FDM-TDMA Cellular Networks: Principles and Inter-cell Hand-off Procedure

شبکه‌های بی‌سیم (628-40)شبکه‌های سلولی FDM-TDMA نیمسال دوّم 98-97 افشین همّتیار دانشکده مهندسی کامپیوتر

Introduction • The FDM-TDMA technique for allocating spectrum and time resources to calls is the most classical one, and systems based on this technique carry a substantial majority of the cellular telephony traffic around the world. • Many of the basic techniques of cellular telephony emerged from the design of such systems; for example, techniques such as frequency reuse management, cell sectorization, power control, and handover management.

Overview Inter-cell hand-off procedure • Spatial reuse: • partitioning the available FDMcarriers into reuse groups • Allocating reuse groups to cells (acceptable co-channel interference.) • Co-channel interference constraint  D/R ratio constraint • Analysis based on Signal-to-Interference Ratio (SIR). • Power attenuation model including path loss and shadowing. • Effects of partitioning on spectrum efficiency. • Various channel allocation strategies • Call blocking analysis • Inter-cell handover mechanism. • Handover blocking Analysis. Inter-cell handover

Principles of FDM-TDMA Cellular Systems (1) GSM multiple access scheme: FDMA + TDMA • GSM: A commercial implementation of FDM-TDMA sytem. • The system bandwidth, Wsystem, is partitioned into several non-overlapping FDM channels, each of which is then digitally modulated, and then time slotted to yield FDM-TDMA channels. • A guard band is left vacant at either end of an operator’s spectrum allocation to prevent power from one operator’s system from interfering with another system. • In the GSM system the FDM channel spacing is 200 KHz. After digital modulation, each such channel is time slotted to provide 8 TDMA channels, each of which can carry one direction of a digitized voice call. uplink spectrum

Principles of FDM-TDMA Cellular Systems (2) TDD and FDD • Each voice call has two directions, and hence for each call we need two links to be established, one from the MS to the BS, and one from the BS to the MS. • The way these two links are established is called the duplexing technique. • In FDM-TDMA systems the common duplexing mechanism employed is Frequency Division Duplexing(FDD);that is, two separate FDM carriers are used to carry the two directions of a call. • The FDM channels fujand fdjare paired, as are the channels fukand fdk.

Principles of FDM-TDMA Cellular Systems (3) • Let us denote the FDM channel, bandwidth by W, the number of FDM channels in the system bandwidth by C , and the number of traffic carrying TDM slots per FDM channel by s. • The number of calls that can be carried simultaneously is: N = C × s • For example, if the operator leases a Wsystem of 5 MHz, and the number of FDM channels is 24, and s is 8, then the system can be used to carry 24×8 = 192 simultaneous calls.

Spatial Reuse (1) • Assume the system may be set up as shown below. • We observe that a BS should to serve the MSs at the periphery of the coverage area, in order to ensure that the SNR exceeds the minimum required for the desired bit error rate. • MSs at the periphery should communicate with the BS at high transmission power. Such MSs will quickly drain their batteries, and will also cause interference to the systems in neighboring coverage areas. • Also, the maximum number of users that can be simultaneously served in this simple system is N.

Spatial Reuse (2) • Now consider the idea of spatial reuse as shown below. • The MS-BS communication can be done using smaller powers. • Suppose that, by exploiting spatial reuse, each channel could be reused, say, five times in the same coverage area; then we would have effectively multiplied the number of calls that can be simultaneously handled by a factor of 5. • The problem is that if an MS is not close to any BS, in order to serve it from some BS, on any channel, a large transmission power will need to be used. This will cause co-channel interference. Co-channel Interference

Spatial Reuse (3) e.g.: Network layout and spatial reuse pattern • In order to address this problem, the cellular FDM-TDMA approach is to tessellate the coverage area into cells, each of which has a BS. • The set of FDM carriers is partitioned into subsets called reuse groups. • These channel groups are then assigned to the cells in such a way that cells with the same group of channels (called co-channel cells) are not close together. • How close co-channel cells can be depends on the SINR required for reliable communication. a centralized downlink OFDMA scenario in a multicellular network enhanced with six fixed relays per cell is considered. The proposed scheme considers efficient use of subcarriers via opportunistic spatial reuse within the same cell. Read More: Article: An Overview of Radio Resource Management in Relay-Enhanced OFDMA-Based Networks.

Trunking (1) Concept of Channel Reuse in a 7-Cell Cluster What is Blocking? • Trunking: serving m users by n servers when m>n • If the arrival rate of new calls into the system is λ per second, and the mean holding time of a call is h seconds, then for this system, the load is: ρ = λh • The blocking probability B(ρ,n) denotes the probability of blocking in a system with a load of ρ Erlangs and n servers. • The number of servers that would be required to obtain a specified blocking probability for a given traffic intensity can be obtained from an Erlangtable (next slide). • We know that, if the SINR required is large, then the co-channel cells will need to be kept far apart. This will require more channel reuse groups, and hence fewer channels per reuse group. • On the other hand, If the reuse groups have only a small number of carriers in them, then trunking efficiency is lost.

Trunking (2)

Trunking (3) • For a fixed probability of blocking, ɛ, let ρɛ(n) denote the Erlangs that can be carried when the number of servers is n, then B(ρɛ(n), n) = ɛ. • Now define gɛ(n) = ρɛ(n)/n , that is the Erlangsper server that can be offered, when the number of servers is n and the target blocking probability is ɛ. n = 1 then gɛ(1) = ɛ/1−ɛ << 1 , n  ∞ then gɛ(n)  1 • Thus, it is beneficial to not partition the set of carriers into small groups (which reduces trunking efficiency).

Tradeoff between SINR and Trunking Efficiency • A larger target SINR results in smaller reuse groups, which results in lower trunking efficiency. • There is a trade-off between keeping the SINR above a required threshold and keeping the trunking efficiency high. • The SINR requirements, the spatial reuse, and the system efficiency are intimately linked, and some analysis is required to evaluate the trade-offs.

SINR Analysis (1) • We depict uplink (right panel) and downlink (left panel) co-channel interference in a configuration in which a channel is reused at the five BSs shown. • The circular boundaries (with radius R) indicate the coverage of each BS. • The distance between the centers of each of the outer BSs and the one in the middle is D. • A large D/R ratio will be required if the co-channel interference has to be kept very small.

SINR Analysis (2) • The MS whose signal performance is being analyzed is considered to be in the most unfavorable position (at the periphery of its BS’s coverage area). • The interferers are also assumed to be in the most unfavorable position, as close as possible to the receiver of the desired transmission. • Let H denote the channel power gain between the transmitter of the desired signal and its receiver, and let Hidenote the power gain from the i-th co-channel interferer to the receiver of the desired signal. • Assuming NIinterferers, SINR at the receiver is given by:

SINR Analysis (3) • Assuming the noise power is much less than the received signal power, we neglect the first term in the denominator. • Let d be the distance between the transmitter and its receiver, and dithe distance between the i-th interfering transmitter and the receiver. Then: Here the ξ, ξ0, and ξi (1≤i≤NI), are i.i.d. normally distributed, zero mean, and with variances σ2/2; thus, the log-normal shadowing standard deviation on any path is σ dB. Hence, we can write the SINR as:

SINR Analysis (4) • In the numerator we have a log-normally distributed random variable of the form 10−1/10Q, where Q has units of dB. • In the denominator we have a sum of NIlog-normally distributed random variables of the form 10−1/10Qi. • We have: • The ratio of two independent log-normal random variables is obviously log-normal, thus: Q and QIarenormally distributed with means mand mIand variances v and vI respectively.

SINR Analysis (5) • For an outage probability (Pr((QI −Q)<γ)<ɛ ) there is a τɛ, such that we need to ensure that: • The Fenton-Wilkinson analysis yields Where α= ln10/10 ≈ 0.23026

SINR Analysis (6) • We thus get the requirement: • Above inequality places a constraint on the ratio between D, the distance between co-channel cells, and R, the cell radius.

SINR Analysis Numerical Example • First consider v= 0, that is, there is no log-normal shadowing, just path loss and η= 4. The D/R constraint reduces to: • For NI= 6, we find: • Now consider shadow fading with standard deviation of 8 dB, and outage probability of 0.01, we find: • Conclusion: Shadow fading has a significant effect on D/R ratio.

Hexagonal Cell Layout • It is not practical to use the cell configuration as shown (slide 14), because of this leaves large portions of the service area uncovered. • So the service area is tessellated with cells. • It is convenient to take the cells to be hexagons of equal size. • Set of FDM carriers is partitioned into disjoint sets in such a way that co-channel cells respect the D/R ratio. Cell Shape

Co-channel Cell Groups (1) • Tessellation of the coverage area by hexagonal cells:

Co-channel Cell Groups (2) • The FDM channels are partitioned into reuse groups. • One of these groups is assigned to Cell “0” which will be our reference cell in the following discussion. • Which other cells in the layout should use the same group of carriers? • A coordinate system with axes inclined at 60◦ to each other, as shown by the axes u and v in the figure (previous slide), is needed. • A third axis, walso pass through the center of the reference cell. • Moving a cell width, C, along any of the axes takes us to the center of a neighboring cell. Thus C is our unit length on each axis.

Co-channel Cell Groups (3) • Moving i units along the u axis and then j units along the v axis, brings us to the cell “1”. • Euclidean distance between the centers is: • Fixing i = 3 and j = 2, we can identify cells “2”, “3”, “4”, “5”, and “6”. These are the co-channel cells in relation to cell “0”.

Co-channel Cell Groups (4) Same Cells Adjacent Cells R j.d d = 2Rcos60o = R√3 i.d D = d√(i2 - 2ijcos120o + j2) = d√(i2 + ij + j2) = R√3 √(i2 + ij + j2)

Co-channel Cell Groups (5) • Carrying out the same procedure and use the same(i, j), for Cell “1” , then we will obtain Cells “2”, “0”, and “6”, and three other cells, above and to the right of Cell “1”. • Starting from any element of this subset yields the same set of cells. • Starting from one of the cells adjacent to Cell “0”, and use the same (i, j), results a subset of cells that is disjoint from the previous one.

Calculating Nreuse(1) • For each cell in the large dashed hexagon with Cell “0” at its center, there is a different subset of hexagons. • All of these subsets (19 for (i,j)=(3,2)) are mutually disjoint and together they form a partition of the tessellation. • We can call each of these subsets of cells a co-channel group. Two different channel groups: 1) Cells 0, 1, 2, 3, 4, 5, and 6. 2) Cells A0, A1, A2, A3, A4, A5, and A6.

Calculating Nreuse(2) • The number of co-channel groups, Nreuse, depends on choice of (i,j). • With (i,j)=(1,0), ther is only one co-channel group. • The area of the large dashed hexagon is (√3/2)D2 (recall C is our unit length on each axis). • In this large hexagon, there are as many co-channel groups as the number of cells. • Thus in a large area, A, the number of cells in a co-channel group is A/(√3/2 D2). • Recall the area of a cell is √3/2, thus the total number of cells is A/(√3/2) and the number of co-channel groups is D2.

Calculating Nreuse(3) • Number of co-channel groups is the number of groups into which the set of FDM carriers is partitioned, so: • For a given (i, j):

Calculating Nreuse(4) • The values of Nreuseand D(i, j)/R for different values of (i, j) are:

D/R Ratio (1) • Accounting only for path loss, and taking the transmitter powers, in the worst-case transmitter-receiver configurations, to be equal, the following is the general expression for the SIR: R is the cell radius, and NIis the number of first tier interferers. Di, 1 ≤ i ≤ NI, is the distance of the interferer from the receiver in the reference cell.

D/R Ratio (2) • For forward channel (downlink) worst-case situation, with approximated distances between the interferers and the receiver, we have:. Nreuse= 9  D/R = 5.20, and η = 4, then Ψ= 95.09 = 19.78 dB

D/R Ratio (3) • For reverse channel (uplink) worst-case situation we have: Nreuse= 9  D/R = 5.20, and η = 4, then Ψ= 51.86 = 17.14 dB • For the same D/R ratio, the performance of uplink is more than 2.5 dB worse than downlink.

Sectorization (1) cell sectorizationusing three-120° directional antennas • There were six first tier interferers at a receiver in considered cases. • The number of interferers can be reducedwhen directional antennas are used in the BSs. • This is achieved by a technique called sectorization.

Sectorization (2) • Suppose each cell is divided into three 120◦ sectors, each with a directional antenna whose angular coverage is designed to coincide with the angular spread of the sector. • So an MS in a given sector of a cell is served by the antenna in that sector. • The channels are reused only in the corresponding sectors of the cell reuse groups (carrier f).

Sectorization (3) • Worst-case configurations of first tier downlink interferers (left) and uplink interferers (right) are shown:

Sectorization (4) • Sectorization advantage (left side): • MS in Cell “0” sees only two first tier interferers, the two BSs in Cells “2” and “3.” • Sectors in Cells “4”, “5”, and “6” could be using the same channel, but their antenna main lobe is not visible to the MS in Cell “0”, thus SIR is given by: • Taking η = 4, Nreuse= 9 ( D/R = 5.20) we find that: Ψ= 489.13 = 26.89 dB, (7.1 dB improvement)

Sectorization (5) • Sectorization advantage (right side): • The SIR is given by: • Taking η = 4, and Nreuse= 9 ( D/R = 5.20), we find that: Ψ= 365.58 = 25.63 dB (8.5 dB improvement) • Sectorization disadvantage: it implies smaller sets of channels in each sector, thus reducing the trunkingefficiency.

Spectrum Efficiency (1) • We consider the simplest approach of partitioning the C carriers into Nreuse subsets. • Each subset of carriers is partitioned into K sets, each of which is allocated to the same sector in all the cells in a reuse group of cells. • Each slot in each carrier in a sector can carry one call. • Assuming a call that is initiated in a sector stays in the same sector for its entire duration (there are no handovers in the system). • For a target blocking probability of ∈, the number of Erlangs that can be offered to a cell is given by: The first term is the number of Erlangs per slot in a sector.

Spectrum Efficiency (2) • Erlangcapacity of the system, denoted by Λ, is given by: where A denote the coverage area of the system, and a denote the area of a cell. • Spectrum efficiency of the system is the Erlang capacity per unit area per Hz of system bandwidth, and denote this by ν:

Spectrum Efficiency (3) • The term sC/(Wsystem) is fixed for a given system bandwidth, and depends on the FDM-TDMA modulation scheme being used. • The term decreases with increasing Nreuse or K, but we need to set Nreuse and K so that the SIR requirements are met while keeping this trunking efficiency term as large as possible. • It also increases with Wsystem, but having leased a certain amount of the spectrum, the operator will want to work within this leased amount. • The Erlang capacity of the system can be increased by reducing the cell size.

Cell size decrease issues • As the cell size decreases, we need to consider handovers, and the handover rate increases with decreasing cell size. This will impact the Erlangcapacity, as resources need to be reserved for handovers. • The signaling load increases due to the increased handover rate. This means that higher capacity call handling systems need to be installed. • Reducing cell size requires the installation of more base stations, which can be expensive. The design of any given system will have to balance these trade-offs.

Channel allocation Fixed Channel Assignment • Apart from reducing the cell size, another way to increase the efficiency is to improve the channel utilization. • Uniform fixed assignment of the FDM carriers: it is possible that channels are idle in one cell, whereas another cell is overloaded. • Trunkingefficiency can be improved if the channels are viewed as being in various common pools, from which allocations are made as needed. • Dynamic channel allocation must respect the co-channel SIR constraints as the channels are allocated, released, and re-allocated to various cells.

Reuse Constraint Graph (1) • Simple model for designing and analyzing dynamic channel allocation strategies is to specify pairwise reuse constraints. • This model specify which pairs of cells cannot use the same FDM carrier at the same time given a set of cells. • In the following figure, a linear array of 10 cells (top), and two sets of pairwise reuse constraints (middle and bottom), are shown as constraint graphs.

Reuse Constraint Graph (2) An example of the hypergraph modeling Illustration of the hypergraph interference model • In a constraint graph, each cell is represented by a vertex. • There is an edge between two vertices if an FDM carrier cannot be simultaneously used in both of the corresponding cells. • Constraint graph in the middle: only constrains neighboring cells from reusing the same channel; channels can be simultaneously used in alternate cells. • Constraint graph at the bottom: permits a channel to be reused only in cells that are separated by at least two other cells. • In general, representation of reuse constraints by pairwise constraints is conservative. • Hyper-graphs: a generalization of graphs in which edges are subsets of nodes with cardinality greater than two. Read More: Radio Resource Allocation for Device-to-Device Underlay Communication Using Hypergraph Theory.

Reuse Constraint Graph (3) • Let B = {1, 2, . . . ,N} denote the set of cells. • Let (B, C) denote the constraint graph with C being the edge set (for i ∈ B and j ∈ Band (i, j) ∈ C the same carrier cannot be used in Cell i and Cell j simultaneously). • Constraint graph is undirected; that is (i, j) ∈ C if and only if (j, i) ∈ C. • Let F = {f1, f2, . . . , fM} be the set of FDM carriers that need to be assigned to the cells. • Let xjdenote the number of carriers required in Cell j, 1 ≤ j ≤ N. • The vector x = (x1, x2, . . . , xN) is feasible if there exists an allocation of xkcarriers to Cell k such that the reuse constraints are respected.

Reuse Constraint Graph (4) • Define X = {x : x feasible} • According to some standard concepts from graph theory, a cliqueof (B, C) is a fully connected sub-graph. • So a carrier can only be used in exactly one of the cells that form a clique. • A maximal clique is one that is not contained in any other clique. • We refer to maximal cliques also as cliques. • So in the bottom diagram the cliques are {1, 2, 3}, {2, 3, 4}, and so on.

Feasible Carrier Requirements (1) • Let Q be the number of cliques in (B,C). • Consider the Q×N matrix A with: • A necessary condition for x∈ X is: Mis the number of carriers, and 1is the Q × 1 vector of 1s. • The expression on the left of above inequality is the number of carriers needed in Clique i in order to achieve the carrier allocation given by x.

Feasible Carrier Requirements (2) • Having ,the suffix CPA expands to Clique Packing Allocation. • XCPA is a convenient characterization of X. • Since every carrier allocation must satisfy the clique constraints, we see that . • In general, X is a strict subset of XCPA. It can be that • but . • If constraint graph shown in next slide is a sub-graph of a constraint graph, then .

Feasible Carrier Requirements (3) • A pentagon reuse constraint graph for five nodes is shown on the left. • With M=2, the vector x=(1,1,1,1,1) satisfies the clique constraints, but there is no feasible allocation of carriers to cells, as seen in the diagram on the right.

FDM-TDMA Cellular Networks: Principles and Inter-cell Hand-off Procedure