Hi Daniel,
On 04.02.2016 22:49, Daniel Pocock wrote:
> To give a more specific example:
>
> a) SDR device sampling the 2 meter band (144 - 148 MHz), this input
> range is locked and can't be changed by users
>
> b) using something like the USRP B200
> - it can do 61 Million samples/sec, 12 bit samples, 732 Mbit/sec
> - but maybe that sample rate is not needed for a band that is 4 MHz wide...
No, 4MS/s should suffice (if you can live with the filter roll-off at
the band edges).
Still, not wasting too much signal quality: for 8bit samples in I and Q,
4MS/s * 2B/S = 8MB/s = 64Mb/s
>
> c) an instance of GNU Radio taking all the samples and encapsulating
> them into packets
>
> d) transmitting to local users
> layer 1/physical: 23cm or 13cm, using 8 - 10 MHz bandwidth
So, since AX.25 doesn't specify a modulation (and if we used the AX.25
that seem to be dominant, we'd end up with a data rate whopping three to
four orders of magnitude too low), let's look at the data rate here to
determine a minimal modulation order and SNR:
Shannon Channel Capacity says that our bitrate $r$ is bound, if we want
to achieve transmission with arbitrarily low bit error rate over a
channel of bandwidth $b$ and given $\text{SNR}$:
\documentclass{article} \usepackage[utf8x]{inputenc}
\usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb}
\usepackage{trfsigns} \DeclareMathOperator*{\argmin}{arg\,min}
\DeclareMathOperator*{\SNR}{SNR} \usepackage{tikz}
\usepackage{circuitikz} \usepackage[binary-units=true]{siunitx}
\sisetup{exponent-product = \cdot} \DeclareSIUnit{\dBm}{dBm}
\newcommand{\imp}{\SI{50}{\ohm}} \newcommand{\wrongimp}{\SI{75}{\ohm}}
\pagestyle{empty} \begin{document} \begin{align*} r &\le b\cdot
\log_2\left(1+\SNR\right)\\ \frac rb &\le \log_2\left(1+\SNR\right)\\
2^{\frac rb} &\le 1+\SNR\\ \SNR &\ge 2^{\frac rb}\\ \intertext{in our
case} \SNR &\ge 2^{\frac{\num{64e6}}{\num{10e6}}} \\ &= 2^{6.4} \\ &=
\left[10\cdot \log_{10}\left(2^{6.4 }\right)\right]\,\text{dB}\\ &=
\left[6.4\cdot3\right]\,\text{dB}\\ &= 19.2\,\text{dB}\,\text{.}
\end{align*} \end{document}
I'd say, wow, for a wide-range 10MHz link, that's a pretty good minimum SNR!
Now, for the modulation:
$\frac{\SI{64}{\mega\bit\per\second}}{\SI{10}{\mega\hertz}}=\SI{6.4}{\bit}$;
i.e. our modulation would have to have at least that many bits per
symbol, which means at least 85 different states.
Effectively, this calls for something like a 128 QAM, or a 256 QAM (from
a gut feeling, this makes sense if SNR is in fact quite a bit higher
than 19.2 dB) ; more likely the latter, because it's a square number,
making the constellation easier to implement, and also, because we'll
definitely want some bitrate headroom to add redundancy for channel
coding/forward error correction, and $\frac{256}{85}=\frac13$, which is
pretty handy for code implementation.
Whether to send those symbols in time-domain or over a set of OFDM or
filterbank carriers would be up for discussion; from an equalization
point of view, using multiple carriers seems to make a lot of sense;
those 10MHz will probably not be nicely flat.
> layer 2: AX.25 (with repeater callsign)
As calculated above, not that much room in those 10 MHz for framing
overhead, the relatively ineffective CRC32 and the 5-bit-stuffing, to be
honest... I don't think AX.25 is the optimum choice here. I'd rather go
for something that has a usuable preamble for equalization, and a more
compressed header, and complements the FEC used more nicely.
> layer 3: IP multicast (UDP packets)
Why that? If we're going to be fully utilizing the link with sample
packets, anyway, it's not really necessary to have different logical
endpoints, right?
>
> e) Receiving stations would receive the UDP multicast packets and feed
> them as input to a flow graph in a local instance of GNU Radio
>
> I can imagine there may be risks with packet loss and the receiving
> users may need directional antennas. As it would be a licensed amateur
> repeater, it would be able to legally put out more watts than a wifi
> router though.
The point is that I lack knowledge about typical SNRs for the 13cm
(2.4GHz) or the 23cm (1.3GHz) bands; problematic for me sounds that free
space loss for 23cm over a distance of 10km would be around
$l=\SI{115}{dB}$.
So with an minimum SNR of $\text{SNR}_\text{min} \approx \SI{20}{dB}$
and a thermal noise floor of $N = N_0 b =
\SI{-174}{\dBm\per\hertz}\cdot \SI{70}{dB\hertz}=\SI{-104}{\dBm}$, and
assuming a relatively nice receiver with a noise figure
$\text{NF}=\SI{3}{dB}$ :
You'd need a transmit power of
$P_\text{TX} = N + \text{NF} + l + \text{SNR}_\text{min} \approx \left[
-104 + 3 + 115 + 20 \right]\si{\dBm} = \SI{4}{dB\watt}$.
So, you can't reliably talk to someone further away than 10km with a
10kW TX, assuming you have no antenna gains. Sure, a very nicely aligned
dish with low losses can achieve almost 30dB, but that effectively only
means that 1kW is enough for 10km.
Best regards,
Marcus
>
> Regards,
>
> Daniel
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