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) Spatial reuse

Spatial reuse is the decision to let the same radio channel carry different conversations in different places at the same time, accepting a controlled amount of co-channel interference in exchange. The service area is divided into cells, each with a base station whose transmit power is deliberately limited so that its signal has decayed into the noise a few cell radii away. The total bandwidth is divided once, into $N$ channel groups, and those $N$ groups are assigned to the $N$ cells of a cluster; the cluster pattern then tiles the whole service area, so every cell in the same relative position uses the same group.

The reason this multiplies capacity is that the bandwidth is divided once but used many times. If the operator holds $B$ hertz and each user needs $B_u$ hertz, a single transmitter covering the whole territory serves $B/B_u$ users, full stop. Break the territory into $M$ cells with cluster size $N$ and each cell holds only $B/(N B_u)$ channels — fewer per cell — but there are $M$ cells, so the system serves $M/N$ times as many users as the single transmitter did, and the multiplier grows without bound as cells are made smaller. Capacity is thus set by the number of cells, not by the amount of spectrum. What limits the shrinking is interference, not spectrum: for hexagonal cells the co-channel reuse ratio is $D/R=\sqrt{3N}$, so a smaller cluster puts co-channel cells closer together and degrades the carrier-to-interference ratio, which for a path-loss exponent $n$ and six first-tier interferers goes as $(3N)^{n/2}/6$. The engineering compromise is to pick the smallest $N$ that still meets the required carrier-to-interference ratio — historically 7 for analogue systems, 3 or 1 for modern systems that tolerate interference by coding.

(b) OFDM and MIMO

Orthogonal frequency division multiplexing splits one wideband channel into many narrow subcarriers spaced exactly $1/T_s$ apart, where $T_s$ is the symbol duration, so that each subcarrier’s spectral nulls fall on all the others — they overlap yet remain separable, which is what the word orthogonal buys. A high-rate stream is demultiplexed across the subcarriers, so each carries a long, slow symbol; adding a cyclic prefix longer than the channel delay spread then makes inter-symbol interference from multipath disappear and turns frequency-selective fading into a set of independent flat-fading subchannels, each correctable by a single complex coefficient. LTE uses 15 kHz subcarrier spacing and implements the whole scheme with an inverse fast Fourier transform, which is why it is cheap.

Multiple-input, multiple-output transmission uses several antennas at both ends of the link. Because the paths between each transmit and each receive antenna fade almost independently, the receiver can invert the resulting channel matrix and separate signals that share the same time and frequency. Used for spatial multiplexing this multiplies throughput by roughly $\min(N_t,N_r)$ at no cost in bandwidth; used for diversity the same antennas instead carry redundant copies and lower the outage probability; used for beamforming they steer energy towards one user and away from others. Multipath, the enemy of the single-antenna link, becomes the resource that makes MIMO work, which is why OFDM and MIMO are paired in LTE: OFDM presents MIMO with the flat per-subcarrier channels its matrix algebra needs.

Given. The GSM multiplexing parameters of part (c) and the spectrum plan of parts (d) and (e), collected below.

Find. The peak (burst) bit rate on the carrier including control bits and the average data-only rate per user; the bandwidth allocated to each cell; and the number of FDMA users supported per cell and across the whole system.

Approach. Divide the frame into slots and remove the guard time to get the interval in which bits are actually radiated (peak rate); divide the data bits by the whole frame to get the long-run average per user; then divide the spectrum once by the cluster size, never by the cell count, and count channels.

1. Part (c) — find the length of one user’s time slot. The frame is shared equally among the eight users, so $$T_{\text{slot}} = \frac{T_f}{n} = \frac{4615\ \mu\text{s}}{8} = 576.875\ \mu\text{s}$$
2. Remove the guard time to get the burst duration. The guard time is dead air that lets the transmitter ramp up and absorbs propagation-delay differences between mobiles; no bits are carried in it, so $$T_b = T_{\text{slot}} - T_g = 576.875 - 30.5 = 546.375\ \mu\text{s}$$
3. Divide the burst by its duration to get the peak rate. All 148 bits, control included, are radiated inside $T_b$, and while a burst is on the air the carrier is being driven at its full instantaneous rate. Because the eight users share one carrier in time, this single figure is also the peak rate for all users together: $$R_{\text{peak}} = \frac{b}{T_b} = \frac{148}{546.375 \times 10^{-6}\ \text{s}} = \boxed{270.9\ \text{kbit}\,\text{s}^{-1}}$$

That figure is a useful sanity check on the whole calculation, because it reproduces the published GSM channel rate of 270.833 kbit/s; the 43 bit/s discrepancy is only the rounding of the true 8.25-bit guard period to the 30.5 $\mu$s quoted in the question. If the guard times are instead treated as part of the overhead rather than excluded, the frame-averaged gross rate is $8\times148/4615\ \mu\text{s} = 256.6$ kbit/s, which is the rate a link budget would use.

4. Average the data bits over the whole frame, per user. A given user gets one slot per frame and no more, so its data bits must be spread over the full 4615 $\mu$s, guard times and other users’ slots included: $$R_{\text{user}} = \frac{b_d}{T_f} = \frac{114}{4615 \times 10^{-6}\ \text{s}} = \boxed{24.7\ \text{kbit}\,\text{s}^{-1}}$$ which is the familiar GSM full-rate figure. Across eight users the carrier delivers $8 \times 24.70 = 197.6$ kbit/s of payload, so the payload efficiency of the burst is $114/148 = 77.0\%$.
5. Part (d) — divide the spectrum by the cluster size, not the cell count. The whole bandwidth is consumed exactly once per cluster, and the cluster tiles the territory, so $$B_{\text{cell}} = \frac{B}{N} = \frac{42\ \text{MHz}}{7} = \boxed{6\ \text{MHz per cell}}$$ The 56 cells form $56/7 = 8$ complete clusters; the cell count does not enter the per-cell allocation at all, and dividing 42 MHz by 56 is the classic error.
6. Part (e) — count the FDMA channels in one cell. Each user occupies a fixed 20 kHz slice for the duration of the call, so $$U_{\text{cell}} = \frac{B_{\text{cell}}}{B_u} = \frac{6 \times 10^{6}}{20 \times 10^{3}} = \boxed{300\ \text{users per cell}}$$
7. Scale to the whole system. Every cell carries its own 300 channels simultaneously, which is spatial reuse doing its work, so $$U_{\text{sys}} = U_{\text{cell}} \times M = 300 \times 56 = \boxed{16\,800\ \text{users}}$$ Equivalently, the 42 MHz supports $42\times10^{6}/20\times10^{3} = 2100$ distinct channels, of which each cell receives $2100/7 = 300$; reuse then multiplies that by the 56 cells.