Question 10 of 10: Matched Filtering and Decoding of Split-Phase Manchester Data

Question 10: Matched Filtering and Decoding of Split-Phase Manchester Data (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. Split-phase pulse $p(t) = +1$ on $(0, T_b/2)$ and $-1$ on $(T_b/2, T_b)$; symbol amplitudes $a_k = \pm 1$ V; bit rate $r_b = 1$ kbps, hence $T_b = 1$ ms; the eye diagram and matched-filter output waveform of Figure 8.

Find. The causal matched-filter impulse response, the pulse response by superposition, the optimum sampling instants and decision rule, and the first six decoded symbols.

Approach. Build $h(t)$ by time-reversing and shifting $p(t)$, obtain the pulse response as a superposition of rectangle-on-rectangle convolutions (each a triangle), then read the eye diagram to fix the sampling phase and apply a zero-threshold decision.

1. Part (a): construct the causal matched filter. The filter matched to a pulse of duration $T_b$, made causal by delaying the time-reversed pulse by $T_b$, is $$h(t)=p(T_b-t)$$ Evaluating: for $0 \lt t \lt T_b/2$ the argument $T_b - t$ falls in the second half of the pulse, where $p = -1$; for $T_b/2 \lt t \lt T_b$ it falls in the first half, where $p = +1$. Hence $$\boxed{h(t)=\begin{cases}-1,&0\lt t\lt T_b/2\\ +1,&T_b/2\lt t\lt T_b\end{cases}}$$ Because $p$ is antisymmetric about $t = T_b/2$, time reversal is equivalent to a polarity flip: $h(t) = -p(t)$. The filter is causal (zero for $t \lt 0$) and has finite duration $T_b$.

For part (b) the superposition is done on rectangles, because the convolution of two rectangles is a triangle and every piece of this problem is built from half-bit rectangles.

2. Decompose both signals into half-bit rectangles. Let $r_a(t)$ denote a unit rectangle occupying $(a, a + T_b/2)$. Then $p(t) = r_0(t) - r_{T_b/2}(t)$ and $h(t) = -r_0(t) + r_{T_b/2}(t)$, so their convolution expands into four terms: $$p*h = -\,r_0*r_0 \;+\; 2\,r_0*r_{T_b/2} \;-\; r_{T_b/2}*r_{T_b/2}$$ the two cross terms being identical.
3. Convolve rectangle with rectangle. Two unit rectangles of width $T_b/2$ starting at $a$ and $b$ convolve to an isosceles triangle of peak $T_b/2$ located at $t = a + b + T_b/2$, spanning $a+b$ to $a+b+T_b$. Applying this to the three terms gives a negative triangle peaking at $T_b/2$, twice a positive triangle peaking at $T_b$, and a negative triangle peaking at $3T_b/2$ — all of height $T_b/2$.
4. Superpose and shift. Summing at the breakpoints, $p*h$ is the piecewise-linear function with values $0,\,-T_b/2,\,+T_b,\,-T_b/2,\,0$ at $t = 0,\,T_b/2,\,T_b,\,3T_b/2,\,2T_b$. Scaling by $A$ and delaying by $T_b$ as the question specifies, $$\boxed{Ap(t-T_b)*h(t)\ \text{peaks at}\ AT_b\ \text{when}\ t=2T_b}$$ with half-bit side lobes of $-AT_b/2$ at $t = 1.5T_b$ and $2.5T_b$, and zeros at $t = T_b$ and $3T_b$.
5. Confirm the peak independently. By the definition of the matched filter, the response at the sampling instant is the pulse energy: $\int_0^{T_b}p^2(t)\,dt = T_b$, which matches the peak $AT_b$ for $A = 1$. This is the defining property — the matched filter maximises the output signal-to-noise ratio precisely by delivering the full pulse energy at one instant.

Crucially, the pulse response is zero at every other bit instant ($t = T_b$ and $t = 3T_b$), so at the correct sampling phase there is no intersymbol interference: each sample sees only its own symbol. This is what produces the wide-open eye of Figure 8(a).

6. Part (c): fix the sampling phase from the eye diagram. With $r_b = 1$ kbps the bit period is $T_b = 1/r_b = 1$ ms. The eye diagram of Figure 8(a) spans two bit intervals and shows a single fully-open eye centred at $t = 1.0$ ms, where the vertical opening between the $+1$ and $-1$ trace bundles is widest and the traces are furthest from the zero axis. The optimum sampling instants are therefore $$\boxed{t_k = kT_b = k\ \text{ms},\qquad k=1,2,3,\ldots}$$ consistent with part (b), where the pulse response of the $k$-th symbol peaks exactly one bit after that symbol's interval ends.
7. State the decision rule. The two hypotheses produce matched-filter peaks of $+AT_b$ and $-AT_b$, which are symmetric about zero, and the noise is symmetric, so the minimum-error threshold sits midway at zero: $$\boxed{\hat a_k = +1\ \text{if}\ y(kT_b)\gt 0,\qquad \hat a_k = -1\ \text{if}\ y(kT_b)\lt 0}$$ No other threshold is optimal here because the two symbols are equiprobable and equal in energy.
8. Read the samples and decode. Reading Figure 8(b) at $t = 1, 2, \ldots, 6$ ms gives the sample values in the table below; applying the zero-threshold rule yields the first six symbols $$\boxed{a_1\ldots a_6 = -1,\,-1,\,+1,\,+1,\,-1,\,+1\ \text{V}}$$ or equivalently the bit pattern $0,0,1,1,0,1$ under the usual mapping $+1\to 1$, $-1\to 0$.

[Figure not reproduced: Figure 8(b) (redrawn) with the decision process overlaid. The six optimum sampling instants t = 1…6 ms give −1.18, −0.55, +0.59, +0.95, −0.88 and +0.72 V, decoding to −1, −1, +1, +1, −1, +1. See the official exam paper.]