Symmetry and Asymmetry of MIMO Fading Channels

Transcription

1 Symmery and Asymmery of MIMO Fadng Channels mmanuel Abbe, mre Telaar, Member, I, and Lzhong Zheng, Member, I Absrac We consder ergodc coheren MIMO channels, and characerze he opmal npu dsrbuon under general dsrbuons of he fadng marx. We descrbe how symmeres n he fadng marx dsrbuon and he consran se, descrbed naurally wh group srucures, mply he symmerc properes of he opmal npu dsrbuon. We use hs approach o gve a general condon under whch he whe Gaussan npu s opmal, and o compue he opmal npu for MISO Rcean channel. In conras, we also nvesgae he Kronecker model, n whch case we show how an asymmerc srucure n he problem s also preserved n he opmal npu, leadng o a new waerfllng soluon. I. INTRODUCTION Vecor fadng channel of he form () has been suded horoughly n he recen years. The opmal npu dsrbuon of x s, however, only characerzed n some specal cases. In hs paper, we ry o srenghen he known resuls on he capacy achevng npu dsrbuons by explong he symmerc properes of he channel. y = Hx + n () We assume n () ha x X = C, and y Y = C r, where and r are he number of ransm and receve anennas, respecvely. We assume ha a each me, he r marx H s drawn from an ergodc process havng margnal probably measure µ H. In hs paper, we wll focus on he case where H s perfecly known a he recever,.e., he coheren communcaon model. We assume he addve nose n s drawn..d from a complex crcularly-symmerc gaussan (C.C.S.G.) random varable of covarance marx K, ndependenly from he H s. Moreover, he npus {x } are consraned n he followng way. If we wan o use a code C = {c(),...,c(m)} X n, n, hen he code has o sasfy ha for every m M: n n = c(m) c(m) D, where D H + () s a gven compac se we use H + (n) o denoe he se of herman posve defne marces of sze n. A parcular example of such a consran s when one consder D o be {A H + () ra P }, for a gven P R. Ths s equvalen o ask for X X P and s called he oal power consran. An ndvdual power consran can also be consdered,.e. when X 2 P, for gven P R,, hen he se D would be. Abbe and L. Zheng are wh he Massachuses Insue of Technology, Cambrdge MA 0239, USA, and. Telaar s wh he PFL, Lausanne 05, CH. (emal: eabbe@m.edu,lzhong@m.edu,emre.elaar@epfl.ch). {A H + () A P, }. Le C be he capacy of hs channel under hs general consran. Then, denong by X a random vecor (r.ve.) n C, we know from sandard nformaon heorec argumens ha C(µ H, D ) = max I(X; Y, H) X:XX D Under he coheren assumpon, he opmal npu of x s alway Gaussan. The opmal covarance marx of x, s however no characerzed for he general cases. In he specal case ha H has..d. C.C.S.G. enres, wh a oal power consran of P, s shown n [] ha he opmal npu covarance marx s whe, XX = P I, and he capacy ncrease lnearly wh mn, r. The proof of hs saemen uses he concavy of he muual nformaon funcon, as well as he symmery of he channel. In a nushell, snce any roaon of he npu dsrbuon would no change he resulng muual nformaon, he opmal npu has o be soropcally dsrbued. The man quesons addressed n he curren paper s how o address he symmery of a vecor channel, and how does he symmery n he channel affec he opmal npu dsrbuon. We argue ha group srucures are he naural ool for hs ask. Usng he noon of groups, we can charaerze a number of dfferen symmeres offered by he vecor channel, mos of whch are weaker han he specal case n []. We explore how does such symmerc srucures ranslae no he symmery n he opmal npu. In parcular, we prove a more general condon, han [], for he opmal npu dsrbuon o be whe Gaussan. Moreover, we also presen some paral resuls when such symmerc propery s absen from he channel. II. GNRAL XPRSSION OF TH CAPACITY Defnon : We defne he opmal npus by X op (µ H, D ) = arg max X C I(X; HX + N, H), where arg max X C f(x), for a real funcon f, denoes he se of he elemens x sasfyng f(x) f(y), y C. We now use he assumpons we made on he channel o gve a more specfc expresson for he capacy and he opmal npus. The fac ha he gaussan dsrbuon maxmzes he enropy under a covarance consran lead o he followng resul.

2 2 Proposon : Le ψ : Q D µh log de(i + K HQH ) R, (2) whch we call he muual nformaon funcon. Then, accordng o prevous defnons and assumpons, we have where X op (µ H, D ) N C (Q op ), Q op (µ H, D ) = arg max Q D ψ(q) and C(µ H, D ) = max ψ(q). Q D n D. We can also argue ha we can resrc he doman of maxmzaon o D, snce he dervave of log de(i + K HQH ) wh respec o any enres of Q s monoone (by concavy of Q log de(i + K HQH )). A. Quanfyng symmeres III. SYMMTRIS Assume ha he channel has he same oupu dsrbuon when sendng any npu X or a permued verson of, say, PX, where P s a permuaon marx. Y = HX + N Y = H(PX) + N. Then, we alk abou a symmery of he channel wh respec o ha ransformaon P. Bu from prevous equvalence, hs s o say ha HP H. Remarks: ) Ths ype of nvarance has a naural group srucure: assume you have he nvarance HP H for a se of marces P, hen clearly HP P j H and f P s nverble HP H. Thus hs nvarance sll holds for he group generaed by hs se. 2) In order o compare X and PX, we need o ensure ha P X s sasfyng he consdered conran oo,.e. s covarance marx P(XX )P has o belong o D as well. 3) Oher group han he permuaons mgh be of neres, for example f we wan o consder suaon where he symmery s expressed by keepng he channel equvalen wheher we send an npu X or a modfed verson of where some componen s sgns have been flpped, hen he group of dagonal marces wh and s he approprae group. These remarks movae he followng defnons. Defnon 2: Le G be a group n M n (C). ) A random marx s G-nvaran (on he rgh) f Hg H, g G. 2) A se of marces D M n (C) s nvaran n G- conjugaon f gqg D, Q D, g G. 3) A funcon Ψ : D R s G-nvaran f D s nvaran n G-conjugaon and f Ψ(gQg ) = Ψ(Q), Q D, g G. Noe ha only subgroups of unary marces are of our neres regardng our MIMO channel seng, because he muual nformaon funcon evaluaed a Q depends on he dsrbuon of HQH. xamples of funcons whch are nvaran n G- conjugaon for unary subgroups are all funcons of he form x f(mxm ) where f s any measurable funcon and M s a random marx ha s G-nvaran on he rgh. The reason for whch a conjugaon nvarance for unary subgroups s relevan n our MIMO sengs s a consequence of he fac ha we are workng wh second order momen consran, whch mples ha he muual nformaon has precsely he above descrbed form (cf. (2)). Fnally, examples of groups n M n (C) are U(n), whch s he larges group we wll consder, and s subgroups Σ(n) and Π(n) (wh he usual marx mulplcaon), defned as: ) U(n): he unary group of sze n, 2) Π(n): he group of permuaon marces of sze n, 3) Σ(n): he dagonal marces group wh and of sze n. We now gave a defnon o quanfy symmeres n he problem, hrough he group of nvarance of H and D, or equvalenly of ψ, he queson s hen: how do we use hs nvarance n order o ge knowledge on he opmal npu? In he nex secon we wll see ha hs s done hrough he commuan. B. xplong symmeres Defnon 3: The commuan of G s defned by he algebra Comm(G) = {A M n (C) Ag = ga, g G} = {A M n (C) A = gag, g G}. We sar wh a rval observaon lnkng he commuan and G-nvaran funcons. Lemma : Le G M n (C) be a group and D M n (C). Le Ψ : D R a G-nvaran funcon havng a unque maxmzer Q op. Then Q op Comm(G) D. Proof: We have ψ(q op ) = ψ(gq op g ), g G. We conclude by he unqueness of he maxmzer. Noe ha he bgger s he group he smaller s he commuan, whch s wha we expec n order o explo symmeres. C. Symmeres n MIMO We rewre one las me he prevous observaons for our MIMO channel conex. Proposon 2: Le a MIMO channel as defned n he nroducon and le G be a subgroup of U(). If he consran se D s nvaran n G-conjugaon, he fadng marx dsrbuon µ H s G-nvaran, hen Q op Comm(G) D. Also noe ha f G, G 2 are wo groups n U() and f D s nvaran n G -conjugaon whereas µ H s G 2 -nvaran, hen Q op Comm(G G 2 ) D.

3 3 We wll now see some specfc applcaons of prevous proposon. The cases ha we wll consder are dealng wh he followng commuans: Comm(Σ(n)) s he se of dagonal marces n M n (C), Comm(Π(n)) = {αi n + βj n α, β C}, where J n = n n, Comm(U(n)) = {αi n α C}. Corollary : Toal power consran For a gven P R +, we consder Q D = {Q H + () r(q) P }. If µ H s nvaran n G-conjugaon for a subgroup G of U(), hen Q c Comm(G) D. Smply observe ha D s nvaran n U()-conjugaon. Two neresng cases of subgroups of U() are Π() and Σ(). From wha we saw n he examples of he commuan, f we consder a dsrbuon µ H nvaran under Σ(), hen Q c s dagonal and f s nvaran under Π(), hen Q c wll have he same value for all componens nsde he dagonal ( P f one works n D ) and also he same value for all elemens ousde he dagonal, as long as says a posve defne marx. xamples of Σ()-nvaran random marces are marces wh ndependen symmerc enres (symmerc means ha H j H j ) and examples of Π()-nvaran ones are marces wh..d. enres or jonly gaussan enres havng a covarance marx of he form αi r + βj r. Corollary 2: Sll consderng Q D, f H s Π()Σ()- nvaran, whch s for example he case when H j are..d. and H j H j, r, j, hen Q c = P I. Noe ha we dd no assume ha he enres of H are gaussan (whch would be a parcular case of hs) n order o ge P I as a maxmzer. Also noe ha he group Π() could be replaced by C(), he group of cyclng permuaons, and we would ge he same concluson. Generally, hs wll be rue as long as we have a group of nvarance G such ha Comm(G) s reduced o he mulple s of he deny. Corollary 3: Local power consran If X s consraned by X 2 P for gven P R +,, and f H s Σ()-nvaran, hen Q c = dag(p,..., P ). The consran X 2 P mples ha Q D = {Q H + () Q P, }, now we no longer have ha D s nvaran n U()-conjugaon, bu we sll have, for example, nvarance n Σ()-conjugaon. Concluson: As has been llusraed n prevous example, he problem of symmeres should be generally approached n he followng way: frs denfy he nvarance propery of he doman D n whch we are workng (we saw examples of oal and local power consran (see corollares 2 and 3), several nermedary cases are possble), hen denfy he nvarance propery of he fadng marx dsrbuon µ H, once we have hese wo groups of nvarance, we know ha we can resrc our search of Q c o marces commung wh hese groups and sayng n D. Whch means ha he commuan s summarzng he nformaon gven by he symmeres n he problem. IV. ASYMMTRIS IN MIMO Le us consder now he followng specfc channel. Defnon 4: The Kronecker channel We consder he consran se D = {Q H + () r(q) P }, and H = AWB, where A M r (C) non-zero, B M (C) non-zero, W s a r random marx beng U()-nvaran on he rgh. Queson: do we sll have some symmeres for such a channel? Any marx mulplcaon on he rgh of H wll be frs acng on he fxed marx B, herefore our nvarance properes seem o be compromsed. Bu before facng he asymmerc srucure of hs channel, le us brng back some symmeres n he problem. A. Brngng back he symmeres As we jus saw, n some suaons, a symmerc srucure s no clearly exsng. Bu wh approprae ransformaon, one can brng some symmeres back no he problem. We now gve an example of how o carry ou such a procedure for he Kronecker channel. One can also ackle he Rcean channel defned by H = A+W, where A s a deermnsc marx, n a smlar way. For boh cases, he followng smple observaon s used. Lemma 2: Le Ψ : D R wh a unque maxmzer Q op and such ha D s nvaran n G-conjugaon. Then, for any M G, we have M Q op M = argmax Q D Ψ(MQM ). Because we are now dealng wh he Kronecker channel, he muual nformaon funcon ψ s gven by Q D ψ(q) = log de(i + K AWBQB W A ). We now denoe he SVD of B by B = U B dag(b)v B, where U B, V B U() and b R +. Usng our prevous lemma, we can choose M = V B, n order o ge ha V BQ op V B = argmax Q D ψ(v BQV B) (3) The advanage of geng o he las maxmzaon problem s ha Q ψ(v B QV B ) s a Σ()-nvaran funcon and hus we can resrc our maxmzaon o marces beng dagonal (wh race smaller han P ). In oher words, we showed he followng observaons: Remark : he egenvecors of Q op are he rgh-egenvecors of B and s egenvalues q op = (q op,...,q op ) are gven by q op = arg max P q R + s.. = log de(i q P +K AW dag(q b 2,...,q b 2 ))W A ).

4 4 B. Asymmeres for he Kronecker channel Suppose ha he marx B n he Kronecker channel s dagonal,.e. B = dag(b), wh b R +. Then we know ha he opmal covarance marx s dagonal bu we do no know wha are he value of he dagonal elemens. Now assume ha b... b n, can we hen expec ha he opmal covarance marx should preserve hs orderng n some sense? We wll now analyze hese knds of quesons. We wll presen wo proposons ha wll help descrbe he opmal npu for he Kronecker channel. We know ha f he random marx H were replaced by he deermnsc marx B, he opmal npu covarance has egenvalues q op gven va waer-flng on he sngular values of B (cf. []). Two parcular properes of he waer-flng soluon are he followng. ) Monooncy: f b b j hen q op q op j (wh equaly f b = b j ) 2) On/Off hreshold: f b + s suffcenly bgger han b, hen we mgh end up by sharng he whole power P on he bgges componens of b. We wll see n he nex wo proposons ha hese wo properes are preserved n he Kronecker model. We sar wh he monooncy resul. Proposon 3: We have where q op sasfes q op q op j Q op = V B dag(q op )V B q op = qop 0, = P, f b > b j, and q op = q op j f b = b j. Noe: If B = I and µ W s G-nvaran on he rgh wh G U(), hen Q op Comm(G) D. Remark: Ths proposon says ha he egenvecors of Q op are he rgh-egenvecors of B (whch has been shown n prevous secon) and ha s egenvalues are monooncally dsrbued wh respec o he sngular values of B. In order o prove hs resul, we need a prelmnary lemma. Le λ (M)... λ n (M) denoe he ordered egenvalues of any marx M H(n) we use H(n) o denoe he se of herman marces of sze n. Lemma 3: Le n, P H + (n) and H H(n). We hen have, λ k (H + P) > λ k (H), k =,...,n. Proof of proposon 3. The nal expresson of he muual nformaon funcon for hs channel s ψ(q) = log de(i + K AWBQB W A ). Frs noe ha A affecs he funcon ψ n he same way as K, n oher words, we could consder one of hs wo marx o be he deny, for example, assume..d. componens for he nose and se à = K 2A. If B = I r any nvarance properes on he rgh for µ W wll be preserved for AW, hus he noe afer he proposon s a drec consequence of proposon 2. The frs par of he proposon s proved n he prevous secon, le us now look a he egenvalues. We have q op = arg max P q R + s.. log de(i = q P +AW dag(q b 2,..., q b 2 ))W A ). Thus we wll consder from now on ψ : q log de(i + AW dag(q b 2,..., q b 2 ))W A ). Now observe ha f b = b j hen ψ s Π() j -nvaran, where Π() j s he subgroup of permuaons keepng he dagonal elemens dfferen han and j nvaran (ransposon), hus we ge from proposon 2 ha q op = q op j. Now, le P = P =3 qop, such ha q op + q op 2 = P. We wll show ha f b > b 2, hen for any 0 P P, q ψ(q) ( P 2, P 2,qop 3,...,q op whch, by he concavy of ψ, mples ha > ) q 2 ψ(q) ( P 2, P, 2,qop 3,...,q op ) q op > q op 2. By symmery of he problem, hs clearly mples he resul for any componens and j (oher han and 2). We have ψ(q) = log de(i + q b 2 Aw (Aw ) ) = where w s he -h column of W. For an nverble marx M, we have he formula mj log de(m) = (M ) j, herefore we have qj ψ(q) = b 2 j r ( I + = q b 2 Aw (Aw ) ) Awj (Aw j ). Le us denoe X = Aw (Aw ), whch ( are herman posve semdefne marces, as well as I + ) = q b 2 X whch s n addon posve defne and nverble. We defne Z = =3 q b 2 X and Z op = b 2 X, we hen rewre =3 qop q ψ(q) = b 2 r ( I + q b 2 X + q 2 b 2 2X 2 + Z ) X (4) q2 ψ(q) = b 2 2 r( I + q b 2 X + q 2 b 2 2 X 2 + Z ) X2 = b 2 2r ( I + q b 2 X 2 + q 2 b 2 2X + Z ) X (5) where n he las lne we nerchanged he random marces X and X 2, as W s Π()-nvaran. To conclude he proof, we have o show ha f b > b 2 (I + P b 2 r > b 2 2 r ( I + P 2 b2 X + P 2 b2 X 2 + P 2 b2 2X 2 + Z op ) X 2 b2 2 X + Z op ) X, for any 0 P. Ths s clearly verfed n he scalar case (r = ). In he marx case, more seps (usng he prevous lemma) are requred o show ha he resul hold. We now presen an On/Off hreshold resul.

5 5 Proposon 4: Le b b 2... b. We assume ha..d. r =, w j N C (), j. Then, for all j =,...,, here exss b(b j ) 0 such ha f b j+ > b(b j ) hen q op = 0, =,...,j. ā(pb 2 j ) Commens: We wll see ha one can ake b(b j ) = P, where ā s gven by he recprocal of he funcon F, wh F(a) = +ax, whch s also known as he or exponenal funcon. The prevous resul says he followng, f here s a value b j+ such ha Pb 2 j+ s bgger han ā(pb2 j ), we hen know ha he opmal q op are zero for =,...,j. In oher words, f some of hese gans (b s) are wo small compared o some ohers, we swch off he correspondng anennas. Proof: In hs seng we have ψ(q) = log( + P q d X ), wh b 2 = d..d., X (), and q Θ() = {x R + = x = }. Le Z j = + P j,j+ q X. We hen have Pd j X j qj ψ(q) =. Z j + Pd j X j + Pd j+ X j+ = Le 0 < T and p (j) be a vecor wh p (j) j+ = T, p(j) j = 0 and hus j,j+ p(j) = T. From he concavy of ψ, f qj ψ(q) q=p (j) < qj+ ψ(q) q=p (j), 0 < T, (6) hen q op = 0, =,...,j. Now, (6) becomes Z + TPd jx j <, 0 < T, Z + TPd j+ X j+ so f for all z and 0 < T we have z + TPd jx j = z/t + Pd jx j <, z + TPd j+ X j+ z/t + Pd j+ X j+ we are done. Las nequaly s equvalen o < z + Pd j+ X j+ z + Pd j, z. Le F(a) = +ax, a j+ = Pd j+ and a j = Pd j, we now wonder when F(a j+ /z) <, z. + a j /z For a gven β R +, le α(β) be he smalles number sasfyng F(α(β)) +β. Then f for any possble a j, ā(a j ) = sup z zα(a j /z) < +, we deduce ha for a j > ā(a j ), we sasfy F(a j+ /z) < +a j/z, z. One can check ha α s convex and ha s a connuous ncreasng funcon wh α(0) = 0. Therefore whch mples ha ā(a j ) = sup z zα(a j /z) = α(a j ). And we conclude by seng b(b ā(pb 2 j ) = j ) P. The funcon b 2 ( ) s connuous convex and ncreasng wh b(0) = 0, a dervave of a 0 and of 0 a nfny. In he followng graphc, he nverse of he funcon b 2 s ploed. Le us now look a some numercal example, we Fg.. Inverse of b 2, P =. assume ha we have r = 6 recevng anennas and P =. If for example B s such ha s sngular values have squared gven by (b 2,...,b 2 6) = (,, 2, 3, 4, ), can be seen from fgure ha b 6 exceeds he On/Off hreshold of b 5 and hus he opmal power allocuon s (q max,..., q6 max ) = (0,...,0, ),.e. n hs case we solved he problem. If we had (b 2,...,b2 6 ) = (,, 2, 5, 6, 8), hen he prevous suaon does no hold anymore, bu b 4 exceeds he On/Off hreshold of b 3 and we are reduced o a half-dmensonal opmzaon problem for he values of b 4, b 5 and b 6. Concluson: The opmal power allocaon s no acheved n he same way han for he case of a deermnsc fadng marx B, bu sll we preserve he same properes, ndeed he monooncy and he on/off hreshold (wh more specfc hypoheses). RFRNCS []. Telaar, Capacy of Mul-anenna Gaussan Channels, uropean Trans. on Telecommuncaons, Vol. 0, No 6, 999, pp [2] V. Marchenko and L. Pasur, Dsrbuon of egenvalues n ceran ses of random marces, Mah. USSR-Sbornk (967), no. 4, pp and hus α(a j /z) = α(a j /z + 0( /z)) α(a j )/z + 0 zα(a j /z) α(a j ), z