22-Elec-B4 Information Technology Networks · May 2018

Question 1 of 5: Cellular Telephony — OFDM Orthogonality, PRB Peak Rate and FDMA Capacity

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

Question 1: Cellular Telephony — OFDM Orthogonality, PRB Peak 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 built from $K$ subcarriers, $s(t)=\sum_{i=1}^{K} X(i) e^{j2\pi i t/T_s}$ on $0 \le t \le T_s$, in which the $i$th subcarrier 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 $r$ in the range $1 \le r \le K$.

Find. Show that the correlator output for subcarrier index $k$ returns exactly the transmitted symbol on that subcarrier, $d_k = X(k)$, 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. Because the sum over $i$ is finite, the integral of the sum is the sum of the integrals: $$d_r = \frac{1}{T_s}\int_0^{T_s}\left[\sum_{i=1}^{K} X(i)e^{j2\pi i t/T_s}\right]e^{-j2\pi r t/T_s}dt = \sum_{i=1}^{K} X(i)\underbrace{\frac{1}{T_s}\int_0^{T_s} e^{j2\pi (i-r)t/T_s}dt}_{\textstyle I_{ir}}$$ Everything now rests on the single elementary integral $I_{ir}$, which depends only on the integer difference $n = i-r$.
2. Evaluate the kernel for 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 matched 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 for 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, $e^{j2\pi n}=\cos 2\pi n + j\sin 2\pi n = 1$, so the numerator vanishes and $I_{ir}=0$. Each unwanted subcarrier completes a whole number of cycles inside the observation window, so its contribution averages exactly to zero.
4. Collapse the sum and read off the result. Steps 2 and 3 together say $I_{ir}=\delta_{ir}$, the Kronecker delta, so $$d_r=\sum_{i=1}^{K}X(i)\,\delta_{ir}=X(r)$$ Setting $r=k$ gives the required statement: $$\boxed{\,d_k = X(k)\,}$$

The essential engineering content is that the subcarriers are spaced by exactly $\Delta f = 1/T_s$, which is the smallest spacing for which the correlation between distinct subcarriers over one symbol period is zero. The subcarrier spectra overlap heavily — each is a $\mathrm{sinc}$ centred on its own carrier — yet every one of them has a null at the centre of every other, so the receiver separates them without any guard band at all. That is precisely why OFDM achieves near-ideal spectral efficiency where a classical FDM system would need guard bands between channels.

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, counting only the resource elements available to user data.

Approach. Count the resource elements in the grid, deduct the reference elements, multiply by the bits carried per 16-QAM symbol, and divide by the PRB duration.

1. Count the resource elements in the block. A resource element is one subcarrier during one OFDM symbol, so $$N_{\text{RE}} = N_{\text{sym}}\times N_{\text{sc}} = 7 \times 12 = 84\ \text{resource elements}$$
2. Deduct the reference elements. The question states that four of these elements over the whole PRB carry pilots for channel estimation and cannot carry data: $$N_{\text{data}} = N_{\text{RE}} - N_{\text{ref}} = 84 - 4 = 80\ \text{elements}$$
3. Convert elements to bits. A 16-QAM constellation has 16 points, so each transmitted symbol carries $$b = \log_2 16 = 4\ \text{bits/symbol}\quad\Longrightarrow\quad N_{\text{bits}} = N_{\text{data}}\,b = 80 \times 4 = 320\ \text{bits per PRB}$$
4. Divide by the block duration to obtain the rate. The block occupies 0.5 ms, so $$R_{\text{peak}} = \frac{N_{\text{bits}}}{T_{\text{PRB}}} = \frac{320\ \text{bits}}{0.5\times10^{-3}\ \text{s}} = 6.40\times10^{5}\ \text{bit/s}$$ $$\boxed{\,R_{\text{peak}} = 640\ \text{kbit/s per PRB}\,}$$

It is worth noting the cost of the pilots explicitly, because examiners often ask for it in the same breath. Had all 84 elements carried data the block would deliver $84\times4/0.5\ \text{ms} = 672\ \text{kbit/s}$, so the four reference symbols cost $4/84 = 4.8\,\%$ of the raw throughput. That is a cheap price for coherent detection: without channel estimates the receiver would be restricted to differential modulation and would give up far more than five per cent in required signal-to-noise ratio.

Part (c) — Simultaneous users in an FDMA cellular network (5 marks)

Given. A city to be covered by an FDMA cellular system, with the data printed in the question:

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

Approach. Divide the system band into user channels, allocate the band across the cells of one cluster (never across all cells), then scale by the number of cells in the city.

1. Find the number of channels in the whole system band. Each user occupies 25 kHz including its guardband, so the band supports $$C_{\text{tot}} = \frac{B_{\text{sys}}}{B_u} = \frac{35\times10^{6}\ \text{Hz}}{25\times10^{3}\ \text{Hz}} = 1400\ \text{channels}$$
2. Divide the band among the cells of one cluster. Frequency re-use works by splitting the band into $N$ disjoint groups, one per cell of a cluster; the same group then reappears in the corresponding cell of every other cluster. The bandwidth available to any single cell is therefore $$B_{\text{cell}} = \frac{B_{\text{sys}}}{N} = \frac{35\ \text{MHz}}{7} = 5\ \text{MHz} \quad\Longrightarrow\quad C_{\text{cell}} = \frac{B_{\text{cell}}}{B_u} = \frac{5\times10^{6}}{25\times10^{3}}$$ $$\boxed{\,C_{\text{cell}} = 200\ \text{simultaneous users per cell}\,}$$ The divisor here is the cluster size, seven, and not the number of cells in the city. Dividing by 42 would allocate each cell only 833 kHz and understate the answer sixfold; it is the classic error on this question.
3. Count the cells and scale to the whole city. The cells tile the city exactly, so $$M = \frac{A_{\text{city}}}{A_{\text{cell}}} = \frac{42\ \text{km}^2}{1\ \text{km}^2} = 42\ \text{cells}\ \ (= 6\ \text{complete clusters of }7)$$ Every cell reuses its own 200 channels, so the system-wide simultaneous capacity is $$C_{\text{sys}} = C_{\text{cell}}\times M = 200\times 42$$ $$\boxed{\,C_{\text{sys}} = 8400\ \text{simultaneous users in the city}\,}$$ The same figure follows from the cluster view as a check: six clusters, each consuming the full 1400 channels once, gives $6\times1400 = 8400$.