Question 4 of 5: Signal Modulation and Detection

Question 4: Signal Modulation and Detection (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. An on-off keyed binary scheme with $s_1(t) = \sin(2\pi t/T)$ on $0\le t\le T$ (one full cycle of the carrier in the bit interval) and zero elsewhere, $s_0(t) = 0$, a filter matched to $s_1$, output sampled at $t = T$, additive zero-mean Gaussian noise of variance $\sigma^{2}$ at that instant, and equiprobable symbols. The identity $\sin^{2}x = (1-\cos 2x)/2$ is supplied for part (b), and the erfc integral identity above for part (d).

Find. (a) sketches of $s_0$, $s_1$ and the matched-filter impulse response $m(t)$; (b) the noise-free output at $t = T$ when a 1 is sent; (c) the optimal threshold rule; (d) $\Pr(\text{error}\mid 0\ \text{sent})$ in terms of erfc.

Approach. Form the matched filter by time-reversing and delaying the signal it is matched to, evaluate the convolution at the sampling instant to show it returns the signal energy, place the decision threshold midway between the two noise-free output levels because the symbols are equiprobable and the noise is symmetric, and finally read the error probability off the supplied Gaussian-tail identity.

1. Part (a) — Construct the matched-filter impulse response. A filter matched to $s_1(t)$ over the observation interval $[0,T]$ and sampled at $t = T$ has $$m(t) = s_1(T-t), \qquad 0\le t\le T,$$ and zero elsewhere — the signal reversed in time and shifted so the correlation peak lands at $t = T$ rather than at $t = 0$ (which would make the filter non-causal). Substituting the given waveform, $$m(t) = \sin\!\left(\frac{2\pi (T-t)}{T}\right) = \sin\!\left(2\pi - \frac{2\pi t}{T}\right) = -\sin\!\left(\frac{2\pi t}{T}\right).$$ Because $s_1$ is a whole cycle, time reversal happens to reproduce the negative of the original; the three waveforms are sketched in Figure 4.1. The filter matched to $s_0(t) = 0$ is trivial, which is why a single filter suffices for on-off keying.
2. Part (b) — Set up the output at the sampling instant. The filter output is the convolution $y(t) = \int s_1(\tau)\,m(t-\tau)\,d\tau$, and at $t = T$ the argument becomes $m(T-\tau) = s_1(T-(T-\tau)) = s_1(\tau)$, so the convolution collapses to an autocorrelation at zero lag: $$y(T) = \int_{0}^{T} s_1(\tau)\,s_1(\tau)\,d\tau = \int_{0}^{T} \sin^{2}\!\left(\frac{2\pi\tau}{T}\right) d\tau = E,$$ the energy of $s_1$. This is the defining property of a matched filter: at the sampling instant it returns the signal energy, and it is exactly what a correlator multiplying by $s_1$ and integrating would produce.
3. Evaluate the energy using the supplied identity. With $\sin^{2}x = (1-\cos 2x)/2$ and $x = 2\pi\tau/T$, $$E = \int_{0}^{T}\frac{1}{2}\left[1 - \cos\!\left(\frac{4\pi\tau}{T}\right)\right] d\tau = \frac{T}{2} - \frac{1}{2}\cdot\frac{T}{4\pi}\left[\sin\!\left(\frac{4\pi\tau}{T}\right)\right]_{0}^{T}.$$ The sine term vanishes at both limits because $4\pi\tau/T$ passes through an exact whole number of cycles, leaving $$\boxed{y(T) = E = \frac{T}{2}}$$ So a “1” produces $T/2$ at the sampler and a “0” produces 0.
4. Part (c) — Identify the two hypotheses at the sampler. The sampled output is $$r = \begin{cases} 0 + n, & \text{symbol 0 sent},\\[2pt] E + n, & \text{symbol 1 sent},\end{cases} \qquad n\sim\mathcal{N}(0,\sigma^{2}).$$ Both hypotheses carry the same Gaussian noise with the same variance, so their conditional densities are identical bells centred at 0 and at $E$.
5. Place the optimal threshold. With equiprobable symbols the maximum a posteriori rule reduces to maximum likelihood, and for two equal-variance Gaussians the likelihood ratio crosses unity exactly midway between the means. The threshold is therefore $\lambda = E/2$, and the rule is $$\text{decide }\; \hat{b} = \begin{cases} 1, & r > \lambda,\\ 0, & r < \lambda,\end{cases} \qquad \boxed{\lambda = \frac{E}{2} = \frac{T}{4}}$$ (the boundary $r = \lambda$ may be assigned either way; it has probability zero). Had the symbols not been equiprobable the threshold would shift towards the less likely symbol by $(\sigma^{2}/E)\ln[\Pr(0)/\Pr(1)]$.
6. Part (d) — Express the error event when a 0 is sent. Given symbol 0, the sampled output is $r = n$ with mean $\mu = 0$ and variance $\sigma^{2}$, and the decoder errs precisely when that sample rises above the threshold: $$\Pr(\varepsilon\mid 0) = \Pr(n > \lambda) = \int_{\lambda}^{\infty}\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\!\left(-\frac{x^{2}}{2\sigma^{2}}\right) dx.$$ This is the supplied identity with $\mu = 0$ and $t = \lambda$.
7. Apply the identity. Substituting $\mu = 0$ and $t = \lambda = E/2 = T/4$, $$\boxed{\Pr(\varepsilon\mid 0) = \frac{1}{2}\,\mathrm{erfc}\!\left(\frac{E}{2\sqrt{2\sigma^{2}}}\right) = \frac{1}{2}\,\mathrm{erfc}\!\left(\frac{T}{4\sqrt{2}\,\sigma}\right)}$$ By the symmetry of the two hypotheses about the threshold, $\Pr(\varepsilon\mid 1)$ is identical, so the overall bit-error probability is the same expression. As a numerical illustration, normalising $T = 1$ s gives $\lambda = 0.25$, and with $\sigma = 0.1$ in the same units the error probability is $\tfrac{1}{2}\,\mathrm{erfc}(1.7678) = 6.21\times 10^{-3}$ — and every halving of $\sigma$ from there drives it down by orders of magnitude, which is the exponential steepness characteristic of Gaussian-channel detection.