Olivier HOUOT wrote:
Using Shannon's channel capacity formula i would say that in a 27 Mhz bandwidth, assuming noise is the same, going from 58.8 Mbps to 6 Mbps reduces power requirement by a factor of more than 21, so the power would be a bit under 100 Kw. Still in the same ballpark, then.
Indeed. I want to add more modules to my RF computations tool first, before doing more "what if" games, but you did bring up one tradeoff. The channel width. A narrow channel requires less power, mais en revanche, to keep channel capacity the same, the narrow channel requires a higher spectral efficiency (b/s/Hz). This translates to a higher marginal C/N ratio, which against you in the path loss computations.
Another effect is that the wider channel makes the receiver noise figure higher, all else equal, which also can work against you.
And then there is the modulation. When you compute transmit power required, that power applies to the weakest of the symbols transmitted. Not to the average. So if you use something that wastes power, like n-VSB, QAM, or COFDM, the actual transmitter has to be bigger. How much bigger depends on the modulation. For example, n-VSB and n^2-QAM should be close to identical in terms of efficiency, and COFDM would be worse. The best is n-PSK, either in a regular or spread spectrum scheme, because each symbol is transmitted at full power (as Frank Eory pointed out some years ago). So DVB-T2, which includes QAM, is not the best here.
So you actually need more info than we have, to make very accurate tradeoffs, but it does look like we're in the same ballpark anyway.
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