Question 4 of 5: Cellular telephony (25 marks)

Question 4 — 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.

Given. The data for the two calculative sub-parts are collected below; parts (a), (d) and (e) are descriptive.

Find. Why cellular partitioning multiplies capacity, the peak data rate of one LTE physical resource block, the number of simultaneous calls system-wide and per cell, a summary of MIMO and its behaviour in fading, and an explanation of frequency division duplexing.

Part (a) — Why dividing space into cells increases capacity

A radio channel is exhausted by one user only within the region where that user's signal is strong enough to matter. Beyond that region the signal has decayed — path loss on a terrestrial link runs at roughly $1/d^{\,\alpha}$ with $\alpha$ between 3 and 4 in a built-up area — and the same frequency can carry a completely different conversation without either interfering with the other. Cellular systems are built on exactly that observation: instead of covering a city with one high-power transmitter, cover it with many low-power base stations, each serving a small cell, and re-use every frequency in every cluster of $N$ cells. Capacity then scales with the number of cells rather than with the bandwidth, and the bandwidth is a fixed, licensed and expensive resource whereas cells are simply capital equipment.

The price is paid in three currencies. Co-channel interference sets a floor on how tightly the pattern may be packed, since the signal-to-interference ratio in a cell depends on the re-use ratio $Q=D/R=\sqrt{3N}$ and not on transmit power. Mobility must now be managed, because a user crossing a cell boundary requires a handover. And the number of base stations, backhaul links and sites multiplies.

Example. Take the network of part (c). A single transmitter covering the whole 28 km² city with 42 MHz in 25 kHz channels supports $42\times10^{6}/25\times10^{3}=1680$ simultaneous calls in the entire city, and no more, however many masts are added. Split the same area into 56 cells of 0.5 km² with a seven-cell re-use cluster and each cell receives $42/7=6$ MHz, that is 240 channels — and, crucially, those 240 channels reappear in every one of the 56 cells. The city now supports $240\times56=13\,440$ simultaneous calls from the same 42 MHz, a factor of eight improvement, which is exactly the number of clusters that fit into the city ($56/7=8$).

Part (b) — Peak data rate of an LTE physical resource block

Approach. Count the resource elements in the block, remove the ones spent on channel estimation, multiply by the bits each modulation symbol carries, and divide by the block duration.

1. Count the resource elements in one PRB. A resource element is one subcarrier during one symbol period, so $$N_{\mathrm{RE}}=N_{\mathrm{sc}}\times N_{\mathrm{sym}}=12\times7=84\ \text{resource elements}.$$
2. Remove the reference symbols. Four of the 84 elements carry pilots for channel estimation and are unavailable to the user: $$N_{\text{data}}=84-4=80\ \text{resource elements}.$$
3. Convert constellation order to bits per symbol. A 64-QAM constellation has 64 distinguishable points, so each symbol carries $$b=\log_{2}M=\log_{2}64=6\ \text{bits}.$$
4. Find the payload of the block. Multiplying the usable elements by the bits each carries, $$N_{\text{bits}}=N_{\text{data}}\times b=80\times6=480\ \text{bits per PRB}.$$
5. Divide by the block duration to obtain a rate. The block occupies one 0.5 ms slot, so $$R=\frac{N_{\text{bits}}}{T_{\mathrm{PRB}}}=\frac{480\ \text{bits}}{0.5\times10^{-3}\ \text{s}}$$ $$\boxed{R=9.60\times10^{5}\ \text{bit/s}=960\ \text{kbit/s}}$$ For comparison, the gross rate before the pilots are deducted would be $84\times6/0.5\ \text{ms}=1.008\ \text{Mbit/s}$, so channel estimation costs about 4.8 % of the block — a modest price for the coherent demodulation that 64-QAM demands.

Part (c) — Simultaneous calls in the system and per cell

Approach. Divide the system bandwidth among the cells of one cluster, convert the per-cell bandwidth into channels, and then multiply by the number of cells in the city — because every cell re-uses its group of channels.

1. Count the cells covering the city. With cells fitting the area exactly and without overlap, $$n_{\text{cells}}=\frac{A_{\text{city}}}{A_{\text{cell}}}=\frac{28\ \text{km}^{2}}{0.5\ \text{km}^{2}}=56\ \text{cells}.$$
2. Convert the whole spectrum into channels. Each user occupies 25 kHz including its guardband, so the band as a whole contains $$n_{\text{ch,total}}=\frac{B_{\text{sys}}}{B_{u}}=\frac{42\times10^{6}\ \text{Hz}}{25\times10^{3}\ \text{Hz}}=1680\ \text{channels}.$$ These 1680 channels are the entire national resource for this network; the cellular structure is about to let each of them be used 8 times over.
3. Share the spectrum across one cluster. The defining rule of frequency re-use is that the whole band is divided among the $N$ cells of a cluster, and the pattern then repeats. Dividing by the cluster size, never by the total number of cells, $$B_{\text{cell}}=\frac{B_{\text{sys}}}{N}=\frac{42\ \text{MHz}}{7}=6\ \text{MHz per cell}.$$
4. Find the channels available in each cell. Either divide the per-cell bandwidth by the channel spacing, or divide the total channel count by the cluster size — the two must agree, and checking that they do is a free check on the work: $$n_{\text{ch,cell}}=\frac{B_{\text{cell}}}{B_{u}}=\frac{6\times10^{6}}{25\times10^{3}}=\frac{1680}{7}=240.$$ $$\boxed{240\ \text{simultaneous calls per cell}}$$
5. Scale up to the whole system. Every one of the 56 cells carries its own 240 conversations at the same time, because the re-use pattern guarantees that cells sharing a group are far enough apart: $$n_{\text{system}}=n_{\text{ch,cell}}\times n_{\text{cells}}=240\times56$$ $$\boxed{n_{\text{system}}=13\,440\ \text{simultaneous calls}}$$
6. Check the result against the single-transmitter case. Without cellular re-use the city would support only the 1680 channels of step 2. The gain is $$\frac{13\,440}{1680}=8=\frac{n_{\text{cells}}}{N}=\frac{56}{7},$$ the number of complete clusters tiling the city — which is precisely the physical meaning of the improvement and confirms the arithmetic.

For completeness, the geometry implied by these numbers: a hexagonal cell of area $A=1.5\sqrt{3}R^{2}=0.5\ \text{km}^{2}$ has radius $R=\sqrt{0.5/(1.5\sqrt{3})}=0.439\ \text{km}$, the re-use ratio is $Q=\sqrt{3N}=\sqrt{21}=4.58$, and co-channel cells are therefore $D=QR=2.01\ \text{km}$ apart.

Part (d) — MIMO and why it performs well in fading

What it is. MIMO — multiple input, multiple output — places $N_{t}$ antennas at the transmitter and $N_{r}$ at the receiver, so that the link is no longer a scalar channel but a matrix one, $\mathbf{y}=\mathbf{H}\mathbf{x}+\mathbf{n}$, where $\mathbf{H}$ is the $N_{r}\times N_{t}$ matrix of complex path gains. Having a matrix rather than a number buys three distinct things, and a good answer names all three:

• Spatial multiplexing. If $\mathbf{H}$ is well conditioned it has several strong singular values, and the receiver can separate several independent data streams sent simultaneously on the same frequency. Capacity grows as $C\approx\min(N_{t},N_{r})\log_{2}(1+\mathrm{SNR})$ — linearly in the number of antennas rather than logarithmically in power.
• Spatial diversity. The same information is carried over $N_{t}N_{r}$ largely independent paths, so the receiver can combine them.
• Array and beamforming gain. With channel knowledge the transmitter can precode so that the signals add coherently at the intended receiver and cancel elsewhere, raising received power and suppressing interference.

Why it thrives in fading. The answer is slightly paradoxical and that is what the marks are for. On a single-antenna link, Rayleigh fading is destructive because the received power occasionally drops into a deep null and the link fails; outage probability falls only as $1/\mathrm{SNR}$. With $N_{t}N_{r}$ paths that fade more or less independently, the probability that all of them are in a null at once is far smaller, and the error probability falls as $\mathrm{SNR}^{-N_{t}N_{r}}$ — the diversity order. More than that, the very thing that causes fading, a rich scattering environment, is what decorrelates the entries of $\mathbf{H}$ and makes it full rank; a clean line-of-sight channel gives a rank-one matrix and no multiplexing gain at all. MIMO therefore converts multipath from an impairment into the resource it exploits, which is why the technique is standard in LTE, 5G NR and every recent WiFi amendment.

Part (e) — Frequency division duplexing

Duplexing is the question of how a two-way conversation shares one radio link. Frequency division duplexing answers it by giving the uplink and the downlink permanently separate frequency bands, offset by a fixed duplex spacing large enough that the handset's own transmitter does not desensitise its receiver. Each channel is thus a pair of frequencies, and both directions operate continuously and simultaneously. A duplexer — a pair of sharp filters — lets the transmitter and receiver share a single antenna.

GSM-900 is the textbook illustration: mobiles transmit on 890–915 MHz, base stations on 935–960 MHz, a duplex spacing of 45 MHz across two 25 MHz bands. Note that the 25 kHz channel of part (c) is really a 25 kHz channel in each direction.

The engineering trade is worth stating. FDD gives genuine full duplex with low, constant latency and no need to synchronise base stations to one another, and it tolerates large cells because no guard time has to be reserved for propagation — which is why it dominates wide-area cellular. Against that, it demands paired spectrum, which regulators must allocate and which is scarce; the duplexer is bulky and lossy; and the split is fixed, so an asymmetric traffic mix of the kind data services generate leaves uplink capacity idle. Time division duplexing, which alternates the two directions in time on a single band, inverts every one of those properties: flexible and adjustable asymmetry and channel reciprocity that makes MIMO precoding easy, at the cost of guard periods, network-wide synchronisation and higher latency.