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 (5 marks). Spatial reuse is the observation that radio power falls off steeply with distance, so the same carrier frequency may be used again at another location provided the two users are far enough apart that each sees the other only as weak interference. A cellular system turns this into a design rule. The service area is divided into small cells, the available spectrum is divided into N channel groups, and those N groups are assigned to the N cells of a repeating cluster so that no two adjacent cells share a group. The whole cluster pattern then tiles the service area: every group reappears once per cluster, and the co-channel cells are separated by the reuse distance $D=R\sqrt{3N}$, where R is the hexagon radius. The system capacity is no longer the number of channels the spectrum contains, but that number multiplied by the number of clusters laid down.

The figure shows the pattern used in part (d) of this question, a cluster size of $N=13$. The example worth quoting is the one the question then computes: 39 MHz of spectrum at 25 kHz per call is only 1560 simultaneous calls if used once, but tiled over a 52-cell city as four clusters of thirteen it supports 6240 simultaneous calls — a fourfold gain purchased entirely by geography. Shrink the cells and the gain grows again, which is why urban microcells are a few hundred metres across.

Part (b) — multipath and signal loss (5 marks). In one sentence: the transmitted wave reaches the receiver by several paths of different length, so the copies arrive with different phases and add vectorially, and when they arrive close to anti-phase they cancel, producing a deep fade in which the received power drops by tens of decibels even though the transmitter power has not changed. A concrete example is a handset carried along a street lined with buildings: the direct ray and a ray reflected from the opposite wall differ in path length by a fraction of a metre, so at 1.8 GHz (wavelength about 167 mm) moving the handset a few centimetres swings the relative phase through 180° and the call quality collapses and recovers over a walking pace — the classic Rayleigh fading pattern. Multipath also spreads the arrival times of one symbol, so at high symbol rates the delayed copies fall into the next symbol interval and cause intersymbol interference, which is a loss of usable signal quite apart from the fading nulls.

Part (c) — MIMO (5 marks). Multiple-input, multiple-output transmission puts several antennas at the base station and several at the terminal and treats the resulting set of paths as a matrix channel rather than a single link. The receiver estimates that matrix from pilot symbols and inverts it, so the transmitter can send several independent data streams on the same frequency at the same time and have them separated at the far end by their distinct spatial signatures — spatial multiplexing. The benefit is twofold: capacity grows roughly in proportion to the smaller of the transmit and receive antenna counts without any extra spectrum or power, and, when the streams are used redundantly instead (diversity or beamforming), the very multipath that caused the fading of part (b) becomes the mechanism that makes the link robust, because it is unlikely that all antenna pairs are in a fade at once.

Given. For part (d): a service area of 52 km² tiled by cells of 1 km² each, a reuse cluster size of 13 cells, a total system bandwidth of 39 MHz, and FDM with 25 kHz per user including the guardband. For part (e): a GSM TDM frame of 4.615 ms shared by 8 users, a 148-bit burst per user per frame, and a guard time of 0.030 ms.

Find. The number of simultaneous calls the whole system supports and the number supported per cell (part d), and the peak bit rate of one GSM user (part e).

Approach. Divide the spectrum into channels, allocate those channels across one cluster rather than across the whole city, and multiply by the number of cells; then for the GSM slot, divide the frame into slots, subtract the guard time to obtain the interval over which the burst is actually transmitted, and divide the burst length by it.

1. Part (d) — count the channels the spectrum contains. With FDM, each simultaneous call occupies one 25 kHz channel including its guardband, so the total number of channels in the system bandwidth is $$N_{\text{ch}}=\frac{B_{\text{sys}}}{B_{u}}=\frac{39\times10^{6}\ \text{Hz}}{25\times10^{3}\ \text{Hz}}=1560\ \text{channels}.$$ These 1560 channels are the entire spectral resource; spatial reuse is what will let them be used more than once.
2. Divide the channels among the cells of ONE cluster. The defining property of a reuse pattern is that the whole band is shared out across the N cells of a cluster and then the pattern repeats. The bandwidth and the channel count per cell are therefore $$B_{\text{cell}}=\frac{B_{\text{sys}}}{N}=\frac{39\ \text{MHz}}{13}=3\ \text{MHz},\qquad N_{\text{cell}}=\frac{N_{\text{ch}}}{N}=\frac{1560}{13}=\boxed{120\ \text{simultaneous calls per cell}}.$$ Note carefully that the divisor is the cluster size 13, not the 52 cells in the city; dividing by the cell count would give 30 channels per cell and understate the allocation fourfold.
3. Count the cells and hence the clusters. The cells tile the city exactly, so $$M=\frac{A_{c}}{A_{\text{cell}}}=\frac{52\ \text{km}^{2}}{1\ \text{km}^{2}}=52\ \text{cells},\qquad \frac{M}{N}=\frac{52}{13}=4\ \text{complete clusters}.$$ The city is covered by exactly four repetitions of the thirteen-cell pattern, which is the reuse factor.
4. Multiply to obtain the system capacity. Every one of the 52 cells runs its own 120 channels simultaneously, so $$N_{\text{sys}}=N_{\text{cell}}\times M=120\times52=\boxed{6240\ \text{simultaneous calls in the system}}.$$ The same number arrives by the other route, $N_{\text{ch}}\times(M/N)=1560\times4=6240$, which is a useful arithmetic check: the whole band is reused once per cluster and there are four clusters.
5. Part (e) — find the slot duration. Eight users share the frame in TDM, so each user owns one slot of $$T_{s}=\frac{T_{f}}{n}=\frac{4.615\ \text{ms}}{8}=0.576875\ \text{ms}=576.875\ \mu\text{s}.$$ This is the interval the user owns, but it is not the interval over which bits flow.
6. Subtract the guard time to obtain the burst interval. The guard time is deliberately silent, protecting against timing-advance errors and the finite ramp-up and ramp-down of the transmitter, so the 148 bits must be squeezed into what remains: $$T_{b}=T_{s}-T_{g}=576.875-30.000=546.875\ \mu\text{s}.$$
7. Divide to obtain the peak bit rate. The peak (instantaneous) rate is the burst length divided by the burst interval: $$R_{\text{peak}}=\frac{L_{b}}{T_{b}}=\frac{148\ \text{bits}}{546.875\times10^{-6}\ \text{s}}=270\,629\ \text{bit/s}=\boxed{270.6\ \text{kbit/s}}.$$ This is a strong self-check: the published GSM channel rate is 270.833 kbit/s, and the 0.08 per cent discrepancy is only because the question rounds the true 30.46 µs guard period to 0.030 ms. A candidate who forgets to subtract the guard time gets 256.6 kbit/s and misses the standard's own figure.
8. Contrast the peak rate with the average rate. The user only transmits in one slot out of eight, so averaged over the whole frame the rate is $$R_{\text{avg}}=\frac{L_{b}}{T_{f}}=\frac{148}{4.615\times10^{-3}}=32.1\ \text{kbit/s},$$ about one eighth of the peak. Both figures are correct answers to different questions, and the distinction — peak measured over the burst, average measured over the frame — is where marks are most often lost.