Optimal Bandwidth and Power Allocation for

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1 Optmal Bandwdth and Power Allocaton for 1 Sum Ergodc Capacty under Fadng Channels n Cogntve Rado Networks arxv: v1 [cs.it] 25 Jun 2010 Xaowen Gong, Student Member, IEEE, Sergy A. Vorobyov, Senor Member, IEEE, and Chntha Tellambura, Senor Member, IEEE Abstract Ths paper studes optmal bandwdth and power allocaton n a cogntve rado network where multple secondary users (SUs) share the lcensed spectrum of a prmary user (PU) under fadng channels usng the frequency dvson multple access scheme. The sum ergodc capacty of all the SUs s taken as the performance metrc of the network. Besdes all combnatons of the peak/average transmt power constrants at the SUs and the peak/average nterference power constrant mposed by the PU, total bandwdth constrant of the lcensed spectrum s also taken nto account. Optmal bandwdth allocaton s derved n closed-form for any gven power allocaton. The structures of optmal power allocatons are also derved under all possble combnatons of the aforementoned power constrants. These structures ndcate the possble numbers of users that transmt at nonzero power but below ther correspondng peak powers, and show that other users do not transmt or transmt at ther correspondng peak power. Based on these structures, effcent algorthms are developed for fndng the optmal power allocatons. Index Terms Bandwdth and power allocaton, cogntve rado, fadng channels, frequency dvson multple access, sum ergodc capacty Ths work was supported n parts by research grants from the Natural Scence and Engneerng Research Councl (NSERC) of Canada and Alberta Ingenuty New Faculty Award. The authors are wth the Department of Electrcal and Computer Engneerng, Unversty of Alberta, St., Edmonton, Alberta, T6G 2V4 Canada. The contactng emals are {xgong2, vorobyov, chntha}@ece.ualberta.ca. Correspondng author: Sergy A. Vorobyov, Dept. of Electrcal and Computer Engneerng, Unversty of Alberta, St., Edmonton, Alberta, T6G 2V4, Canada; Phone: +1 (780) , Fax: +1 (780)

2 2 I. INTRODUCTION Cogntve rado s a promsng technology for mprovng spectrum utlzaton n wreless communcatons systems [1]. A secondary user (SU) n a cogntve rado network s allowed to access the lcensed spectrum allocated to a prmary user (PU) f the spectrum s not utlzed by the PU. Such a spectrum sharng strategy, whch s referred to as spectrum overlay or opportunstc spectrum access (OSA) [2], requres correct detecton of spectrum opportuntes by the SU. Exstng works on spectrum overlay have manly studed spectrum sensng and access polces at the medum access control (MAC) layer [3]- [6]. An alternatve strategy, whch s known as spectrum underlay [7] [12], enables the PU and the SU to transmt smultaneously, provded that the receved nterference power by the PU s below a prescrbed threshold level. A number of works have recently studed nformaton theoretc lmts for resource allocaton n the context of spectrum underlay. In [13], the optmal power allocaton whch ams at maxmzng the ergodc capacty acheved by an SU s derved for varous channel fadng models subject to the peak nterference power (PIP) constrant or average nterference power (AIP) constrant mposed by a PU. In [14], the authors derve the optmal power allocaton for the ergodc capacty, outage capacty, and mnmum-rate capacty of an SU under both the PIP and AIP constrants from a PU. The ergodc capacty, delay-lmted capacty, and outage capacty of an SU s studed n [15] under dfferent combnatons of the peak transmt power (PTP) constrant or average transmt power (ATP) constrant at the SU and the PIP constrant or AIP constrant from a PU. However, all the papers mentoned above consder the setup of a sngle SU. The most recent work [16] studes a cogntve rado network of multple SUs under multple access channel and broadcast channel models, where the optmal power allocaton s derved to acheve the maxmum sum ergodc capacty of the SUs subject to varous mxed transmt and nterference power constrants. The optmalty condtons for the dynamc tme dvson multple access scheme are also derved. In ths paper, we focus on a cogntve rado network where multple SUs share the lcensed spectrum of a PU usng the frequency dvson multple access (FDMA) scheme. The sum ergodc capacty of the SUs, whch s a relevant network performance metrc for delay-tolerant traffcs, s studed. Besdes the transmt power constrants at the SUs and the nterference power constrant mposed by the PU, whch are also consdered n [13]- [16], we also take nto account the total

3 3 bandwdth constrant of the shared spectrum. Such study s motvated by the fact observed for a number of dfferent applcatons that jont bandwdth and power allocaton can sgnfcantly mprove the performance of systems wth lmted both ndvdual (power) and publc (bandwdth) resources [17] [23]. Thus, n ths paper, nstead of conventonal fxed and equal bandwdth allocaton used n FDMA, we nvestgate dynamc and unequal bandwdth allocaton, where the bandwdth allocaton vares for dfferent SUs at dfferent channel fadng states. Moreover, dfferent from the exstng works [13]- [16], all combnatons of the transmt power constrants and the nterference power constrants are consdered, ncludng both PTP and ATP constrants combned wth both PIP and AIP constrants. We frst derve the optmal bandwdth allocaton for any gven power allocaton n any channel fadng state, whch results n equvalent problems that only nvolve power allocaton. Usng the convexty of the resultant power allocaton problems, we apply dual decomposton whch transforms these problems nto equvalent dual problems, where each dual functon nvolves a power allocaton subproblem assocated wth a specfc channel fadng state. The dual problems can be solved usng standard subgradent algorthms. For the power allocaton subproblem under all combnatons of the power constrants, we derve the structures of the optmal power allocatons. These structures ndcate the possble numbers of users that transmt at nonzero power but below ther correspondng peak powers, and show that other users do not transmt or transmt at ther correspondng peak power. Based on these structures, we develop algorthms for fndng the optmal power allocatons n each channel fadng state. The rest of the paper s organzed as follows. Secton II summarzes the system model and formulates correspondng sum ergodc capacty maxmzaton problems. Secton III derves the optmal bandwdth allocaton for the problems formulated n Secton II subject to the bandwdth constrant. Secton IV obtans the optmal power allocatons from the resultant problems n Secton III under all combnatons of the transmt power constrants and nterference power constrants. Numercal results for the maxmum sum ergodc capacty under dfferent combnatons of the power constrants and the bandwdth constrant are shown n Secton V. Secton VI concludes the paper.

4 4 II. SYSTEM MODEL Consder a cogntve rado network of N SUs and one PU. The PU occupes a spectrum of bandwdth W for ts transmsson, whle the same spectrum s shared by the SUs. The spectrum s assumed to be dvded nto dstnct and nonoverlappng flat fadng channels wth dfferent bandwdth, so that the SUs share the spectrum through FDMA to avod nterferences wth each other. The channel power gans between the th SU transmtter (SU-Tx) and the th SU recever (SU-Rx) and between the th SU-Tx and the PU recever (PU-Rx) are denoted by h and g, respectvely. The channel power gans,.e., g [g 1 g 2 g N ] and h [h 1 h 2 h N ], are assumed to be drawn from an ergodc and statonary vector random process. We further assume that full channel state nformaton (CSI),.e., the jont probablty densty functon (PDF) of the channel power gans and the nstantaneous channel power gans, are known at the SUs. 1 The nose at each SU-Rx plus the nterference from the PU transmtter (PU-Tx) s assumed to be addtve whte Gaussan nose (AWGN) wth unt power spectral densty (PSD). We denote the transmt power of the th SU-Tx and the channel bandwdth allocated to the th SU-Tx as p (g,h) and w (h,g), respectvely, for the nstantaneous channel power gans g and h. Then the total bandwdth constrant can be expressed as The PTP constrants are gven by w (h,g) W, h,g. (1) p (h,g) P pk,,h,g (2) where P pk gven by denotes the maxmum peak transmt power of the th SU-Tx. The PIP constrant s g p (h,g) Q pk, h,g (3) where Q pk denotes the maxmum peak nterference power allowed at the PU-Rx. The ATP constrants are gven by E{p (h,g)} P av, (4) 1 Note that the full CSI assumpton s typcal n the context of cogntve rado and s also made n other works such as [13]- [16]. Indeed, under ths assumpton we am at nvestgatng the nformaton-theoretc lmts on the sum ergodc capacty.

5 5 where the expectaton s taken over h and g, and P av denotes the maxmum average transmt power of the th SU-Tx. The AIP constrant s gven by { N } E g p (h,g) Q av (5) where Q av denotes the maxmum average nterference power allowed at the PU-Rx. The objectve s to maxmze the sum ergodc capacty of the SUs, whch can be wrtten as { N ( max E w (h,g)log 1+ h ) } p (h,g) (6) {w (h,g),p (h,g)} F w (h,g) where F s a feasble set specfed by the bandwdth constrants (1) and a partcular combnaton of the transmt power constrants{(2), (4)} and the nterference power constrants{(3), (5)}. Note that the constrants on nonnegatvty of the bandwdth and power allocatons,.e., w (h,g) 0 and p (h,g) 0,,h,g, are natural and, thus, omtted through out the paper for brevty. It can be shown that the objectve functon of the problem (6) s concave wth respect to {w (h,g),p (h,g)},,h,g. It can also be seen that the bandwdth and power constrants (1) (5) are lnear and, thus, convex. Therefore, the sum ergodc capacty maxmzaton problem (6) under dfferent combnatons of the constrants (1) (5) s a convex optmzaton problem. III. OPTIMAL BANDWIDTH ALLOCATION Gven that the power allocaton p (h,g),,h,g, s fxed, the maxmum sum ergodc capacty can be expressed as E{f 0 (h,g)}, where f 0 (h,g) s gven by f 0 (h,g) max {w (h,g)} s.t. G (w (h,g)) w (h,g) W where G (w (h,g)) w (h,g)log(1+h p (h,g)/w (h,g)) s an ncreasng and concave functon of w (h,g). The problem (7a) (7b) s smlar to the classcal water-fllng power allocaton problem. Thus, the optmal soluton of the problem (7a) (7b), denoted by {w (h,g)}, must satsfy G (w (h,g)) w (h,g) (7a) (7b) (7c) = G j(w j (h,g)) w (h,g)=w (h,g) w j (h,g), j. (8) wj (h,g)=w j (h,g)

6 6 Snce we have G (w (h,g)) w (h,g) = log w (h,g)=w (h,g) ( 1+ h ) p (h,g) w (h,g) h p (h,g) w (h,g)+h p (h,g) =Y ( ) h p (h,g) w (h,g) where Y(x) log(1 + x) x/(1 + x) s a monotoncally ncreasng functon, we can obtan from (8) that h p (h,g) w (h,g) = h jp j (h,g), j (10) (h,g) w j It follows from (7b) that at optmalty we have N w (h,g) = W. Furthermore, usng (10), we can obtan that w (h,g) = W h p (h,g) N h p (h,g). (11) Substtutng the optmal w (h,g) gven by (11) nto (6), we can equvalently rewrte (6) as { ( )} N max E W log 1+ h p (h,g) (12) {p (h,g)} F W where F s a feasble set specfed only by a partcular combnaton of the power constrants {(2), (3), (4), (5)}. Therefore, the optmal power allocaton obtaned from the problem (6) and denoted by {p (h,g)}, can also be obtaned by solvng the equvalent problem (12). Then the optmal bandwdth allocaton obtaned from the problem (6) and denoted by {w (h,g)}, can be found as w(h,g) = W h p (h,g) N h (13) p (h,g). IV. OPTIMAL POWER ALLOCATION In ths secton, we study the optmal power allocaton obtaned from the problem (12) wth F specfed by dfferent combnatons of the power constrants. A. Peak transmt power wth peak nterference power constrants Consder F = {the constrants (2) and (3)}. Then the optmal value of the problem (12) can be expressed as E{f 1 (h,g)}, where f 1 (h,g) s gven by ( ) N f 1 (h,g) max W log 1+ h p (h,g) {p (h,g)} W s.t. p (h,g) P pk, g p (h,g) Q pk. (9) (14a) (14b) (14c)

7 7 For brevty, we drop the dependence on h and g that specfes nstantaneous channel power gans. Also let {p } denote the optmal soluton of the problem (14a) (14c). Introducng q g p, the problem (14a) (14c) can be equvalently rewrtten as max {q } h g q s.t. q g P pk, q Q pk. (15a) (15b) (15c) Let {q } denote the optmal soluton of the problem (15a) (15c) and (s 1,s 2,,s N ) denote a permutaton of the SU ndexes such that h s1 /g s1 > h s2 /g s2 > > h sn /g sn. It s assumed that h /g h j /g j, j, snce h, g, h j, and g j are drawn from a contnuous-valued random process. Then the followng lemma s n order. Lemma 1: There exsts k, 1 k N, such that q s = g s P pk s,, 1 k 1, 0 < q s k g sk P pk s k, and q s = 0,, k +1 N. Proof: Let q s j > 0 for some j and let l < j for some l. Frst we prove that q s l = g sl P pk s l by contradcton. If q s l < g sl P pk s l, then we can always fnd q > 0 and defne a feasble soluton {q s } of the problem (15a) (15c) q s j q s j q,q s l q s l + q,q s q s,, j, l such that the objectve functon n (15a) acheves larger value for {q s } than for the optmal soluton {q }, snce we have N h s g s q s ( h s q hsl s g = h ) s j q > 0. (16) s g sl g sj Therefore, t contradcts the fact that {q s } s the optmal soluton of the problem (15a) (15c). Let q s j < g sj P pk s j for some j and let l > j for some l. Usng the result obtaned above, t can be proved also by contradcton that q s l = 0. Ths completes the proof. Lemma 1 shows that for the optmal power allocaton under the constrants (2) and (3), there exsts at most one user that transmts at nonzero power and below ts peak power, whle any other user ether does not transmt or transmts at ts peak power. Note that ether the constrants (15b) or the constrant (15c) must be actve at optmalty. Usng the structure of {q } gven n Lemma 1, k can be found by Algorthm 1.

8 8 Algorthm 1 Algorthm for fndng k n Lemma 1 Intalze: k = 1 whle k g s P pk s < Q pk and k N 1 do k = k +1 end whle Output: k Snce p s = qs /g s, we obtan Ps pk, 1 k 1 p s = mn{ps pk,(q pk k 1 g s Ps pk )/g s }, = k 0, k +1 N. (17) Note that for brevty, we say n ths paper that n x = 0 f n = 0 wth a lttle abuse of notaton. B. Average transmt power wth average nterference power constrants Consder F = {the constrants (4) and (5)}. Then the dual functon of the problem (12) can be wrtten as f 2 ({λ },µ) E{f 2(h,g)}+ λ P av +µq av (18) where {λ 1 N} and µ are the nonnegatve dual varables assocated wth the correspondng constrants n (4) and (5) and f 2 (h,g) s gven by ( ) N f 2 (h,g) max W log 1+ h p (h,g) {p (h,g)} W γ p (h,g) (19) wth γ λ +µg. Let {p } denote the optmal soluton of the problem (19), where we drop the dependence on h and g for brevty. Also let F({p }) denote the objectve functon n (19). If p > 0 for some, the followng must hold F({p }) h p = {p }={p } 1+ N h p /W γ = 0. (20) Then the followng lemma s of nterest. Lemma 2: If h γ for some, then p = 0.

9 9 Proof: If p j = 0, j, then p = 0. If p j 0 for some j, t can be seen that (20) can not be satsfed snce h γ. Thus, p = 0. If p = 0 for some, the followng must hold F({p }) h p = {p }={p } 1+ N h p /W γ 0. (21) Then the next lemma s n order. Lemma 3: p = 0,, f and only f h γ,. Proof: It can be seen from Lemma 2 that f h γ,, then p = 0,. Moreover, t can be seen from (21) that f p = 0,, then h γ,. Let (s 1,s 2,,s N ) denote a permutaton of the SU ndexes such that h s1 /γ s1 > h s2 /γ s2 > > h sn /γ sn. Then we can also prove the followng lemma. Lemma 4: There exsts at most one k such that p k > 0. Moreover, k = s 1. Proof: We prove t by contradcton. It can be seen from (20) that f p > 0 and p j > 0 for some j, the followng must hold h γ = h j γ j. (22) Snce h, γ, h j, and γ j are ndependent constants gven n the problem (19), (22) can not be satsfed. Let p k > 0 and p must hold Therefore, we must have k = s 1. = 0,, k. Then t follows from (20) and (21) that the followng h k γ k h γ, k. (23) Lemma 4 shows that for the optmal power allocaton under the constrants (4) and (5), there exsts at most one user that transmts at nonzero power, whle any other user does not transmt. Case 1: Consder the case when h γ,. It follows from Lemma 3 that p Case 2: = 0,. Consder the case when h γ does not hold for some. Usng Lemma 4, let p k > 0 and p = 0,, k. Substtutng {p } nto (20), we have p s 1 = W(1/γ s1 1/h s1 ). Therefore, we obtan W (1/(λ p s1 +µg s1 ) 1/h s1 ), = 1 s = 0, 2 N. (24)

10 10 C. Peak transmt power wth average nterference power constrants Consder F = {the constrants (2) and (5)}. Then the dual functon of the problem (12) can be wrtten as f 3 (µ) E{f 3 (h,g)}+µqav (25) where µ s the nonnegatve dual varable assocated wth the constrant (5), and f 3 (h,g) s gven by f 3 (h,g) max W log {p (h,g)} s.t. p (h,g) P pk,. ( ) N 1+ h p (h,g) µ W g p (h,g) (26a) (26b) Let {p } denote the optmal soluton of the problem (26a) (26b) after droppng the dependence on h and g for brevty. The followng cases are of nterest. Case 1: Consder the case when h µg,. Snce the problem (26a) (26b) wthout the constrants (26b) has the same form as the problem (19), and p = 0,, satsfes the constrant (26b), t can be seen from Lemma 3 that p = 0,. Case 2: Consder the case when h µg does not hold for some. The problem (26a) (26b) s equvalent to maxw log {q } ( ) N 1+ h q /µg W s.t. q µg P pk, where q µg p. Let {q } denote the optmal soluton of the problem (27a) (27b) and q (27a) (27b) (s 1,s 2,,s N ) denote a permutaton of the SU ndexes such that h s1 /µg s1 > h s2 /µg s2 > > h sn /µg sn. Then the followng lemma s n order. Lemma 5: There exsts k, 1 k N, such that q s = g s P pk s,, 1 k 1, 0 < q s k g sk P pk s k, and q s = 0,, k +1 N. Proof: Consder the followng ntermedate problem h max q (28a) {q } µg s.t. q µg P pk, q = Q (28b) (28c)

11 11 where Q N q and t s unknown snce {q } s unknown. Let {q } denote the optmal soluton of the problem (28a) (28c). If {q } {q }, we have N h q /µg N h q /µg snce {q } s a feasble soluton of the problem (28a) (28c). Then we have ( ) ( N F ({q }) F ({q}) = W log 1+ h ) q /µg N W log 1+ h q/µg 0 (29) W W where F({q }) denotes the objectve functon n the problem (27a) (27b). Snce {q } s a feasble soluton of the problem (27a) (27b), t contradcts the fact that {q } s the optmal soluton of the problem (27a) (27b). Therefore, t must be true that {q } = {q }. It can be seen from the constrants (27b) that N q = N q = Q N µg P pk. Then the problem (28a) (28c) s equvalent to the followng problem max {q } h µg q (30a) s.t. q µg P pk, q Q (30b) (30c) snce the constrant (30c) s actve at optmalty. Therefore, the problem (27a) (27b) s equvalent to the problem (30a) (30c). Snce the problem (30a) (30c) s smlar to the problem (15a) (15c) n Secton IV-A, we conclude that {q } has the same structure as that gven n Lemma 1. The result of Lemma 5 s smlar to that of Lemma 1. Specfcally, t shows that for the optmal power allocaton under the constrants (2) and (5), there exsts at most one user that transmts at nonzero power and below ts peak power, whle any other user ether does not transmt or transmts at ts peak power. Usng Lemma 5, let q s = µg s P pk s,, 1 k 1, 0 < q s k µg s P pk s, and q s = 0,, k +1 N. Then we only need to fnd k and q s k to determne {q }. Consder the case when 0 < qs k < µg sk Ps pk k, 1 k N. Then the followng must be true H(q sk ) h sk /µg q sk = ( sk N 1 = 0 (31) qsk =qs k 1+, k h s qs /µg s +h sk qs k /µg sk )/W where H(q sk ) W log ( 1+ N, k h s q s /µg s +h sk q sk /µg sk W ), k q s q sk. (32)

12 12 Substtutng {q s } nto (31), we obtan q s k = W(1 µg sk /h sk ) µg sk k 1 h s P pk s /h sk. Snce qs k must satsfy 0 < qs k < µg s Ps pk, t must be true that k 1 ( ) h s Ps pk hsk < W 1 < µg sk k h s P pk s. (33) Consder the case when qs k = µg sk Ps pk k, 1 k N 1. Then the followng must hold H(q sk ) h sk /µg q sk = ( sk N 1 0 (34) qsk =qs k 1+, k h s qs /µg s +h sk qs k /µg sk )/W and H(q sk+1 ) q sk+1 = qsk+1 =qs k+1 1+ ( N Substtutng {q } nto (34) and (35), we obtan ( ) hsk+1 k W 1 µg sk+1 h sk+1 /µg sk+1, k+1 h s q s /µg s +h sk+1 q s k+1 /µg sk+1 )/W h s P pk s W 1 0. (35) ( ) hsk 1, 1 k N 1. (36) µg sk If qs k = µg sk Ps pk k, k = N, then only (34) must be true and t follows that k ( ) h s Ps pk hsk W 1, k = N. (37) µg sk Lemma 6: There exsts only one set of values for {q } that satsfes only one of the necessary condtons (31), (34) or (35). Proof: It s equvalent to prove that there exsts only one k that satsfes only one of (33), (36) or (37). Let L j j h s P pk s and M j W(h sj /µg sj 1) for brevty. Then t must be true that L 0 < L 1 < < L N, M 1 > M 2 > > M N and L 0 < M 1. It can be seen that f (37) holds,.e., f L < M,, 1 N, then (33) and (36) do not hold. If (37) does not hold, then these exsts such l that L < M,, 1 l 1 and L > M,, 1 N. The followng two cases should be consdered. () If L l 1 < M l < L l, (33) holds for k = l. Snce L < M,, 1 l 1, (33) does not hold for k < l as well. Snce M < M l < L l L 1,, l +1, (33) does not hold for k > l. Snce L < L +1 < M +1,, 1 l 2, (36) does not hold for k < l 1. Snce L l 1 < M l, (36) does not hold also for k = l 1. Moreover, snce M < L,, l, (36) does not hold for k > l 1. Therefore, only (33) holds for only k = l. () If M l < L l 1 < M l 1, (36) holds for k = l 1. Smlar to the case (), t can be proved that only (36) holds for only k = l 1.

13 13 Algorthm 2 Algorthm for fndng k n Lemma 5 Intalze: k = 0, c = 0 whle c = 0 do k = k +1 f k 1 h s P pk s < W(h sk /µg sk 1) < k h s P pk s then c = 1 end f f {W(h sk+1 /µg sk+1 1) k h s P pk s W(h sk /µg sk 1) and k N 1} or { k h s P pk s W(h sk /µg sk 1) and k = N} then c = 2 end f end whle Output: k, c Usng Lemma 6, Algorthm 2 s developed to fnd the unque k n Lemma 5. Note that k satsfes (33) and (36) or (37) f the output of Algorthm 2 s c = 1 and c = 2, respectvely. Snce p s = qs /µg s, when c = 1, we obtan Ps pk, 1 k 1 p s = W(1/µg sk 1/h sk ) k 1 h s Ps pk /h sk, = k 0, k +1 N, 1 N (38) and when c = 2, we obtan P p s pk s =, 1 k 0, k +1 N, 1 N. (39) D. Average transmt power wth peak nterference power constrants Consder F = {the constrants (3) and (4)}. Then the dual functon of the problem (12) can be wrtten as f 4 ({λ }) E{f 4 (h,g)}+ N λ P av (40)

14 14 where {λ 1 N} are the nonnegatve dual varables assocated wth the correspondng constrants (4) and f 4 (h,g) s gven by f 4 (h,g) max W log {p (h,g)} s.t. ( ) N 1+ h p (h,g) W g p (h,g) Q pk. λ p (h,g) (41a) (41b) Let {p } denote the optmal soluton of the problem (41a) (41b) where the dependence on h and g s dropped for brevty. The followng three cases are of nterest. Case 1: Consder the case when h λ,. Smlar to Case 1 n Secton IV-C, t can be seen from Lemma 3 that p = 0,. Case 2: Consder the case when h λ does not hold for some and the constrant (41b) s nactve at optmalty. Let (s 1,s 2,,s N ) denote a permutaton of the SU ndexes such that h s1 /λ s1 > h s2 /λ s2 > > h sn /λ sn. Snce the problem (41a) (41b) wthout the constrant (41b) has the same form as the problem (19), t can be seen from (24) that p s 1 = W(1/λ s1 1/h s1 ) and p s = 0,, 2 N, f t satsfes the constrant (41b),.e., N g s p s = g s1 W(1/λ s1 1/h s1 ) < Q pk. Case 3: Consder the case when h λ does not hold for some and the constrant (41b) s actve at optmalty,.e., g s1 W(1/λ s1 1/h s1 ) Q pk. The dual functon of the problem (41a) (41b) can be wrtten as f 4 (µ) f 4 +µqpk, where µ s the nonnegatve dual varable assocated wth the constrant (41b), and f 4 s gven by ( ) N f 4 max W log 1+ h p {p } W λ p µ g p. (42) Let µ denote the optmal dual varable. Also let F({p }) denote the objectve functon n the problem (42). If p > 0 for some, the followng must hold F({p }) h p = {p }={p } 1+ N h p /W λ µ g = 0. (43) If p = 0 for some, the followng must hold F({p }) h p = {p }={p } 1+ N h p /W λ µ g 0. (44) Note that snce the problem (41a) (41b) s convex, the necessary condtons (43) and (44) for the optmal soluton {p } are also suffcent condtons.

15 15 p k Lemma 7: There exsts at most two j k such that p j > 0 and p k > 0. Proof: We prove t by contradcton. It can be seen from (43) that f p > 0, p j > 0 for some j, j k, k, the followng must hold > 0, and h λ +µ g = h j λ j +µ g j = h k λ k +µ g k. (45) Snce h, λ, g, h j, λ j, g j, h k, λ k, and g k are ndependent constants gven n the problem (41a) (41b), and only µ s a varable, (45) can not be satsfed. Lemma 7 shows that for the optmal power allocaton under the constrants (3) and (4), there exsts at most two users that transmt at nonzero power, whle any other user does not transmt. Then Case 3 can be further dvded nto the followng two subcases. Case 3.1: Consder the subcase when p k > 0 and p = 0, k. Snce the constrant (41b) s actve at optmalty,.e., N g p = g k p k = Qpk, we obtan that p k = Qpk /g k. Then substtutng {p } nto (43) we have µ = 1 g k /h k +Q pk /W λ k g k. (46) Note that µ gven n (46) must satsfy µ 0. Substtutng {p } nto (44), we can see that µ gven n (46) also must satsfy µ h /g 1+h k Q pk /g k W λ g,, k. (47) Then Algorthm 3 can be used to fnd k. Note that {p } does not exst n Case 3.1 f the output Algorthm 3 Algorthm for fndng k n Case 3.1 ( ) k = argmax {} W log 1+ h Q pk g W µ = 1 g k /h k +Q pk /W λ k f µ < max { k} k = 0 end f Output: k g k λ Q pk g h /g 1+h k Q pk /g k W λ g or µ < 0 then of Algorthm 3 s k = 0. Case 3.2: Consder the subcase when p j > 0, p k > 0, j k and p follows from (43) that h j λ j +µ g j = = 0,, j, k. It h k λ k +µ g k. (48)

16 16 Therefore, we obtan that µ = λ j/h j λ k /h k g k /h k g j /h j. (49) Note that µ gven n (49) must satsfy µ 0. Usng (43) and the fact that the constrant (41b) s actve at optmalty, we have h j p j +h kp k = Wh j/(λ j +µ g j ) W g j p j +g kp k = Qpk. Solvng the system of equaton (50), we obtan (50) p j = Qpk /g k a/h k g j /g k h j /h k, p k = a/h j Q pk /g j h k /h j g k /g j (51) where a Wh j /(λ j +µ g j ) W. Note that p j and p k gven n (51) must satsfy p j > 0 and p k > 0. Substtutng {p } and µ nto (44), we can see that j and k must satsfy λ j /h j λ k /h k g k /h k g j /h j λ j/h j λ /h g /h g j /h j,, j, k. (52) Then Algorthm 4 can be used to fnd j and k. Note that {p } does not exst f the output of Algorthm 4 s j = 0 and k = 0. E. Combnatons of more than two power constrants ConsderF = {the constrants (2), (4), and (5)} orf = {the constrants (3), (4), and (5)}. It can be shown that the correspondng dual functons of the problem (12) under these two combnatons of the power constrants have the same form as those n Subsectons IV-C and IV-D, respectvely. Therefore, optmal solutons can be found smlarly theren and, thus, are omtted here. ConsderF = {the constrants (2), (3), and (4)} or F = {the constrants (2), (3), and (5)} or F = {the constrants (2), (3), (4), and (5)}. It can be shown that the correspondng dual functons of the problem (12) under the frst two combnatons of the power constrants have the same form as that under the thrd combnaton. Therefore, we only focus on the case F = {the constrants(2), (3), (4), and (5)}. Then the dual functon of the problem (12) can be wrtten as f 5 ({λ },µ) E{f 5(h,g)}+ λ P av +µq av (53)

17 17 Algorthm 4 Algorthm for fndng j and k n Case 3.2 Intalze: I = for j = 1,,N 1 do for k = j +1,,N do µ = λ j/h j λ k /h k g k /h k g j /h j f µ 0 then a = Wh j /(λ j +µ g j ) W p j = Qpk /g k a/h k g j /g k h j /h k, p k = a/h j Q pk /g j h k /h j g k /g j f p j > 0 and p k > 0 then I = I {(j,k)} ( ) v j,k = W log 1+ h jp j +h kp k λ W j p j λ kp k end f end f end for end for (j,k) = argmax {(,l) I} v,l f λ j/h j λ k /h k g k /h k g j /h j < max { j,k} λ j /h j λ /h g /h g j /h j then j = 0, k = 0 end f Output: j, k where {λ 1 N} and µ are the nonnegatve dual varables assocated wth the correspondng constrants n (4) and (5) and f 5 (h,g) s gven by ( ) N f 5 (h,g) max W log 1+ h p (h,g) λ p (h,g) µ {p (h,g)} W s.t. g p (h,g) Q pk p (h,g) P pk,. g p (h,g) (54a) (54b) (54c) Let {p } denote the optmal soluton of the problem (54a) (54c) where the dependence on h and g s dropped for brevty. The followng cases are of nterest.

18 18 Case 1: Consder the case when h λ +µg,. Smlar to Case 1 n Subsectons IV-C and IV-D, t can be seen from Lemma 3 that p = 0,. Case 2: Consder the case when h λ +µg does not hold for some and the constrant (54b) s nactve at optmalty. Snce the problem (54a) (54c) wthout the constrant (54b) has the same form as the problem (26a) (26b), {p } can be found usng Algorthm 2 and (38) or (39) f t satsfes the constrant (54b). Case 3: Consder the case when h λ +µg does not hold for some and the constrant (54b) s actve at optmalty. The dual functon of the problem (54a) (54c) can be wrtten as f 5 (β) f 5 + βqpk, where β s the nonnegatve dual varable assocated wth the constrant (54b) and f 5 s gven by f 5 maxw log {p } s.t. p P pk,. ( ) N 1+ h p W γ p β g p (55a) (55b) where γ λ +µg. Let β denote the optmal dual varable and F({p }) stands for the objectve functon n the problem (55a). If P pk > p > 0 for some, the followng must hold F({p }) h p = {p }={p } 1+ N h p /W γ β g = 0. (56) If p = Ppk Moreover, f p for some, the followng must hold F({p }) h p = {p }={p } 1+ N h p /W γ β g 0. (57) = 0 for some, the followng must hold F({p }) h p = {p }={p } 1+ N h p /W γ β g 0. (58) Note that snce the problem (54a) (54c) s convex, the necessary condtons (56), (57) and (58) for the optmal soluton {p } are also suffcent condtons. p j Lemma 8: There exsts at most two j and k, j k such that P pk j > p j > 0 and Ppk k > p k > 0. Proof: We prove t by contradcton. It can be seen from (56) that f P pk > p > 0, P pk j > > 0, and Ppk k > p k > 0 for some j, j k, k, the followng must be true h γ +β g = h j γ j +β g j = h k γ k +β g k. (59)

19 19 Snce h, γ, g, h j, γ j, g j, h k, γ k, and g k are ndependent constants gven n the problem (54a) (54c), and only β s a varable, (59) can not be satsfed. Lemma 8 shows that for the optmal power allocaton under the constrants (2), (3), (4) and (5), there exsts at most two user that transmt at nonzero power and below ther peak power, whle any other user ether does not transmt or transmts at ts peak power. Then Case 3 can be further dvded nto the followng two subcases. Case 3.1: Consder the subcase when P pk k > p k > 0 and p {P pk,0}, k. Let N 1 and N 0 denote the sets of SU ndexes such that p = P pk, N 1 and p = 0, N 0. Snce the constrant (54b) s actve at optmalty,.e., N g p = g k p k + N 1 g P pk obtan p k = (Qpk N 1 g P pk )/g k. Note that p k substtutng {p } nto (56) we obtan = Q pk, we gven here must satsfy Ppk k > p k > 0. Then β = 1+ h k /g ( k h k (Q pk N 1 g P pk )/g k + ) N 1 h P pk /W γ k g k. (60) Note that β gven by (60) must satsfy β 0. Substtutng {p } nto (57) we can see that β gven by (60) must satsfy β 1+ h /g ( h k (Q pk N 1 g P pk )/g k + ) N 1 h P pk /W Substtutng {p } nto (58), we can see that β gven n (60) also must satsfy γ g, N 1. (61) β 1+ h /g ( h k (Q pk N 1 g P pk )/g k + ) N 1 h P pk /W γ g, N 0. (62) Let S (1),S (2),,S (2N 1 ) denote all the subsets of the set N\{} where \ denotes the set dfference operator. Then Algorthm 5 can be used to fnd k, N 1, and N 0. Note that {p } does not exst f the output of Algorthm 5 s k = 0. Case 3.2: Consder the subcase when P pk j > p j > 0, Ppk k > p k > 0 and p {P pk,0}, j,k. Let N 1 and N 0 denote the sets of SU ndexes such that p = P pk, N 1 and p = 0, N 0, respectvely. It follows from (56) that Therefore, we obtan that h j γ j +β g j = h k γ k +β g k. (63) β = γ j/h j γ k /h k g k /h k g j /h j. (64)

20 20 Algorthm 5 Algorthm for fndng k, N 1, N 0 n Case 3.1 Intalze: I = for k = 1,2,,N do for l = 1,2,,2 N 1 do N 1 = S (l) k p k = (Qpk N 1 g P pk )/g k f P pk k > p k > 0 then I = I {l} ( ) r l = W log 1+ h kp k + N h P pk 1 W end f end for v k = max { I} r, t = argmax { I} r S k = S(t) k I = end for k = argmax {} v N 1 = S k N 0 = N\N 1 \{k} β = f β < 0 or β > or β < γ k p k N 1 γ P pk h k /g k 1+(h k (Q pk N g P pk 1 )/g k + γ N h P pk k 1 )/W g k h /g 1+(h k (Q pk N g P pk 1 )/g k + γ N h P pk 1 )/W g, N 1 h /g 1+(h k (Q pk N 1 g P pk )/g k + N 1 h P pk )/W γ g, N 0 then k = 0 end f Output: k, N 1, N 0 Note that β gven n (64) must satsfy β 0. Followng (56) and the fact that the constrant (54b) s actve at optmalty, we have h j p j +h kp k = Wh j/(γ j +β g j ) W N 1 h P pk g j p j +g k p k = Qpk N 1 g P pk. (65)

21 21 Solvng the system of equaton (65), we obtan p j = a/g k b/h k g j /g k h j /h k, p k = b/h j a/g j h k /h j g k /g j (66) where a Q pk N 1 g P pk and b Wh j /(γ j +β g j ) W N 1 h P pk. Note that p j and p k gven n (66) must satsfy Ppk j > p j by (64) nto (57), we obtan > 0 and Ppk k > p k > 0. Substtutng {p } and β gven γ j /h j γ k /h k g k /h k g j /h j γ j/h j γ /h g /h g j /h j, N 1. (67) Moreover, substtutng {p } and β gven by (64) nto (58), we also obtan Let S (1),j,S(2),j,,S(2N 2 ),j γ j /h j γ k /h k g k /h k g j /h j γ j/h j γ /h g /h g j /h j, N 0. (68) denote all the subsets of the set N\{,j}. Then Algorthm 6 can be used to fnd j, k, N 1, and N 0. Note that {p } does not exst f the output of Algorthm 6 s j = 0 and k = 0. V. SIMULATION RESULTS Consder a cogntve rado network whch conssts of one PU and four SUs. For smplcty, we assume that only Raylegh fadng s present n all lnks. The varance of the channel power gan s set to σ 2 = 1. We also set W = 1, P pk = 10,, P av = 10,, Q pk = 1, and Q av = 1 as default values f no other values are specfed otherwse. The AWGN wth unt PSD s assumed. We use 1000 randomly generated channel power gans for h and g n our smulatons. The results are compared under the followng fve combnatons of the power constrants: the PTP wth PIP constrants (PTP+PIP), the PTP wth AIP constrants (PTP+AIP), the ATP wth PIP constrants (ATP+PIP), the ATP wth AIP constrants (ATP+AIP), the PTP and ATP wth PIP and AIP constrants (PTP+ATP+PIP+AIP). Frst, we am at showng by Fg. 1 that the nformaton-theoretc lmt for the sum ergodc capacty s ndeed sgnfcantly hgher when bandwdth s allocated optmally as compared to the case when t s allocated equally among SUs. In ths fgure, OBPA stands for optmal bandwdth and power allocaton, whle EBPA stands for equal bandwdth and power allocaton. The case of PTP+PIP s only depcted n Fg. 1, but the concluson about the superorty of optmal bandwdth and power allocaton holds true for other combnatons of power constrants. Then

22 22 Algorthm 6 Algorthm for fndng j, k, N 1, N 0 n Case 3.2 Intalze: I = for j = 1,2,,N 1 do for k = j +1,,N do for l = 1,2,,2 N 2 do N 1 = S (l) j,k β = γ j/h j γ k /h k g k /h k g j /h j f β 0 then a Q pk N 1 g P pk, b Wh j /(γ j +β g j ) W N 1 h P pk p j = a/g k b/h k g j /g k h j /h k, p k = b/h j a/g j h k /h j g k /g j f P pk j > p j > 0 and Ppk k > p k > 0 then I = I {l} ( ) r l = W log 1+ h jp j +h kp k + N h P pk 1 W end f end f end for v j,k = max { I} r, t = argmax { I} r S j,k = S(t) j,k I = end for end for (j,k) = argmax {(,l)} v,l N 1 = S j,k N 0 = N\N 1 \{j,k} γ j p j γ k p k N 1 γ P pk f γ j/h j γ k /h k g k /h k g j /h j > γ j/h j γ /h g /h g j /h j, N 1 or γ j/h j γ k /h k g k /h k g j /h j < γ j/h j γ /h g /h g j /h j, N 0 then j = 0, k = 0 end f Output: j, k, N 1, N 0

23 23 Fg. 2 shows and compares the maxmum sum ergodc capacty under PTP+PIP, PTP+AIP and PTP+ATP+PIP+AIP constrants versus P pk where P pk = P pk, s assumed. It can be seen from the fgure that the maxmum sum ergodc capacty acheved under PTP+AIP s larger than that acheved under PTP+PIP for any gven P pk. Ths s due to the fact that the AIP constrant s more favorable than the PIP constrant from SUs perspectve, snce the former allows for more flexblty for SUs to allocate transmt power over dfferent channel fadng states. It s also observed that the performance under PTP+ATP+PIP+AIP s very close to that under PTP+PIP that s because the PTP constrant domnates over the ATP, PIP, and AIP constrants for all values of P pk. Fg. 3 shows the maxmum sum ergodc capacty under ATP+PIP, ATP+AIP and PTP+ATP+ PIP+AIP constrants versus P av where P av = P av, s assumed. The maxmum achevable sum ergodc capacty acheved under ATP+AIP s larger than that acheved under ATP+PIP for all values of P av snce the PIP constrant s strcter than the AIP constrant. The sum ergodc capacty under PTP+ATP+PIP+AIP s much smaller than that under ATP+PIP and ATP+AIP due to the fact that the PTP constrant s domnant over other constrants for all values of P av. Fg. 4 shows the maxmum sum ergodc capacty under PTP+PIP, ATP+PIP and PTP+ATP+ PIP+AIP constrants versus Q pk. It can be seen from the fgure that the maxmum sum ergodc capacty acheved under ATP+PIP s larger than that acheved under PTP+PIP for any gven Q pk. Ths s because the power allocaton s more flexble for SUs under the ATP constrant than under the PTP constrant. The sum ergodc capacty under PTP+ATP+PIP+AIP saturates earler than that under PTP+PIP and ATP+PIP, because t s restrcted by the AIP constrant. Fg. 5 shows the maxmum sum ergodc capacty under PTP+AIP, ATP+AIP and PTP+ATP+ PIP+AIP constrants versus Q av. Due to the same reasons as for the results n Fg. 4, the sum ergodc capacty acheved under ATP+AIP s larger than that acheved under PTP+AIP. The sum ergodc capacty under PTP+ATP+PIP+AIP saturates earler than that for PTP+AIP and ATP+AIP because of the presence of the PIP constrant. Fnally, Fg. 6 shows the maxmum sum ergodc capacty under PTP+PIP, PTP+AIP, ATP+PIP, ATP+AIP and PTP+ATP+PIP+AIP versus W. Smlar performance comparson results as n the prevous fgures can be observed. One dfference s that the sum ergodc capactes do not saturate wth the ncrease of W.

24 24 VI. CONCLUSION A cogntve rado network where multple SUs share the lcensed spectrum of a PU usng the FDMA scheme has been consdered. The maxmum achevable sum ergodc capacty of all the SUs has been studed subject to the total bandwdth constrant of the lcensed spectrum and all possble combnatons of the peak/average transmt power constrants at the SUs and nterference power constrant mposed by the PU. Optmal bandwdth allocaton has been derved n each channel fadng state for any gven power allocaton. Usng the structures of the optmal power allocatons under each combnaton of the power constrants, algorthms for fndng the optmal power allocatons have been developed too. REFERENCES [1] J. Mtola, Cogntve rado: An ntegrated agent archtecture for software defned rado,, Ph.D. dssertaton, KTH, Stockholm, Sweden, Dec [2] S. Haykn, Cogntve rado: Bran-empowered wreless communcatons, IEEE J. Sel. Areas Commun., vol. 23, pp , Feb [3] Q. Zhao, L. Tong and S. Ananthram, Decentralzed cogntve MAC for opportunstc spectrum access n ad hoc networks: A POMDP framework, IEEE J. Sel. Areas Commun., vol. 25, pp , Apr [4] Y. Chen, Q. Zhao, and S. Ananthram, Jont desgn and separaton prncple for opportunstc spectrum access n the presence of sensng errors, IEEE Trans. Inform. Theory, vol. 54, pp , May [5] S. Ahmad, M. Lu, T. Javd, Q. Zhao, and B. Krshnamachar, Optmalty of myopc sensng n multchannel opportunstc access, IEEE Trans. Inform. Theory, vol. 55, pp , Sep [6] N. B. Chang and M. Lu, Optmal channel probng and transmsson schedulng for opportunstc spectrum access, IEEE Trans. Networkng, vol. 55, pp , Sep [7] Q. Zhao and B. M. Sadler, A survey of dynamc spectrum access, IEEE Sgnal Process. Mag., vol. 24, pp , May [8] K. T. Phan, S. A. Vorobyov, N. D. Sdropoulos, and C. Tellambura, Spectrum sharng n wreless networks va QoS-aware secondary multcast beamformng, IEEE Trans. Sgnal Process., vol. 57, pp , June [9] R. Zhang and Y.-C. Lang, Exploatng mult-antennas for oportunstc spectrum sharng n cogntve rado networks, IEEE J. Sel. Topcs Sgnal Process., vol. 2, no. 1, pp , Feb [10] R. Zhang, On peak versus average nterference power constrants for protectng prmary users n cogntve rado networks, IEEE Trans. Wreless Commun., vol. 8, pp , Apr [11] R. Zhang and Y.-C. Lang, Investgaton on multuser dversty n spectrum sharng based cogntve rado networks, IEEE Commun. Letters, vol. 14, pp , Feb [12] G. Zheng, K.-K. Wong, and B. Ottersten, Robust cogntve beamformng wth bounded channel uncertantes, IEEE Trans. Sgnal Process., vol. 57, pp , Dec [13] A. Ghasem and E. S. Sousa, Fundamental lmts of spectrum-sharng n fadng envronments, IEEE Trans. Wreless Commun., vol. 6, pp , Feb

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26 OBPA under PTP+PIP EBPA under PTP+PIP Sum Ergodc Capacty(nats/s) P pk (db) Fg. 1. Sum ergodc capacty vs P pk PTP+AIP PTP+PIP PTP+ATP+PIP+AIP Sum Ergodc Capacty(nats/s) P pk (db) Fg. 2. Sum ergodc capacty vs P pk.

27 Sum Ergodc Capacty(nats/s) P av (db) ATP+AIP ATP+PIP PTP+ATP+PIP+AIP Fg. 3. Sum ergodc capacty vs P av ATP+PIP PTP+PIP PTP+ATP+PIP+AIP Sum Ergodc Capacty(nats/s) Q pk (db) Fg. 4. Sum ergodc capacty vs Q pk.

28 ATP+AIP PTP+AIP PTP+ATP+PIP+AIP Sum Ergodc Capacty(nats/s) Q av (db) Fg. 5. Sum ergodc capacty vs Q av. Sum Ergodc Capacty(nats/s) ATP+AIP PTP+AIP ATP+PIP PTP+PIP PTP+ATP+PIP+AIP W(Hz) Fg. 6. Sum ergodc capacty vs W.