Somewhat OT: MythTV / DVB in Israel

[Erez D]

2012/1/14 Udi Finkelstein <Linux-IL at udif.com>

>
>
> On Fri, Jan 13, 2012 at 11:50 AM, Shachar Shemesh <shachar at shemesh.biz>wrote:
>
>>  On 01/12/2012 02:27 PM, Udi Finkelstein wrote:
>>
>> yielding about 30% higher bitrate for the same bandwidth
>>
>> Complete and utter nitpicking.
>>
> If you nitpick, make sure your are correct first...
>
>
>> Bitrate is the number of bits per second (usually measured in kilo bits
>> per second, or kbps, or sometimes mbps). This means that bit rate and
>> bandwidth are, for all practical purposes, one and the same.
>>
>
> I meant every word I said.
> And your assumption that bitrate and bandwidth is the same is definitely
> wrong!
> Every heard the of the distinctions between baud and bit/s?
> Just look at the evolution on modems from the 110bps half duplex to the
> 53600 full duplex (57600 is cheating because it relies on a digital line,
> so its not fail to compare it with earlier standards).
>
> As for DVB-T2, I will not go into the technical details , but feel free to
> look at:
>
> http://en.wikipedia.org/wiki/DVB-T2#System_differences_with_DVB-T
>
>
>>
>> What you (probably) meant to say was that the new encoding allows
>> transferring the same video quality for 30% less bit rate.
>>
>
> Sorry, wrong again...
>
> Unlike earlier DVB-T efforts in the world (e.g. UK's Freeview) that used
> MPEg2, The Israeli standard already used H.264 over the DVB-T physical
> layer from day one. The new transmissions will keep H.264 but will use the
> new DVB-T2 encoding for the HD channels.
>
> Udi
>
>
>
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>


In general. the maximum bit rate which can go through a channel is its
analog bandwidth, multiplied by log2 of the ratio of signal to noise (plus
one, but usually that's negligible).
i.e if you have 1Mhz analog bandwidth and your signal energy is 128 times
your noise energy (what we call 7x3=21 DB), you get to a maximum
theoretical of 1Mhz x 7 = 7 Mbps.
why is that: if you take a signal and change it 1 million times a second -
you will be transferring 1 million symbols a second, and will be using 1MHZ
bandwidth.
now lets say a symbol can be any amplitude between 0 and 127 (because the
difference between levels must be bigger then the noise), then we can
encode 7 bits per symbol, for a total of 7Mbit per second.

i do not know of DVB-T. but if i compare DVB-S (2 bit per symbol) and
DVB-S2 (usually 3 bits per symbol) - you see that DVB-S2 is more sensitive
to noise but have x1.5 bitrate.

there are other considerations: we need error correction. for that we pay,
a simple example: we can send every bit three times, so if we have error,
we still can take a majority vote and know the right value for the bit.
this is of course a very not effective error correction scheme, it is a
rate 1/3 (we use triple bandwidth).
btw DVB-S2 has also a better error correction scheme.

another consideration is the gap, either time domain, or frequency domain.
for example: i have two channels. as analog filters has a transition
region, between what they let pass and what they block, we need a gap which
takes some analog bandwidth.

just my 2c ( and 1$ experience ;-))
erez.
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