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) — why cellular subdivision multiplies capacity. A radio channel is a shared resource only within the region in which it is audible. If one transmitter serves a whole city, every hertz of the licensed band is consumed city-wide: 42 MHz divided into 25 kHz channels yields 1680 conversations for the entire city, and no amount of extra base-station power changes that number. Cellular architecture attacks the problem geometrically rather than spectrally. The service area is tiled into small cells, each with a low-power base station whose signal decays as roughly the fourth power of distance in a built-up environment, so the same frequency can be re-used in another cell far enough away that the co-channel interference arriving there is small compared with the wanted signal. Capacity is therefore no longer set by the bandwidth alone but by the bandwidth multiplied by the number of times it is re-used across the coverage area.

The band is first divided among the N cells of a re-use cluster, and that cluster pattern is then repeated across the map. Example. With the 42 MHz and 25 kHz channelisation of part (c) and a cluster size of seven, each cell receives 6 MHz — 240 channels — and a 35-cell city contains five complete clusters, so 8400 users can be served simultaneously against the 1680 a single-transmitter system would allow: a five-fold gain, exactly the number of clusters. Shrinking the cells further multiplies capacity again, at the cost of more base stations, more handovers and more backhaul.

Part (b) — peak data rate of one LTE physical resource block.

Given.

Find. The peak user data rate, in bits per second, that one PRB can deliver once the reference symbols have been removed.

Approach. Count the resource elements in the block, subtract the reference symbols, convert the remaining symbols to bits through the constellation order, and divide by the block duration.

1. Count the resource elements in the block. A PRB is a rectangle of subcarriers by symbols, so $$N_{sym} = N_{sc}\,N_{s} = 12 \times 7 = 84 \text{ resource elements.}$$ Each resource element carries exactly one modulation symbol.
2. Remove the reference symbols. Four of the 84 elements are pilots used by the receiver to estimate the channel and carry no user data, so $$N_{data} = N_{sym} - N_{ref} = 84 - 4 = 80 \text{ data symbols.}$$ That is an overhead of 4.76 percent, which is the price of coherent demodulation.
3. Convert symbols to bits. A 64-QAM constellation has 64 distinct points, so each symbol resolves $$b = \log_2 M = \log_2 64 = 6 \text{ bits/symbol,}$$ and the block therefore carries $N_{data}\,b = 80 \times 6 = 480$ bits.
4. Divide by the block duration. The 480 bits are delivered in one PRB interval of $T = 0.5 \text{ ms} = 0.5 \times 10^{-3} \text{ s}$, so $$R_{PRB} = \frac{N_{data}\,b}{T} = \frac{480}{0.5 \times 10^{-3}} = \boxed{9.60 \times 10^{5}\ \text{bit/s} = 960\ \text{kbit/s}}$$ This is a peak figure: it assumes the highest modulation order is usable, that no coding redundancy is applied and that no control channel steals resource elements.

For comparison, had all 84 elements carried data the rate would have been 1.008 Mbit/s, so the pilots cost about 48 kbit/s per block. Aggregate rates are obtained by multiplying by the number of PRBs allocated: a 20 MHz LTE carrier offers 100 PRBs, giving 96 Mbit/s per 0.5 ms slot pair before coding.

Part (c) — simultaneous users in the cellular network.

Given.

Find. The number of users the whole system can carry simultaneously, and the number carried by one cell.

Approach. Convert the system bandwidth to a channel count, divide that count among the cells of one cluster to obtain the per-cell allocation, count the cells that tile the city, and multiply.

1. Convert the licensed band into FDM channels. Each user occupies 25 kHz including its guardband, so the band supports $$N_{ch} = \frac{B_{sys}}{B_{u}} = \frac{42 \times 10^{6}}{25 \times 10^{3}} = 1680 \text{ channels.}$$ This is the total pool available to the operator, not the number available anywhere in particular.
2. Share the pool across one re-use cluster. Co-channel interference forbids neighbouring cells from using the same frequency, so the pool is divided among the $N = 7$ cells of a cluster: $$N_{cell} = \frac{N_{ch}}{N} = \frac{1680}{7} = \boxed{240 \text{ channels (users) per cell}}$$ equivalently $B_{sys}/N = 42/7 = 6$ MHz of spectrum in every cell.
3. Count the cells tiling the city. The cells fit the city exactly and without overlap, so $$n_{cells} = \frac{A_{city}}{A_{cell}} = \frac{35}{1} = 35 \text{ cells},$$ which is $35/7 = 5$ complete re-use clusters.
4. Multiply to obtain system capacity. Every cell independently supports its own 240 conversations, so $$U = n_{cells} \times N_{cell} = 35 \times 240 = \boxed{8400 \text{ simultaneous users}}$$ or, equivalently, $5 \times 1680$: the licensed band is re-used once per cluster.

Part (d) — why multipath fading causes signal loss. In a built-up mobile environment the receiver almost never sees a single wave. The transmitted signal reaches the antenna by many routes — directly, and after reflection from buildings, vehicles and terrain, or diffraction over rooftops — and each route imposes its own delay, and therefore its own carrier phase. The antenna sums these copies as phasors. Where the copies happen to arrive in phase the sum is larger than any one of them; where they arrive in antiphase the sum can very nearly cancel, and the received power collapses by 20 to 30 dB over a distance of a fraction of a wavelength (a few centimetres at 2 GHz). This is small-scale or Rayleigh fading, and it is a loss of signal-to-noise ratio, not of transmitted power: raising the transmitter power does not remove the nulls, it only moves the depth at which they matter.

Two consequences follow for a mobile network. First, a moving handset sweeps through the interference pattern, so its received level fluctuates rapidly (at 100 km/h and 2 GHz the fading rate is of order 200 Hz), producing bursts of errors rather than isolated ones — which is exactly the behaviour that damages TCP in Question 2(c). Second, when the spread of path delays becomes comparable with the symbol period the copies of one symbol overlap the next, giving frequency-selective fading and intersymbol interference. The standard countermeasures follow directly from the mechanism: diversity in space (multiple antennas), in frequency (spread spectrum, OFDM subcarriers) or in time (interleaving plus coding), equalisation, and a link margin of some 10 to 20 dB reserved for fading.

Part (e) — frequency division duplexing. Duplexing is the arrangement that lets a subscriber transmit and receive at the same time on what the user perceives as one connection. In frequency division duplexing (FDD) the operator is allocated two separate blocks of spectrum, one for the uplink (handset to base station) and one for the downlink, permanently separated by a fixed duplex spacing. A handset transmits continuously on its assigned uplink channel and receives continuously on the paired downlink channel; a duplexer — a pair of sharp filters sharing one antenna — keeps the handset's own transmitter, which may be 80 dB stronger than the wanted incoming signal, out of its own receiver. In Canada the pairing and spacing are fixed by ISED's Standard Radio System Plans; the classic cellular example is the 850 MHz band, with uplink at 824–849 MHz, downlink at 869–894 MHz and a 45 MHz duplex spacing.

FDD suits cellular telephony because voice traffic is symmetric and continuous: both directions need the channel all the time, and a permanently open path minimises delay and avoids the switching transients of the alternative. The alternative, time division duplexing, sends both directions on one frequency in alternating time slots; it needs no paired spectrum and can be made asymmetric to favour the downlink, which is why it is preferred for bursty data services, but it requires network-wide slot synchronisation and adds a guard period for propagation. Note that duplexing answers a different question from multiple access: FDD says how one user's two directions are separated, while FDMA, TDMA or CDMA say how many users share the same direction.