Question 1 of 5: Cellular Telephony — OFDM Symbols, Resource Blocks and Frequency Re-use

Question 1: Cellular Telephony — OFDM Symbols, Resource Blocks and Frequency Re-use (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) — identifying the three quantities. The expression is one OFDM symbol written in the time domain as a sum of $K$ complex exponentials. Everything that carries information sits in the coefficients; the exponentials themselves are fixed by the system design.

$K$ — the number of subcarriers. $K$ counts the orthogonal complex exponentials (subcarriers) that are added together to form the symbol, and therefore also the number of modulation symbols carried in parallel by one OFDM symbol. It is the size of the transform in part (b). In LTE one physical resource block occupies $K = 12$ subcarriers, and a full carrier stacks many such blocks: a 20 MHz LTE carrier populates 1200 subcarriers inside a 2048-point transform, the unused ones acting as guard bands at the edges of the channel.

$X(i)$ — the complex modulation symbol on subcarrier $i$. Each $X(i)$ is one point of the modulation constellation — QPSK, 16-QAM or 64-QAM in LTE — so its magnitude and phase encode the coded data bits destined for that subcarrier. Because the subcarriers are independent, the modulation order of $X(i)$ can in principle be chosen per subcarrier according to the channel quality that subcarrier experiences, which is what adaptive modulation and coding exploits. A minority of the $X(i)$ carry no user data at all: they are known reference (pilot) symbols used by the receiver to estimate the channel, which is exactly the deduction part (c) asks about.

$T_s$ — the useful symbol duration. $T_s$ is the length of the interval over which the sum is transmitted and over which the receiver integrates. It fixes the subcarrier spacing, because subcarrier $i$ sits at frequency $f_i = i/T_s$ and adjacent subcarriers are therefore separated by

$$\Delta f = \frac{1}{T_s}.$$

This is the whole point of the construction: any two distinct subcarriers are orthogonal over one symbol period,

$$\frac{1}{T_s}\int_{0}^{T_s} e^{\,j2\pi \frac{i}{T_s}t}\; e^{-j2\pi \frac{m}{T_s}t}\,dt \;=\; \begin{cases} 1, & i = m \\ 0, & i \neq m \end{cases}$$

so the spectra of neighbouring subcarriers overlap yet do not interfere — each one's spectral null falls on every other one's peak. In LTE $\Delta f = 15\ \text{kHz}$, giving $T_s = 66.7\ \mu\text{s}$; a cyclic prefix of about $4.7\ \mu\text{s}$ is prepended for transmission, so seven symbols occupy the $0.5\ \text{ms}$ slot of part (c).

Part (b) — the FFT as the modulator and demodulator. Building $s(t)$ literally — $K$ oscillators, $K$ mixers and a summing network — is not practical for $K$ in the hundreds or thousands. Instead the symbol is generated digitally. Sample $s(t)$ at the $K$ instants $t = nT_s/K$, $n = 0, 1, \ldots, K-1$:

$$s[n] \;=\; s\!\left(\frac{nT_s}{K}\right) \;=\; \sum_{i=1}^{K} X(i)\,e^{\,j2\pi in/K} \;=\; K \cdot \mathrm{IDFT}\{X(i)\}.$$

The sampled OFDM symbol is exactly the inverse discrete Fourier transform of the constellation vector, to within the constant $K$. So the modulator is an inverse FFT: load the $K$ constellation points into the frequency-domain input, transform, prepend the cyclic prefix, and feed the result to a single digital-to-analogue converter and one RF up-converter.

The demodulator reverses this. After down-conversion and sampling, the receiver discards the cyclic prefix and applies a $K$-point forward FFT to the remaining block, which recovers the coefficients $Y(i) = H(i)X(i) + N(i)$, where $H(i)$ is the channel response at subcarrier $i$. Provided the cyclic prefix is longer than the channel's delay spread, the linear convolution with the channel becomes a circular convolution, so multipath acts as a single complex scalar on each subcarrier rather than as inter-symbol interference across many samples. Equalisation then costs one complex division per subcarrier, $\hat{X}(i) = Y(i)/\hat{H}(i)$, with $\hat{H}(i)$ interpolated from the reference symbols. The saving is decisive: the FFT costs $O(K \log_2 K)$ operations against $O(K^2)$ for a bank of correlators.

Part (c) — peak data rate of one resource block. The rate is fixed by how many resource elements carry data, how many bits each carries, and how long the block lasts.

Given.

Find. The peak data rate delivered by one PRB, in bits per second.

Approach. Count the resource elements, subtract the reference elements, multiply by the bits each 16-QAM symbol carries, and divide by the block duration.

1. Count the resource elements in the block. One resource element is one subcarrier during one OFDM symbol, so $$N_{\text{RE}} = N_{\text{sym}} \times N_{\text{sub}} = 7 \times 12 = 84.$$
2. Remove the reference elements. Four of the 84 are known pilots used to estimate $H(i)$ and are unavailable to the user: $$N_{\text{data}} = 84 - 4 = 80.$$
3. Bits per resource element. A 16-QAM constellation has 16 points, so each surviving element carries $$b = \log_2 16 = 4\ \text{bits}.$$
4. Bits per resource block. Multiplying the two, $$B = N_{\text{data}}\, b = 80 \times 4 = 320\ \text{bits per PRB}.$$
5. Divide by the block duration. Those 320 bits are delivered in $T = 0.5\ \text{ms}$, so $$R = \frac{B}{T} = \frac{320\ \text{bits}}{0.5\times 10^{-3}\ \text{s}} = \boxed{640\ \text{kbit/s per PRB}}.$$

The figure above shows the block: 84 squares, four of them shaded as reference elements. Had all 84 elements carried data the rate would have been $84 \times 4 / 0.5\ \text{ms} = 672\ \text{kbit/s}$, so channel estimation costs about 4.8 per cent of the block here — a realistic LTE configuration spends rather more than that on pilots.

Part (d) — system and per-cell capacity. This is a frequency-reuse count. The trap is the denominator: the band is divided among the cells of one cluster, not among all the cells in the city, because the cluster pattern repeats and every group of channels is used again in every cluster.

Given.

Find. The number of users that can be served simultaneously across the whole city, and the number per cell.

Approach. Divide the system band into channels, share those channels among the $N$ cells of one cluster, count the cells and hence the clusters covering the city, and multiply.

1. Total channels in the system band. With FDM and 25 kHz per user, guard band already included, $$N_{\text{ch}} = \frac{B_{\text{sys}}}{B_u} = \frac{35 \times 10^{6}}{25 \times 10^{3}} = 1400\ \text{channels}.$$
2. Channels available in each cell. The whole band is split across the $N = 3$ cells of a cluster so that no two adjacent cells share a channel: $$N_{\text{cell}} = \frac{N_{\text{ch}}}{N} = \frac{1400}{3} = 466.7 \;\rightarrow\; 466\ \text{channels per cell}.$$ So about 466 users can be served simultaneously in any one cell (1400 does not divide by 3, so a real plan gives two cells 467 and one 466).
3. Count the cells and the clusters. The cells tile the city exactly, so $$N_{\text{cells}} = \frac{A_{\text{city}}}{A_{\text{cell}}} = \frac{42}{0.2} = 210\ \text{cells}, \qquad N_{\text{clusters}} = \frac{210}{3} = 70\ \text{clusters}.$$
4. Total simultaneous users. Every cluster re-uses the entire band, so the city-wide capacity is the number of clusters times the channels per cluster: $$M = N_{\text{clusters}} \times N_{\text{ch}} = 70 \times 1400 = \boxed{98\,000\ \text{simultaneous users}}.$$ Counting the other way round gives the same figure, $210 \times 466.7 = 98\,000$; with the integer split of step 2 it is $210 \times 466 = 97\,860$, i.e. the same answer to three significant figures.

For completeness, the geometry that goes with the count: a hexagonal cell of area $0.2\ \text{km}^2$ has radius $R = \sqrt{A/(1.5\sqrt{3})} = 0.277\ \text{km}$, and the co-channel re-use distance is $D = R\sqrt{3N} = 0.83\ \text{km}$, a re-use ratio of $Q = D/R = \sqrt{3N} = 3$ exactly. A ratio that small is aggressive — it is why $N = 3$ buys capacity but demands tight power control and good sectorisation to keep co-channel interference tolerable.

Final results.