## A note about non-matched pulse filtering

This is a short note about the losses cause by non-matched pulse filtering in the demodulation of a PAM waveform. Recently I’ve needed to come back to these calculations several times, and I’ve found that even though the calculations are simple, sometimes I make silly mistakes on my first try. This post will serve me as a reference in the future to save some time. I have also been slightly surprised when I noticed that if we have two pulse shapes, let’s call them A and B, the losses of demodulating waveform A using pulse shape B are the same as the losses of demodulating waveform B using pulse shape A. I wanted to understand better why this happens.

Recall that if $$p(t)$$ denotes the pulse shape of a PAM waveform and $$h(t)$$ is a filter function, then in AWGN the SNR at the output of the demodulator is equal to the input SNR (with an appropriate normalization factor) times the factor$\begin{equation}\tag{1}\frac{\left|\int_{-\infty}^{+\infty} p(t) \overline{h(t)}\, dt\right|^2}{\int_{-\infty}^{+\infty} |h(t)|^2\, dt}.\end{equation}$This factor describes the losses caused by filtering. As a consequence of the Cauchy-Schwarz inequality, we see that the output SNR is maximized when a matched filter $$h = p$$ is used.

To derive this expression, we assume that we receive the waveform$y(t) = ap(t) + n(t)$with $$a \in \mathbb{C}$$ and $$n(t)$$ a circularly symmetric stationary Gaussian process with covariance $$\mathbb{E}[n(t)\overline{n(s)}] = \delta(t-s)$$. The demodulator output is$T(y) = \int_{-\infty}^{+\infty} y(t) \overline{h(t)}\, dt.$The output SNR is defined as $$|\mathbb{E}[T(y)]|^2/V(T(y))$$. Since $$\mathbb{E}[n(t)] = 0$$ due to the circular symmetry, we have$\mathbb{E}[T(y)] = a\int_{-\infty}^{+\infty} p(t)\overline{h(t)}\,dt.$Additionally,$\begin{split}V(T(y)) &= \mathbb{E}[|T(y) – \mathbb{E}[T(y)]|^2] = \mathbb{E}\left[\left|\int_{-\infty}^{+\infty} n(t)\overline{h(t)}\,dt\right|^2\right] \\ &= \mathbb{E}\left[\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} n(t)\overline{n(s)}\overline{h(t)}h(s)\,dtds\right] \\ &= \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \mathbb{E}\left[n(t)\overline{n(s)}\right]\overline{h(t)}h(s)\,dtds \\ &= \int_{-\infty}^{+\infty} |h(t)|^2\, dt. \end{split}$Therefore, we see that the output SNR equals$\frac{|a|^2\left|\int_{-\infty}^{+\infty} p(t) \overline{h(t)}\, dt\right|^2}{\int_{-\infty}^{+\infty} |h(t)|^2 dt.}.$

The losses caused by using a non-matched filter $$h$$, in comparison to using a matched filter, can be computed as the quotient of the quantity (1) divided by the same quantity where $$h$$ is replaced by $$p$$. This gives$\frac{\frac{\left|\int_{-\infty}^{+\infty} p(t) \overline{h(t)}\, dt\right|^2}{\int_{-\infty}^{+\infty} |h(t)|^2\, dt}}{\frac{\left|\int_{-\infty}^{+\infty} |p(t)|^2\, dt\right|^2}{\int_{-\infty}^{+\infty} |p(t)|^2\, dt}}=\frac{\left|\int_{-\infty}^{+\infty} p(t) \overline{h(t)}\, dt\right|^2}{\int_{-\infty}^{+\infty} |p(t)|^2\, dt\cdot \int_{-\infty}^{+\infty} |h(t)|^2\, dt}.$

We notice that this expression is symmetric in $$p$$ and $$h$$, in the sense that if we interchange $$p$$ and $$h$$ we obtain the same quantity. This shows that, as I mentioned above, the losses obtained when filtering waveform A with pulse B coincide with the losses obtained when filtering waveform B with pulse A. This is a clear consequence of these calculations, but I haven’t found a way to understand this more intuitively. We can say that the losses are equal to the cosine squared of the angle between the pulse shape vectors in $$L^2(\mathbb{R})$$. This remark makes the symmetry clear, but I’m not sure if I’m satisfied by this as an intuitive explanation.

As an example, let us compute the losses caused by receiving a square pulse shape, defined by $$p(t) = 1$$ for $$0 \leq t \leq \pi$$ and $$p(t) = 0$$ elsewhere, with a half-sine pulse shape filter, defined by $$h(t) = \sin t$$ for $$0 \leq t \leq \pi$$ and $$h(t) = 0$$ elsewhere. This case shows up in many different situations. We can compute the losses as indicated above, obtaining$\frac{\left(\int_0^\pi \sin t \, dt\right)^2}{\int_0^\pi \sin^2t\,dt\cdot \int_0^\pi dt} = \frac{2^2}{\frac{\pi}{2}\cdot\pi}= \frac{8}{\pi^2}\approx -0.91\,\mathrm{dB}.$

## WSPR with the LimeRFE

A few days ago, I received a LimeRFE from Andrew Back of Lime Microsystems. He was kind enough to send me a unit so that I can test it and make some usage demos during the ongoing crowdfunding campaign at Crowd Supply.

The LimeRFE is intended to work as an RF frontend for the LimeSDR family, although it can work coupled with any other SDR or conventional radio. As such, it has power amplifiers, filters and LNAs designed to cover the huge frequency range of these SDRs. It is designed to cover all the Amateur radio bands from HF up to 9cm, and a few cellular bands.

As anyone will know, designing broadband RF hardware is often quite difficult or expensive (Amateur radio amplifiers and LNAs are usually designed for a single band), so packing all this into a single unit is a considerable feat. The output power on most bands is around a couple watts, which is already enough for many experiments and applications. The block diagram of the LimeRFE can be seen below.