Network Beamforming Using Relays with Perfect Channel Information

Netw ork Beamforming Using Relays with Perfect Channel Information Y I N D I J I N G A N D H A M I D J A FA R K H A N I ∗ Uni v ersity of California, Irvine, Irvine, CA, 92697 Abstract This paper is on be amforming in wireless relay net works with perfect chan nel informat ion at relays, the recei ver , a nd the transmitter if there is a direct link between the transmitter and recei ver . It is assumed that e ver y node in the network has its o wn power constrai nt. A two-step amplify- and-forw ard pr otocol is used, in w hich the transmitt er and relays not only use match filters to form a beam at the recei ve r b ut also adapti vely adjust their transmit power s accord ing to th e chann el strength infor mation. For a netwo rk with an y numb er of rela ys and no direc t link, the optimal po wer control is solved analytic ally . The comple xity of finding the exact solution is linear in the number of relays. Our results sho w that the transmitter should always use its maximal power and the optimal po wer used at a relay is not a binary function. It can take any v alue between zero and its maximum transmit po wer . Also , surprisingly , this valu e depend s on the quality of all other channels in addition to the relay’ s own chann els. Despite this coupling fact , distrib uti ve strategi es are proposed in w hich, w ith the aid of a low-ra te broadcas t from the recei ver , a rel ay n eeds only it s o wn ch annel i nformation to implement the optimal po wer control. Simulated performance sho ws that netwo rk beamforming achie ves the maximal div ersity and outper forms other existi ng schemes . ∗ This work was suppor ted in part by AR O un der the Multi-University Research Initiative (MURI) grant #W91 1NF- 04-1- 0224. 1 Then, beamformin g in networ ks with a direct link are consid ered. W e show that when the direct lin k exists d uring the first ste p only , the opt imal power control at the tran smitter and relays is the same as t hat of network s with no direct link. For netw orks with a direct link durin g the second step only and both steps, recursi ve numerical algorith ms are proposed to solv e the po wer control problem. Simulation sho ws that by adjust ing the transmitter and relays’ po wers adapti vely , networ k performance is significantly improve d. 1 Introduction It is wel l-known that due to t he fading effe ct, the transmissio n over wireless channels suffers from sev ere attenuatio n in si gnal strength . Performance of wireless comm unication is m uch worse than that o f wired communication. For the simplest point-to-point comm unication system , which is com- posed of one transmitter and one recei ver only , the use of multipl e antennas can improve the capacity and reli ability . Space-time coding and beamformi ng are amon g the mos t successful techni ques de- veloped for multipl e-antenna sys tems during the last decades [1, 2]. Ho w e ver , in many situations, due to the limited size and processing po wer , it i s not practical for some users, especially small wireless mobile de vices, to implement multiple antennas. Thus, recently , wireless network commu- nication is attracting more and m ore at tention. A large amount o f effort has been given to improve the communication by having differ ent users i n a network cooperate. This im provement is con ven- tionally addressed as cooperativ e di versity and t he techniques cooperativ e schemes. Many cooperative schemes have been proposed in l iterature [3–21]. Som e assume channel in- formation at the receiv er but not the t ransmitter and relays, for example, the n oncoherent am plify- and-forward protocol in [8, 9] and distributed space-time coding in [10]. Some ass ume channel information at the receiving side o f each t ransmission , for example, the decode-and-forward proto - col in [8, 12] and the coded-cooperation i n [13]. S ome assume no channel informatio n at any node, for example, th e differe ntial transm ission schemes proposed in dependently in [14– 16]. The coher- ent amplify-and-forward scheme i n [9, 11] assumes full channel informati on at both relays and th e recei ver . But only channel di rection information is u sed at relays. In all these cooperativ e schemes, 2 the relays always coop erate on their hi ghest powers. None of the above p ioneer work allow relays to adjust their transmi t powers adaptively according to channel m agnitude information, and t his is exactly the concern of this paper . There have been s e veral papers on relay networks with adaptiv e power control. In [22, 23], outage capacity of networks with a single relay and perfect channel information at all nodes were analyzed. Both w ork ass ume a total power constraint on the relay and the transmitter . A decode-and- forward protocol is used at the relay , wh ich resul ts in a binary p ower all ocation between the relay and the t ransmitter . In [24], performance of networks with multiple ampl ify-and-forward relays and an a ggregate power cons traint was analyzed. A distributi ve scheme for the optimal power allo cation is proposed, i n whi ch each relay only needs to know its own channels and a real number that can be broadcasted by the receive r . Another related work on networks with one and two amplify-and- forward relays can be found in [25]. In [26], outage min imization of single-relay netw orks with limited chann el-information feedback is performed. It is assumed that there i s a lon g-term power constraint on the tot al power of the transmit ter and the relay . In t his paper , we consider n etworks with a general number of amplify-and-forward relays and we assume a separate po w er constraint on each relay and th e transmitter . D ue to the di ff erence in the power assum ptions, compared to [24], analysis of this new m odel is more dif ficult and totally diffe rent results are obtained. For m ultiple-antenna systems , when there is no channel information at the transmitter , space - time coding can achiev e ful l di versity [1]. If the t ransmitter has perfect or partial channel infor- mation, performance can be further i mproved through beamforming sin ce it takes advantage of the channel information (both direction and strength) at t he transm it side to obtain higher recei ve SNR [2]. W ith perfect channel informatio n or high qualit y chann el in formation feedback from th e recei ver at the transmitter , one-dimension al beamform ing is proved optimal [2, 27, 28]. The more practical m ultiple-antenna system s with partial channel informati on at th e transm itter , channel statis- tics o r quanti zed inst antaneous channel information, are als o analyzed e xtensively [29– 33]. In m any situations, appropriate com bination of beamforming and space-time coding outperforms eith er on e of the two schemes alone [34–37]. In this paper , we will see similar performance improvement in networks using network beamforming over d istributed space-tim e coding and other existing schemes 3 such as best-relay selection and coherent amplify-and-forward. W e consider networks with one pair of transmitter a nd recei ver but multiple relays. The recei ver knows all channels and e very relay kno ws i ts o wn channels per fectly . In networks with a direct link (DL) between th e transmitter and the receiver , we als o assume th at the transmitter knows t he DL ful ly . A two-step ampli fy-and-forward protocol is used, where in the first step, the transmi tter sends informat ion and in the second step, the transmit ter and relays, if t here is a DL, transm it. W e first s olve the power control problem for networks with no DL analytically . The exact solution can be obtained with a complexity that i s linear in the number of relays. Then, to perform network beamforming, we propose two distributive s trategies in which a relay needs only its own channel information and a low-rate broadcast from the receiv er . Simulation shows th at the optim al power control or network beamformin g outperforms o ther existing schemes. W e then consider networks with a DL during the first transm ission step, the second transmi ssion step, and both. For the first case, the power con trol p roblem is proved to be the same as the one in networks wit hout the DL. For the other two cases, recursiv e n umerical algorithms are provided. Simu lation s hows that they hav e much better performance compared to networks without power c ontrol. W e shou ld clarify that only amplify-and-forward is considered here. For decode-and-forward, the result m ay be different and it depends on the details of the coding schemes. The paper is organized as follows. In the next section, t he relay netw ork model and the main problem are int roduced. Section 3 works on the power control probl em in relay networks with n o DL and Section 4 consi ders networks wi th a DL. Section 5 contains t he conclusion and several future directions. 2 W i r eless Re lay Network M odel and Problem St atement Consider a relay network with one transm it-and-receiv e pair and R relays as depicted in Fig. 1. Every r elay has only one single antenna which can be used for both transmi ssion and reception. Denote the channel from the transmit ter to the i th relay as f i and the channel from the i th relay to the recei ver as g i . If the DL between t he transmit ter and the receiver e xists, we denote it as 4 f 0 . W e assume that t he transm itter knows f 0 , the i th relay k nows its own channels f i and g i , and the recei ver kn ows all channels f 0 , f 1 , . . . , f R and g 1 , . . . , g R . The channels can have b oth fading and path -loss effects. Actually , our results are valid for any channel st atistics. W e assum e that for each transm ission, the p ower s used at the transmitter and the i th relay are no larger than P 0 and P i , respectiv ely . Note that in this paper , only short-term power constraint is considered, that is, there is an upp er bound on the av erage transmit power of each nod e for each transmiss ion. A node cannot sav e its power to fa vor transmissions with better channel realizations. W e use a two-step amplify-and-forward protocol. During the first step, the transmitter sends α 0 √ P 0 s . The information symbol s is selected randoml y from t he codebo ok S . If we normali ze it as E | s | 2 = 1 , the a verage power used at t he t ransmitter i s α 2 0 P 0 . The i th relay and the recei ver , if a DL exists during the first step, rec eive r i = α 0 p P 0 f i s + v i and x 1 = α 0 p P 0 f 0 s + w 1 , (1) respectiv ely . v i and w 1 are the n oises at the i th relay and the receiver at Step 1. W e assume that they are C N (0 , 1) . During the second step, the transmi tter sends β 0 √ P 0 e j θ 0 s , i f a DL exists during this step. At the same tim e, the i th relay sends t i = α i s P i 1 + α 2 0 | f i | 2 P 0 e j θ i r i . The a verage tra nsmit po wer of the i th relay can be calculated to be α 2 i P i . If we assum e that f 0 keeps constant for the two steps, the recei ver gets x 2 = β 0 p P 0 f 0 e j θ 0 s + R X i =1 g i t i + w 2 = p P 0 β 0 f 0 e j θ 0 + α 0 R X i =1 α i f i g i e j θ i √ P i p 1 + α 2 0 | f i | 2 P 0 ! s + R X i =1 α i g i e j θ i √ P i p 1 + α 2 0 | f i | 2 P 0 v i + w 2 . (2) w 2 is the noise at the receiv er at Step 2, which is also assum ed to be C N (0 , 1) . Not e that if th e transmitter sends du ring both st eps, we assum e that th e total av erage power it uses is no larger than P 0 . W ith this, the total aver age powe r i n transmitting one symbol is no lar g er t han P R i =0 P i . Clearly , the coefficients α 0 , α 1 , . . . , α R are in troduced i n the model for power control. The po wer constraints at the transmitter and relays require that α 2 0 + β 2 0 ≤ 1 and 0 ≤ α i ≤ 1 . 5 Our ne twork beamforming design is thus the design of θ 0 , θ 1 , · · · , θ R and α 0 , β 0 , α 1 , · · · , α R , such that the error rate of the network i s th e smallest. This is equiv alent to maximi ze the recei ve SNR, or the tot al receive SNR of both branches if a DL exists durin g the first step. From (2), we can easily prov e that an optimal choice of the angles a re θ 0 = − arg f 0 and θ i = − (arg f i + arg g i ) . That is, match filters s hould be used at relays and the transm itter during the second st ep to cancel the phases of their channels and form a beam at the recei ver . W e thus hav e x 2 = p P 0 β 0 | f 0 | + α 0 R X i =1 α i | f i g i | √ P i p 1 + α 2 0 | f i | 2 P 0 ! s + R X i =1 α i | g i | √ P i p 1 + α 2 0 | f i | 2 P 0 e − j arg f i v i + w 2 . (3) What is lef t is the optimal power control, i.e., the choice of α 0 , β 0 , α 1 , . . . , α R . This is also t he main contribution of our w ork. 3 Optimal Relay P ower Contr ol In thi s section, we in vestigate th e opti mal adaptiv e power control at the transmitter and relays in net- works without a DL. Section 3.1 presents the analytical power cont rol result. Section 3.2 comments on the result and gives distributi ve schemes for the optimal power cont rol. Section 3.3 provides simulated performance. 3.1 Analytical Result W ith no DL, we have β 0 = 0 and x 1 = 0 . From (3), t he recei ve SNR can be calculated to be α 0 P 0  P R i =1 α i | f i g i | √ P i √ 1+ α 2 0 | f i | 2 P 0  2 1 + P R i =1 α 2 i | g i | 2 P i 1+ α 2 0 | f i | 2 P 0 . It is an increasing function of α 0 . Therefore, the transmitter should always u se its maxim al power , i.e., α ∗ 0 = 1 . The receive S NR is thus : P 0  P R i =1 α i | f i g i | √ P i √ 1+ | f i | 2 P 0  2 1 + P R i =1 α 2 i | g i | 2 P i 1+ | f i | 2 P 0 . 6 Before going i nto detail s of the SNR optimi zation, we first introduce some notation to help the presentation. h· , · i ind icates the inner product. k · k in dicates the 2-norm. P indicates the probability . a i denotes the i th coordinate of vector a and a i 1 ,...,i k denotes the k -di mensional vector h a i 1 · · · a i k i T , where · T represents the transpose. If a , b are two R -dim ensional vectors, a  b means a i ≤ b i for all i = 1 , . . . , R . 0 R is the R -dim ensional vector with all zero entries. Denote the set 0 R  y  a or equi valently , 0 ≤ y i ≤ a i for i = 1 , . . . , R , as Λ . For 1 ≤ k ≤ R − 1 , denote the set 0 k  y i 1 ,...,i k  a i 1 ,...,i k as Λ i 1 ,...,i k , where { i 1 , . . . , i k } is a k -subset of { 1 , . . . , R } . Define x =      α 1 . . . α R      , b =       | f 1 g 1 | √ P 1 √ 1+ | f 1 | 2 P 0 . . . | f R g R | √ P R √ 1+ | f R | 2 P 0       , a =       | g 1 | √ P 1 √ 1+ | f 1 | 2 P 0 . . . | g R | √ P R √ 1+ | f R | 2 P 0       , and A = diag { a } , where diag { a } indicates th e diagonal matrix who se i th d iagonal entry is a i . W ith the transform ation y = A x , or equiv alently , x = A − 1 y , we have S N R = P 0 h b , x i 2 1 + k A x k 2 = P 0 h c , y i 2 1 + k y k 2 , where c = A − T b =      √ 1+ | f 1 | 2 P 0 | g 1 | √ P 1 · · · 0 . . . . . . . . . 0 · · · · · · √ 1+ | f R | 2 P 0 | g R | √ P R            | f 1 g 1 | √ P 1 √ 1+ | f 1 | 2 P 0 . . . | f R g R | √ P R √ 1+ | f R | 2 P 0       =      | f 1 | . . . | f R |      . The recei ve SNR optimization problem is thus equi valent to max y h c , y i 2 1 + k y k 2 s.t. y ∈ Λ . (4) The di ffi culty of the problem lies i n the sh ape of the feasibl e set. If y is con strained on a hypersph ere, that is, k y k = r , the solutio n is obvious at least geomet rically . Given that k y k = r , h c , y i 2 1 + k y k 2 = r 2 k c k 2 1 + r 2 cos 2 ϕ, 7 where ϕ is the angle between c and y . The optimal s olution sh ould be the vector which has t he smallest angle with c . Thus , we decompose (4) as max r 1 1 + r 2  max k y k = r h c , y i  2 s.t. y ∈ Λ and 0 ≤ r ≤ k a k . (5) Since P( a i > 0 ) = 1 and P( c i > 0 ) = 1 , we assum e that a i > 0 and c i > 0 . Define φ j = φ ( f j , g j , P j ) = c j a j = | f j | p 1 + | f j | 2 P 0 | g j | p P j , (6) for i = 1 , . . . , R and, for the sake of presentation, define φ R +1 = 0 . Order φ j as φ τ 1 ≥ φ τ 2 ≥ · · · ≥ φ τ R ≥ φ τ R +1 . (7) ( τ 1 , τ 2 , . . . , τ R , τ R +1 ) is thus an ordering of (1 , 2 , . . . , R , R + 1) and τ R +1 = R + 1 . D efine r 0 =0 , r 1 = φ − 1 τ 1 k c k = q φ − 2 τ 1 k c τ 2 ,...,τ R k 2 + a 2 τ 1 , r 2 = q φ − 2 τ 2 k c τ 2 ,...,τ R k 2 + a 2 τ 1 = v u u t φ − 2 τ 2 k c τ 3 ,...,τ R k 2 + 2 X i =1 a 2 τ i , . . . r R − 1 = v u u t φ − 2 τ R − 1 k c τ R − 1 ,τ R k 2 + R − 2 X i =1 a 2 τ i = v u u t φ − 2 τ R − 1 | c τ R | 2 + R − 1 X i =1 a 2 τ i , r R = v u u t φ − 2 τ R | c τ R | 2 + R − 1 X i =1 a 2 τ i = k a k . Since φ τ j − 1 ≥ φ τ j , we h a ve r j − 1 ≤ r j for j = 1 , . . . , R . Thus, the feasible interval o f the radi us, [0 , k a k ] , can be decompos ed into the follo wing R intervals: [0 , k a k ] = [ r 0 , r 1 ] ∪ [ r 1 , r 2 ] ∪ · · · ∪ [ r R − 2 , r R − 1 ] ∪ [ r R − 1 , r R ] . W e denote Γ i = [ r i , r i +1 ] for i = 0 , . . . , R − 1 . Thus, (5) is equiv alent to max i =1 ,...,R max r ∈ Γ i 1 1 + r 2  max k y k = r ∈ Γ i , y ∈ Λ h c , y i  2 . 8 W e have decompos ed the optim ization problem into R subp roblems. W e now work on the i th subproblem: max r ∈ Γ i 1 1 + r 2  max k y k = r ∈ Γ i , y ∈ Λ h c , y i  2 . (8) Denote the solution of the inner optimization problem, max k y k = r ∈ Γ i , y ∈ Λ h c , y i , (9) as z ( i ) . W e have t he following two lemmas. Lemma 1. z ( i ) j = a j for j = τ 1 , . . . , τ i . Pr oo f. W e pro ve t his lemma by cont radiction. Assume that z ( i ) j < a j for some j ∈ { τ 1 , . . . , τ i } . W e first show that there exists an l ∈ { τ i +1 , . . . , τ R } such that z ( i ) j c j < z ( i ) l c l . Assume that z ( i ) j c j ≥ z ( i ) m c m for all m ∈ { τ i +1 , . . . , τ R } . W e ha ve z ( i ) m ≤ c m z ( i ) j c j < c m a j c j = c m φ − 1 j . Thus , k z ( i ) k = v u u t i X m =1  z ( i ) τ m  2 + R X m = i +1  z ( i ) τ m  2 < v u u t i X m =1 a 2 τ m + R X m = i +1 c 2 τ m φ − 2 j = v u u t φ − 2 j k c τ i +1 ,...,τ R k 2 + i X m =1 a 2 τ m ≤ v u u t φ − 2 τ i k c τ i +1 ,...,τ R k 2 + i X m =1 a 2 τ m because of (7) = r i . This contradicts k z ( i ) k ∈ Γ i . thu s, there exists an l ∈ { τ i +1 , . . . , τ R } such t hat z ( i ) j c j < z ( i ) l c l . Define another vector z ′ as z ′ j = z ( i ) j + δ , z ′ l = r  z ( i ) l  2 − 2 δ z ( i ) j − δ 2 , and z ′ m = z ( i ) m for m 6 = i, l , wh ere 0 < δ < min ( 2 c j  1 + c 2 j c 2 l  − 1 z ( i ) l c l − z ( i ) j c j ! , r  z ( i ) j  2 +  z ( i ) l  2 − z ( i ) j , a j − z ( i ) j ) . Since we hav e assum ed that z ( i ) j < a j and have just proved that z ( i ) j c j < z ( i ) l c l , such δ is achiev able. T o contradict the assumption that z ( i ) is the optimal, it is enough to prov e the following t wo items: 9 1. z ′ is a feasible point: k z ′ k = r and z ′ ∈ Λ , 2. h c , z ( i ) i < h c , z ′ i . From the definition of z ′ , we hav e k z ′ k 2 =  z ′ j  2 +( z ′ l ) 2 + X m 6 = j,l ( z ′ m ) 2 =  z ( i ) j + δ  2 +  z ( i ) l  2 − 2 δ z ( i ) j − δ 2 + X m 6 = j,l  z ( i ) m  2 = k z ( i ) k 2 = r 2 . Since 0 < δ < a j − z ( i ) j , we have 0 < z ′ j < a j . Also, since 0 < δ < r  z ( i ) j  2 +  z ( i ) l  2 − z ( i ) j , we can easily prove that z ′ l = r  z ( i ) l  2 − 2 δ z ( i ) j − δ 2 > 0 and z ′ l < z ( i ) l ≤ a l . Thus, z ′ ∈ Λ . The first item has been proved. For the second item, since δ < 2 c j  1 + c 2 j c 2 l  − 1  z ( i ) l c l − z ( i ) j c j  , we hav e  1 + c 2 j c 2 l  δ 2 < 2 c j z ( i ) l c l − z ( i ) j c j ! δ = 2  z ( i ) l c j c l − z ( i ) j  δ ⇒  z ( i ) l  2 + c 2 j c 2 l δ 2 − 2 c j c l z ( i ) l δ <  z ( i ) l  2 − 2 z ( i ) j δ − δ 2 = z ′ 2 l ⇒ c l z ( i ) l − c j δ < c l z ′ l ⇒ c l z ( i ) l + c j z ( i ) j − ( c l z ′ l + c j z ′ j ) < 0 ⇒ h c , z ( i ) i < h c , z ′ i . Lemma 2. z ( i ) j = √ r 2 − P i m =1 a 2 τ m k c τ i +1 ,...,τ R k c j for j = τ i +1 , . . . , τ R . Pr oo f. F rom Lemm a 1, z ( i ) j = a j for j = τ 1 , . . . , τ i . Thus, (9) can be written as max k y k = r ∈ Γ i , y ∈ Λ i X m =1 a τ m c τ m + h c τ i +1 ,...,τ R , y τ i +1 ,...,τ R i = i X m =1 b τ m + max k y τ i +1 ,...,τ R k = √ r 2 − P i m =1 a 2 τ m , r ∈ Γ i , y τ i +1 ,...,τ R ∈ Λ τ i +1 ,...,τ R h c τ i +1 ,...,τ R , y τ i +1 ,...,τ R i . Define λ = √ r 2 − P i m =1 a 2 τ m k c τ i +1 ,...,τ R k . It i s obvious that h c τ i +1 ,...,τ R , y τ i +1 ,...,τ R i ≤ h c τ i +1 ,...,τ R , λ c τ i +1 ,...,τ R i for all k y τ i +1 ,...,τ R k = q r 2 − P i m =1 a 2 τ m . I n o ther words, to maxi mize the i nner product, y τ i +1 ,...,τ R should 10 hav e the same direction as c τ i +1 ,...,τ R . Thus, we only need t o show that t his direction is feasible for r ∈ Γ i . This is equiv alent to show that λ c τ i +1 ,...,τ R ∈ Λ τ i +1 ,...,τ R for any r ∈ Γ i . W e can easil y prove that r ∈ Γ i ⇔ λ ∈ Ω i , where Ω i = h φ − 1 τ i , φ − 1 τ i +1 i for i = 0 , . . . , R − 1 . Thus, for any r ∈ Γ i and j = τ i +1 , . . . , τ R , we have 0 ≤ λc j ≤ φ − 1 τ i +1 c j ≤ φ − 1 j c j = a j . Hence, λ c τ i +1 ,...,τ R ∈ Λ τ i +1 ,...,τ R . Combining Lemma 1 and Lemma 2, we ha ve z ( i ) j =    a j j = τ 1 , . . . , τ i λc j j = τ i +1 , . . . , τ R (10) and thus max k y k = r ∈ Γ i , y ∈ Λ h c , y i = i X m =1 b τ m + λ k c τ i +1 ,...,τ R k 2 . (11) W e hav e solved the inner optimization of Subproblem i . Th e solution of the R subproblems can thus be obtained. Lemma 3. F or i = 1 , . . . , R , define λ i = 1 + P i m =1 a 2 τ m P i m =1 b τ m . The so lution of Su bpr oblem 0 is y (0) = φ − 1 τ 1 c . The solution of Subpr oblem i for i = 1 , . . . , R − 1 is y ( i ) that is defined as y ( i ) j =    a j j = τ 1 , . . . , τ i min n λ i , φ − 1 τ i +1 o c j j = τ i +1 , . . . , τ R . (12) Pr oo f. F rom (11), Subproblem i is e quiv alent to the foll owing 1-dimensi onal optimization problem: max λ ∈ Ω i  P i m =1 b τ m + k c τ i +1 ,...,τ R k 2 λ  2 1 + P i m =1 a 2 τ m + k c τ i +1 ,...,τ R k 2 λ 2 . (13) When i = 0 , (13) i s equiva lent to max λ ∈ Ω i k c k 4 λ 2 1+ k c k 2 λ 2 . Since k c k 4 λ 2 1+ k c k 2 λ 2 is an increasing function of λ , its maximum is at λ = φ − 1 τ 1 . 11 For i = 1 , . . . , R − 1 , Define ξ i ( λ ) =  P i m =1 b τ m + k c τ i +1 ,...,τ R k 2 λ  2 1 + P i m =1 a 2 τ m + k c τ i +1 ,...,τ R k 2 λ 2 . W e hav e, ∂ ξ i ∂ λ = 2  P i m =1 b τ m + k c τ i +1 ,...,τ R k 2 λ  k c τ i +1 ,...,τ R k 2  1 + P i m =1 a 2 τ m + k c τ i +1 ,...,τ R k 2 λ 2  2 1 + i X m =1 a 2 τ m − i X m =1 b τ m λ ! . Thus, ∂ ξ i ∂ λ > 0 if λ < λ i and ∂ ξ ∂ λ < 0 if λ > λ i . So, if λ i ≤ φ − 1 τ i +1 , the optim al solut ion is reached at λ = λ i . Otherwis e, the optimal solut ion is reached at λ = φ − 1 τ i +1 . From (10), Subprobl em i is solved at y ( i ) as defined in (12). Now , we can work on the relay po wer control problem presented in (4). Theor em 1. De fine x ( i ) as x ( i ) j =    1 j = τ 1 , . . . , τ i λ i φ j j = τ i +1 , . . . , τ R . (14) The solution of the SNR optimizatio n is x ( i 0 ) , wher e i 0 is the smallest i such that λ i < φ − 1 τ i +1 . Pr oo f. F irst, since φ R +1 = 0 , we ha ve λ R < φ − 1 τ R +1 = φ − 1 R +1 = ∞ . Thus, i 0 exists. Also, since λ i 0 < φ − 1 τ i 0 +1 , and φ τ j decreases with j , we have x ( i 0 ) j ≤ 1 for j = τ i 0 +1 , . . . , τ R . This means t hat x ( i 0 ) is in the feasible region of the optimization problem. Denote η ( y ) = h c , y i 2 1 + k y k 2 . Note that k y (0) k = r 1 . Since r 1 ∈ Γ 1 , y (0) is also a feasible point of S ubproblem 1. T hus, η  y (0)  ≤ η  y (1)  due the optimality of y (1) in Subprob lem 1. This means that t here is no need to consid er Subproblem 0. For i = 1 , . . . , R − 2 , if λ i ≥ φ − 1 τ i +1 , y ( i ) j =    a j j = τ 1 , . . . , τ i φ − 1 τ i +1 c j j = τ i +1 , . . . , τ R . 12 and k y ( i ) k = v u u t φ − 2 τ i +1   c τ i +1 ,...,τ R   2 + i X j =1 | a τ j | 2 = r i +1 . Since r i +1 ∈ Γ i +1 , y ( i ) is a feasible p oint of Subp roblem i + 1 . Thus, η  y ( i )  ≤ η  y ( i +1)  due the optimalit y of y ( i +1) in Subprobl em i + 1 . This m eans that there is no need to con sider Subproblem i + 1 . Thus, we only need to check thos e y ( i ) ’ s wit h λ i < φ − 1 τ i +1 , and find the one that results in the largest recei ve SNR. From the definition i n (14), th is is t he same as to check t hose x ( i ) ’ s wi th λ i < φ − 1 τ i +1 . Now , we prove that λ i +1 < φ − 1 τ i +2 if λ i < φ − 1 τ i +1 . First, from λ i < φ − 1 τ i +1 , we hav e 1 + P i m =1 a 2 τ m P i m =1 b τ m < φ − 1 τ i +1 . Since a 2 τ i +1 b τ i +1 = φ − 1 τ i +1 , we can prove easily that λ i +1 = 1 + P i +1 m =1 a 2 τ m P i +1 m =1 b τ m = 1 + P i m =1 a 2 τ m + a 2 τ i +1 P i m =1 b τ m + b τ i +1 < φ − 1 τ i +1 < φ − 1 τ i +2 . Thus, we only need to check those x ( i ) ’ s for i 0 ≤ i ≤ R and find th e one causing the largest receive SNR. From pre vious discussio n, i 0 ≥ 1 . Define S N R i = h b , x ( i ) i 2 1+ k A x ( i ) k 2 . No w , we prove that S N R i > S N R i +1 for i 0 ≤ i ≤ R . From the proof of Lemma 3, we hav e S N R i =  P i m =1 b τ m + k c τ i +1 ,...,τ R k 2 λ i  2 1 + P i m =1 a 2 τ m + k c τ i +1 ,...,τ R k 2 λ 2 i = R X m = i +1 c 2 τ m +  P i m =1 b τ m  2 1 + P i m =1 a 2 τ m = S N R i +1 + b 2 τ i +1 a 2 τ i +1 +  P i m =1 b τ m  2 1 + P i m =1 a 2 τ m −  P i +1 m =1 b τ m  2 1 + P i +1 m =1 a 2 τ m = S N R i +1 +  1 + P i m =1 a 2 τ m  a 2 τ i +1 1 + P i +1 m =1 a 2 τ m  φ i +1 − λ − 1 i +1  2 >S N R i +1 . Thus, the optimal power control vector that maximizes the recei ve SNR is x ( i 0 ) . 13 3.2 Discussion It is natu ral to expect the p ower cont rol at relays to undergo an on -or -off scenario: a relay uses its maximum power if its channels are good enough and otherwise not t o coop erate at all. Our result shows otherwise. The opt imal power used at a relay can be any value between 0 and its m aximal power . In many situatio ns, a relay should use partial of its power , whose value is determined not only by it s own channels but all o thers’ as well. This is because ev ery relay has t wo ef fects on t he transmissio n. For o ne, it h elps the transm ission by forwarding the informatio n, whil e for the other , it harms t he transmissi on by forwarding noise as w ell. Its t ransmit power has a non -linear ef fect on the powers of both the signal and the no ise, which makes the optim ization solution not an on-or-of f one, not a decoupled one, and, in general, not e ven a dif ferentiable function of channel coef ficients. As shown in Theorem 1 and Lem ma 3, t he fraction of power used at relay j sati sfies α j = 1 for j = τ 1 , . . . , τ i 0 and α j = λ i 0 φ j for j = τ i 0 +1 , . . . , τ R . Thus, th e i 0 relays whose φ ’ s are the l ar gest use t heir maximal powers. Since i 0 ≥ 1 , there is at l east one relay th at us es its maxim um power . This tells us that the relay with the largest φ always uses its maximal power . The remaining R − i 0 relays whose φ ’ s are smaller on ly use parts of their powers. For j = τ i 0 +1 , . . . , τ R , the p ower u sed at the j th relay is α 2 j P j = λ 2 i 0 φ 2 j P j = λ 2 i 0 | f j /g j | 2 (1 + | f j | 2 P 0 ) , which is propo rtional to | f j /g j | 2 (1 + | f j | 2 P 0 ) since λ i 0 is a constant for each channel realization. Alt hough P j does not appear explicitly in the formula, it affe cts th e decis ion of whether the j th relay shoul d use its m aximal p ower . Actually , i n determining whether a relay should use its maximal power , not only do the channel coef ficients and power constraint at thi s relay account, but also all other chann el coefficients and power constraints. The power const raint of the transmitter , P 0 , plays a roll as well. Due t o these special properti es of t he optim al power control solut ion, it can be implem ented distributiv ely with each relay knowing only its own channel informati on. In the foll owing, we propose two distrib uted strategies. One is for networks with a small number of relays, and the other is more economical in networks with a lar ge number of relays. The re ceiv er , which kn ows all channels, can solve the power control problem. Wh en the number of relays, R , is small , the recei ver broadcasts the index es of the relays that use their full po wers and 14 the coefficient λ i 0 . If relay j hears its own index from the receiv er , it will use its m aximal power to transmi t durin g t he second st ep. Otherwis e, it will use power λ 2 i 0 | f j /g j | 2 (1 + | f j | 2 P 0 ) . The bits needed for the feedback is i 0 log R + B 1 < R log R + B 1 , where i 0 is the number of relays that use their maximal power s and B 1 is the number of bits needed in broadcasting the real number λ i 0 . Instead, the receiv er can also broadcast two real n umbers: λ i 0 and a real num ber d that s atisfies φ τ i 0 > d > φ τ i 0 +1 . Relay j calculates its own φ j . If φ j > d , relay j uses its maximal power . O therwise, it uses power λ 2 i 0 | f j /g j | 2 (1 + | f j | 2 P 0 ) . The numb er of bits needed for the feedback is 2 B 1 . Thus , when R is large, this strategy needs less bits of feedback compared to the first one. Networks with an aggregate power constraint P on relays were analyzed in [24]. In th is case, with the same notation in Section 3.1, P j = P and P R j =1 α 2 j ≤ 1 . T he optimal solution is α j = | f j g j | √ 1+ | f j | 2 P 0 | f j | 2 P 0 + | g j | 2 P +1 q P R m =1 | f m g m | 2 (1+ | f m | 2 P 0 ) ( | f m | 2 P 0 + | g m | 2 P +1) 2 . α j is a functi on of its own channels f j , g j only and an extra coeffi cient c = q P R m =1 | f m g m | 2 (1+ | f m | 2 P 0 ) ( | f m | 2 P 0 + | g m | 2 P +1) 2 , which is the s ame for all relays. Therefore, this powe r allocation can b e done distrib utively wit h the extra knowledge of one single coeffi cient c , which can be broadcasted by the receiv er . In our case, e very relay has a s eparate po wer constraint. Thi s is a more practical assumption in sens or networks since ev ery sensor or wi reless device has its own battery power l imit. The power control sol utions of the two cases are totally dif ferent. If relay selection is used and o nly one relay is all owed to cooperate, it can be proved easily that we should choose the relay with the highest h j = h ( f j , g j , P j ) = P j | f j g j | 2 1 + | f j | 2 P 0 + | g j | 2 P j . W e call h the relay s election function si nce a relay with a l ar ger h j results in a higher receive S NR. While all relays ar e allowed to c ooperate, the concepts of the best relay and relay selection function are not clear . Since the power control problem is a coupl ed one, it is hard to measure how much 15 contribution a relay has. As discussed before, in network beamforming, a relay wit h a lar ger φ j does not necessarily use a larger power o r has more contribution. But we can conclude th at if φ k > φ l , the fraction o f power used at relay k , α k , i s no less than the fraction of power used at relay l , α l . It is worth t o mention that in network beamformi ng, relays with larger enough φ ’ s u se their maximal powers no matter what t heir m aximal powers are. Actually , it is not hard to see t hat i f at one time channels of all relays are good , e very relay should use its maximum po wer . 3.3 Simulation Results In this section, we show simulated performance of network beamforming and compare it wit h per - formance of other existing schemes. Figures 2(a) and 2(b) show performance of networks with Rayleigh fading channels and the same power constrain t on the transm itter and relays. In ot her words, f i , g i are C N (0 , 1 ) and P 0 = P 1 = · · · = P R = P . The horizont al axis of the figures indicates P . In Fig. 2(a), simulated block error rates of network beamformi ng with opt imal power control are compared t o those of best -relay selection, Larsson’ s scheme in [24] wit h total relay power P , distributed s pace-time coding in [10 ], and amplify-and-forwa rd without powe r control (e very re- lay uses its maximal po wer) in a 2-re lay network. The information symbol s is modulated as BPSK. W e can see that net work beamforming wit h op timal power control outperforms all other schemes. It is about 0.5dB and 2dB b etter than Larsson’ s scheme and best-relay selection, respectively . W i th perfect channel knowledge at relays, it is 7 dB b etter than Alamouti distributed space-time coding , which needs no chann el informati on at relays. Ampli fy-and-forward wi th no power control only achie ves div ersity 1, distributed space-time coding achie ves a div ersity slightly less than two, whil e best-relay selection, network beamform ing, and Larsson’ s s cheme achie ve diver sity 2. Fig. 2(b) shows simulated performance of a 3-relay network under differ ent schemes. Similar d iv ersity re- sults are obtained. But for the 3-relay case, n etwork beamformi ng is about 1 .5dB and 3.5dB better than Larsson’ s scheme and best-relay selection, respecti vely . In Fig. 3(a), we s how performance of a 2-relay network in which P 0 = P 1 = P and P 2 = P / 2 . That is, the t ransmitter and the first relay hav e the same p ower constraint while the second relay has 16 only h alf the power of t he first relay . T he channels are assumed t o be Rayleigh fading channels. In Fig. 3(b), we show performance of a 2-relay network whose channels have both f ading and path-loss ef fects. W e assume that t he distance between the first relay and th e transmitt er/recei ver is 1, wh ile the distance between the second relay and the transmitt er/recei ver is 2. The path-loss exponent [38] is assume to be 2. W e also assume that t he transm itter and relays hav e t he sam e power constraint, i.e., P 0 = P 1 = P 2 = P . In bot h cases, distributed space -time codi ng does not apply , and Larsson’ s scheme applies for th e second case only . So, we compare network b eamforming with best-relay selection and amp lify-and-forward wi th no power control only . Performance of Larsson’ s scheme is shown in Fig. 3(b) as well. Both figures show the sup eriority of network beamforming to o ther schemes. 4 Networks with a Dir ect Link The previous section is on po wer control of relay networks with no DL between the transmitter and recei ver . In th is section, we dis cuss networks with a DL. As in [8], there are sev eral scenarios, which we discuss separately . 4.1 Dir ect Link During the First Step Only In thi s subsection, we consid er relay networks with a DL durin g the first step only . This happens when the receiv er knows that the transmitter is in vicin ity and listens during th e first step, while the transmitter is not aware of the DL o r is unwil ling to do the opt imization because of it s p ower and delay const raints. It can also happen when the t ransmitter is in the listening or sleeping mod e during the second step. In this case, β 0 = 0 . From (1) and (3), the syst em equations can be written as   x 1 x 2   =    α 0 √ P 0 f 0 α 0 √ P 0 P R i =1 α i | f i g i | √ P i √ 1+ α 2 0 | f i | 2 P 0    s +    w 1 w 2 + P R i =1 α i | g i | √ P i √ 1+ α 2 0 | f i | 2 P 0 e − j arg f i v i    . 17 Using maximum ratio combining, the ML decoding is arg max s    x 1 − α 0 p P 0 f 0 s    2 + 1 + R X i =1 α 2 i | g i | 2 P i 1 + α 2 0 | f i | 2 P 0 ! − 1      x 2 − α 0 p P 0 R X i =1 α i | f i g i | √ P i p 1 + α 2 0 | f i | 2 P 0 s      2 . The op timization probl em i s t hus the m aximization of t he total receive SNR of both transmissi on branches, which equals α 2 0 P 0 | f 0 | 2 + α 2 0 P 0  P R i =1 α i | f i g i | √ P i √ 1+ α 2 0 | f i | 2 P 0  2 1 + P R i =1 α 2 i | g i | 2 P i 1+ α 2 0 | f i | 2 P 0 . First, both term s in the SNR formula increase as α 0 increases. Thus, α ∗ 0 = 1 , i.e., the transmit ter should use it s maximum power . The SNR optimizati on problem becomes the one in Section 3. 1, in which there is no DL. Therefore, the power c ontrol of networks with a DL during the first step only is exactly the same as that of networks wi thout a DL. T his result i s intuitiv e. Since w ith a DL during the first step only , operations at both the transmitter and relays keep the same as networks w ithout the DL. The only di ff erence is that the receiver obtains some e xtra informatio n from the transmitter during the first s tep, and i t can use the inform ation to improve the performance without any extra cost. For the s ingle-relay case, it can be proved easil y that to maximize the receive SNR, the relay should use its maximal power as well, t hat is, α ∗ 1 = 1 . 4.2 Dir ect Link During the Second Step Only In this subsection, we consider re lay networks with a DL during the second step only . Thi s happens when t he transm itter knows that the receiv er is at vicinit y and determ ines to do more opt imization to allocate its power between the two transmis sion steps. Ho wev er , th e receiv er is unaware of the DL and i s not li stening during the first step. It can also happen when t he receiver is i n transm itting or sleeping mode during the first step. In this case, x 1 = 0 and x 2 is giv en in (3). The recei ve SNR can be calculated to be P 0  β 0 | f 0 | + α 0 P R i =1 α i | f i g i | √ P i √ 1+ α 2 0 | f i | 2 P 0  2 1 + P R i =1 α 2 i | g i | 2 P i 1+ α 2 0 | f i | 2 P 0 18 First, we show t hat α 2 0 + β 2 0 should take it s m aximal value 1, i.e., the transmitt er should use all its power . Assume th at ˆ α 2 0 + ˆ β 2 0 < 1 i s the optim al solution. Define ˜ β 0 = p 1 − ˆ α 2 0 . W e hav e ˜ β 0 > ˆ β 0 . Therefore, S N R ( ˆ α 0 , ˆ β 0 ) < S N R ( ˆ α 0 , ˜ β 0 ) . This contradicts the assumpt ion that ( ˆ α 0 , ˆ β 0 ) is optimal. Define ˆ a i = | g i | √ P i p 1 + α 2 0 | f i | 2 P 0 , ˆ b i = α 0 | f i g i | √ P i p 1 + α 2 0 | f i | 2 P 0 , ˆ c i = ˆ b i ˆ a i , ˆ A = diag { ˆ a } , and ˆ y i = ˆ a − 1 i α i . The recei ver SNR can be calculated to be ψ ( α 0 , x ) = P 0  p 1 − α 2 0 | f 0 | + h ˆ b , x i  2 1 + k ˆ A x k 2 = P 0  p 1 − α 2 0 | f 0 | + h ˆ c , ˆ y i  2 1 + k ˆ y k 2 . For any fixed α 0 , we can optimi ze α 1 , . . . , α R following the analy sis in Section 3. 1. The foll owing theorem can be proved. Theor em 2. De fine ˆ φ j = ˆ c j ˆ a j for j = 1 , . . . , R and ˆ φ R +1 = 0 . F or any fixed α 0 ∈ (0 , 1) , or der ˆ φ j as ˆ φ ˆ τ 1 ≥ · · · ≥ ˆ φ ˆ τ R ≥ ˆ φ ˆ τ R +1 . F or i = 0 , . . . , R , let ˆ λ i = 1+ P i m =1 ˆ a 2 τ m √ 1 − α 2 0 | f 0 | + P i m =1 ˆ b τ m and define ˆ x ( i ) is defined as ˆ x ( i ) j =    1 j = ˆ τ 1 , . . . , ˆ τ i ˆ λ i ˆ φ j j = ˆ τ i +1 , . . . , ˆ τ R The r eceive SNR is ma ximized at ˆ x ( ˆ i 0 ) , wher e ˆ i 0 is the smallest i such that ˆ λ i < ˆ φ − 1 ˆ τ i +1 . Pr oo f. The proof of this theorem follo ws the one of Theorem 1 and the lemmas it uses. As discuss ed in Section 3 .1, for networks w ith no DL, there is no need to cons ider t he s olution of S ubproblem 0. Here it is differ ent. Define ˆ r 1 = ˆ φ − 1 ˆ τ 1 k ˆ c k . If we deno te the solution o f S ubproblem 0, max | ˆ y |∈ [0 , ˆ r 1 ] , 0 R  ˆ y  ˆ a ( a + h ˆ c , ˆ y i ) 2 1+ k ˆ y k 2 , as ˆ y (0) , b ecause of the existence of the DL d uring t he s econd s tep, it is possible that   ˆ y (0)   < ˆ r 1 . Now we discuss the optimization of α 0 . W e first consi der the case of α 0 ∈ (0 , 1) . For any gi ven x = h α 1 · · · α R i T , the α 0 that maximizes t he receiv e SNR satis fies ∂ ψ ∂ α 0 = 0 . Thus, t he op timal α 0 19 can be found numerically by so lving ∂ ψ ∂ α 0 = 0 . It can be p roved easily that ∂ ψ ∂ α 0 > 0 when α 0 → 0 + and ∂ ψ ∂ α 0 < 0 when α 0 → 1 − . Thus, the maximu m of ψ is reached inside (0 , 1) . When the power at t he transmitter is high ( P 0 ≫ 1 ), the receiv e SNR can be approximated by ψ ( α 0 , x ) ≈ d ( α 0 ) = P 0  p 1 − α 2 0 | f 0 | + d 1  2 1 + d 2 /α 2 0 , where d 1 = 1 √ P 0 P R i =1 α i | g i | √ P i and d 2 = 1 P 0 P R i =1 α 2 i | g i /f i | 2 P i . It can be calculated straightfor- wardly that for α 0 ∈ (0 , 1) , ∂ d ∂ α 0 = 4 P 0  p 1 − α 2 0 | f 0 | + d 1  α 3 0 p 1 − α 2 0 (1 + d 2 /α 2 0 ) 2  −| f 0 | α 4 0 − 2 b | f 0 | α 2 0 + b | f 0 | + ab q 1 − α 2 0  . and ∂ d ∂ α 0 = 0 ⇔ | f 0 | 2 α 8 0 − 4 d 2 | f 0 | 2 α 6 0 + 2 d 2 | f 0 | 2 α 4 0 + d 2 2 (4 | f 0 | 2 − d 2 1 ) α 2 0 + d 2 2 ( d 2 1 − | f 0 | 2 ) = 0 . This is a quartic equ ation of α 2 0 , whose so lutions can be calculated analytically . Note that ∂ d ∂ α 0 > 0 when α 0 → 0 + and ∂ d ∂ α 0 < 0 when α 0 → 1 − . Thus th e maximum of d is reached inside (0 , 1) . An approximate solution of α 0 can thus be obtained analytically at high transmit power s. Now we consider the cases of α 0 = 0 and α 0 = 1 . If α 0 = 0 , the system degrades to a point- to-point one since only the DL works. Thus, the recei ve SNR i s | f 0 | 2 P 0 . For α 0 = 1 , we can obtain the optim al x us ing Theorem 2. Thu s, we obt ain th ree sets of α and x for the th ree cases: α 0 ∈ (0 , 1 ) , α 0 = 0 , and α 0 = 1 , respectively . The o ptimal solut ion of t he sy stem is the set of α 0 and x corresponding t o the largest recei ve SNR. The power control problem in networks wit h a DL during the second step only can thus be solved using the follo wing recursi ve algorithm. Algorithm 1. 1. Initializa tion: S et x ( pr ev ious ) 1 = 1 R , the R -dimensi onal vector of al l ones, S N R ( pr ev ious ) 1 = 0 , and count = 0 . Set the maximal number of iterations i ter and the thr eshold thre . 2. Optimize α 0 with x = x ( pr ev ious ) 1 . Denote the solution as α (1) 0 . W e can either do this numeri- cally or calculate the high SNR appr oximation. 20 3. W ith α 0 = α (1) 0 , find the x that maximizes th e rec eive SNR us ing Theor em 2. Denote it as x 1 . Calculate S N R 1 = ψ ( α (1) 0 , x 1 ) . 4. Set count = co unt +1 . If count < iter and    S N R 1 − S N R ( pr ev ious ) 1    > thr e , set x ( pr ev ious ) 1 = x 1 , S N R ( pr ev ious ) 1 = S N R 1 , and go to step 2. 5. F in d the solution of x with α 0 = 1 using Theor em 2. Denote this solu tion as x 2 . 6. The optimal solutio n is: ( α ∗ 0 , x ∗ ) = arg max n ψ ( α (1) 0 , x 1 ) , ψ (1 , x 2 ) , ψ (0 , 0 R ) o . Similarly , the distributi ve strategies proposed in Sec tion 3 .2 can be applied here. 4.3 Dir ect Link During Both Steps In this subs ection, we consider relay n etworks with a DL d uring both th e first and the second steps. This happens when b oth the t ransmitter and the receiver k now that they are not too far away from each other and decide to communi cate during bo th steps with the help of relays during the second step. From (1) and (3), the system equation can be written as   x 1 x 2   =    α 0 √ P 0 f 0 β 0 | f 0 | + α 0 √ P 0 P R i =1 α i | f i g i | √ P i √ 1+ α 2 0 | f i | 2 P 0    s +    w 1 w 2 + P R i =1 α i | g i | √ P i √ 1+ α 2 0 | f i | 2 P 0 e − j arg f i v i    . Similar to the networks discussed in Section 4.1, the maximum ratio com bining results in the following M L decoding: arg max s    x 1 − α 0 p P 0 f 0 s    2 +     x 2 − √ P 0  β 0 | f 0 | + α 0 P R i =1 α i | f i g i | √ P i √ 1+ α 2 0 | f i | 2 P 0  s     2  1 + P R i =1 α 2 i | g i | 2 P i 1+ α 2 0 | f i | 2 P 0  − 1 . The total recei ve SNR of both transmission branches can be calculated to be α 2 0 P 0 | f 0 | 2 + P 0  β 0 | f 0 | + α 0 P R i =1 α i | f i g i | √ P i √ 1+ α 2 0 | f i | 2 P 0  2 1 + P R i =1 α 2 i | g i | 2 P i 1+ α 2 0 | f i | 2 P 0 = α 2 0 P 0 | f 0 | 2 + ψ ( α 0 , x ) . 21 The same as the networks in Section 4.2, α 2 0 + β 2 0 should tak e its m aximal v alue, which is 1. That is, β 0 = p 1 − α 2 0 . Similar to the SNR optim ization in Section 4.2, for any giv en α 0 ∈ (0 , 1) , the SNR maximization i s the s ame as th e m aximization of ψ , which is so lved by Theorem 2 . But du e t o t he diffe rence in the receiv e SNR f ormula, the op timal α 0 giv en α 1 , . . . , α R is different. It is the so lution of 2 α 0 P 0 | f 0 | 2 + ∂ ψ ∂ α 0 = 0 . When t he DL exists during bot h steps, the case of α 0 = 0 , whose receiv e SNR is | f 0 | 2 P 0 will never outperform the ca se of α 0 = 1 , whose receive SNR is | f 0 | 2 P 0 + ψ (1 , x ) for some x . Th us, the case α 0 = 0 n eeds not to be considered. The power control problem in networks with a DL during both steps can thus be solved using the follo wing recursi ve al gorithm. Algorithm 2. 1. Initializa tion: Set x ( pr ev ious ) 1 = 1 R , S N R ( pr ev ious ) 1 = 0 , and cou nt = 0 . Set the maximal number of iterations ite r and t he thr eshold thr e . 2. Optimize α 0 with x = x ( pr ev ious ) 1 . Denote t he solution as α (1) 0 . W e can do this numerically . 3. W ith α 0 = α (1) 0 , find th e x that ma ximizes ψ usin g Theor em 2. Denote it a s x 1 . Calculate S N R 1 =  α (1) 0  2 | f 0 | 2 P 0 + ψ ( α (1) 0 , x 1 ) . 4. Set count = co unt +1 . If count < iter and    S N R 1 − S N R ( pr ev ious ) 1    > thr e , set x ( pr ev ious ) 1 = x 1 , S N R ( pr ev ious ) 1 = S N R 1 and go to step 2. 5. F in d the solution of x with α 0 = 1 using Theor em 2. Denote this solu tion as x 2 . 6. The optimal sol ution is: ( α ∗ 0 , x ∗ ) = arg max   α (1) 0  2 | f 0 | 2 P 0 + ψ ( α (1) 0 , x 1 ) , | f 0 | 2 P 0 + ψ (1 , x 2 )  . Again, the distributiv e strategies proposed in Section 3.2 can be applied here. 4.4 Pe rformance Com parison In this subsection, we compare single-relay networks in which t he power cons traints at the transmi t- ter and the relay are same, i.e., P 0 = P 1 = P . The channels are assumed to have both t he fading 22 and path-loss effect. There are four cases: no DL, a DL d uring the first step on ly , a DL during the second step only , and a DL during both steps. In Fig. 5, we compare networks in wh ich th e dist ance of every l ink is t he same, i.e., the three nodes are v ertexes of an equi lateral triangl e with uni t-length edges as sho wn in Fig. 4(a) . W e c an s ee that the network with no DL has diversity 1 while networks with a DL and power control achie ve div ersity 2. Th e network wi th a DL during the first step performs less th an 0.5dB better than the network with a DL during t he s econd step only , whi le the network with a DL during b oth st eps performs the best (about 1dB better than th e network with a DL during the first s tep only). T o illuminat e t he effe ct of power con trol, we show performance of networks whose transmit power at the relay and transm itter are fixed. For the network wi th a DL during the first step onl y , there is n o power control problem since it is optimal for bo th the transmitter and t he relay to use their maximal powers. For the ot her two cases, we let t he transm itter uses half of its po wer , P / 2 , to each of the tw o steps and the relay al ways uses i ts maximum power P . W e can see that, if the DL only e xists during the second step, w ithout power control, the achiev able div ersity is 1. At bl ock error rates of 10 − 2 and 10 − 3 , it performs 3 and 6dB worse, respectively . For networks wi th a DL du ring both st eps, power control resul ts in a 1.5dB improvement. In Fig. 6(a) and 6(b), we show p erformance o f li ne networks wi th path-loss exponents 2 and 3 respectiv ely . As shown in Fig. 4(b), the three nodes are o n a line and the relay is in the m iddle of the transmit ter and receive r . The dis tance between t he transmit ter and receive r is assumed to be 2. The same phenom enon as in the equilateral triangle networks can be observed. The network with a DL during bo th steps performs the best (about 1dB better th an t he network with a DL duri ng the first step only). The network wi th a DL at first step only performs sl ightly better than th e one with a DL during the second step only . But the differe nce is smaller than that in Fig. 5. The performance diffe rence b etween line networks with and wi thout DLs is smaller than tho se in equilateral tri angle networks, and it gets even sm aller for larger path-loss exponents. This is because as th e distance between the transmitter and recei ver or the path-loss exponent is l ar ger , the qualit y of the DL is lower . Therefore, the i mprovement d ue to this lin k is s maller . For both cases, power control results in a 1.5dB imp rovement when the DL li nk exists for bot h steps and a higher div ersity when the D L 23 exists for the second step only . Then we work on the random network in Fig. 4 (c), in which the relay locates random ly and uniformly within a circle in the midd le of th e transmitter and th e recei ver . The distance between the transmitter and th e receiv er is assu med to be 2 . The radius of the circle is denoted as r . W e assume that r < 1 . This i s a reasonable m odel for ad hoc wireless networks since if comm unications between two nodes is allowed to be h elped b y one other relay , o ne sh ould choose a relay t hat is around th e middle of the two no des. In other words, the distance between the relay and th e transm itter or recei ver should be shorter than that between the transmitter and recei ver . W e work out th e geometry first. As in Fig. 4(c), we denot e the posit ions of the transmitter , the receiv er , the relay , and t he mi ddle point of t he transmitt er and the receiv er as A, C , D , and B , respectiv ely . D enote the angle of AB and B D as θ and the length of B D as ρ . The length s of AD and C D are thus p 1 + ρ 2 − 2 ρ cos θ and p 1 + ρ 2 + 2 ρ cos θ . Since D is uniformly distributed within the circle, θ is uniform in [0 , π ) and the pdf and cdf of ρ can be calculated to be p ( ρ ) = 2 ρ r 2 and P ( ρ < x ) = x 2 r 2 , respectiv ely . Define Y = r √ X . If X is uniform on (0 , 1 ) , it can be prov ed that P( Y ≤ x ) = P( r √ X ≤ x ) = P  X ≤ x 2 r 2  = x 2 r 2 . Thus, Y has th e same distribution as ρ . Therefore, we generate Y to represent ρ . Fig. 7 s hows performance of random networks with p ath-loss exponent 2 and r = 1 / 2 . W e can see that the same phenomenon as in line networks can be obs erved. W ith a DL at bot h steps, th e random network performs about 1dB w orse than the line network. 5 Conclusio ns and Futur e W ork In this paper , we propose the novel id ea of beamforming in wireless relay networks to achiev e both d iv ersity and array gain. The s cheme i s based on a two-step ampl ify-and-forward protocol. W e assum e that each relay knows its own channels perfectly . Unl ike previous works in network 24 div ersity , the scheme dev eloped here uses not only t he channels’ phase information but also th eir magnitude. Match filt ers are applied at t he transmitter and relays during th e second step to cancel t he channel ph ase ef fect and th us form a coherent beam at the receiv er , i n t he m ean whil e, optimal power control is p erformed based on the channel magnitude t o decide the power used a t the transmit ter and relays. T he power cont rol problem for networks wi th any numbers of relays and no direct link is solved analy tically . The solution can be obt ained with a complexity that i s linear i n the num ber of relays. The p ower used at a relay depends on not o nly its o wn channels nonlinearly but also all other channels in the network. In general, it is not even a diffe rentiable functi on of channel coeffi cients. Simulation with Rayleigh fading and path -loss channels show that net work beamforming achiev es the maximum di versity while amplify-and-forwar d without powe r control achie ves di versity 1 only . Network beamforming also o utperforms oth er cooperativ e strategies. For example, i t is about 4dB better than best-relay selection. Relay n etworks with a direct link b etween the transmitter and receiver are also considered i n thi s paper . For net works with a d irect l ink d uring th e first st ep only , the power control at relays and the transmitter is exactly the s ame as t hat of networks with no direct link. F o r networks wit h a direct link during the second st ep only and networks with a di rect link during b oth steps, the solut ions are diffe rent. Recursiv e num erical algorithms for the power control at both the transmi tter and relays are giv en. Simulated per formance of singl e-relay networks with dif ferent topologies sh ows that optimal power control results in abou t 1.5dB improv ement in networks with a direct link at both steps and a higher div ersity in netw orks with a direct lin k at the second step only . W e hav e just scratched the surface of a brand-new area. There are a lot of ways to extend and generalize thi s work. First, it is assumed in this work th at relays and someti mes the t ransmitter know thei r channels p erfectly , which i s not practical i n many networks. Netw ork beamformi ng with li mited and delayed feedback from the recei ver is an im portant issue. In mu ltiple-antenna systems, beamforming wit h limited and delayed channel information feedback has been widely probed. Howe ver , beamforming in networks di f fers from beamforming in multiple-antenna systems in a couple o f ways. In networks, i t i s d iffi cult for relays to cooperate whi le in a m ultiple-antenna system, di f ferent ant ennas of the transmitter can cooperate fully . Th ere are tw o transmi ssion steps in 25 relay networks while o nly one in multi ple-antenna systems, which leads t o different error rate and capacity calculation and thus different desig ns. Second, the relay network probed in t his paper has only one pair of transmitter and recei ver . 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P rentice Hall, 2nd ed., 2002. 29 2 1 t 2 t R 1 r r 2 r R 1 g g 2 g R R f f 1 f f 0 0 0 1 1 0 0 1 1 00 00 11 11 00 00 11 11 00 00 11 11 transmitter relays receiver . . . . . . . . . s x t Figure 1: W ireless relay network. 10 12 14 16 18 20 22 24 10 −5 10 −4 10 −3 10 −2 10 −1 P (dB) Block error rate Alamouti DSTC Network beamforming Larsson’s scheme Best relay selection AF without power control (a) 2-relay network 6 8 10 12 14 16 18 10 −4 10 −3 10 −2 10 −1 P (dB) Block error rate AF without power control Larsson’s scheme Network beamforming Best relay selection (b) 3-relay network Figure 2: Networks with fading channels and same po wer constraint for all nodes. 30 10 12 14 16 18 20 22 24 10 −5 10 −4 10 −3 10 −2 10 −1 P (dB) Block error rate AF without power control Network beamforming Best relay selection (a) Network with different relay powers 10 12 14 16 18 20 22 24 26 28 30 10 −4 10 −3 10 −2 10 −1 Block error rate P (dB) AF without power control Network beamforming Larsson’s scheme in Best relay selection (b) Network with path-loss plus fading channels Figure 3: 2-relay networks with different relay p ower s and pass-loss plus fading channels. 1 transmitter Receiver Relay 1 1 (a) T riangle network 2 1 1 transmitter Relay Receiver (b) Line network 00 00 11 11 transmitter Receiver B C A D r Relay Region 1 1 ρ θ (c) Random network Figure 4: Network topology . 31 6 8 10 12 14 16 18 20 10 −4 10 −3 10 −2 10 −1 P (dB) Block error rate No DL DL 1st step DL 2nd step DL both steps DL 2nd step no power control DL both steps no power control Figure 5: Eq uilateral triangle network. 6 8 10 12 14 16 18 20 10 −3 10 −2 10 −1 P (dB) Block error rate No DL DL 1st step DL 2nd step DL both steps DL 2nd step no power control DL both steps no power control (a) Path-loss e xpo nent 2 6 8 10 12 14 16 18 20 10 −3 10 −2 10 −1 P (dB) Block error rate No DL DL 1st step DL 2nd step DL both steps DL 2nd step no power control DL both steps no power control (b) Path-loss exponent 3 Figure 6: Single-relay line network. 32 6 8 10 12 14 16 18 20 10 −2 10 −1 P (dB) Block error rate No DL DL 1st step DL 2nd step DL both steps DL both steps line network DL 2nd step no power control DL both steps no power control Figure 7: Single-relay random network with pass-loss exponent 2 and r = 1 / 2 . 33