverage bit rate, so if we multiply the burst duration for each earth station by the burst bit rate forthe transponder, 31.12 Mbps, we must have the correct number of bits/frame for each station.

Station A:Station B:Station C:

482.0 ps Nu = 482"0 p.s x 31.12 Mbps = 15,00032L3 p.s Nv = 321.3 p,s X 31.12l"Ibps = 10,000160.7 p.s Nu = 160.7 p.s x 31"12 Mbps = 5,000

Each station must ransmit at the same symbol rate of 15.56 Msps, regardless of the number ofbis dent per frame. In the previous FDMA example, a transponder oulput power of,20 W = 13 dBWwas achieved with a total earth gation power of 250 W = 24 dBW and a transponder gain of 105 dB.With mMA, we are using a I dB tansponder output backoff, and a transponder gain of 104 dB, sothe transponder output power is now 16 - 1 : 15 dBV/, an increase of,2 dB, and we have lost I dBof gain in the transponder. This requires an earth station oulput power, frbm each earth station, of

Station A;Station B:Station C:

"f"r A -

I B -

t c -

P,", = 24 + 2 + I = 27 W : 500 W

Time diaision multiple access (mMA) 305THROUGHPUT(/ .)

NUMBER OF ACCESSES

Figure 6.21 The efficiency of the INTELSAT/EUTELSAT TDMA system; the 100% valueindicated for a single access corresponds to the capacity of the single carrier which passes throughthe transponder and is transmitted continuously.

5.5.6 Conclusion

Time division multiple access (TDMA) is characterised by access to the channelduring a time slot. This has certain advantages:

-At each instant the satellite repeater channel amplifies only a single carrier whichoccupies all of the repeater channel bandwidth; there are no intermodulationproducts and the carrier benefits from the saturation power of the channel.

-Transmission throughput remains high for a large number of accesses.-There is no need to control the transmitting power of the stations.-All stations transmit and receive on the same frequency whatever the origin or

destination of the burs| this simplifies tuning.TDMA, however, has certain disadvantages:

-The need for synchronisation which implies complex procedures and theprovision of two reference stations. Fortunately, these procedures can beautomated and computer driven.

-The need to increase power and bandwidth as a result of high burst bit rate,compared to continuous access, as with FDMA for instance.

Consider a station-to-station link. The quality objective is specified in terms of errorprobability. The imposed value determines the required value of the ratio E/No.The ratio C/No for the overall link is determined by the relation established inChapter 3 and recalled here:

C/No : (E/NO)R (6.17)It can be seen that C/N6 is proportional to R for which the expression is given byequation (6.8). For a capacity Rtr, a station must be dimensioned in power andbandwidth to transmit a bit rate R which is high when the duty cycle TslTe is low,

100