Question 2 of 5: LTE, OFDM and Duplexing

Question 2: LTE, OFDM and Duplexing (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 parameters (10 marks)

The expression $s(t) = \sum_{i=1}^{K} X(i)\,e^{\,j2\pi (i/T_s)t}$ is the complex baseband waveform of a single OFDM symbol: a sum of $K$ complex exponentials, each carrying one data symbol, transmitted simultaneously for a duration $T_s$.

$K$ — the number of subcarriers. It is the count of orthogonal complex exponentials the transmitter sums, and therefore the number of independent narrowband channels into which the wideband channel has been divided. In LTE, $K$ is set by the carrier bandwidth: 12 subcarriers make one physical resource block, and a 1.4, 3, 5, 10, 15 or 20 MHz carrier carries 6, 15, 25, 50, 75 or 100 resource blocks respectively, so $K$ runs from 72 to 1200 active subcarriers. The reason for choosing $K$ large is the central purpose of OFDM: dividing a wideband channel into $K$ narrow ones makes each subcarrier's bandwidth much smaller than the channel's coherence bandwidth, so each subcarrier experiences flat fading — a single complex gain — and can be equalised with one complex multiplication instead of a long time-domain equaliser.

$X(i)$ — the complex modulation symbol on subcarrier $i$. Each $X(i)$ is one point drawn from the constellation in use on that subcarrier: QPSK (2 bits), 16-QAM (4 bits) or 64-QAM (6 bits) in LTE, with 256-QAM added in later releases. Its magnitude sets the amplitude and its argument the phase of that subcarrier for the duration of the symbol, so $X(i)$ is precisely the payload — the whole of the information the symbol carries lies in the set $\{X(1), \ldots, X(K)\}$. LTE assigns the constellation per user and per subframe by adaptive modulation and coding, so a user near the cell centre may carry 64-QAM on its subcarriers while a user at the edge carries QPSK on the same carrier at the same instant.

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

In LTE $\Delta f = 15\ \text{kHz}$, so $T_s = 1/(15\times10^{3}) = 66.67\ \mu\text{s}$. This spacing is what makes the subcarriers orthogonal: over one symbol period,

because $i-m$ is a non-zero integer and the exponential completes a whole number of cycles in $T_s$. The subcarriers overlap in frequency — each is a sinc whose peak falls on the nulls of all the others — yet they do not interfere, which is why OFDM is spectrally about twice as efficient as a guard-banded FDM system carrying the same symbols.

Two practical points complete the picture. First, the transmitted symbol is actually longer than $T_s$: a cyclic prefix of $4.69\ \mu\text{s}$ (normal) or $16.67\ \mu\text{s}$ (extended) is prepended by copying the tail of the symbol to its front, so that multipath echoes shorter than the prefix fall inside it and the receiver still integrates over exactly one clean period. Seven symbols of $66.67\ \mu\text{s}$ occupy $466.7\ \mu\text{s}$ of the $500\ \mu\text{s}$ slot, and the remaining $33.3\ \mu\text{s}$ is precisely the cyclic-prefix budget. Second, the choice of $\Delta f = 15$ kHz is a compromise: narrower subcarriers would resist delay spread better but would be more sensitive to Doppler shift and oscillator offset, which destroy the orthogonality above.

Part (b) — Modulating and demodulating with the FFT (5 marks)

Sample $s(t)$ at $N$ points spaced $T_s/N$ apart, so $t = nT_s/N$. The sum becomes

which is exactly the inverse discrete Fourier transform of the sequence $X(i)$ (zero-padded from $K$ to $N$ points). The OFDM modulator is therefore not $K$ oscillators and $K$ mixers but a single $N$-point IFFT: the transmitter loads the constellation points into the frequency-domain input bins, takes the IFFT, prepends the cyclic prefix, and clocks the resulting time samples out through one digital-to-analogue converter. The receiver reverses this — discard the cyclic prefix, take the $N$-point FFT of the remaining samples, and bin $i$ of the output is $H(i)X(i)$, the transmitted symbol scaled by the channel's complex gain on that subcarrier, from which $X(i)$ is recovered by dividing by the pilot-estimated $H(i)$ (single-tap equalisation) and slicing to the nearest constellation point.

The reason this matters is cost. A direct implementation of the sum needs $O(K^{2})$ complex multiplications per symbol and $K$ separate analogue chains that must be kept phase-coherent; the FFT needs $O(N\log_2 N)$ and one chain. For a 20 MHz LTE carrier with $N = 2048$ and 1200 active subcarriers, that is roughly 11 000 butterflies instead of well over a million multiplications — the difference between a practical handset and an impossible one. It is no exaggeration to say OFDM became deployable only because the FFT exists.

One structural detail makes the scheme work end to end: because the cyclic prefix makes the received block a circular convolution of the transmitted block with the channel impulse response, and circular convolution in time is pointwise multiplication in the DFT domain, the frequency-selective channel is diagonalised. That is the formal statement of “one complex multiply per subcarrier” equalisation.

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

Given.

Find. The peak data rate carried by one physical resource block, 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 $N_{\text{RE}} = N_{\text{sym}} \times K = 7 \times 12 = 84$ resource elements per block.
2. Bits per resource element. A 16-QAM constellation has 16 points, so each carries $b = \log_2 16 = 4$ bits.
3. Bits per resource block. Multiplying, $N_{\text{bits}} = N_{\text{RE}} \times b = 84 \times 4 = 336$ bits in every 0.5 ms block.
4. Convert to a rate. Dividing by the block duration, $$R = \frac{N_{\text{bits}}}{T_{\text{PRB}}} = \frac{336\ \text{bits}}{0.5\times10^{-3}\ \text{s}} = \boxed{672\times10^{3}\ \text{bit/s} = 672\ \text{kbit/s}}$$
5. Cross-check by symbol rate. The block delivers $7/(0.5\ \text{ms}) = 14\,000$ OFDM symbols per second on each of 12 subcarriers, i.e. $168\,000$ resource elements per second, and $168\,000 \times 4 = 672\ \text{kbit/s}$ — the same figure by a different route.

This is a peak rate in the strict sense the question intends: it counts every resource element as user data and assumes uncoded transmission. A live carrier reserves some elements for reference (pilot) signals, control channels and synchronisation, and applies a turbo code of rate roughly one-third to three-quarters, so the delivered rate per block is materially lower. Scaling the peak figure is nonetheless the standard sanity check on an LTE link budget: a 20 MHz carrier holds 100 resource blocks and two 0.5 ms slots per millisecond, giving $100 \times 672\ \text{kbit/s} \times 2 = 134.4\ \text{Mbit/s}$ on a single antenna stream — which is exactly the published single-stream 16-QAM figure for LTE Category 3.

Part (d) — Frequency division duplexing (5 marks)

Frequency division duplexing separates the two directions of a call in frequency rather than in time. The regulator allocates the operator two disjoint blocks of spectrum — in LTE Band 7, for example, 2500–2570 MHz for the uplink and 2620–2690 MHz for the downlink — and every base station transmits continuously in the downlink block while every handset transmits continuously in the uplink block. The constant separation between a paired uplink and downlink channel is the duplex spacing, 120 MHz in that band.

Because both directions are active at the same instant, the handset's own transmitter is radiating tens of milliwatts into an antenna a few millimetres from a receiver trying to detect a signal below a picowatt. A duplexer — a pair of sharply tuned filters sharing one antenna port — is therefore mandatory: it passes the downlink block to the receiver, passes the transmitter's output to the antenna, and provides 50 dB or more of isolation between them at the duplex spacing. The duplex gap between the two blocks exists to make that filter realisable, and it is the main cost of FDD: the duplexer is bulky, lossy and band-specific, which is why multi-band handsets need a bank of them.

The advantages are equally concrete. Continuous transmission in both directions means no switching guard time, no waiting for a turnaround, and the lowest achievable latency, which suits symmetric traffic such as voice. It also makes power control and interference management simpler, because uplink and downlink never interfere with each other. The contrast is time division duplexing, which uses one block for both directions and alternates between them; TDD needs only a switch instead of a duplexer, needs no paired spectrum, and can shift the uplink/downlink split to match asymmetric traffic, but pays for it in switching guard periods and in the requirement that neighbouring cells be time-synchronised. LTE supports both; FDD dominates in the paired spectrum most operators hold, and TDD dominates in the unpaired bands (LTE Band 40, and much of 5G NR at 3.5 GHz).