Question 2 of 5: Matched-Filter Detection of a Sinusoidal Pulse

Question 2: Matched-Filter Detection of a Sinusoidal Pulse (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. On-off keying over one symbol interval of duration $T$: $s_1(t) = \sin(2\pi t/T)$ for $0 \le t \le T$ and zero elsewhere; $s_0(t) = 0$. The receiver is a filter matched to $s_1(t)$, sampled once at $t = T$. The sample is corrupted by additive Gaussian noise $n$ of zero mean and variance $\sigma^2$, and the two symbols are equiprobable.

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 (minimum-error-probability) decision rule; (d) $\Pr(\text{error} \mid 0 \text{ sent})$ expressed with erfc.

Approach. Build $m(t)$ from the matched-filter definition $m(t) = s(T-t)$; evaluate the convolution at $t = T$, which reduces to the signal energy; place the equiprobable-symbol threshold midway between the two conditional means; and integrate the Gaussian tail beyond that threshold using the identity the paper supplies.

1. Part (a) — construct the matched-filter impulse response. A filter matched to a signal $s(t)$ of duration $T$ and sampled at $t = T$ has the time-reversed, delayed impulse response $$m(t) = s_1(T-t), \qquad 0 \le t \le T .$$ Substituting the given waveform and using $\sin(2\pi - \theta) = -\sin\theta$, $$m(t) = \sin\!\left(\frac{2\pi (T-t)}{T}\right) = \sin\!\left(2\pi - \frac{2\pi t}{T}\right) = \boxed{-\sin\!\left(\frac{2\pi t}{T}\right)}, \quad 0 \le t \le T,$$ and zero elsewhere. The three sketches are drawn in Figure 2.1: $s_1$ is one complete sine cycle in $[0,T]$, $s_0$ is identically zero, and $m$ is that same cycle turned upside down — time reversal about the midpoint of a full cycle is equivalent to a sign change. Note that no filter can be matched to $s_0(t) = 0$; the receiver needs only the one filter, and the decision is made on its output.
2. Part (b) — evaluate the noise-free output at the sampling instant. The filter output is the convolution $y(t) = \int_0^{t} s_1(\tau)\, m(t-\tau)\, d\tau$. At $t = T$ the reversal in $m$ undoes itself and the convolution becomes the autocorrelation of $s_1$ at zero lag — that is, the signal energy: $$y(T) = \int_0^{T} s_1(\tau)\, s_1(\tau)\, d\tau = \int_0^{T} \sin^2\!\left(\frac{2\pi \tau}{T}\right) d\tau .$$ Applying the identity the paper supplies, $\sin^2 x = (1-\cos 2x)/2$ with $x = 2\pi\tau/T$, $$y(T) = \int_0^{T} \frac{1}{2}\left[1 - \cos\!\left(\frac{4\pi \tau}{T}\right)\right] d\tau = \frac{T}{2} - \frac{T}{8\pi}\Big[\sin\!\left(\tfrac{4\pi\tau}{T}\right)\Big]_0^{T} .$$ The cosine integrates over exactly two whole periods, so its contribution vanishes ($\sin 4\pi = \sin 0 = 0$) and $$\boxed{y(T) = E = \frac{T}{2}}$$ This is the general result that a matched filter sampled at the end of the symbol delivers the signal energy $E$, here numerically half the symbol duration because the sine has unit amplitude and mean-square value $1/2$.
3. Part (c) — place the decision threshold. With noise present the sample is $Y = y(T) + n$, where $n \sim \mathcal{N}(0,\sigma^2)$. The two hypotheses therefore produce Gaussian samples of the same variance and different means: $$Y \mid 1 \sim \mathcal{N}(E,\ \sigma^2), \qquad Y \mid 0 \sim \mathcal{N}(0,\ \sigma^2)$$ because $s_0(t) = 0$ contributes nothing. With equiprobable symbols the minimum-error-probability rule is the maximum-likelihood rule, $\Pr(Y \mid 1) \gtrless \Pr(Y \mid 0)$; for equal-variance Gaussians the likelihood ratio crosses unity exactly midway between the means. Hence the optimal rule is $$\boxed{\text{decide } 1 \text{ if } Y > \gamma = \frac{E}{2} = \frac{T}{4}, \quad \text{otherwise decide } 0}$$ (the boundary $Y = \gamma$ may be assigned either way; it has zero probability). Had the symbols not been equiprobable the threshold would shift by $\sigma^2 \ln[\Pr(0)/\Pr(1)] / E$, so equiprobability is precisely what puts it at the midpoint.
4. Part (d) — integrate the error tail. Given that a 0 was sent, an error occurs whenever the sample exceeds the threshold, so $$\Pr(\varepsilon \mid 0) = \Pr\big(Y > \gamma \;\big|\; 0 \big) = \int_{\gamma}^{\infty} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\!\left(-\frac{x^2}{2\sigma^2}\right) dx .$$ This is exactly the supplied integral with $\mu = 0$ (the conditional mean under a transmitted 0) and lower limit $t = \gamma$. Substituting directly, $$\Pr(\varepsilon \mid 0) = \frac{1}{2}\,\mathrm{erfc}\!\left(\frac{\gamma - 0}{\sqrt{2\sigma^2}}\right) = \frac{1}{2}\,\mathrm{erfc}\!\left(\frac{E/2}{\sigma\sqrt{2}}\right),$$ and inserting $E = T/2$ from part (b), $$\boxed{\Pr(\varepsilon \mid 0) = \frac{1}{2}\,\mathrm{erfc}\!\left(\frac{T}{4\sigma\sqrt{2}}\right) = \frac{1}{2}\,\mathrm{erfc}\!\left(\frac{E}{2\sqrt{2}\,\sigma}\right)}$$ By symmetry the conditional error probability given a 1 is identical — the threshold is equidistant from both means — so the overall bit error probability for this on-off keyed system is the same expression. A sanity check on the trend: as $\sigma \to 0$ the argument grows without bound and $\mathrm{erfc} \to 0$, while as $\sigma \to \infty$ the argument tends to zero, $\mathrm{erfc}(0) = 1$, and the error probability approaches $1/2$ — the coin-flip a receiver drowned in noise is reduced to.