On Single-User Interactive Beam Alignment in Millimeter Wave Systems: Impact of Feedback Delay
Abbas Khalili, Shahram Shahsavari, Mohammad A. Amir Khojastepour, Elza Erkip
OOn Single-User Interactive Beam Alignment inMillimeter Wave Systems: Impact of Feedback Delay
Abbas Khalili † , Shahram Shahsavari ‡ , Mohammad A. (Amir) Khojastepour (cid:5) , Elza Erkip † . † NYU Tandon School of Engineering, ‡ University of Waterloo, (cid:5)
NEC Laboratories America, Inc..Emails: † { ako274, Elza } @nyu.edu, ‡ [email protected], (cid:5) [email protected]. Abstract —Narrow beams are key to wireless communicationsin millimeter wave frequency bands. Beam alignment (BA) allowsthe base station (BS) to adjust the direction and width of the beamused for communication. During BA, the BS transmits a numberof scanning beams covering different angular regions. The goalis to minimize the expected width of the uncertainty region (UR)that includes the angle of departure of the user. Conventionally,in interactive BA, it is assumed that the feedback correspondingto each scanning packet is received prior to transmission ofthe next one. However, in practice, the feedback delay couldbe larger because of propagation or system constraints. Thispaper investigates BA strategies that operate under arbitraryfixed feedback delays. This problem is analyzed through a sourcecoding prospective where the feedback sequences are viewedas source codewords. It is shown that these codewords forma codebook with a particular characteristic which is used todefine a new class of codes called d − unimodal codes. By analyzingthe properties of these codes, a lower bound on the minimumachievable expected beamwidth is provided. The results revealpotential performance improvements in terms of the BA durationit takes to achieve a fixed expected width of the UR over the state-of-the-art BA methods which do not consider the effect of delay. Index Terms —Millimeter wave, Analog beam alignment, In-teractive beam alignment, Non-interactive beam alignment, Con-tiguous beams.
I. I
NTRODUCTION
Millimeter wave (mmWave) communication greatly im-proves throughput of wireless networks by using the widebandwidths available at high frequencies [1]. In order toestablish a viable communication link in highly directionalmmWave links and mitigate the high path-loss and intenseshadowing, it is necessary to perform beamforming [2]. Beam-fomring methods concentrate the transmit power in a desireddirection by utilizing narrow beams [3].It is known that mmWave channels are sparse and consistof only a few spatial clusters [4]. Therefore, to reduce beam-forming overhead and maximize system throughput, beamalignment (BA) (a.k.a. beam training and beam search) isused to find narrow beams aligned with the direction of thechannel clusters [5]. In BA, the wireless transceiver searchesover the angular space through a set of scanning beams tolocalize the direction of the channel clusters, i.e., namely,the angle of arrival (AoA) and angle of departure (AoD)of the channel clusters at the receiver and transmitter sides,
This work is supported by National Science Foundation grants EARS-1547332, SpecEES-1824434, and NYU WIRELESS Industrial Affiliates. respectively. Moreover, due to high power consumption inmmWave systems it is often assumed that the transceivers onlyuse one RF-chain during BA, a method known as analog BA.There is a large body of work on BA methods in theliterature [6]–[16]. In general, BA strategies can be classifiedas interactive
BA and non-interactive
BA. To elaborate, let usconsider the BA procedure at the transmitter whose objectiveis to localize the AoD of the channel. The transmitter sendsa set of BA packets through a set of scanning beams to scanthe angular space. In non-interactive BA, the transmitter doesnot receive any feedback from the receiver until all the BApackets are sent. In interactive BA, however, the transmitterreceives feedback during the transmission of the BA packetswhich allows it to refine the scanning beams and better localizethe AoD of the channel compared to non-interactive BA.The problem of multi-user non-interactive BA is consid-ered in [6] where we analyzed the BA problem through aninformation theoretic perspective and provided bounds onthe minimum average expected beamwidth of data beamsallocated to the users given a fixed BA duration along withachievablity schemes. A more challenging problem is to con-sider the interactive case which necessitates optimally utilizingthe feedback information during the BA. Prior research oninteractive BA methods [7]–[15], [17]–[19] consider no delayfor the receiver’s feedback on the scanning packets. However,this might not always be the case due to practical reasons suchas processing delay at the transceivers.In this paper, we consider the problem of interactive analogBA at the base station (BS) in a single-user downlink systemwhere the channel has one dominant cluster and the feedbackto each transmitted BA packet is received after a fixed knowndelay. Due to practical constraints, we only look at the casewhere the beams are contiguous [6]–[8]. Similar to [6], weassume that the BA packets and feedback at the user and BSare received error free. As a result, at the end of the BA phasethe BS can allocate a beam for the data communication whichincludes the AoD of the channel with probability one. We referto the angular region of this beam as uncertainty region (UR)on the channel AoD. Our objective is to minimize the expectedwidth of the UR similar to [6]. Overview of the contributionsof this paper is as follows: • We view the BA with feedback delay as a source codingproblem in which the BS needs to ask b yes/no questionswhere each question is an angular interval. We showthat the resulting source codewords (feedback sequences) a r X i v : . [ c s . I T ] F e b = 1 t = b t = b + d Scanning
Waiting t = T Transmision
Phase
CoherenceInterval
PhasePhase
Fig. 1: Time-slotted system.have a special characteristic using which we define a newfamily of codes that we name d − unimodal (Section III). • We provide a lower bound on the minimum expectedwidth of the UR given any arbitrary prior on the AoD byanalyzing properties of d − unimodal codes. Through nu-merical evaluations, we show the potential improvementin terms of the number of required time-slots to achievea certain angular resolution for the expected width of theUR when compared with state-of-the-art (Section IV).II. S YSTEM M ODEL AND P RELIMINARIES
In this section, we outline general system assumptions (II-Aand II-B) and then provide the mathematical formulation ofthe problem (II-C and II-D).
A. Network Model
We consider a single-user downlink communication in asingle-cell mmWave system scenario. Motivated by previousworks [9], [10], [19], [20] and experimental results [4], weassume that the channel has only a single dominant cluster. Wedenote the AoD corresponding with this cluster by ψ which isunknown to the BS. In our setup, motivated by [6], [7], [12],we consider that the BS performs analog BA during whichit is able to search one beam at a time while the user hasan omnidirectional reception pattern. The goal is to find asmall angular interval (i.e., UR) which contains ψ . We assume Ψ ∼ f Ψ ( ψ ) for ψ ∈ (0 , π ] which accounts for the priorknowledge on the AoD (e.g., corresponding to the history ofpreviously localized AoD in beam tracking applications).Due to practical constraints, we only consider use of con-tiguous beams as in [6]. Similar to [6], [7], [12], we assumethat the beams are ideal and use the sectored antenna modelfrom [21]. In this model, each beam is characterized bya constant main-lobe gain and an angular coverage region(ACR). In the case of contiguous beams, this ACR is anangular interval inside (0 , π ] that is covered by the main-lobe. This model is often used in the literature (e.g. [22], [23])and is justified as the BSs are envisioned to use large antennaarrays which allows for beams close to ideal shape [1]. B. Frames and Feedback
We consider an interactive BA scenario in which the BSreceives feedback form the user during the transmission ofBA packets and can change the subsequent scanning beamsbased on the feedback. Unlike conventional interactive BAin which the feedback to each transmitted packet is available instantaneously, we consider a fixed known delay of d time-slots for each feedback. This delay accounts for practicalconstraints such as processing delay at the transceivers. If thisdelay cannot be accurately measured, an upper bound can beutilized for our analysis. We assume that the feedback to eachpacket is either an acknowledgement (ACK) that the packetwas received by the user or a negative acknowledgement(NACK) which indicates the user did not receive the packet.Similar to [7], we consider that the feedback is receivedthrough a control channel and is error free [1]. Also, as in[6], [9], we assume that the BA packets are detected at theuser without error.Motivated by the above discussion, we consider a time-slotted system in which the user has fixed AoD over coherenceintervals of duration T time-slots. We assume that the commu-nication spanning a coherence interval includes three phasesas shown in Fig. 1. We first have the scanning phase in whichthe BS transmits b BA packets through a set of scanning beamsto scan the angular space. Since the response to each packettakes d time-slots, we consider a waiting phase in which theBS waits to receive the feedback to the scanning beams. Thisphase lasts for d time-slots and can be used for example, fordata transmission to other users for which the BS has alreadyperformed BA. The rest of the coherence interval, i.e., the last T − b − d time slots, is called transmission phase which isused for data transmission.Our main focus in this paper is the design of the beamsto be used in the scanning phase and the resulting expectedbeamwidth for data transmission beam. C. Scanning Beam Set and Data Beam
The objective of BA is to maximize the beamforminggain which in turn maximizes the data communication rate.Towards this goal, we consider minimizing the expected widthof the UR for the AoD of the user’s channel.The BS uses b scanning beams { Φ i } i ∈ [ b ] to transmit the b BA packets . Let a i ∈ { , } denote the feedback received forthe i th BA packet (i.e., a i = 1 if ACK was received for Φ i and a i = 0 otherwise). Based on the received feedback sequenceby the i th time-slot (i.e., ( a , a , . . . , a i − d ) ), there are multiplechoices for Φ i . To model this, we use a hierarchical beam set S = {S i } i ∈ [ b ] , where S i = { S i,m } m ∈ [ M ( i )] denotes the setof all possible scanning beams given that there are a total of M ( i ) ≤ i − d possible feedback sequences. To elaborate, notethat by the i th time-slot, the BS has received the feedbacksequence corresponding to the scanning beams Φ j , j ≤ i − d .We design the set S i = { S i,m } m ∈ [ M ( i )] to contain a beamfor each of the possible feedback sequences. Therefore, uponreception of a particular feedback sequence at the i th time-slot, the BS selects the beam S i,m which was designed forthat feedback sequence and uses it for transmission of the i th BA packet (i.e., Φ i = S i,m ).Given an AoD realization ψ , let us denote the ACR that theBS chooses for data transmission (i.e., UR) by Beam ( S , ψ ) . We use the notation [ n ] to represent the set { , , . . . , n } . nder the assumption of single dominant path channel anderror free system, the minimum length ACR which includesthe user AoD is Beam ( S , ψ ) = ∩ bi =1 Θ(Φ i , a i ) , (1)where Θ(Φ i , a i ) = Φ i if a i = 1 which corresponds to ψ ∈ Φ i ,and Θ(Φ i , a i ) = (0 , π ] − Φ i otherwise. D. Problem Formulation
We formulate the problem of minimizing the expected widthof the UR for the AoD as S ∗ = arg min S E Ψ [ | Beam ( S , Ψ) | ] , (2)where expectation is taken over the distribution f Ψ ( ψ ) .Based on (1), given S , we get an UR for each possiblefeedback sequence ( a , a , . . . , a b ) . Let us denote the set ofpossible URs for the AoD of the user by U = { u m } m ∈ M ( b ) ,where M ( b ) ≤ b is the number of possible feedback se-quences. It is easy to see that Beam ( S , Ψ) = u m for Ψ ∈ u m .Hence, we can write the expectation in (2) as E Ψ [ | Beam ( S , Ψ) | ] = M ( b ) (cid:88) m =1 | u m | (cid:90) ψ ∈ u m f Ψ ( ψ ) dψ. (3)The notation | u m | is the Lebesgue measure of u m , which isequal to the total width of the intervals in the case where u m is the union of a finite number of intervals.In the next sections, we will show that the feedback se-quences can be viewed as source codewords with a specialcharacteristic. This characteristic lets us define a new classof codes which we refer to by d − unimodal codes. Then, bystudying these codes, we lower bound the minimum expectedbeamwidth in Sec. IV. Explicit construction of optimal scan-ning beam set is reported elsewhere due to space constraint.III. B EAM A LIGNMENT AND U NIMODAL C ODES
We view the discussed BA problem as a source codingproblem in which the BS asks b questions whose answers (thefeedback sequences) represent the source codewords. Unlikea finite alphabet source coding problem, here the alphabet iscontinuous and the questions are intervals inside (0 , π ] . Inthis section, we examine the properties of the aforementionedsource code in our BA problem and define a new class of codescalled d − unimodal . We also establish the connection betweenthe BA schemes and the design of d − unimodal codes.To define d − unimodal codes, we need the following Definition 1 ( Unimodal Loop ) . A binary loop is called unimodal iff the location of ones (if any) are consecutive .As an example, the loop (cid:12){ , , , } is unimodal but theloop (cid:12){ , , , } is not . As we will elaborate later, unimodalloops represent the scanning beams in our BA problem. Now,we can define d − unimodal codes as follows: A loop is a cyclically ordered set of elements [24] (i.e., the elements canbe arranged on a circle). The notation (cid:12){ . . . } indicates the loop of the ordered set { . . . } . Definition 2 ( d − unimodal Code ) . A binary code (collectionof codewords) with codewords of length b is called d -unimodal and is denoted by C ( b, d ) , if there exists an ordered set ofits codewords which could also include repetition of somecodewords whose associated loop satisfies:1) For i ≤ d , the loop created by the i th bits of thecodewords in the loop is unimodal.2) For i > d , for each sub-loop of the loop consisting onlyof codewords with same prefix of length i − d , the binaryloop of the bits in the i th position is unimodal.We refer to such loop as characteristic loop of the code. The code cardinality is the number of codewords in C , denoted by |C| . For example, the code C = { , , } is a − unimodalcode with a characteristic loop of L = (cid:12){ , , } . Moreexamples are provided later in the paper.Characteristic loop of a code is not unique and may containrepetition of the codewords. For example, consecutively re-peating a codeword in a characteristic loop generates anothervalid characteristic loop. A minimal characteristic loop (MCL) is defined as one which does not contain any consecutiverepetitions. Yet, an MCL may still contain repetitions thatare not consecutive. For example, consider C = { , , } with the characteristic loop L = (cid:12){ , , , } which isminimal but contains repetition.As part of our first main result (Thm. 1), we show that thefeedback sequences in our BA problem form a d − unimodalcode. Moreover, one can also find a construction that givena d − unimodal code, generates a scanning beam set S whosefeedback sequences are that code. This second part forms thefoundation of our explicit construction of optimal BA schemesand will be pursued elsewhere due to space constraint.Before providing the theorem statement, we first providethe necessary definitions and show through a set of exampleshow a scanning beam relates to a unimodal loop and how agiven scanning beam set S leads to d − unimodal code.Suppose we are given a scanning beam set S . The scanningbeams inside this set, partition the interval (0 , π ] into a setof angular intervals which we call component beams . Wedefine these component beams as follows. Each scanning beamis an angular interval with two endpoints. After sorting theendpoints of all the scanning beams in S and removing therepetitions, each angular interval in between two consecutiveendpoints is a component beam. Since the component beamsare contiguous and partition the interval (0 , π ] , one can usetheir positions on the circle and form a loop of the componentbeams. We denote this loop using I . To better understand thenotation and the relation between S and I , let us consider thefollowing example which we will build upon in the paper. Example 1.
Fig. 2 illustrates a possible set of scanning beamsfor b = 4 and d = 3 . In this case, S = {S , S , S , S } , with S i = { Φ i } for i ∈ { , , } each consisting of a single possiblescanning beam as no feedback is received prior to fourth time-slot. However, at the fourth time-slot, we receive the feedbackto the beam Φ and so there are two possibilities for Φ .Here, we have S = { S , , S , } . As shown in Fig. 2, the set Φ Φ I I I I I I I I I I S , Φ Φ Φ = ∪ { I , I , I , I , I } Φ = ∪ { I , I , I , I } Φ = ∪ { I , I , I , I , I } Φ = I or ∪ { I , I } S , Fig. 2: An example set of scanning beams for b = 4 and d = 3 and the corresponding component beams. S creates the component beam loop I = (cid:12){ I , I , . . . , I } which includes ten component beams.It is easy to see that each of the scanning beams in S can bewritten as union of subsets of component beams in I . Considerone of these beams. By replacing the elements of the loop I with if the component beam is included in the beam and otherwise, we will have a binary loop. As a result, we canuniquely determine any beam in S using a binary loop given I . Note that since the scanning beams are contiguous, theyeach include adjacent component beams and so the positionof ones inside their binary loops are consecutive. Therefore,these binary loops are unimodal. To elaborate, consider thesetup of Example 1. The beam S , in Fig. 2 is partitioned bycomponent beams I and I . Hence, its corresponding binaryloop is (cid:12){ , , , , , , , , , } which is unimodal.Next, we show how these unimodal binary loops corre-sponding to the scanning beams lead to d − unimodal codes.For this purpose, let us first consider the following example. Example 2.
Consider the setup in Example 1. If we replacethe component beams in the loop I with their correspondingfeedback sequences, we get the loop L = (cid:12){ , , , , , , , , , } . Let uslook at the second bit of the codewords in L which leads tothe binary loop (cid:12){ , , , , , , , , , } . This loopis unimodal and is the same binary loop representing thecontiguous scanning beam Φ = ∪{ I , I , I , I } . Next,consider the loop of the fourth bits (cid:12){ , , , , , , , , , } . This is not a unimodal loop. Here, this loop correspondsto the beam S , ∪ S , which is not contiguous. Notice thateach of these beams are contiguous. So if we can separatetheir binary loops, we should get unimodal loops. Note thatthe decision that which one of these beams is used for Φ isbased on the feedback sequences received at the fourth time-slot which is a in our case. Now, if we look at the sub-loopsof the loop L whose feedback sequences have same value for a , we will get the binary loops (cid:12){ , , , , } and (cid:12){ , , , , } . These are both unimodal loops where the formerand the latter correspond to S , and S , , respectively whichare contiguous.Let us form a loop of binary codewords by replacing eachcomponent beam in loop I with its corresponding feedbacksequence (i.e., ( a , a , . . . , a b ) ) and denote it by L . Following A sub-loop of a loop is a loop in which some of the elements of theoriginal loop are removed.
Example 2, if we look at the loop consisting of the i th bit ofeach codeword, we have a binary loop which relates to thescanning beams in S i . For i ≤ d , since we have not receivedany feedback, S i consists of only one contiguous scanningbeam and the binary loop becomes unimodal. However, for i > d , this is no longer the case since there are multiplescanning beams in S i . Yet, similar to Example 2, if we createsub-loops of L that include feedback sequences of same prefixof ( a , a , . . . , a i − d ) and then look at the loop of the i th bitsfor each sub-loop, we will get unimodal loops. These claimsare proved rigorously in the next theorem were we show thatthe loop L is in fact an MCL for the d − unimodal code whosecodewords are the feedback sequences resulting from S .An interesting observation from the MCL created using thefeedback sequences and the component beams loop is thatwhen it has repetition, one or more of the URs are non-contiguous. The repetition of a codeword means there aremultiple component beams with same feedback sequence andadjacent component beams of different feedback sequences. onthe other hand, from Sec. II-D, we know that each feedbacksequence corresponds to an UR. Therefore, there is an URthat includes these component beams but not their adjacentwhich makes it non-contiguous. This is important since as wediscussed in Sec. II, each UR is a possible data beam andthe data beams are preferred to be contiguous. As an exampleof this observation, consider Example 2. The MCL has therepetition of the codeword . On the other hand, if weform the set of URs, we get U = { u m } m =1 , where u m = I m for m ∈ [10] \ and u = ∪{ I , I } . The non-contiguous URis u whose feedback sequence is the repeated codeword . Theorem 1 ( Beam Alignment and Unimodal Codes ) . Con-sider the BA problem introduced in Section II where thenumber of BA scanning packets is set to b and the delayis set to d . Given any scanning beam set S , the feedbacksequences form a d − unimodal code C whose MCL L is theloop of binary codewords resulted from replacing the elementsof the component beams I with their corresponding feedbacksequences.Proof. The proof is provided in Appendix A. (cid:3)
This theorem shows that the collection of feedback se-quences of any possible scanning beam set is a d − unimodalcode. We will use this to lower bound the performance ofthe considered BA problem in terms of minimum expectedbeamwidth in the next section.V. L OWER B OUND ON E XPECTED B EAMWIDTH
In this section, we investigate the properties of d − unimodalcodes to lower bound the optimal performance in terms ofexpected beamwidth for our BA problem. To this end, wedefine a parent-child hierarchy between the codes C ( b, d ) and C ( b − , d ) which we will use in our proofs. This hierarchy isformally defined below. Definition 3 ( Parent Code ) . For a C ( b, d ) code with an MCL L ( b, d ) , the loop containing the prefix of length b − i of allthe codewords in the loop is an MCL that defines a parentcode of order i , i.e., C ( b − i, d ) . The parent code of order issimply called the parent code .It can be inferred that given a code, its corresponding parentcode is unique and d -unimodal. However, a parent code canresult in different child codes. Note that based on Thm. 1,given a scanning beam set, the resulting collection of feedbacksequences is a d − unimodal code. Also, from Sec. II-D, weknow that the number of possible URs is the same as thenumber of possible feedback sequences. As a result, wecan upper bound the number of URs (number of feedbacksequences) by finding an upper bound for the cardinality of d − unimodal codes. In the next theorem, we use the parent-child hierarchy to bound the cardinality of d − unimodal codeswhich also gives us a bound on the number of URs. Theorem 2 ( Maximum Code Cardinality ) . Let M ( b, d ) denote the maximum cardinality for the code C ( b, d ) . Then,for d = 1 , M ( b, d ) = 2 b and for d > , M ( b, d ) ≤ (cid:40) M ( b − , d ) + 2 M ( b − d, d ) b > d, b b ≤ d. (4) Proof.
The proof is provided in Appendix B. A sketch ofwhich is as follows. From Def. 2, we know that the cardinalityof a d − unimodal code is less than or equal to the length of itsMCL. Moreover, it is easy to see that the length of any MCLfor C (1 , d ) is at most . On the other hand, using parent-childhierarchy, we bound the difference between the cardinality ofa child’s MCL with its parent’s MCL. Based on these, wecalculate the upper bound in the theorem. (cid:3) Using the above results, we can bound the minimum ex-pected beamwidth for UR as in the next theorem.
Theorem 3 ( Minimum Expected Beamwidth ) . The minimumexpected beamwidth i.e., the objective function in optimizationproblem (2) when contiguous scanning beams are used isbounded as h (Ψ) M ( b, d ) ≤ E Ψ [ | Beam ( S , Ψ) | ] (5) Proof.
From Thm. 1 and Thm. 2, we observe that the maxi-mum number of URs is bounded by M ( b, d ) . Using this with[6, Prop. 2], will give us the lower bound. (cid:3) We conclude this section, by providing a comparison of thetotal (i.e., scanning phase + waiting phase) BA duration that Fig. 3: Total BA duration for a given fixed expected databeamwidth resolution of / ≈ ◦ for different BAmethods and feedback delays for ψ ∼ Uniform(0 , π ] .different BA methods and the derived lower bound require,given a fixed expected UR width for different values offeedback delay and Ψ ∼ Uniform(0 , π ] . The result is plottedin Fig. 3. In the modified exhaustive search method, for agiven b , we divide the (0 , π ] into b + 1 equal width URs, andat each time-slot, scan one UR. Since the system is error free,we can find the UR including the AoD by only searching b of b + 1 URs. We observe that as the delay increases,the performance of bisection method which is optimal forcase of d = 1 rapidly degrades and after delay of d = 8 time-slots, even the modified exhaustive search outperformsbisection method. This figure also shows that as the delayincreases the lower bound becomes closer to the performanceof the optimal non-interactive method [6]. In fact, if we allowfor more delays, they become exactly the same. The reasonis that the optimal non-interactive method in [6] is a specialcase of our problem for d > b . This plot also suggests thatthere is potential of improving the performance of the state-of-the-art methods using an appropriate BA scheme. In fact, theproposed framework can also be used to construct the optimalBA method achieving the lower bound in the Fig. 3. Detailsand the derivation of optimal BA solution are left for futurepublication due to space constraint.V. C ONCLUSION
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Czechoslovak Mathematical Jour-nal , vol. 32, no. 3, pp. 460–473, 1982. A PPENDIX AP ROOF OF T HEOREM I with its corresponding feedback sequencesby L . To prove the theorem statement, we show that L firstly,does not have any consecutive repetitions and secondly, is acharacteristic loop for a code consisting of all the feedbacksequences of S . We prove the second part by showing L satisfies each of the conditions in Def. 2. No consecutive repetitions:
We show this by contradiction.Assume L has a consecutive repetition. This means twoconsecutive component beams lets say I and I have samefeedback sequences. If so, all the scanning beams in S eitherinclude both I and I or none. Therefore, if we form thecomponent beams of S , we should get I ∪ I as a componentbeam instead of two separate component beams I and I . Thisis a contradiction and so L does not have any repetitions.To prove that L is a characteristic loop of a code consistingof all the feedback sequences. First, note that the loop L includes all the possible feedback sequences as the componentbeams of I partition the entire (0 , π ] . Next, we show that L satisfies the conditions in Def. 2. Def. 2, first condition:
Assume i ≤ d , since no feedback isreceived, the set S i includes only one scanning beam. Let usdenote this beam by Φ i . By construction of component beamsloop I , Φ i can be written as union of subset of componentbeams in I . It is easy to see that if a component beam isincluded in Φ i , the feedback sequence corresponding to thatcomponent beam would have one in the i th position andotherwise zero. Now, Let us consider the binary loop derivedfrom the i th elements of the feedback sequences in L anddenote it by L i . Based on our discussions, in L i , at the positionof the component beams which are included in Φ i , we shouldhave one and otherwise we should have zero. Since Φ i iscontiguous, the component beams included in Φ i are adjacentand so the positions of ones in L i are consecutive. Therefore, L i is by definition unimodal and the condition holds. Def. 2, second condition:
Suppose that we are at the i th time-slot of BA where i ≥ d . Without loss of generalityassume that the BS has received the feedback sequence ( a , a , . . . a i − d ) for which it uses the scanning beam S i,m from S i for some m ∈ [ M ( i )] . We need to show that if weform the sub-loop of the feedback sequences in L that havethe prefix ( a , a , . . . , a i − d ) , the binary loop of the i th bits isunimodal. Let us denote this binary loop by (cid:101) L i .ote that the scanning beams Φ j , j ≤ i − d and theconsidered feedback sequence ( a , a , . . . , a i − d ) determine anangular region that includes the AoD of the user at the i th time-slot. By construction of the component beam loop I ,we can write this angular region as a union of subset ofthe component beams in I . Given this, let us form a sub-loop of I by removing the component beams which are notincluded in this angular region and name it (cid:101) I . Consider oneof the component beams inside this sub-loop and denote itby (cid:101) I . It is straight forward to see that the i th bit of thefeedback sequence ( a , a , . . . , a b ) corresponding to (cid:101) I is oneif (cid:101) I ∈ S i,m and zero otherwise. Therefore, if we replace thecomponent beams inside (cid:101) I with their feedback sequences andlook at the loop of i th bits it would be unimodal since S i,m is contiguous. Next, we show that this loop is the loop (cid:101) L i .To show this, note that (cid:101) I is a sub-loop of I and includes alland only component beams of I whose feedback sequenceshave the prefix ( a , a , . . . , a i − d ) . Therefore, if we replace thecomponent beams inside (cid:101) I with their corresponding feedbacksequences and form the binary loop of the i th bits by definitionit will be the loop (cid:101) L i . A PPENDIX BP ROOF OF T HEOREM d = 1 , we know that the maximum numberof feedback sequences using b yes/no questions is b which isalso achievable (e.g., bisection method). Therefore, M ( b,
1) =2 b .For d > , we first bound the cardinality of the MCL L ( b, d ) , and then bound M ( b, d ) using Def. 2 which indicates M ( b, d ) ≤ |L ( b, d ) | .Looking at the Def. 2, by grouping the codewords in theMCL L ( b, d ) into sub-loops whose codewords have sameprefix of length b − d , the loop of the last bit of the codewordsin each sub-loop becomes unimodal. On the other hand, weknow that if we remove the last bit of all the codewords in theMCL L ( b, d ) and eliminate the consecutive repetitions of thecreated codewords, by definition of the parent code, we willget an MCL L ( b − , d ) . Note that if no consecutive repetitionsare caused by removing the last bit of the codewords, wewould have |L ( b, d ) | = |L ( b − , d ) | . However, if there werecodewords to be eliminated due to the consecutive repetitions,we would get |L ( b, d ) | > |L ( b − , d ) | . So, by findingthe maximum possible reduction of codewords due to theconsecutive repetitions, we can find the maximum possibledifference between |L ( b, d ) | and |L ( b − , d ) | . To count this,observe that the MCL L ( b, d ) cannot have any consecutiverepetitions by definition. Therefore, if the bits removed fromtwo consecutive codewords in L ( b, d ) are the same, theycannot lead to consecutive repetitions in L ( b − , d ) . Also,removing the last bit of the consecutive codewords in L ( b, d ) which have different prefixes of length b − d does not leadin consecutive repetitions either since the codewords don’thave the same first b − d bits. As a result, the only way thatconsecutive repetitions might happen is when the last bits arenot the same and the codewords have same prefix of length b − d . Suppose we group the codewords into sub-loops thathave same prefix of length b − d . We know that the loop of thelast bits in each sub-loop is unimodal. Moreover, the maximumnumber of times that two consecutive bits in a unimodal loopcan be different is . Therefore, |L ( b, d ) | ≤ |L ( b − , d ) | + 2( number of sub-loops ) . (6)The number of sub-loops of codewords with same prefix oflength b − d for any possible MCL of C ( b, d ) is by definitionof a parent code equal to | C ( b − d, d ) | . Combining this with(6) and using the fact that M ( b − d, d ) ≤ |L ( b − d, d ) | , we get M ( b, d ) ≤ |L ( b, d ) | ≤ |L ( b − , d ) | + 2 |L ( b − d, d ) | . (7)When b ≤ d , we can conclude form in [6, Prop. 6] that M ( b, d ) = |L ( b, d ) | = 2 b . Therefore, M ( b, d ) = |L ( b, d ) | = 2 b. for b ≤ dd