Question 2 of 5: Modulation Schemes and Channel Capacity

Question 2: Modulation Schemes and Channel 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.

Given. A rectangular 16QAM constellation ($M = 16$, hence $\log_2 16 = 4$ bits per symbol); available system bandwidth $B = 1$ MHz; transmit pulses that just satisfy the Nyquist criterion for zero intersymbol interference; capacity formula $C = B\log_2(1+S)$ with $S$ the signal-to-noise ratio as a power ratio.

Find. A description of how bits are mapped onto and recovered from the 16QAM constellation; the maximum bit rate the 1 MHz channel supports; the signal-to-noise ratio at which Shannon capacity just equals that bit rate; and a professional judgement on the vendor’s claim.

(a) Transmission using the rectangular 16QAM constellation (7 marks)

Quadrature amplitude modulation transmits two independent amplitude-modulated streams on the same carrier frequency by exploiting the orthogonality of the sine and cosine over a symbol interval. The transmitted waveform in the $k$-th symbol interval is

where $g(t)$ is the baseband pulse shape, $T_s$ the symbol period, and the pair $(a_k, b_k)$ is one point of the constellation. For rectangular 16QAM each of $a_k$ and $b_k$ is drawn independently from the four levels $\{-3, -1, +1, +3\}$ (in units of a scaling amplitude), and the sixteen resulting combinations form a $4 \times 4$ square lattice in the in-phase/quadrature plane.

Transmission proceeds as follows. The incoming bit stream is grouped into blocks of four bits, since $\log_2 16 = 4$. Each block is mapped to one constellation point by a Gray-coded assignment: the first two bits select one of the four in-phase levels and the last two select one of the four quadrature levels, arranged so that horizontally or vertically adjacent points differ in exactly one bit. The chosen amplitudes $(a_k, b_k)$ scale a band-limited baseband pulse, the in-phase branch multiplies $\cos(2\pi f_c t)$ and the quadrature branch multiplies $-\sin(2\pi f_c t)$, and the two branches are summed and amplified for transmission. Because the two carriers are orthogonal over a symbol interval, they share the same spectrum without interfering, and the composite waveform for each symbol has a distinct amplitude and phase — which is why QAM is described as joint amplitude and phase modulation.

At the receiver the signal is coherently downconverted against locally generated $\cos(2\pi f_c t)$ and $-\sin(2\pi f_c t)$ references, low-pass filtered, passed through a matched or receive filter and sampled once per symbol to recover noisy estimates of $a_k$ and $b_k$. The detector then chooses the nearest constellation point — which for a rectangular constellation reduces to two independent four-level threshold comparisons, one on each axis — and the Gray map is inverted to return four bits. Coherent carrier recovery, symbol timing recovery and, on a fading channel, amplitude and phase equalisation are all prerequisites, because 16QAM carries information in the absolute amplitude as well as the phase and cannot be detected non-coherently.

(b) Maximum data rate over a 1 MHz channel (7 marks)

Approach. Find the highest symbol rate a Nyquist-shaped passband pulse can support in 1 MHz, then multiply by the number of bits each 16QAM symbol carries.

1. Part (b) — apply the Nyquist criterion to fix the symbol rate. A pulse satisfying the Nyquist criterion for zero intersymbol interference at symbol rate $R_s$ needs a minimum baseband bandwidth of $R_s/2$. Quadrature modulation translates that baseband spectrum to the carrier and mirrors it about $f_c$, so the transmitted passband signal occupies $$B = 2 \times \frac{R_s}{2} = R_s .$$ With the whole 1 MHz available for the modulated signal, $$R_s = B = 1 \times 10^{6}\ \text{symbols/s}.$$
2. Part (b) — convert symbols to bits. Each rectangular 16QAM symbol conveys $$\log_2 M = \log_2 16 = 4\ \text{bits/symbol},$$ so the maximum information rate is $$R_b = R_s \log_2 M = (1 \times 10^{6})(4),$$ $$\boxed{R_b = 4 \times 10^{6}\ \text{bit/s} = 4\ \text{Mbit/s}}$$ Equivalently, the link achieves a spectral efficiency of 4 bit/s per hertz, which is the defining figure of merit of 16QAM.

(c) Signal-to-noise ratio required to reach that rate (7 marks)

Approach. Set the Shannon capacity equal to the bit rate from part (b) and invert the capacity formula for $S$.

1. Part (c) — set capacity equal to the required rate. The rate of part (b) is achievable only if the channel capacity is at least as large, so at the break-even point $$C = B\log_2(1+S) = R_b \quad\Longrightarrow\quad \log_2(1+S) = \frac{R_b}{B} = \frac{4\times 10^{6}}{1 \times 10^{6}} = 4 .$$ The ratio $R_b/B$ is exactly the spectral efficiency in bit/s/Hz, so the capacity formula is being asked to deliver 4 bit/s/Hz.
2. Part (c) — invert to obtain the signal-to-noise ratio. Exponentiating, $$1 + S = 2^{4} = 16 \quad\Longrightarrow\quad S = 16 - 1,$$ $$\boxed{S = 15 \;\;(\text{a power ratio}) = 10\log_{10}(15) = 11.76\ \text{dB}}$$ This is the minimum signal-to-noise ratio at which 4 Mbit/s is information-theoretically possible in 1 MHz. A real 16QAM link needs appreciably more — roughly 20 dB for an uncoded bit error rate near $10^{-6}$ — because Shannon’s bound assumes ideal Gaussian-like coding of unbounded block length, not a fixed sixteen-point constellation with a hard decision device.

(d) Assessing the vendor’s claim (4 marks)

The claim is not reasonable and should be rejected. The capacity $C = B\log_2(1+S)$ is not an engineering rule of thumb or a limitation of present-day hardware; it is a theorem. Shannon’s channel-coding theorem states that for a band-limited channel with additive white Gaussian noise, reliable communication — meaning an error probability that can be driven arbitrarily close to zero by increasing the code block length — is possible at every rate below $C$ and at no rate above it. The converse half of the theorem is what matters here: at any rate exceeding $C$ the error probability is bounded away from zero no matter how the transmitter and receiver are designed. No modulation format, coding scheme, signal-processing algorithm or manufacturing advance can evade it, because the bound follows from the definition of mutual information rather than from any assumption about the equipment.

The professionally appropriate response is therefore to ask what the vendor is actually measuring, because plausible explanations usually exist for the numbers being quoted. The vendor may be counting the gross line rate including framing and error-control overhead rather than the net payload rate; may be quoting the rate after data compression, which increases the useful information per transmitted bit but does not raise the channel’s capacity in bits per second; may be operating over a larger bandwidth or at a higher signal-to-noise ratio than the one in the comparison, both of which legitimately raise $C$; may be exploiting spatial multiplexing across multiple antennas, where a multiple-input multiple-output channel genuinely has a larger capacity than the single-input single-output formula gives, but still obeys its own Shannon limit; or may simply be reporting peak burst throughput rather than a sustained error-free rate. A demonstration under controlled conditions, with the bandwidth, signal-to-noise ratio, payload definition and residual error rate all measured independently, will resolve which of these applies. As the engineer of record one should require that evidence before endorsing the product, and should record that a claim of transmission above capacity, taken literally, is physically impossible.