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From: Mark on 5 Aug 2010 10:36
Probably due to poor math skills, I am having a problem reconciling
two aspects of the Shannon limit.

see:
http://www.digitalsignallabs.com/shannonlimit.pdf

written by Randy who I see here frequently..

Aspect 1)
Using Eq 3 and Fig 1, one would conclude that -1.6 dB Eb/No is the
worst channel that is usable. With a 0 dB Eb/No channel, one could
send 1 bit per second / Hz but if the channel degrades to worse then
-1.6 dB, then no matter how slowly one sends the info , it cannot be
received error free.

Aspect2)
Using Eq 1 as in example #2 on Wikipedia here:
http://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem

one can conclude that if you are willing to send the data at a very
slow rate. (50 kbps over a 1MHz BW channel for the given example) the
channel can be as bad as -14.5 dB.

I cannot see how both of these can be true....

What am I missing?

thanks

Mark


From: Jason on 5 Aug 2010 11:03
On Aug 5, 10:36 am, Mark <makol...(a)yahoo.com> wrote:
> Probably due to poor math skills, I am having a problem reconciling
> two aspects of the Shannon limit.
>
> see:http://www.digitalsignallabs.com/shannonlimit.pdf
>
> written by Randy who I see here frequently..
>
> Aspect 1)
> Using Eq 3 and Fig 1, one would conclude that  -1.6 dB  Eb/No is the
> worst channel that is usable.  With a 0 dB Eb/No channel,  one could
> send 1 bit per second / Hz but if the channel degrades to worse then
> -1.6 dB, then no matter how slowly one sends the info , it cannot be
> received error free.
>
> Aspect2)
> Using Eq 1 as in example #2 on Wikipedia here:http://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem
>
> one can conclude that if you are willing to send the data at a very
> slow rate. (50 kbps over a 1MHz BW channel for the given example)  the
> channel can be as bad as -14.5 dB.
>
> I cannot see how both of these can be true....
>
> What am I missing?
>
> thanks
>
> Mark

Your confusion comes from the different terms that are used to
describe "signal-to-noise ratio." Your first reference concludes that
you need an Eb/No of -1.6 dB for reliable communication (with an
infinite block size in your coding scheme). This ratio is the energy
per bit divided by the noise power spectral density (i.e. power per
Hz) over the transmission bandwidth. This is one quantity that is
often used to express signal to noise ratio in digital communications.

Your second reference explicitly uses SNR, or S/N. This is the ratio
of the signal power level to the noise power level, both measured
across the transmission bandwidth. This is a different measure of
signal quality.

The two values are closely related. Recall that Eb = S*T, where T is
the bit period. Also, assuming the noise is white over the band of
interest, No = N / B, where B is the transmission bandwidth. So, Eb/No
can be written as:

Eb/No = S*T / (N / B) = (S/N) * B * T

In your second example, you have:

S/N = -14.5 dB
B = 1 MHz
T = 1 / (50000) sec
Eb/No = (S/N) * B * T
Eb/No (dB) = (S/N) (dB) + 10*log10(B*T) = -14.5 dB +
10*log10(1000000 / 50000000) = -1.5 dB

which, apart from rounding errors in the calculations that yielded
-14.5 dB, gives you the ~-1.6 dB bound for Eb/No that you had in the
first reference.

Jason
From: Mark on 5 Aug 2010 21:27

>
> > I cannot see how both of these can be true....
>
> > What am I missing?
>
> > thanks
>
> > Mark
>
> Your confusion comes from the different terms that are used to
> describe "signal-to-noise ratio."
> Jason

Jason,
got it...
thank you
Mark

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