22-Elec-B4 Information Technology Networks · December 2018

Question 2 of 5: Cellular Telephony — OFDM Orthogonality, PRB Rate and FDMA Capacity

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Cellular Telephony — OFDM Orthogonality, PRB Rate and FDMA Capacity (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) — Orthogonality of the OFDM subcarriers (15 marks)

Given. A noiseless OFDM symbol assembled from $K$ subcarriers, $s(t)=\sum_{i=1}^{K} X(i)e^{j2\pi i t/T_s}$ on the interval $0 \le t \le T_s$, in which subcarrier $i$ carries the complex data symbol $X(i)$ and the subcarrier spacing is $\Delta f = 1/T_s$. The receiver forms the correlation $d_r=(1/T_s)\int_0^{T_s}s(t)e^{-j2\pi r t/T_s}\,dt$ for an integer $r$ with $1 \le r \le K$.

Find. Show that the correlator output on subcarrier $r$ returns exactly the symbol transmitted on that subcarrier, $d_r = X(r)$, with no contribution from any other subcarrier.

Approach. Substitute the sum into the integral, interchange the finite sum with the integral, and show that the resulting kernel $(1/T_s)\int_0^{T_s}e^{j2\pi(i-r)t/T_s}dt$ is the Kronecker delta $\delta_{ir}$; the sum then collapses to the single term $i=r$.

1. Substitute the transmitted signal into the detector. The sum over $i$ is finite, so the integral of the sum is the sum of the integrals: $$d_r = \frac{1}{T_s}\int_0^{T_s}\left[\sum_{i=1}^{K}X(i)e^{j2\pi i t/T_s}\right]e^{-j2\pi r t/T_s}dt = \sum_{i=1}^{K}X(i)\,I_{ir}, \qquad I_{ir} \equiv \frac{1}{T_s}\int_0^{T_s}e^{j2\pi (i-r)t/T_s}dt$$ Everything now rests on the single elementary integral $I_{ir}$, which depends only on the integer difference $n = i-r$.
2. Evaluate the kernel on the matched subcarrier, $i = r$. With $n = 0$ the exponential is identically unity, so $$I_{rr} = \frac{1}{T_s}\int_0^{T_s}e^{0}\,dt = \frac{1}{T_s}\,T_s = 1$$ The wanted term therefore passes through the correlator with unit gain, which is what makes the detector an unbiased estimator of the data symbol.
3. Evaluate the kernel on every other subcarrier, $i \ne r$. For $n \ne 0$ the exponential integrates in closed form: $$I_{ir} = \frac{1}{T_s}\left[\frac{T_s}{j2\pi n}e^{j2\pi n t/T_s}\right]_0^{T_s} = \frac{e^{j2\pi n}-1}{j2\pi n}$$ Since $n = i-r$ is a non-zero integer, $e^{j2\pi n} = \cos 2\pi n + j\sin 2\pi n = 1$, the numerator vanishes and $I_{ir} = 0$. Each unwanted subcarrier completes a whole number of cycles inside the observation window, so its contribution averages to exactly zero.
4. Collapse the sum and read off the result. Steps 2 and 3 together state that $I_{ir} = \delta_{ir}$, the Kronecker delta, so $$d_r = \sum_{i=1}^{K}X(i)\,\delta_{ir} = X(r)$$ which is the required result: $$\boxed{\,d_r = X(r)\,}$$

The engineering content behind the algebra is that the subcarriers are spaced by exactly $\Delta f = 1/T_s$, the smallest spacing at which the correlation between two distinct subcarriers over one symbol period is zero. Their spectra overlap heavily — each is a $\mathrm{sinc}$ centred on its own carrier — yet every one has a spectral null at the centre of every other, so the receiver separates them with no guard band whatsoever. That is why OFDM reaches a spectral efficiency a classical FDM system, which must insert guard bands, cannot match. It is also why the scheme is unforgiving of frequency offset: a carrier error of even a few per cent of $\Delta f$ moves the nulls off the neighbouring carriers and the orthogonality above fails, producing inter-carrier interference.

Part (b) — Peak data rate of one physical resource block (5 marks)

Given. One LTE physical resource block, with the parameters printed in the question:

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

Approach. Count the resource elements in the block, multiply by the bits each constellation point carries, and divide by the block duration.

1. Count the resource elements. One resource element is one subcarrier during one OFDM symbol, so $$N_{RE} = N_{sym}\,N_{sc} = 7 \times 12 = 84\ \text{resource elements}$$ The question reserves none of them for channel-estimation reference symbols, so all 84 are available for user data.
2. Convert the constellation to bits per element. A 4-QAM (equivalently QPSK) constellation has four points, and each transmitted point therefore resolves $$b = \log_2 M = \log_2 4 = 2\ \text{bits per resource element}$$
3. Total the bits in the block. Multiplying the two previous results, $$B_{PRB} = N_{RE}\,b = 84 \times 2 = 168\ \text{bits per PRB}$$
4. Divide by the block duration to get the rate. The block occupies 0.5 ms, so $$R_{peak} = \frac{B_{PRB}}{T_{PRB}} = \frac{168\ \text{bits}}{0.5 \times 10^{-3}\ \text{s}} = \boxed{3.36\times10^{5}\ \text{bit/s} = 336\ \text{kbit/s}}$$ The same figure follows from the per-subcarrier view as a check: each subcarrier delivers $7/0.5\ \text{ms} = 14\,000$ symbols per second, so $14\,000 \times 2 \times 12 = 336$ kbit/s.

This is a peak figure in two senses. It assumes every element carries data, whereas a live LTE downlink spends several elements per PRB on reference, control and synchronisation signals; and it counts channel bits rather than payload bits, so the rate seen by an application is lower still once the forward-error-correction and header overheads are removed.

Part (c) — Simultaneous users in the city and per cell (5 marks)

Given. The FDMA cellular plan printed in the question:

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

Approach. Divide the band by the cluster size to get the bandwidth available in one cell, divide that by the per-user allocation to get the channels per cell, then multiply by the number of cells covering the city.

1. Count the cells needed to cover the city. The cells tile the city exactly, so $$n_{cells} = \frac{A_{city}}{A_{cell}} = \frac{36\ \text{km}^2}{0.5\ \text{km}^2} = 72\ \text{cells}$$ which is $72/3 = 24$ complete re-use clusters.
2. Count the channels the whole band supports. Each user occupies 25 kHz including its guardband, so the system band supports $$n_{ch} = \frac{B_{sys}}{B_u} = \frac{21\times10^{6}\ \text{Hz}}{25\times10^{3}\ \text{Hz}} = 840\ \text{channels}$$ These 840 channels are the entire spectral resource; the re-use plan decides how they are distributed, not how many there are.
3. Share the band among the cells of one cluster. The defining property of a cluster of size $N$ is that its $N$ cells between them use the whole band once, so each cell receives $$B_{cell} = \frac{B_{sys}}{N} = \frac{21\ \text{MHz}}{3} = 7\ \text{MHz}, \qquad n_{ch,cell} = \frac{B_{cell}}{B_u} = \frac{7\times10^{6}}{25\times10^{3}} = \boxed{280\ \text{simultaneous users per cell}}$$
4. Scale to the whole city. Every one of the 72 cells carries its own 280 channels, because the plan guarantees that co-channel cells are never adjacent: $$n_{users} = n_{ch,cell}\times n_{cells} = 280 \times 72 = \boxed{20\,160\ \text{simultaneous users in the city}}$$ The same number arrives by the cluster route, $840 \times 24 = 20\,160$, which is a useful arithmetic check.

The frequency-re-use gain is the whole point of a cellular architecture: a single high-power transmitter with 21 MHz would support 840 users across the entire city, whereas subdividing the same band over 72 low-power cells serves 24 times as many. The price is 72 base-station sites, the handover machinery needed to move a call between them, and a co-channel interference floor set by the cluster size — which is why a designer raises $N$ when the signal-to-interference ratio is inadequate and lowers it when capacity is short.