Question 1 of 5: Cellular Telephony

Question 1: 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. Radio spectrum is a fixed, licensed and expensive resource, so a cellular operator cannot give every subscriber a private frequency. Spatial reuse is the observation that a radio signal attenuates with distance — roughly as $d^{-n}$ with a path-loss exponent $n$ between 2 and 4 in real terrain — so the same carrier frequency can be used again at another location provided the two transmitters are far enough apart that the co-channel interference each causes at the other is below the receiver threshold. The service area is therefore divided into small cells, each served by a low-power base station, and the total channel set is divided into $N$ groups. Cells are gathered into clusters of $N$ cells; each cell in a cluster uses a different group, and the whole cluster pattern is then tiled across the coverage area.

The reuse factor $N$ cannot be arbitrary: for the standard hexagonal lattice it must satisfy $N = i^{2} + ij + j^{2}$ for non-negative integers $i$ and $j$, giving the familiar values 1, 3, 4, 7, 9, 12, 13 and so on. The co-channel reuse ratio then follows as $D/R = \sqrt{3N}$, where $R$ is the cell radius and $D$ the distance between co-channel cell centres, so a larger cluster buys a better signal-to-interference ratio at the cost of fewer channels per cell.

A concrete example: this question's own system carries 49 MHz over 70 cells with $N = 7$. Without reuse, 70 cells would each need their own slice and every cell would receive only 0.7 MHz. With sevenfold reuse each cell receives 7 MHz — a tenfold gain in per-cell capacity from geometry alone, no extra spectrum purchased. Reducing the cell radius and re-tiling the same pattern (cell splitting) multiplies capacity again, which is precisely how operators add capacity in dense downtown cores such as Vancouver or Toronto without acquiring new licences from ISED Canada.

Part (b) — Voice over an entirely packet-switched LTE network. GSM and UMTS kept a circuit-switched domain: a voice call reserved a dedicated 64 kbit/s-equivalent path end to end for its whole duration, and the mobile switching centre guaranteed the resource once the call was admitted. LTE removed that domain entirely. The Evolved Packet Core carries only IP packets, so voice ceases to be a bearer service the network natively understands and becomes just another application riding on the data plane.

The immediate consequence is that a bare LTE deployment cannot place a call by itself. Early networks therefore used circuit-switched fallback (CSFB): when a call arrives the handset is commanded back onto the legacy 2G/3G network for the duration of the call, which costs seconds of set-up delay and drops the data session to the older radio's speed. The engineered answer is Voice over LTE (VoLTE), in which an IMS core signals the call with SIP and the packets themselves ride a dedicated bearer with QCI 1, a guaranteed bit rate class carrying a 100 ms packet-delay budget and a $10^{-2}$ loss target. Robust header compression squeezes the 40-byte IP/UDP/RTP header down to a few bytes, and semi-persistent scheduling avoids re-granting the uplink every 20 ms speech frame.

The engineering impact is therefore mixed and worth stating plainly. Voice gains spectral efficiency, wideband AMR quality and instant interworking with any other IP service, and the operator no longer maintains two parallel core networks. Against that, delivered quality now depends on scheduler policy rather than on a reservation: jitter must be absorbed in a de-jitter buffer at the receiver, lost speech frames are concealed rather than retransmitted (retransmission would arrive too late to be played out), and emergency-call location and lawful-intercept functions have to be rebuilt in IMS rather than inherited from the circuit core.

Part (c) — MIMO wireless networks. A multiple-input multiple-output (MIMO) system places several antennas at both the transmitter and the receiver — $N_{t}$ and $N_{r}$ respectively — and treats the resulting $N_{r} \times N_{t}$ matrix channel $\mathbf{H}$ as the object to be exploited rather than as a nuisance. Because scatterers in a rich multipath environment give each antenna pair an effectively independent fading coefficient, the matrix has full rank and can be decomposed into up to $\min(N_{t}, N_{r})$ parallel spatial streams occupying the same frequency at the same time.

Three distinct features follow, and a well-designed system trades between them dynamically. Spatial multiplexing sends independent data streams on each eigenmode, so ergodic capacity grows roughly as $C \approx \min(N_{t},N_{r}) \, B \log_{2}(1 + \mathrm{SNR})$ — a linear rather than logarithmic return on antenna count, which is the headline result. Spatial diversity instead sends the same information redundantly across antennas, for example by an Alamouti space-time block code, raising the diversity order to $N_{t}N_{r}$ and flattening deep fades; this is the right choice at low SNR or high mobility. Beamforming weights the antenna elements in phase so that energy is steered toward the wanted user and nulled toward others, which raises received SNR and enables multi-user MIMO, where one base station serves several handsets on the same resource block.

MIMO is what carries 802.11n onward and every LTE release from the first: LTE mandates $2 \times 2$ in the downlink and supports $4 \times 4$ and $8 \times 8$, and massive MIMO in 5G extends the same mathematics to dozens of elements. The practical limits are that the antennas must be separated by roughly half a wavelength to decorrelate, that the transmitter needs channel state information to precode well, and that each chain costs a separate RF front end and its power.

Part (d) — Bandwidth allocated to each cell.

Given. The system parameters printed in the question are collected below.

Find. The bandwidth $B_{\text{cell}}$ available to each individual cell.

Approach. Divide the total bandwidth by the cluster size, because a cluster is the smallest group of cells that together consumes the entire spectrum once, after which the pattern repeats.

1. Identify how many distinct channel groups the spectrum must be cut into. The cluster size is $N = 7$, so the spectrum is divided into seven groups; each cell of a cluster gets exactly one group and no two adjacent cells share one.
2. Divide the total bandwidth among the groups. $$B_{\text{cell}} = \frac{B_{\text{tot}}}{N} = \frac{49 \times 10^{6}\ \text{Hz}}{7}$$ which evaluates to $$\boxed{B_{\text{cell}} = 7.0\ \text{MHz per cell}}$$
3. Check the geometry is consistent. With 70 cells and clusters of 7, the system is tiled by $70/7 = 10$ complete clusters. Each of the seven groups is therefore reused ten times over the coverage area, and the total deployed bandwidth is $10 \times 49 = 490$ MHz of effective capacity from a 49 MHz licence.

Part (e) — Maximum bandwidth per user under FDMA.

Given. The same system as part (d) — 49 MHz total, 70 cells, $N = 7$, hence $B_{\text{cell}} = 7$ MHz — now required to carry at least $U_{\text{tot}} = 21{,}000$ simultaneous users system-wide using frequency-division multiple access.

Find. The largest bandwidth $B_{\text{user}}$ that can be given to each user while still meeting the 21,000-user requirement.

Approach. Users are served by the cell they are in, so convert the system-wide requirement into a per-cell requirement, then split that cell's 7 MHz equally among its users. Under FDMA each user holds a distinct frequency slot for the whole call, so the per-cell bandwidth divides directly.

1. Distribute the users over the cells. Assuming a uniformly loaded system, $$U_{\text{cell}} = \frac{U_{\text{tot}}}{M} = \frac{21{,}000}{70} = 300 \ \text{users per cell.}$$ This is the step where the cell count $M = 70$ legitimately enters, in contrast to part (d).
2. Divide the cell's spectrum among its users. Each cell owns $B_{\text{cell}} = 7$ MHz, so $$B_{\text{user}} = \frac{B_{\text{cell}}}{U_{\text{cell}}} = \frac{7 \times 10^{6}\ \text{Hz}}{300} = 23{,}333\ \text{Hz}$$ and therefore $$\boxed{B_{\text{user}} \approx 23.3\ \text{kHz per user}}$$
3. Confirm the figure is an upper bound and a plausible one. Giving each user more than 23.3 kHz would leave fewer than 300 slots per cell and the system would fall short of 21,000 simultaneous users, so 23.3 kHz is the maximum, not merely a feasible choice. Multiplying back, $300 \times 23{,}333 = 7.0$ MHz recovers the cell allocation exactly. The number is also physically sensible: it sits between the 25 kHz of analogue AMPS and the 30 kHz of IS-136, and any real design would spend part of it on guard bands.