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Improving Broadcast Channel Rate Using Hierarchical Modulation Hugo Me´ric∗†, Je´roˆme Lacan†∗, Fabrice Arnal‡, Guy Lesthievent§ and Marie-Laure Boucheret†∗ ∗Te´SA, Toulouse, France †Universite´ de Toulouse, Toulouse, France ‡Thales Alenia Space, Toulouse, France §CNES, Toulouse, France Email: [email protected], [email protected], [email protected], [email protected], [email protected] 2 1 Abstract—We investigate the design of a broadcast system to merge two different streams at the modulation step. The 0 wheretheaimistomaximisethethroughput.Thistaskisusually High Priority (HP) stream is used to select the quadrant, and 2 challenging due to the channel variability. Modern satellite the Low Priority (LP) stream selects the position inside the communications systems such as DVB-SH and DVB-S2 mainly n quadrant. The HP stream is dedicated to receivers with bad rely on time sharing strategy to optimize throughput. They a J consider hierarchical modulation but only for unequal error channel quality, unlike the LP stream which requires a good protection or backward compatibility purposes. We propose in SNR to be decoded. Today, hierarchical modulation is used 7 thisarticletocombinetimesharingandhierarchicalmodulation mainly for unequal error protection. For instance in [3], SVC 1 togetherandshowhowthisschemecanimprovetheperformance encoded video [4] is protected using hierarchical modulation. in terms of available rate. We present the gain on a simple ] The base layer of the video is transmitted in the HP stream, I channel modeling the broadcasting area of a satellite. Our work N isappliedtotheDVB-SHstandard,whichconsidershierarchical whiletheenhancedlayeriscarriedbytheLPstream.Another . modulation as an optional feature. usage is backward compatibility [5]. The DVB-S2 standard is s called upon to replace DVB-S, but many DVB-S receivers are c [ I. INTRODUCTION already installed. Then the hierarchical modulation helps to In most broadcast applications, the signal-to-noise ratio the migration by allowing the DVB-S receivers to operate. In 2 (SNR) experienced by each receiver varies greatly. For in- [6] , the authors propose to provide local content with hierar- v 5 stance, in satellite communications the channel quality de- chical modulation. The principle is to carry local information 2 creases with the presence of clouds in Ku or Ka band, or that is of interest to a particular geographic area in the LP 4 with shadowing effects of the environment in lower bands. stream, while the HP stream transmits global content. Finally, 2 This leads to many difficulties to design an efficient (in otherworksimprovetheperformanceofrelaycommunication . 1 terms of throughput) broadcast system. The first solution for system [7] or OFDMA-based networks [8]. 1 broadcasting was to design the system for the worst-case Our work focuses on using hierarchical modulation in 1 reception,butthisleadstopoorperformanceasmanyreceivers modernbroadcastsystemstoincreasethetransmissionrate.In 1 do not exploit their full potential. Then, two other schemes satellite standards such as DVB-SH or DVB-S2, time sharing : v have been proposed in [1] and [2]: time division multiplex- and hierarchical modulation are not used simultaneously. This i X ing with variable coding and modulation, and superposition articleinvestigatestheperformance,intermsofthroughput,of coding. Time division multiplexing, or time sharing, allocates a broadcast system where these two schemes work together. r a to each receiver a fraction of time where it can use the full We focus on the DVB-SH standard and show on a simple channel with any modulation and error protection level. This example that the gain can be significant, up to 17%. solution is the most used in practice in standards today. In The paper is organized as follows: Section II computes the superposition coding, the available energy is shared among throughputforthedifferentbroadcastingsolutions,SectionIII several data streams which are sent simultaneously in the studies the performance of each scheme and Section IV same band. This scheme was introduced by Cover in [1] in concludes the paper by summarizing the results. order to improve the transmission rate from a single source to II. THROUGHPUTCOMPUTATION multiple receivers. When communicating with two receivers, Thissectionintroducesthethroughputforthetwofollowing the principle is to superimpose information for the receiver schemes: with the best SNR. This superposition can be done directly at the forward error correction (FEC) encoding level or at • Time sharing strategy with no hierarchical modulation, the modulation level. Hierarchical modulation is a practical referred as classical time sharing. implementation of superposition coding. Figure 1 presents • Time sharing strategy where hierarchical modulation is the principle of hierarchical modulation with a non-uniform allowed, referred as hierarchical modulation time shar- 16-QAM and the mapping used in this paper. The idea is ing. Q is described using the constellation parameter. The DVB- 0100 0101 0001 0000 SH standard considers the hierarchical 16-QAM presented in HP Encoder 00 0110 0111 0011 0010 Figure 1. The constellation parameter α is defined by dh/dl, where2d istheminimumdistancebetweentwoconstellation I h Hierarchical 16−QAM points carrying different HP bits and 2dl is the minimum LP Encoder 1110 1111 1011 1010 distance between any constellation point. Typically, we have 11 2dl α ≥ 1, where α = 1 corresponds to the uniform 16-QAM, 1100 1101 1001 1000 but it is also possible to have α ≤ 1. At a given energy per 2dh symbol (E ), when α grows, the constellation points in each s Fig.1:HierarchicalModulationusinganon-uniform16-QAM quadrant become farther to the I and Q axes. Thus it is easier to decode the HP stream. However, in the same quadrant, the points get closer and the LP stream requires a good channel A. Classical time sharing quality to be decoded. We consider a broadcast system with n receivers, each one We define the QPSK parameter as the minimum distance with a particular signal-to-noise ratio SNR (i = 1,..,n). between two points in a QPSK constellation. As shown in i Given SNR , receiver i has a rate R which corresponds to Figure1,thehierarchical16-QAMisthesuperpositionoftwo i i the best rate it can afford. This rate is the amount of useful QPSK modulations, one with parameter 2(dh +dl) carrying data transmitted on the link and depends on the modulations the HP stream and the other with parameter 2dl carrying the and coding rates available in the system. For instance, if the LP stream. Thus the energy ratio between the two streams is modulationisaQPSKandthecodingrateis1/3,thentherate E equals 2×1/3 bit/symbol. In our study, the physical layer is hp =(1+α)2, (4) E lp based on the DVB-SH standard [9], [10]. Table I resumes the modulations and coding rates used for the classical time where Ehp and Elp corresponds to the energy allocated to sharing scheme. the HP and LP streams respectively [6]. The standard [9] recommends two values for α: 2 and 4. In fact, DVB-SH also TABLE I: Classical time sharing physical layer considers α=1 but only for the Variable Coding Modulation mode. The values 2 and 4 are defined in order to provide Modulations QPSK,16-QAM unequalprotection.Fromanenergypointofview,thisamounts Codingrates 1/5,2/9,1/4,2/7,1/3,2/5,1/2,2/3 to give 90% (α=2) or 96% (α=4) of the available energy totheHPstream.However,ourgoalisnottoprovideunequal The time sharing scheme allocates a fraction of time t to i protection.Itiswhynewvaluesofαareusedinordertoboost receiver i. The performance, in terms of throughput, results the throughput of the system. Our simulations include α=1, from the time allocation. We define the average rate for α = 0.8 and α = 0.5, which provide a better repartition of receiver i as t R . In our study, we are interested to offer i i theenergy:theHPstreamcontains80%,76%and69%ofthe the same average rate to everyone. Then, we need to solve total power. (cid:40) t R =t R ∀i,j To determine the throughput of this scheme, we first inves- i i j j (1) (cid:80) tigate the case with two receivers. We begin by computing t =1. i i the rates offered by all the possible modulations (Table I), It is easy to verify that the set of ti defined as follows, including the hierarchical 16-QAM considered in the DVB- ti = (cid:80)n (cid:81)(cid:16)k(cid:81)(cid:54)=iRkR (cid:17), (2) SpaHirss,tawndhaerrde.RThiuisswtheeobratateinoafsreetceoifvaecrhii,eviab=le1(,R2.1,CRla2s)sricaatel j=1 k(cid:54)=j k modulations (Table I) achieve rate pairs of the form (R1,0) and(0,R ),whilethehierarchical16-QAMallowsratespairs is the unique solution of (1). We remark that increasing R 2 i of the form (R ,R ). When the hierarchical modulation is reduces t , which is a consequence of our rate policy. Finally, 1 2 i used, we assume that the LP stream is dedicated to the the the average rate offered to each receiver is receiver with the largest SNR. Moreover, when two sets of (cid:81) R= (cid:16)kRk (cid:17). (3) rates (R1,R2) and (R1∗,R2∗) are achievable, the time sharing (cid:80)n (cid:81) R achieves any rate pair j=1 k(cid:54)=j k B. Hierarchical modulation time sharing τ(R1,R2)+(1−τ)(R1∗,R2∗)=(τR1+(1−τ)R1∗,τR2+(1−τ)R2∗), (5) We first give some definitions and then compute the where 0 ≤ τ ≤ 1 is the proportion of time allocated to throughput for the hierarchical modulation time sharing (R ,R ). The achievable rate region is the convex hull of the 1 2 scheme.Asmentionedbefore,hierarchicalmodulationsmerge set of achievable (R ,R ) rate pairs. Figure 2 presents the 1 2 severalstreamsinasamesymbol.Theyoftenusenon-uniform achievable rate region for two receivers, one with a SNR of constellation where the points are not uniformly distributed 4 dB and the other 8 dB. We also represent the classical time in the plane. The geometry of non-uniform constellations sharingachievablerateregion.Notethatweonlyrepresentthe TABLE II: Hierarchical 16-QAM decoding thresholds (in dB) rate pairs which are relevant. The points for α=4 and α=2 aregivenin[9,Table7.11]where0.3dBneedstoberemoved Coding α=4 α=2 α=1 α=0.8 α=0.5 due to the pilots. The other points (i.e., α = 1, α = 0.8 rate HP LP HP LP HP LP HP LP HP LP and α = 0.5) are computed using the method described in 1/5 -3.6 10.3 -3.2 6.2 -2.5 3.7 -2.1 3.2 -1.4 2.6 [11]. All the decoding thresholds are given in Table II and 2/9 -3.1 10.8 -2.6 6.8 -1.9 4.1 -1.6 3.6 -0.8 2.9 1/4 -2.5 11.4 -2 7.3 -1.2 4.6 -0.9 4.1 0 3.4 the Appendix details the computation for non standard points 2/7 -1.8 12.1 -1.3 8 -0.4 5.2 0 4.7 0.9 3.9 on one example. As we are interested to offer the same rate 1/3 -0.9 12.9 -0.4 8.8 0.7 6 1.1 5.4 2.2 4.6 to the receivers, we calculate the intersection of the convex 2/5 0.2 14 2 6.8 9.9 6.9 2.5 6.3 3.8 5.4 hull with the curve y =x, which corresponds to the cross in 1/2 1.6 15.3 7.3 11.3 3.7 8.1 4.4 7.5 6.2 6.5 Figure2.Theinterestofusinghierarchicalmodulationisclear 2/3 3.9 17.5 4.9 13.4 7 10.2 8.1 9.5 11 8.4 as the final rate results of time sharing between two operating points where one is obtained using the hierarchical 16-QAM with α=1. sharing. The rate is known for each receiver and (2) can be Figure 2 deserves few more comments. First, the rates applied.Forinstance,considerasystemwhereareceiverwith obtained for the classical time sharing strategy come from aSNRof4dBisinpairwithareceiverhavingaSNRof8dB. the 1/3 16-QAM for the receiver at 4 dB and the 1/2 16- The rate for each receiver is obtained using the hierarchical QAM for the other. As mentioned before, it is the best that 16-QAM (α = 1) a fraction of time a1 and the 16-QAM a each receiver can afford. Concerning α=4, it leads to a zero fractionoftimea2 asshowninFigure2.Whenequalizingthe rate for the receiver who decodes the LP stream. We have rates, (2) gives the same fraction of time t to both receivers. already said that the points in one quadrant get closer when α At last, the broadcast system allocates to our pair of receivers grows (at a given Es), which requires the receiver decoding thehierarchical16-QAM(α=1)foratimeproportion2t×a1 the LP stream to have a good channel quality as observed in and the 16-QAM for 2t×a2. Table II. This is why large values of α are interesting only Finally, the performance obtained here is a consequence of when some receivers experience very good channel quality. our rate policy: we choose to offer the same rate to everyone. Using α = 4 and the smallest coding rate in the standard However, our solution can be adapted to any other policy. For (1/5), the decoding threshold is 10.3 dB and then the receiver instance,ifthesystemcontainspremiumuserswhopaytoget withaSNRof8dBcannotdecodeanything.Wealsoremark a better rate, the fraction of time dedicated to these receivers that α = 0.5 and α = 0.8 give the same rate pairs, which will be computed according to this new policy. is a consequence of the coding rate discretization. In fact, a III. APPLICATIONTOBROADCASTCHANNELSYSTEMS close look to Table II (colored cells) shows the HP and LP streamsusethe2/5and1/2codingratesrespectivelyforboth This section compares the schemes described in Section II α. Finally, the new values of α give the best results in that ontwobroadcastchannels.Thefirstoneconsistsofonesource case. and two receivers, which is the simplest broadcast channel. Then, the performance is evaluated on an advanced example with one transmitter and six receivers. Receiver 1 SNR = 4 dB, receiver 2 SNR = 8 dB 1.4 (R1, R2) = (Rhm, Rhm) 1166−−QQAAMM RR==11//32 A. Channel with two receivers hier. 16−QAM α=4 1.2 hier. 16−QAM α=2 hier. 16−QAM α=1 We consider one source communicating with two receivers. hier. 16−QAM α=0.8 mbol) 1 hTiiemre. 1s6h−aQrinAgM α=0.5 Figure 2 already shows the throughput gain of the combined R (in bit/sy10.8 Convex hull swchheenmeonienrceocmeivpearrishoans atoStNheRcolafs4sicdaBl tainmdethsheaoritnhgersatraSteNgRy Receiver 1 rate 00..46 oathgfrao8iungsdthBpt.uhteFoifgrfeuecrreeeidv3ebrysp’rtheSseeNcnlRtass,stiwhceahletarhnerdoRuhgiteshrpaaurncthdriacRtailohmmRodhaumrlea/tRitohtnes (R1, R2) = (Rts, Rts) timesharingrespectively.Itallowstohaveaglobalideaonthe 0.2 performance.AmoredetailedstudycanbedonewithFigure4, which corresponds to cuts in Figure 3 for various SNR values 00 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Receiver 2 rate R2 (in bit/symbol) of the receiver 1. In general, the gains are more important when the SNR Fig. 2: Achievable rate region difference between the two receivers is large. For instance, in Figure 4a and Figure 4b, when receiver 1 experiences a We now consider a broadcast system with n receivers. The bad channel quality, if the SNR of receiver 2 is large, the first step is to group the receivers in pairs in order to use the gain is almost 25%. Moreover, in all figures, a gain of at hierarchical modulation. Then for each pair, the achievable least 20% is observed. We also remark there is no gain when rate is computed as described previously. Finally, we need to SNR is both small (e.g., Figure 4a) or large (e.g., Figure 4e equalize the rate between each receiver. This is done by time and Figure 4f). The first comment is helpful when we have to B. Advanced example We consider one source communicating with six receivers (Rec ,i = 1..6). The channel is intended to model a satellite 1.3 i spot beam with clear sky conditions. At the center of the 1.25 beam, the channel quality is good and there are few receivers, 1.2 whileawayfromthecenter,theSNRdecreasesbutthenumber /Rmts1.15 of receivers increases. Two parameters describe the channel: Rh 1.1 SNR is the SNR at the center of the beam and ∆ is used max 1.05 to characterize the attenuation due to the distance from the 1 center of the beam. We assume that the SNR distribution is 12 10 12 as follows: 8 10 6 8 • Rec1 has a SNR of SNRmax−∆ 6 Es/No receiver 2 (dB) 4 2 2 4 Es/No receiver 1 (dB) •• RReecc42,aRndecR5eacn3dhRaveec6ahSaNveRaoSfNSRNoRfmSaNx−Rm2∆ax−3∆ Using the DVB-SH guidelines [9] and (3), the classical Fig. 3: Throughput ratio according to receivers’ SNR time sharing strategy is easily treated. For the hierarchical modulation time sharing, we first need to group the receivers in pairs. For instance, we can group Rec with Rec , Rec 1 4 2 groupthereceiversinpairsinalargebroadcastsystem.Infact, with Rec and Rec with Rec . In fact, the performance 5 3 6 the next example will show that the grouping strategy has a only depends on the SNRs of the receivers in pairs. Then greatimpactontheperformanceofthesystem,andthenmust the previous grouping is equivalent in terms of performance be chosen carefully. Finally, the variations of all the curves to the following scheme: Rec with Rec , Rec with Rec 1 6 2 4 in Figure 4 can be explained as we compute the ratio of two and Rec with Rec . These two groupings refer to strategy 3 5 increasing staircase functions, R and R . hm ts A in Figure 5. Thus only three grouping strategies have to be considered. Figure 5 presents the three strategies studied for this example. 1.3 1.3 # of receivers # of receivers # of receivers 1.2 1.2 R/Rhmts R/Rhmts 3 3 3 1.1 1.1 2 2 2 1 1 1 12 4 6 8 10 12 12 4 6 8 10 12 Es/No receiver 2 (dB) Es/No receiver 2 (dB) (a)Receiver1SNR:2dB (b)Receiver1SNR:4dB Es/No (dB) Es/No (dB) Es/No (dB) (a)StrategyA (b)StrategyB (c)StrategyC 1.3 Fig. 5: Grouping strategies 1.2 1.2 Foreachgroupingstrategy,wecomputethebestthroughput R/Rhmts1.1 R/Rhmts available for the receivers when they implement the classical 1.1 or hierarchical modulation time sharing. We illustrate its computation on an example. We consider the classical time sharingstrategywiththechannelparameters(SNR ,∆)= 12 4 6 8 10 12 12 4 6 8 10 12 max Es/No receiver 2 (dB) Es/No receiver 2 (dB) (10 dB,2 dB). Rec has a SNR of 8 dB and can decode the 1 (c)Receiver1SNR:6dB (d)Receiver1SNR:8dB 1/2 16-QAM (best possible rate), Rec and Rec have a SNR 2 3 of 6 dB and can decode the 2/5 16-QAM and finally Rec , 4 1.2 Rec and Rec have a SNR of 4 dB and can decode the 1/3 5 6 16-QAM (or 2/3 QPSK) [9]. The average rate is computed 1.2 with (3). For the hierarchical modulation time sharing, the R/Rhmts1.1 R/Rhmts calculation is done the same way for each strategy. Figure 6 1.1 give the throughputs for the different solutions. The abscissa corresponds to the channel parameters (SNR ,∆) and max 12 4 6 8 10 12 12 4 6 8 10 12 the hierarchical modulation time sharing corresponds to the Es/No receiver 2 (dB) Es/No receiver 2 (dB) histograms labeled Strategy A, B and C. (e)Receiver1SNR:10dB (f)Receiver1SNR:12dB Finally, Table III presents the throughput gain for several Fig. 4: Throughput ratio channel configurations. We obtain a maximum gain of 17% 0.45 capacity is defined by C = 1C , where C is the Classical Time Sharing mod m mod mod Strategy A modulation’s capacity and m denotes the number of bits per ol) 0.4 SSttrraatteeggyy CB symbol.Nowwewouldliketoobtainthedecodingthresholds b bit/sym 0.35 fRoarteth(eBhEiRer)aorcfh1i0c−al51.6T-hQeAHMPawnidthLαPs=tre1amatsautsaergbeotthBtihteE1rr/5o-r ut (in 0.3 turbocodes.Themethodrequirestheperformancecurve(e.g., p ugh 0.25 BER against Es/N0) for one reference modulation, which o hr is the QPSK in the example. The decoding thresholds are T 0.2 computed as follow: 0.15 1) Use the performance curve of the QPSK with rate (12,1)(10,1) (8,1) (6,1) (12,2)(10,2) (8,2) (6,2) Channel configuration (SNRmax, ∆) R = 1/5 to get the operating point (Es/N0)ref cor- responding to the desired performance. In the DVB-SH Fig. 6: Throughput according to channel configuration guidelines [9, Table 7.5], we find, BER (−3.9 dB)=10−5. QPSK withthestrategyAandalargeSNRchannel.Forallstrategies, 2) Compute the normalized capacity R˜ for the QPSK, we have an improvement of the throughput in comparison to the classical time sharing strategy. However, the strategy A R˜ =R˜ =C (−3.9 dB)≈0.2454, HP LP QPSK obtains the best results, sometimes two times better than the 3) For the hierarchical 16-QAM, compute E /N such as otherstrategies.Itshowsthegroupingstrategygreatlyimpacts s 0 thenormalizedcapacityatthisSNRequalsR˜.Then,the on the final performance of the system. Moreover, strategy A decoding threshold for the HP and LP streams are, has the largest average SNR difference between two receivers in pairs. This supports the observation made on the previous (E /N ) = C−1 (cid:16)R˜ (cid:17)=−2.46 dB example and give an idea on how to group the receivers in a s 0 HP HP,α=2 HP (cid:16) (cid:17) large broadcast system. (E /N ) = C−1 R˜ =3.72 dB s 0 LP LP,α=2 LP TABLE III: Throughput gain REFERENCES (SNRmax,∆) StrategyA StrategyB StrategyC [1] T.M.Cover,“Broadcastchannels,”IEEETransactionsonInformation (12dB,1dB) 17% 11% 11% Theory,vol.IT-18,pp.2–14,January1972. [2] P. P. Bergmans and T. M. Cover, “Cooperative broadcasting,” IEEE (10dB,1dB) 9% 9% 9% TransactionsonInformationTheory,vol.IT-20,pp.317–324,May1974. (8dB,1dB) 4% 2% 2% [3] C.Hellge,S.Mirta,T.Schierl,andT.Wiegand,“MobileTVwithSVC (6dB,1dB) 3% 2% 2% andhierarchicalmodulationforDVB-Hbroadcastservices,”Broadband (12dB,2dB) 12% 6% 6% MultimediaSystemsandBroadcasting,2009. (10dB,2dB) 14% 9% 6% [4] H.Schwarz,D.Marpe,andT.Wiegand,“Overviewofthescalablevideo (8dB,2dB) 11% 6% 7% codingextensionoftheH.264/AVCstandard,”CircuitsandSystemsfor (6dB,2dB) 1% 2% 1% VideoTechnology,IEEETransactionson,vol.17,no.9,pp.1103–1120, sept.2007. [5] E. Chen, J. L. Koslov, V. Mignone, and J. Santoru, “DVB-S2 IV. 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COMPUTATIONOFTHEDECODINGTHRESHOLDS The decoding thresholds are computed using the method described in [11]. First, for any modulation, the normalized