Sorry, I had brain rottage.
So, 4 dBW is a little more than 2W (unlike 40dBW, which is 10kW,
indeed). Also, small errors in the SNR formula (suddenly dropped a -1,
but that doesn't really hurt much, there).
On 05.02.2016 11:51, Marcus Müller wrote:
> 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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