An erasure FEC for SSDV

SSDV is an amateur radio protocol that is used to transmit images in packets, in a way that is tolerant to packet loss. It is based on JPEG, but unlike a regular JPEG file, where losing even a small part of the file has catastrophic results, in SSDV different blocks of the image are compressed independently. This means that packet loss affects only the corresponding blocks, and the image can still be decoded and displayed, albeit with some missing blocks.

SSDV was originally designed for transmission from high-altitude balloons (see this reference for more information), but it has also been used for some satellite missions, including Longjiang-2, a Chinese lunar orbiting satellite.

Even though SSDV is tolerant to packet loss, to obtain the full image it is necessary to receive all the packets that form the image. If some packets are lost, then it is necessary to retransmit them. Here I present an erasure FEC scheme that is backwards-compatible with SSDV, in the sense that the first packets transmitted by this scheme are identical to the usual \(k\) packets of standard SSDV, and augments the transmission with FEC packets in such a way that the complete image can be recovered from any set of \(k\) packets (so there is no encoding overhead). The FEC packets work as a fountain code, since it is possible to generate up to \(2^{16}\) packets, which is a limit unlikely to be reached in practice.

About channel capacity and sub-channels

The Shannon-Hartley theorem describes the maximum rate at which information can be sent over a bandwidth-limited AWGN channel. This rate is called the channel capacity. If \(B\) is the bandwidth of the channel in Hz, \(S\) is the signal power (in units of W), and \(N_0\) is the noise power spectral density (in units of W/Hz), then the channel capacity \(C\) in units of bits per second is\[C = B \log_2\left(1 + \frac{S}{N_0B}\right).\]

Let us now consider that we make \(n\) “sub-channels”, by selecting \(n\) disjoint bandwidth intervals contained in the total bandwidth of the channel. We denote the bandwidth of these sub-channels by \(B_j\), \(j = 1,\ldots,n\). Clearly, we have the constraint \(\sum_{j=1}^n B_j \leq B\). Likewise, we divide our total transmit power \(S\) into the \(n\) sub-channels, allocating power \(S_j\) to the signal in the sub-channel \(j\). We have \(\sum_{j=1}^n S_j = S\). Under these conditions, each sub-channel will have capacity \(C_j\), given by the formula above with \(B_j\) and \(S_j\) in place of \(B\) and \(S\).

The natural question regards using the \(n\) sub-channels in parallel to transmit data: what is the maximum of the sum \(\sum_{j=1}^n C_j\) under these conditions and how can it be achieved? It is probably clear from the definition of channel capacity that this sum is always smaller or equal than \(C\). After all, by dividing the channel into sub-channels we cannot do any better than by considering it as a whole.

People used to communications theory might find intuitive that we can achieve \(\sum_{j=1}^n C_j = C\), and that this happens if and only if we use all the bandwidth (\(\sum_{j=1}^n B_j = B\)) and the SNRs of the sub-channels, defined by \(S_j/(N_0B_j)\), are all equal, so that \(S_j = SB_j/B\). After all, this is pretty much how OFDM and other channelized communication methods work. In this post I give an easy proof of this result.

Categorised as Maths Tagged

A note about non-matched pulse filtering

This is a short note about the losses caused 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\[\frac{\left(\int_0^\pi \sin t \, dt\right)^2}{\int_0^\pi \sin^2t\,dt\cdot \int_0^\pi dt} = \frac{2^2}{\frac{\pi}{2}\cdot\pi}= \frac{8}{\pi^2}\approx -0.91\,\mathrm{dB}.\]

An error in the DSN Telecommunications Link Design Handbook description of Reed-Solomon

The DSN Telecommunications Link Design Handbook is a large document describing many aspects pertaining deep space communications and how they are implemented by the NASA Deep Space Network. One of the many things it contains is a description of a Reed-Solomon encoder for the CCSDS code using the Berlekamp bit-serial architecture. While following this description to implement an encoder, I have found an error. In this post, I explain the error and where I think it comes from.

Categorised as Maths, Space Tagged

Voyager 1 and Reed-Solomon

In one of my previous posts about Voyager 1, I stated that the Voyager probes used as forward error correction only the k=7, r=1/2 CCSDS convolutional code, and that Reed-Solomon wasn’t used. However, some days ago, Brett Gottula asked about this, citing several sources that stated that the Voyager probes used Reed-Solomon coding after their encounter with Saturn.

My source for stating that Reed-Solomon wasn’t used was some private communication with DSN operators. Since the XML files describing the configuration of the DSN receivers for Voyager 1 didn’t mention Reed-Solomon either, I had no reason to question this. However, the DSN only processes the spacecraft data up to some point (which usually includes all FEC decoding), and then passes the spacecraft frames to the mission project team without really looking at their contents. Therefore, it might be the case that it’s the project team the one who handles the Reed-Solomon code for the Voyagers. This would make sense specially if the code was something custom, rather than the CCSDS code (recall that Voyager predates the CCSDS standards). If this were true, the DSN wouldn’t really care if there is Reed-Solomon or not, and they might have just forgotten about it.

After looking at the frames I had decoded from Voyager 1 in more detail, I remarked that Brett might be right. Doing some more analysis, I have managed to check that in fact the Voyager 1 frames used Reed-Solomon as described in the references that Brett mentioned. In this post I give a detailed look at the Reed-Solomon code used by the Voyager probes, compare it with the CCSDS code, and show how to perform Reed-Solomon decoding in the frames I decoded in the last post. The middle section of this post is rather math heavy, so readers might want to skip it and go directly to the section where Reed-Solomon codewords in the Voyager 1 frames are decoded.

Computing the symbol error rate of m-FSK

How to compute the symbol error rate rate of an m-FSK modulation is something that comes up in a variety of situations, since the math is the same in any setting in which the symbols are orthogonal (so it also applies to some spread spectrum modulations). I guess this must appear somewhere in the literature, but I can never find this result when I need it, so I have decided to write this post explaining the math.

Here I show an approach that I first learned from Wei Mingchuan BG2BHC two years ago during the Longjiang-2 lunar orbiter mission. While writing our paper about the mission, we wanted to compute a closed expression for the BER of the LRTC modulation used in the uplink (which is related to \(m\)-FSK). Using a clever idea, Wei was able to find a formula that involved an integral of CDFs and PDFs of chi-squared distributions. Even though this wasn’t really a closed formula, evaluating the integral numerically was much faster than doing simulations, specially for high \(E_b/N_0\).

Recently, I came again to the same idea independently. I was trying to compute the symbol error rate of \(m\)-FSK and even though I remembered that the problem about LRTC was related, I had forgotten about Wei’s formula and the trick used to obtain it. So I thought of something on my own. Later, digging through my emails I found the messages Wei and I exchanged about this and saw that I had arrived to the same idea and formula. Maybe the trick was in the back of my mind all the time.

Due to space constraints, the BER formula for LRTC and its mathematical derivation didn’t make it into the Longjiang-2 paper. Therefore, I include a small section below with the details.

Categorised as Maths Tagged ,

ESA NEOCC riddle 1

A few weeks ago, the ESA Nearth Earth Object Coordination Center started a series of NEOCC riddles about Near Earth Object orbits and related topics. The first riddle was about orbits with a peculiar characteristic: they spend 50% of the time inside some fixed radius from the Sun (1.3au in the riddle), and the remaining 50% of the time outside this radius. It was published on June 4. Shortly after that I submitted my solution. The deadline for sending solutions ended yesterday, so today NEOCC has published their solution together with the list of people that solved the riddle correctly. In this post I publish my solution and make some additional comments.

Reverse-engineering the DSCS-III convolutional encoder

One thing I left open in my post yesterday was the convolutional encoder used for FEC in the DSCS-III X-band beacon data. I haven’t seen that the details of the convolutional encoder are described in Coppola’s Master’s thesis, but in a situation such as this one, it is quite easy to use some linear algebra to find the convolutional encoder specification. Here I explain how it is done.

Earth rotation corrections for range and range-rate in GNSS

In GNSS, when considering the propagation of signals from the satellites to a receiver, it is easier to work in an ECI reference frame, since (ignoring the gravitational potential of Earth), light travels in straight lines in ECI coordinates. However, it is often common to do all the calculations in an ECEF frame, as the final goal is to obtain the receiver’s position in ECEF coordinates, and the ephemerides also use ECEF coordinates to describe the satellite positions. Therefore, a non-relativistic correction needs to be applied to account for the fact that light no longer travels in straight lines when one considers ECEF coordinates. Often, the correction is done as some kind of approximation. These types of corrections are known in the GNSS literature as the Sagnac effect.

The goal of this post is to discuss where the corrections arise from, the typical approximations that can be made, and how these corrections affects the calculation of range and range-rate. I didn’t find a good source in the literature where this is described in detail and in a self-contained way, so I decided to write it myself.

Simulating the TED gain for a polyphase matched filter

Trying to improve the performance of the demodulators in gr-satellites, I am switching to the Symbol Sync GNU Radio block, which was introduced by Andy Walls in GRCon17. This block covers the functionality of all the other clock synchronization blocks, such as Polyphase Clock Sync and Clock Recovery MM, while fixing many bugs.

One of the new features of the Symbol Sync block is the ability to specify the gain of the timing error detector (TED) used in the clock recovery feedback loop. All the other blocks assumed unity gain, which simply causes the loop filter taps to be wrong. However, the TED gain needs to be calculated beforehand either by analysis or simulation, as it depends on the choice of TED, samples per symbol, pulse shaping, SNR and other.

While Andy shows how to use the Symbol Sync block as a direct replacement for the Polyphase Clock Sync block in his slides, he leaves the TED gain as one, since that is what the Polyphase Clock Sync block uses. In replacing the Polyphase Clock Sync block by Symbol Sync in gr-satellites, I wanted to use the correct TED gain, but I didn’t found anyone having computed it before. This post shows my approach at simulating the TED gain for polyphase matched filter with maximum likelyhood detector.