Question 2 of 5: Cellular telephony

Question 2: Cellular telephony (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Why dividing space into cells increases the number of simultaneous users

Radio power falls off steeply with distance, roughly as $d^{-n}$ with the path loss exponent $n$ between 3 and 4 in a built-up service area. That decay is a nuisance for coverage but it is the entire basis of cellular capacity: two transmitters far enough apart can use the same frequency without either receiver noticing the other, because the wanted signal arrives so much stronger than the interferer. Spectrum is therefore not a quantity that is spent once; it is a quantity that can be spent again in every cell that is far enough away.

The design variable is the cluster size $N$, the number of cells over which the whole allocation is divided before the pattern repeats. Each cell receives $1/N$ of the spectrum, and cells using the same group are separated by the co-channel reuse distance

$$ \frac{D}{R} = \sqrt{3N} $$

where $R$ is the cell radius. Small $N$ gives each cell more channels but places co-channel cells closer together, so the carrier-to-interference ratio falls; large $N$ does the reverse. $N$ is chosen as the smallest value meeting the required $C/I$.

The paper's own system is the example. With 45 MHz and $N = 9$, each cell gets 5 MHz. The 63 cells form $63/9 = 7$ complete clusters, so the 45 MHz allocation is in use seven times over across the service area: the operator has manufactured $7 \times 45 = 315$ MHz of effective spectrum out of a 45 MHz licence. A single high-power transmitter covering the same area would offer that 45 MHz once, and serve one-seventh of the traffic. Halving the cell radius again quadruples the number of cells in the same area and quadruples capacity, at the price of four times as many sites and far more frequent handovers — which is exactly how operators added capacity to congested downtown areas.

(b) Why directional antennas are used at base stations

An omnidirectional base-station antenna radiates into — and receives interference from — the whole horizon. Replacing it with three 120° sector antennas on the same tower, each serving one third of the cell with its own channel group, changes the interference geometry rather than the transmitted power. In the omnidirectional case a cell is surrounded by six co-channel interferers in the first tier; with 120° sectoring, only two of those six lie within the beam of any given sector, and the rest are attenuated by the antenna's front-to-back and side-lobe rejection. The carrier-to-interference ratio improves by roughly a factor of three, and that margin can be spent on a smaller cluster size — typically $N = 7$ down to $N = 4$ or even 3 — which puts more channels in every cell.

Four further reasons matter in practice. A directional antenna concentrates radiated power, so its gain adds directly to the link budget in both directions and extends range or improves indoor penetration. Narrowing the azimuth beam reduces the spread of multipath arrival times, which lowers delay spread and therefore inter-symbol interference. Directional coverage lets an operator aim capacity where the traffic is — along a highway, at a stadium, into a downtown canyon — instead of wasting it on a lake or an escarpment. And sectorisation is the physical foundation of the later MIMO and beamforming techniques that push this idea to its limit.

The costs should be stated too: three sectors means three sets of radios and feeders, handovers now occur between sectors of the same site as well as between sites, and splitting one channel pool into three smaller pools loses trunking efficiency, since a small pool serves proportionally fewer users at the same blocking probability.

(c) Duplexing, with a TDD and an FDD example

Duplexing is the arrangement that lets one user communicate in both directions — uplink to the base station and downlink from it — on what the user perceives as one continuous, simultaneous connection. It is a distinct question from multiple access, which is how different users are kept apart; a system must answer both. The difficulty is that a handset's transmitter is some 10 orders of magnitude stronger at its own antenna than the signal it is trying to receive, so the two directions must be separated somehow.

Frequency division duplexing gives the two directions permanently separate frequency bands, split by a duplex spacing wide enough for a filter to reject the local transmitter. GSM 1800 is the textbook example: uplink occupies 1710–1785 MHz and downlink 1805–1880 MHz, so each direction has 75 MHz and the duplex spacing is 95 MHz. Both directions are continuously available, which suits symmetric traffic such as voice, and the handset needs a duplexer — a pair of sharp filters — to share one antenna between the two.

Time division duplexing gives the two directions the same frequency and alternates them in time. Bluetooth is the clearest example: the master transmits in even 625 µs slots and the addressed slave replies in the odd slot that follows, so the link is strictly half duplex but alternates fast enough that speech is unaffected. DECT cordless telephones, TD-LTE and the Wi-Fi contention model work the same way. TDD needs no duplexer, so the handset is cheaper; it can be re-tuned to asymmetric traffic simply by giving downlink more slots than uplink, which suits data; and because both directions use the same frequency, the channel is reciprocal, so a measurement made on the uplink is valid for shaping the downlink beam. Its costs are guard time between the two directions and tight synchronisation between neighbouring cells, since an unsynchronised neighbour transmits straight into your receive window.

(d) and (e) Bandwidth per cell, and maximum bandwidth per user

Given. The design starts from one system-wide spectrum licence divided among cells by a repeating reuse pattern, and then subdivided into FDMA channels within each cell.

Find. (d) the bandwidth allocated to each cell, and (e) the largest bandwidth that may be given to one FDMA user while still supporting 14,000 simultaneous users across the whole system.

Approach. Divide the total allocation by the cluster size to get the per-cell bandwidth, then divide that by the number of users each cell must carry to get the per-user channel width.

1. Part (d) — divide the allocation by the cluster size, not by the cell count. The whole allocation is shared out once per cluster, and the pattern then repeats; a cell therefore receives one of $N$ channel groups: $$ B_{\text{cell}} = \frac{B_{\text{total}}}{N} = \frac{45\ \text{MHz}}{9} = \boxed{5\ \text{MHz per cell}} $$
2. Confirm that the cell count is a distractor, and check what it is for. Dividing by 63 would give $45/63 = 0.714$ MHz, seven times too little. The 63 enters nowhere in part (d): its role is to say that the service area holds $63/9 = 7$ complete clusters, so the same 45 MHz is deployed seven times over, for $7 \times 45 = 315$ MHz of effective system bandwidth.
3. Part (e) — convert the system-wide user requirement into a per-cell requirement. Users are spread over all 63 cells, and each cell must carry its share: $$ n_{\text{users/cell}} = \frac{n_{\text{users}}}{n_{\text{cells}}} = \frac{14{,}000}{63} = 222.2\ \text{users per cell} $$
4. Divide the cell's spectrum among that many FDMA channels. In FDMA one simultaneous user occupies one channel for the duration of the call, so the channel width is the cell's bandwidth divided by the number of channels it must provide: $$ B_{\text{user}} = \frac{B_{\text{cell}}}{n_{\text{users/cell}}} = \frac{5 \times 10^{6}}{222.2} = \boxed{22.5\ \text{kHz per user}} $$ Equivalently, in one line from the given data, $B_{\text{user}} = B_{\text{total}}\,n_{\text{cells}} / (N\,n_{\text{users}}) = (45\times10^{6})(63)/(9 \times 14{,}000) = 22.5$ kHz.