FREQUENCY PLANNING & CHANNEL ASSIGNMENT STRATEGIES ECE 2526 - PowerPoint PPT Presentation

FREQUENCY PLANNING & CHANNEL ASSIGNMENT STRATEGIES ECE 2526 – WIRELESS & MOBILE COMMUNICATION Monday, 17 February 2020

CELL COVERAGE- CHOOSING A SHAPE FOR MODELLING The ideal cell shape should be able to: (a) cover the whole coverage area without leaving overlaps and gaps; (b) Support the weakest signal which occurs at the cell boundaries; and (c) Cover the maximum area within the mobile territory. Fig. 1.1 The circular geometry fails to satisfy these conditions. Regular polygons, namely, (i) equilateral triangle, (ii) square and (iii) hexagon can cover the entire area without any overlap and gaps and cover maximum area.

POLYGONS THAT FORM TESSELLATIONS Circles do not tessellate. Three regular polygons that always tessellate are: (a) Equilateral triangle (b) Square (c) Regular Hexagon 3

REGULAR HEXAGON Why is regular hexagon preferred? Idealized coverage area 1. For a given distance between the centre of a polygon and its farthest perimeter points, the hexagon has the largest area. 2. The hexagon closely approximates a circular radiation pattern which would occur for an omni-direction base station (BS). Where on the hexagon is the BS transmitter located? 1. Centre-excited cells employ Omnidirectional antennas. 2. Corner-excited cells employ directional antennas. Centre-excited Corner-excited Uses omni-directional antenna Uses directional antenna

REGULAR HEXAGON CALCULATIONS • Height of each of the equilateral triangle, R 3 𝑆 2 − 𝑆 2 2 = Τ ℎ = 2 𝑆 • Area of one triangle is: 𝐵 = 𝑆 3 4 𝑆 2 2 ℎ = • Area of the hexagon: 4 𝑆 2 × 6 = 3 3 3 2 𝑆 2 5

FREQUENCY REUSE Frequency reuse (or frequency planning), is a technique of reusing frequencies and channels within a communication system to improve capacity and spectral efficiency.

NUMBER OF CELLS IN A CLUSTER, N 1. In practice, the cluster size N cannot take on any value and is given only by the following equation: N = i 2 + ij + j 2 , i ≥ 0 , j ≥ 0 where i and j are integer numbers. 2. This means that N can be 1, 3, 4, 7, 9, 12,

TYPICAL CLUSTER SIZES C = 1, i = 1, j = 0 - Cluster size for CDMA C = 3, i = 1, j = 1 C = 4, i = 2, j = 0 C = 7, i = 2, j = 1 - Cluster sizes for 1 st Generation Systems C = 9, i = 3, j = 0 C = 12, i = 2, j = 2

WORKED EXAMPLE • Find the relationship between any two nearest co-channel cell distance D and the cluster size N using hexagonal pattern shown below.

SOLUTION (1) 2 𝑏 2 + 𝑐 2 = 𝑆 2 𝑆 𝑆 2 − 𝑏 = 2 a 𝑏 = 𝑆 3 b=R/2 4 𝑏 = 𝑆 a 3 2 R R 60 0 2𝑏 = 𝑆 3

SOLUTION (2) The normalized co-channel cell distance D n can be calculated by traveling ’ i ’ cells in one direction and then traveling ’j’ cells in anticlockwise 120 o of the primary direction. Using law of vector addition, 2 = 𝑗 2 𝑑𝑝𝑡 2 30 𝑝 + 𝑗 + 𝑘𝑡𝑗𝑜(30 𝑝 2 = 𝑗 2 + 𝑗𝑘 + 𝑘 2 = N 𝐸 𝑜 or 𝐸 𝑜 = 𝑂 Multiplying the actual distance Multiplying the actual distance between two cells 𝑆 = 3𝑆 , we get 𝐸 𝑜 = 3𝑂 R

FREQUENCY REUSE PATTERNS IN GSM 1. The frequencies used in GSM radio frequency planning are divided among different frequency groups. 2. Common re-use patterns are: a) A cluster of 4 Base Stations with 12 ARFCNs commonly referred to as (4/12) b) A Cluster of 3 base Stations with 9 ARFCNs (Commonly referred to as 3/9)

4/12 GSM FREQUENCY REUSE PATTERN • 4/12 refers to 12 ARFCNs offered on a cluster of 4 base stations. • The available GSM network frequencies are divided into 12 ARFCNs across 4 Base Station sites (cluster)

3/9 FREQUENCY REUSE PATTERN • 3/9 refers to 9 ARFCNs offered on a cluster of 3 base stations. • The available GSM network frequencies are divided into 9 ARFCNs across a cluster of 3 Base Station sites (cells).

ADAPTATION FOR FUTURE GROWTH • Future growth in demand for services generally require a decrease in cell sizes. • Macro-cellular 1 - 30 km · • Micro-cellular 200 - 2000 m · • Pico-cellular 4 - 200 meter

CHANNEL ASSIGNMENT STRATEGIES 1. A common problem in frequency palling is that of allocation frequencies/channels to the various cells. 2. Several channels allocation schemes have been developed and this area still continues which can be classified as i. Fixed Channel Allocation (FCA), ii. Dynamic Channel Allocation (DCA), iii. Hybrid Channel Allocation (HCA).

FIXED CHANNEL ASSIGNMENT (FCA) 1. Fixed Channel Assignment (FCS) is the earliest channel assignment strategy. 2. Each cell is allocated a fixed number of voice channels. 3. Any communication to/from the cell can only be made with the designated unused channels of that particular cell. 4. If all the channels are occupied, then the call is blocked. 5. Later this was improved to allow borrowing of channels from adjacent cell if all of its own designated channels were occupied. 6. This was named as borrowing strategy . In such cases the MSC supervises the borrowing process and ensures that none of the calls in progress are interrupted.

SIMPLE CHANNEL BORROWING (CB) SCHEME 1. In Channel Borrowing (CB) schemes , cell (acceptor cell) that has used all its nominal channels borrows free channels from its neighbouring cell (donor cell) to accommodate new calls. 2. Borrowing is done from an adjacent cell which has largest number of free channels, i.e borrowing from the richest . 3. A search algorithm is used to select the first free channel found. 4. The borrowed channel is returned when channel becomes free in the basic algorithm with reassignment. 5. To be available for borrowing, the channel must not interfere with existing calls.

DYNAMIC CHANNEL ASSIGNMENT (DCA) 1. In Dynamic Channel Assignment (DCA), all channels are kept in a central pool and are assigned dynamically to new calls as they arrive in the system by the MSC. 2. After each call is completed, the channel is returned to the central pool. 3. DCA schemes can be: (i) centralized, or (ii) distributed 4. The centralized DCA scheme involves a single controller (MSC) selecting a channel for each cell. 5. The distributed DCA scheme involves a number of controllers(MSCs) scattered across the network.

CENTRALIZED DYNAMIC CHANNEL ASSIGNMENT (DCA) MSC RAM stores: • Status of all cells-signal strength 1. Using Dynamic Channel Assignment (DCA), the • Status of all Channels-occupancy MSC monitors all cells and all channels. 2. Each time a call request is made, serving BS requests a channel from the MSC. 3. MSC runs an DCA assignment algorithm which assigns a channel only if it is not used and if it will not cause co‐channel interference with any cell in range. 4. Dynamic Channel Assignment(DCA) algorithm provides higher capacity(less blocking). 5. But, it requires higher computational power since the MSC collects real time data of channel occupancy, traffic distribution, and radio signal strengths.

EXAMPLE 2 A total of 33 MHz bandwidth is allocated to a Frequency Division Duplex (FDD) cellular system with two 25 KHz simplex channels to provide full duplex voice and control channels. (i) Compute the number of channels available per cell if the system uses (i) 4 cell, (ii) 7 cell, (iii) 8 cell reuse technique. Assume 1 MHz of spectrum is allocated to control channels. (ii) Give a distribution of voice and control channels.

SOLUTION (1) The Question States: One duplex channel = 2 x 25 = 50 kHz of spectrum. Hence the total available duplex channels are = 33 MHz / 50 kHz = 660 in number. Among these channels, 1 MHz / 50 kHz = 20 channels are kept as control channels. (a) For N = 4 First check that it is a valid cluster number N = i 2 + ij + j 2 , i ≥ 0 , j ≥ 0 for N = 4, i = 2 and j = 0 or j = 2 and i = 0 VALID! Total channels per cell = 660/4 = 165. Among these, voice channels are 640/4 = 160 Control channels are 20/4 = 5 in number.

SOLUTION (2) (b) For N = 7, (i) First check that it is a valid cluster number N = i 2 + ij + j 2 , i ≥ 0 , j ≥ 0 for N = 7, i = 2 and j = 1 or j = 2 and i = 1 VALID! (ii) Total channels per cell are 660/7 ≈ 94. Therefore, we have to make assumptions to get the solution. (iii) From the results in (a), a total of 20 control channels and a total of 640 voice channels are kept. (iv) Therefore, 6 cells can use 3 control channels and the rest (one) can use 2 control channels each. (v) On the other hand, 5 cells can use 92 voice channels and the rest two can use 90 voice channels each. (vi) Thus the total solution for this assumption is: 6 x 3 + 1 x 2 = 20 control channels, and, 5 x 92 + 2 x 90 = 640 voice channels. (vii) This is one solution, there might exist other solutions based on other assumptions.

SOLUTION (3) (c) The option N = 8 is not a valid number of cells in a cluster since it cannot satisfy equation N = 𝑗 2 + ij + 𝑘 2 for i, j ≥ 0