Craig Birkmaier wrote:
First of all, I assume the 704 X 480 analog SD occupying 6 MHz more or less represents NTSC. So in fact, it is 704 X 480 at 60i. And for good, clear reception, in the 6 MHz channel, you need about 40 dB of C/N. Less than that and the image quality suffers very rapidly, compared with an SD 480i.Try a more realistic raster that better reflects the frequency response of NTSC. 480 x 480 interlaced is about right.
I'm not sure what point you're trying to make.Obviously, if the analog is even less capable than I assume Olivier was suggesting, all you're proving is that the analog example is even further away from the Shannon limit compared with HDTV in the 6 MHz channel.
That's fine for me.
Rubbish. Take a look at a spectrum analyzer of ANY MPEG-2 encoded HD source with a bit rate of say 17Mbps (about the max video payload for ATSC). The only stuff you will see above 20 MHz ...
So, did you do the same test with 4.2 MHz in NTSC? I'm baffled by this persistent refrain.
The C/N of the channel is completely irrelevant to this discussion - the frequency response is no longer a function of an analog channel, it is a function of the digital encoder.
I agree that the frequency response of the video signal is no longer directly related to the bandwidth of the RF channel in MHz. But you missed the entire point of Olivier's musings, which I actually found rather interesting.
The Shannon limit certainly does come in play. Bert _________________________________________________________________Dave vs. Carl: The Insignificant Championship Series. Who will win? http://clk.atdmt.com/MSN/go/msnnkwsp0070000001msn/direct/01/?href=http://davevscarl.spaces.live.com/?icid=T001MSN38C07001
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