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.

Part (a) — spatial reuse of frequencies. Spatial reuse is the practice of using the same radio channel simultaneously at two locations that are far enough apart that path loss attenuates each transmitter's signal, at the other's receiver, to below the level that would interfere with it. It is what converts a fixed and legally scarce block of spectrum into a capacity that grows with the number of base stations rather than being fixed by the bandwidth.

The mechanism is a tiling. The service area is modelled as a lattice of hexagonal cells; the cells are grouped into clusters of $N$ cells; the whole spectrum is divided into $N$ channel groups and one group is assigned to each cell of the cluster, so that no two cells of a cluster share a channel; the cluster pattern is then repeated over the coverage area. Every channel group therefore appears once per cluster, and the number of times the whole spectrum is reused equals the number of clusters that fit the service area.

The separation this achieves is fixed by the geometry. For hexagonal cells of circumradius $R$ and cluster size $N$, the distance between the centres of two co-channel cells is

where $Q$ is the co-channel reuse ratio. Taking the numbers this paper supplies in part (d) as the example — a cell of area 1 km² and a cluster size of 7 — the hexagon's circumradius follows from $A = \tfrac{3\sqrt{3}}{2}R^{2}$, giving $R = 0.620$ km, and hence $D = 0.620\sqrt{21} = 2.84$ km with $Q = 4.583$. Two cells only 2.84 km apart therefore transmit on the identical 7 MHz of spectrum at the same instant. The capacity consequence is the point of the exercise: the 28 km² city holds four complete clusters, so the operator's 49 MHz is reused four times over and the system carries four times the traffic that a single high-power transmitter on the same licence could carry. Shrinking the cells (cell splitting) multiplies that gain again, which is why urban capacity is added by adding sites rather than by acquiring spectrum.

Part (b) — co-channel interference and where it is worst. Co-channel interference is the interference suffered by a receiver from transmitters in other cells that are using the same channel group, and it is the price paid for spatial reuse — unlike thermal noise it cannot be overcome by raising transmit power, because raising every cell's power raises the interference in proportion.

It is worst at the cell edge, on the boundary facing the nearest co-channel cell, because that is simultaneously the point of lowest wanted signal (the subscriber is at maximum range $R$ from the serving base station) and of highest unwanted signal (the subscriber is at the minimum possible distance $D-R$ from the nearest co-channel transmitter), so the signal-to-interference ratio takes its minimum there. With a path-loss exponent $n$ and six first-tier co-channel cells the worst-case ratio is approximately

which is the relation used to choose $N$: a larger cluster pushes $Q$ up and fixes an inadequate margin, at the cost of giving each cell fewer channels.

Part (c) — multipath propagation and signal loss. A transmitted signal reaches the receiver over several paths of different lengths — a direct ray plus reflections from buildings, ground and vehicles — and because the path-length differences translate into phase differences, the components can add destructively and produce a deep fade even though the total radiated power has not changed.

The example worth giving is how little movement it takes. At the 1.9 GHz PCS frequency the wavelength is $\lambda = c/f = 15.8$ cm, so a path-length difference of $\lambda/2 = 7.9$ cm inverts the phase of a reflected ray relative to the direct ray; a handset that moves only $\lambda/4 = 3.9$ cm can therefore travel from a fade to a peak, which is exactly the rapid Rayleigh fading heard as a flutter when a subscriber walks past a building. The same mechanism produces a second, distinct impairment: because the echoes are also delayed, the delay spread smears each symbol into its neighbour and causes intersymbol interference once the spread approaches a symbol period, which is why GSM specifies an adaptive equaliser and why LTE uses OFDM with a cyclic prefix.

Part (d) — how many simultaneous calls?

Given.

Find. The number of users who can be in a call at the same instant across the whole system, and the number per cell.

Approach. Divide the spectrum into channels, divide those channels among the $N$ cells of one cluster to obtain the per-cell figure, then multiply the per-cell figure by the number of cells covering the city.

1. Part (d) — count the cells covering the city. The cells tile the city without overlap, so $$n_{\text{cells}} = \frac{A_{\text{city}}}{A_{\text{cell}}} = \frac{28\ \text{km}^{2}}{1\ \text{km}^{2}} = 28\ \text{cells}$$ which is exactly four complete clusters of seven, $n_{\text{clusters}} = 28/7 = 4$.
2. Divide the spectrum into user channels. Each user occupies 25 kHz including its guardband, so the full licence supports $$n_{\text{ch}} = \frac{B_{\text{sys}}}{B_{u}} = \frac{49 \times 10^{6}\ \text{Hz}}{25 \times 10^{3}\ \text{Hz}} = 1960\ \text{channels}.$$ These 1960 channels are the entire pool; they are not all available in any one cell.
3. Share the pool over one cluster, not over the whole city. The seven cells of a cluster must use disjoint channel groups, so the pool is split seven ways: $$B_{\text{cell}} = \frac{B_{\text{sys}}}{N} = \frac{49\ \text{MHz}}{7} = 7\ \text{MHz}, \qquad n_{\text{cell}} = \frac{n_{\text{ch}}}{N} = \frac{1960}{7} = \boxed{280\ \text{channels per cell}}$$ and the same figure follows independently from the per-cell bandwidth, $7\ \text{MHz} / 25\ \text{kHz} = 280$, which confirms the division. Every cell in the city gets these 280 channels, whichever cluster it belongs to, so 280 users can be in a call simultaneously in any given cell.
4. Scale to the whole system. Each of the 28 cells carries its own 280 calls at the same time, because the co-channel cells are far enough apart to reuse the identical frequencies: $$n_{\text{sys}} = n_{\text{cell}} \times n_{\text{cells}} = 280 \times 28 = \boxed{7840\ \text{simultaneous users}}$$ Equivalently, four clusters each reuse the full 1960-channel pool once, $4 \times 1960 = 7840$, which is the same answer reached from the other direction.

Part (e) — peak bit rate of a GSM user.

Given. A TDM frame of duration $T_f = 4.615$ ms shared by $n = 8$ users; each user transmits a burst of $b = 148$ bits in its slot; a guard time of $T_g = 0.030$ ms separates one burst from the next.

Find. The peak bit rate at which a user's transmitter must run — that is, the instantaneous rate during the burst, not the average over the frame.

Approach. Divide the frame into eight equal slots, remove the guard time to obtain the interval in which bits are actually radiated, then divide the burst length by that interval.

1. Part (e) — find the slot duration. The frame is shared equally among the eight users, so $$T_{\text{slot}} = \frac{T_f}{n} = \frac{4.615\ \text{ms}}{8} = 0.576875\ \text{ms} = 576.875\ \mu\text{s}.$$
2. Remove the guard time. The guard interval exists so that bursts from mobiles at different ranges cannot overlap at the base station; no bits are sent during it, so the 148 bits must be squeezed into what is left: $$T_{\text{burst}} = T_{\text{slot}} - T_g = 576.875 - 30.000 = 546.875\ \mu\text{s}.$$
3. Divide burst bits by burst time. The peak rate is the rate sustained while the transmitter is keyed on: $$R_{\text{peak}} = \frac{b}{T_{\text{burst}}} = \frac{148\ \text{bits}}{546.875 \times 10^{-6}\ \text{s}} = \boxed{270.6\ \text{kbit/s}}$$ This is a free self-check, and it is the reason the question is set: the published GSM channel rate is 270.833 kbit/s, and the value obtained here differs from it by only 0.08 per cent — the small discrepancy being the paper's rounding of the true 30.5 microsecond (8.25-bit) guard period to 0.030 ms.
4. Contrast with the average rate, which is not what was asked. Averaged over the whole frame the same user delivers only $$R_{\text{avg}} = \frac{b}{T_f} = \frac{148}{4.615\ \text{ms}} = 32.07\ \text{kbit/s},$$ a factor of eight lower. The transmitter must nevertheless be designed for the 270.6 kbit/s figure, because that is the rate its modulator, power amplifier and channel filters actually see.