[opendtv] Re: "Live" DTV from Mars now possible !
- From: "Manfredi, Albert E" <albert.e.manfredi@xxxxxxxxxx>
- To: <opendtv@xxxxxxxxxxxxx>
- Date: Thu, 12 Aug 2004 18:16:51 -0400
Olivier HOUOT wrote:
> 1) Why should frequency have any influence on propagation
> loss in the vacuum ?
> Isn't a simple inverse square ratio consideration sufficient
> to explain the attenuation of any form of radiative energy,
> whatever the content ?
> Or is it some side effect of the smaller beam angle (i guess
> diffraction laws will allow you to produce a more parallel
> beam at a higher frequency, for a given antenna size) ? But
> that would point to better propagation for high frequencies.
It's caused by the way antenna apertures work at capturing
energy. In essence, the free space path loss equation is
telling you that for an antenna of the same size RELATIVE TO
THE WAVELENGTH, the higher frequency antenna is physically
smaller, and therefore captures less energy (power density).
In other words, the constant here is antenna *gain*.
So for a dish antenna, double the frequency and for the same
antenna size wrt wavelength you cut the radius by 1/2 and the
area by a factor of 1/4. So it follows that you're capturing
less signal.
Of course, if you increase the size of this higher frequency
antenna, you will be increasing its gain. So you can
compensate for the loss of signal level. But you also *need*
to compensate, is the point.
So in fact, it didn't make a lot of sense to make a big
deal about the 8 GHz center frequency, I don't think. What's
to hype? That you need more antenna gain?
> 2) Why the better SNR ? Is it because if you do FM like
> modulation, you can use a bigger frequency deviation for a
> given information because you have more bandwidth available?
> But more bandwidth means more noise...And the modulation
> would rather be QPSK or BPSK. Or is it because natural
> sources of noise are quieter in this frequency domain?
I think you have it in general. In the Shannon equation, one
of the variables is SNR. The way to look at this intuitively
is this:
Assume a channel of X width in Hz. If you want to pump more
and more bits/sec through that fixed width channel, what are
your options?
First, you try to cram as many symbols per second as you
can through the channel. Link them up so there's no gap
between the symbol train, make them beautifully smooth so
they don't slop over beyond your assigned bandwidth, and
you have successfully maximized the symbol rate through
the channel. (Symbols in RF channels typically use a so-
called "raised cosine" shape. Draw a cosine, and raise it
above the X axis. No sharp edges, no wasted bandwidth.)
Now you have the channel all full of these symbols, so you
can't do any better. Right?
Wrong. You can assign more and more bits to each symbol. But
to do so, as you already alluded to, you need to vary
frequency or phase and possibly also amplitude of the
symbols, to be able to distinguish, for example, a 001 symbol
from a 010 symbol. But the more fancy you get with delicate
little variations of the symbol, the more difficult it is to
receive it without errors. Because a noisy channel or noisy
receiver will make it hard to tell the difference between
small changes in amplitude or phase.
So Shannon cleverly takes this into account. Not only that,
his equation also tells us that he leaves it up to you to
decide how much to fix with error correction vs how robust
to make the symbols to begin with. He doesn't care.
And you're right about the wider the channel, the more
the noise energy. Noise levels are typically given as
nV/SQR(Hz), which says that noise power varies with
bandwidth. Which means that these wide band receivers
will have to be very high tech indeed.
So the whole thing is a delicate balancing act. Everything
is a compromise.
Bert
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