Question 10 of 10: Matched Filtering and Optimum Sampling for Split-Phase (Manchester) Signalling

Question 10: Matched Filtering and Optimum Sampling for Split-Phase (Manchester) Signalling (20 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. $p(t)$: $+1$ on $(0,T_b/2)$, $-1$ on $(T_b/2,T_b)$; a noisy received waveform with $a_k=\pm1$ V, $r_b=1$ kbps ($T_b=1$ ms), matched-filtered per part (a); Figure 8 (eye diagram and matched-filter output waveform).

Find. (a) $h(t)$; (b) the isolated pulse response $Ap(t-T_b)*h(t)$; (c) the optimum sampling instants, decoding rule, and the first six decoded symbols.

Approach. Build the causal matched filter as $h(t)=p(T_b-t)$, convolve it with $p(t)$ analytically (as a difference of two rectangular pulses, whose self-convolutions are triangles) to get the isolated pulse response, then read off where that response is zero versus peak to identify the zero-ISI sampling instants.

1. Part (a) — h(t). The causal matched filter for a pulse of duration $T_b$ is $h(t)=p(T_b-t)$ (time-reversed and delayed by the pulse duration, so $h(t)=0$ for $t < 0$). Evaluating piecewise: for $0 < t < T_b/2$, $T_b-t\in(T_b/2,T_b)\Rightarrow p(T_b-t)=-1$; for $T_b/2 < t < T_b$, $T_b-t\in(0,T_b/2)\Rightarrow p(T_b-t)=+1$. So $$h(t)=\boxed{p(T_b-t)=-p(t)}:\quad-1\ \text{on}\ (0,T_b/2),\quad+1\ \text{on}\ (T_b/2,T_b)$$ — the mirror image of $p(t)$ about $T_b/2$, which for this particular antisymmetric pulse equals simply $-p(t)$; it is zero outside $(0,T_b)$, so causal as required.

2. Part (b) — isolated pulse response. Write $p(t)$ as a difference of two unit-height rectangles, each of width $T_b/2$, centred at $T_b/4$ and $3T_b/4$. Since $h(t)=-p(t)$, the isolated response to a pulse occupying $(0,T_b)$ is $q(t)=p(t)*h(t)=-\big[p(t)*p(t)\big]$, and the convolution of two such rectangle-differences is a sum of three shifted triangles (each of peak height $T_b/2$, base $T_b$): $$q(t)=-\Big[\Lambda(t;T_b/2)-2\Lambda(t;T_b)+\Lambda(t;3T_b/2)\Big]$$ where $\Lambda(t;c)$ is the triangle centred at $c$. This is confirmed by direct numerical convolution: $q(t)/T_b$ runs through $(0,-0.25,-0.5,0.25,1,0.25,-0.5,-0.25,0)$ at $t/T_b=(0,\tfrac14,\tfrac12,\tfrac34,1,\tfrac54,\tfrac32,\tfrac74,2)$, i.e. a peak of $T_b$ at $t=T_b$ and EXACT ZEROS at $t=0$ and $t=2T_b$. The requested pulse response, for a received pulse of amplitude $A$ occupying $(T_b,2T_b)$, is this same shape shifted by $T_b$: $$Ap(t-T_b)*h(t)=\boxed{A\,q(t-T_b)}\quad\text{— peaking at } t=2T_b\text{, exactly }T_b \text{ after the pulse's own end.}$$

3. Part (c) — optimum sampling, decoding rule, and decoded symbols. Because the isolated pulse response $q(t)$ from part (b) is EXACTLY ZERO at $t=0$ and $t=2T_b$, and reaches its full value $T_b$ only at $t=T_b$, a running stream of symbols produces, at each bit boundary $t=kT_b$, the $k$-th symbol's own PEAK contribution with exactly zero leakage (zero ISI) from both immediate neighbours — their own pulse-response sidelobes vanish precisely at that instant. So the $$\textbf{optimum sampling instants are } \boxed{t=kT_b\ (k=1,2,3,\dots),\ \text{i.e. every bit boundary}}$$ and, for equiprobable antipodal ($\pm1$ V) signalling, the optimum decoding rule is simply the SIGN of the sample: $$\boxed{\hat a_k=+1\ \text{if the sample at }t=kT_b>0,\quad \hat a_k=-1\ \text{if}<0.}$$ Reading the given matched-filter output waveform at $t=1,2,3,4,5,6$ ms (with $T_b=1/r_b=1$ ms) gives approximately $-1.1,\ -0.6,\ +0.5,\ +0.7,\ -0.9,\ +0.7$ V — every one unambiguously signed despite the noise — so the first six decoded symbols are $$\hat a_1,\dots,\hat a_6=\boxed{-1,\ -1,\ +1,\ +1,\ -1,\ +1\ \text{(V)}}.$$