spiney wrote:
you're still doing it!


Doing WHAT?


Is my assertion that your use of the Shannon formula is wrong correct,
or is it incorrect?


This is Shannon's theorem:

C = B log2 (1 + SNR)

It is a theoretical formula which you rightly say is an upper bound, but
you were the person that appealed to Shannon's theorem when you said
this:

"A smaller antenna is inherently worse, as SNR is lower, therefore
channel capacity is reduced, regardeless of other factors (see
Shannon's formula)."

So, all I'm doing is using that formula.


I don't dispute the arithmetic, it's the science that's wrong.


You need to say what science is wrong.


Is the example I have given correct, or is it wrong?


This is your example:

"For example, 4Khz bandwidth and S/R of 7 gives same C value as 3Khz
bandwidth and S/R of 15, but noise power also decreases by a quarter,
which means a 25% bandwidth reduction needs a 60% S/R increase for same
C value."

No, that is not correct, because you're mixing up SNR and signal power:

A change from an SNR of 7 to an SNR of 15 is a change of 114%. But SNR
is

SNR = signal power / noise power

so a reduction in the noise power by 25% along with the change of SNR
from 7 to 15 shows that it is the SIGNAL POWER that changes:

7 x (1.6 S / 0.75 N) = 15

i.e. the S is increased by 60% and the noise is reduced by 25%.

But if you take the correct change of 114% that does back-up my original
assertion that C is weakly dependent on SNR, because a reduction in
bandwidth of 25% requires an increase in SNR of 60%. That proves that C
is more dependent on changes in bandwidth, which is what I've been
saying all along.

It's obvious that this is true from the formula:

C = B log2 (1 + SNR)

ignore the 1 in the formula (e.g. at highish SNRs):

C = B log2 SNR

for log2 SNR = constant, then

C = B x constant - i.e. C is directly proportional to B; double B
implies a doubling of C

for B = constant:

C = log2 SNR x constant

Here's a table of log2 values

SNR - log2 SNR
2 1
4 2
8 3
16 4
32 5
64 6
128 7
256 8
512 9
1024 10

So, increasing the capacity by a factor of 10 for a given B requires the
SNR to be increased by a factor of 512.

To increase the capacity by a factor of 10 for a given SNR requires the
bandwidth only to be increased by a factor of 10 - direct
proportionality.


Why are you not taking any account of changing noise power?


I am, because I'm using SNR.


But ok, i did another typo, yes QPSK is used in sat mod (90 deg shift,
not 180).


180 degree shift is BPSK, not DPSK. DPSK refers to differential phase
shift keying, which is not used on DVB-S.


Which doesn't change anything I said (but it would have been
wrong if I'd given a numerical example!).

I pointed out that "spectral efficiency" does not give complete
information.


Spectral efficiency is a figure of merit for a digital communications
system. It is probably the most important figure of merit, so ignore it
at your peril Analogue Man.


DTT COFDM is multicarrier, and also operates at much
lower SNR than QPSK,


Required C/N for DVB-T using 64-QAM code rate 2/3 is 19.3 dB.

If you look at Figure 3 on page 4 of this:

http://www.ebu.ch/trev_300-morello.pdf

Required C/N for DVB-S for 33 Mbps with a symbol rate of 22 Msymbols/s
is only 7 dB.

So, you're wrong.


as satellite needs much higher SNR to cope with
rain attenuation etc (wait for rain, watch SNR go right down!).


You're just confusing things AGAIN. SNR is a value AT THE RECEIVER
INPUT. NOT the transmitter power. So SNR is a value that is measured
AFTER RAIN FADE has occurred.


--
Steve - www.digitalradiotech.co.uk - Digital Radio News & Info

Find the cheapest Freeview, DAB & MP3 Player Prices:
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http://www.digitalradiotech.co.uk/mp...e_capacity.htm

Adrian wrote:
DAB sounds worse than FM wrote:
spiney wrote:
Unfortunately, "DAB" is wrong in most of what he says, as I shall
now show .....


Seeing as you've been talking nonsense up to now I doubt what you
will show will prove a thing.

I see that spiney is still posting,


Indeed.


I'd hoped he'd gone as I hadn't
seen any replies to him for acouple of days.


He seems to be one of those people that no matter how wrong he gets
things he always thinks he's right. You'd have thought he'd have learnt
his lesson when he started going on about MP3 standing for MPEG-3, which
is of course ludicrous. But no.


--
Steve - www.digitalradiotech.co.uk - Digital Radio News & Info

Find the cheapest Freeview, DAB & MP3 Player Prices:
http://www.digitalradiotech.co.uk/fr..._receivers.htm
http://www.digitalradiotech.co.uk/da...tal_radios.htm
http://www.digitalradiotech.co.uk/mp...rs_1GB-5GB.htm
http://www.digitalradiotech.co.uk/mp...e_capacity.htm

I'm interested in facts, ie, why technical systems behave as they do.
DAB has said lots of things which are completely and very bizarrely
wrong. I'm interested in why. Whenever I try to find out, he calls me
rude names, then quickly moves on to something else!

However, I have managed to "pin him down" on Information Theory,
where he's shown utter ignorance, and given "examples" which are
complete drivel.

This is an important point. So, although I detest name dropping,
let's go straight to Shannon's original paper on Information
Theory, to see exactly what he says about "that famous formula".

There's a copy at
http://cm.bell-labs.com/cm/ms/what/s...hannon1948.pdf .

"Theorem 18"

" ......If the noise is itself white, N=N1 and the result reduces to
the formula proved previously: C=Wlog(1+P/N). If the noise is Gaussian,
but with a spectrum which is not necessarily flat, N1 IS THE GEOMETRIC
MEAN OF THE NOISE POWER OVER THE VARIOUS FREQUENCIES IN THE BAND W.
Thus ......."

(and an integral is then given).

I've used capital letters, for emphasis, but the words are entirely
Shannon's (because this is a direct quote!).

Ok, then, there's the famous formula. But, as Shannon makes quite
clear, it's only conditionally true. There's a further condition,
ie, a separate relationship between W and N, which must also hold true.
Otherwise, the formula becomes invalid, and any "calculations" done
with it will be meaningless gibberish!

So, there's the formula, relating variables C,W, S, N . Initially, we
can give these any values we like! But then, if we change W, we must
also accordingly re-calculate the corresponding value of N. If we
don't, then the formula no longer holds.

So, each time we change W, we must also re-calculate N. Consequently,
each time, there's a quite different numerical relationship between
C, W, and S.

(So, there we are! Shannon's exact words, and my explanation of what
they mean for his formula. Feel entirely free to compare both versions,
very closely, as many times as you like.).

OK, then, what does "DAB" do? He completely ignores this! He
changes W, but doesn't re-calculate N. And, the result is gibberish.
Complete rubbish. Not Information Theory at all!

Above, I've given an actual example (with "nice" numbers, from a
textbook). A 25% reduction in bandwidth gives a 60% reduction in S/R
(requiring a 60% increase in S, if we want to keep C the same!). But,
DAB thinks this is wrong.

In fact, not only is C very strongly dependent on S/R, but, as W tends
towards infinity, the W term completely disappears, leaving C dependent
only on S/R (a result derived by Shannon, just after the above quote,
and repeated in most text books). It's one of those "funny"
results, completely counter-intuitive, but quite true!

But, we don't really need Shannon to make this point, just some
experience, and common sense!

Have you got a Sky digibox or Freeview receiver? Look at the "signal
quality" (ie, SNR, or S/R in the formula), and try slowly rotating
the dish or aerial. What happens? The SNR very rapidly becomes worse.
In fact, there's a "safety margin" of about 10dB (20dB for
satellite, to allow for rain!). Once you go beyond that, reception
suddenly stops!

But, according to DAB, S/R has only a very tiny unnoticeable effect on
the channel capacity, C.

Let's go further, and unplug the antenna (Why not? It only gives
trouble!). Then, there's something like a drop of 10,000 (let's
say) in S/R. But, according to DAB's method of calculating, this will
reduce channel capacity by approximately 10 bits/s. Not even
measurable, never mind not noticeable! So that's alright, then, since
the reception is entirely unaffected. In fact, we don't really need
antennas at all!

(Oh, I forgot to say how you calculate N. Anyone like to show DAB
how?).

I'm interested in facts, ie, why technical systems behave as they do.
DAB has said lots of things which are completely and very bizarrely
wrong. I'm interested in why. Whenever I try to find out, he calls me
rude names, then quickly moves on to something else!

However, I have managed to "pin him down" on Information Theory,
where he's shown utter ignorance, and given "examples" which are
complete drivel.

This is an important point. So, although I detest name dropping,
let's go straight to Shannon's original paper on Information
Theory, to see exactly what he says about "that famous formula".

There's a copy at
http://cm.bell-labs.com/cm/ms/what/s...hannon1948.pdf .

"Theorem 18"

" ......If the noise is itself white, N=N1 and the result reduces to
the formula proved previously: C=Wlog(1+P/N). If the noise is Gaussian,
but with a spectrum which is not necessarily flat, N1 IS THE GEOMETRIC
MEAN OF THE NOISE POWER OVER THE VARIOUS FREQUENCIES IN THE BAND W.
Thus ......."

(and an integral is then given).

I've used capital letters, for emphasis, but the words are entirely
Shannon's (because this is a direct quote!).

Ok, then, there's the famous formula. But, as Shannon makes quite
clear, it's only conditionally true. There's a further condition,
ie, a separate relationship between W and N, which must also hold true.
Otherwise, the formula becomes invalid, and any "calculations" done
with it will be meaningless gibberish!

So, there's the formula, relating variables C,W, S, N . Initially, we
can give these any values we like! But then, if we change W, we must
also accordingly re-calculate the corresponding value of N. If we
don't, then the formula no longer holds.

So, each time we change W, we must also re-calculate N. Consequently,
each time, there's a quite different numerical relationship between
C, W, and S.

(So, there we are! Shannon's exact words, and my explanation of what
they mean for his formula. Feel entirely free to compare both versions,
very closely, as many times as you like.).

OK, then, what does "DAB" do? He completely ignores this! He
changes W, but doesn't re-calculate N. And, the result is gibberish.
Complete rubbish. Not Information Theory at all!

Above, I've given an actual example (with "nice" numbers, from a
textbook). A 25% reduction in bandwidth gives a 60% reduction in S/R
(requiring a 60% increase in S, if we want to keep C the same!). But,
DAB thinks this is wrong.

In fact, not only is C very strongly dependent on S/R, but, as W tends
towards infinity, the W term completely disappears, leaving C dependent
only on S/R (a result derived by Shannon, just after the above quote,
and repeated in most text books). It's one of those "funny"
results, completely counter-intuitive, but quite true!

But, we don't really need Shannon to make this point, just some
experience, and common sense!

Have you got a Sky digibox or Freeview receiver? Look at the "signal
quality" (ie, SNR, or S/R in the formula), and try slowly rotating
the dish or aerial. What happens? The SNR very rapidly becomes worse.
In fact, there's a "safety margin" of about 10dB (20dB for
satellite, to allow for rain!). Once you go beyond that, reception
suddenly stops!

But, according to DAB, S/R has only a very tiny unnoticeable effect on
the channel capacity, C.

Let's go further, and unplug the antenna (Why not? It only gives
trouble!). Then, there's something like a drop of 10,000 (let's
say) in S/R. But, according to DAB's method of calculating, this will
reduce channel capacity by approximately 10 bits/s. Not even
measurable, never mind not noticeable! So that's alright, then, since
the reception is entirely unaffected. In fact, we don't really need
antennas at all!

(Oh, I forgot to say how you calculate N. Anyone like to show DAB
how?).

spiney wrote:
I'm interested in facts, ie, why technical systems behave as they do.
DAB has said lots of things which are completely and very bizarrely
wrong.


FFS, you are one of the most annoying posters I've seen in a long while.


I'm interested in why. Whenever I try to find out, he calls me
rude names, then quickly moves on to something else!


I've I'd have wanted to move on to something else I wouldn't have
carried on replying to your posts.


However, I have managed to "pin him down" on Information Theory,
where he's shown utter ignorance,


Just for the record, Information Theory was a subject I studied at
university, and IIRC I got about 85% in the exam.


and given "examples" which are
complete drivel.


Only yesterday you claimed the the SNR needs to go up 60% when in fact
it was the signal power that needed to go up 60% and the SNR had to go
up 114%.


This is an important point. So, although I detest name dropping,
let's go straight to Shannon's original paper on Information
Theory, to see exactly what he says about "that famous formula".

There's a copy at
http://cm.bell-labs.com/cm/ms/what/s...hannon1948.pdf .

"Theorem 18"

" ......If the noise is itself white, N=N1 and the result reduces to
the formula proved previously: C=Wlog(1+P/N). If the noise is
Gaussian, but with a spectrum which is not necessarily flat, N1 IS
THE GEOMETRIC MEAN OF THE NOISE POWER OVER THE VARIOUS FREQUENCIES IN
THE BAND W. Thus ......."

(and an integral is then given).

I've used capital letters, for emphasis, but the words are entirely
Shannon's (because this is a direct quote!).

Ok, then, there's the famous formula. But, as Shannon makes quite
clear, it's only conditionally true.


http://en.wikipedia.org/wiki/Shannon%27s_theorem

"In information theory, the Shannon-Hartley theorem states the maximum
amount of error-free digital data (that is, information) that can be
transmitted over a communication link with a specified bandwidth in the
presence of noise interference. The law is named after Claude Shannon
and Ralph Hartley. The Shannon limit or Shannon capacity of a
communications channel is the theoretical maximum information transfer
rate of the channel."

Note the last sentence in particular:

"The Shannon limit or Shannon capacity of a communications channel is
the theoretical maximum information transfer rate of the channel"

That's obviously important in digital communication systems design.


There's a further condition,
ie, a separate relationship between W and N, which must also hold
true. Otherwise, the formula becomes invalid, and any "calculations"
done with it will be meaningless gibberish!


No, it never becomes meaningless gibberish. It is a theoretical limit
above which error-free communication is impossible.


So, there's the formula, relating variables C,W, S, N . Initially, we
can give these any values we like!


No, you cannot give any values you like, because for the assumptions
given this is a mathematical proof, so if you provide values of W, S and
N you get a value for C.


But then, if we change W, we must
also accordingly re-calculate the corresponding value of N. If we
don't, then the formula no longer holds.


For God's sake. The formula is an equality, so if one variable in the
equation changes then the others HAVE TO change, OBVIOUSLY.


So, each time we change W, we must also re-calculate N. Consequently,
each time, there's a quite different numerical relationship between
C, W, and S.

(So, there we are! Shannon's exact words, and my explanation of what
they mean for his formula.


Your explanation is so incredibly stupid that it defies belief. What
you're effectively doing is showing surprise that when you change a
value in an equality other values must also change so that both sides of
the equation balance.

Your explanation is about as profound as saying that 2 x 2 = 4. It
really is as ridiculously stupid as suggesting that we should be
surprised that 2 x 2 = 4.

Here's an analogy to what you're saying:

2 x 2 = 4

Now, keep 4 on the right hand side and change the first value to 1:

1 x 2 = 4 - Incorrect

and notice that the equation doesn't balance, so you have to change the
second value to 4 to make the equation balance:

1 x 4 = 4 - Correct

Give spiney a Nobel Prize for mathematics.


Feel entirely free to compare both
versions, very closely, as many times as you like.).


You make it sound like it is difficult to find fault with your logic. Au
contraire. My 8 year old niece is better at maths than you are.


OK, then, what does "DAB" do? He completely ignores this!


I certainly do work on a slightly higher plane than 2 x 2 = 4, I must
admit.


He
changes W, but doesn't re-calculate N. And, the result is gibberish.
Complete rubbish. Not Information Theory at all!


Total bull****, because I use SNR, and don't separate S from N.


Above, I've given an actual example (with "nice" numbers, from a
textbook). A 25% reduction in bandwidth gives a 60% reduction in S/R


You're doing it again.


(requiring a 60% increase in S, if we want to keep C the same!). But,
DAB thinks this is wrong.


It is wrong. Here's your original assertion:

"For example, 4Khz bandwidth and S/R of 7 gives same C value as 3Khz
bandwidth and S/R of 15, but noise power also decreases by a quarter,
which means a 25% bandwidth reduction needs a 60% S/R increase for same
C value."

(i)

C = B log2 (1 + SNR)

C = 4000 log2 (1 + 7) = 12000 bits/s

(ii)

12000 = 3000 log2 (1 + SNR)

SNR = 15

15 / 7 = 2.14

i.e. SNR has increased by 114%, but 60%. SIGNAL POWER HAS INCREASED BY
60%, NOT SNR.



In fact, not only is C very strongly dependent on S/R,


Here's a table of log2 values

SNR - log2 SNR
2 1
4 2
8 3
16 4
32 5
64 6
128 7
256 8
512 9
1024 10

So, increasing the capacity by a factor of 10 for a given B requires the
SNR to be increased by a factor of 512.

To increase the capacity by a factor of 10 for a given SNR requires the
bandwidth only to be increased by a factor of 10 - direct
proportionality.

Conclusion: Capacity is weakly dependent on SNR, but is strongly
dependent on bandwidth.

QED.



but, as W tends
towards infinity, the W term completely disappears,


It doesn't completely disappear at all. Look at the equation:

C = W log2 (1 + SNR)

split SNR into S / N.W (N.W = noise x bandwidth)

C = infinity log2 (1 + S / infinity) = infinity log2 (1) = 0

What you're doing is confusing SNR with SIGNAL POWER. This is what
you've done from the start.


leaving C
dependent only on S/R (a result derived by Shannon, just after the
above quote, and repeated in most text books). It's one of those
"funny"
results, completely counter-intuitive, but quite true!


I don't see why it's counter-intuitive, because if you have a
finite-valued signal power spread across an infinite bandwidth then the
signal will vanish altogether and all that remains is noise.


But, we don't really need Shannon to make this point, just some
experience, and common sense!

Have you got a Sky digibox or Freeview receiver? Look at the "signal
quality" (ie, SNR, or S/R in the formula),


Could you stop calling it S/R? Signal / Ratio? It doesn't make any
sense. Please call it S/N if you're going to put a divide sign in there.


and try slowly rotating
the dish or aerial. What happens? The SNR very rapidly becomes worse.


Shock horror! You have a capacity of zero when you don't receive
anything. My God, a Nobel prize is being loaded into the DHL van as we
speak.


In fact, there's a "safety margin" of about 10dB (20dB for
satellite, to allow for rain!). Once you go beyond that, reception
suddenly stops!

But, according to DAB, S/R has only a very tiny unnoticeable effect on
the channel capacity, C.


No, what I am saying is that within the normal operating region of a
digital communication system, C is far more dependent on bandwidth than
on ************SNR*************. SNR IS NOT NOT NOT NOT NOT NOT SIGNAL
POWER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!

This is what you said originally, and it's what you've maintained all
along:

"For example, 4Khz bandwidth and S/R of 7 gives same C value as 3Khz
bandwidth and S/R of 15, but noise power also decreases by a quarter,
which means a 25% bandwidth reduction needs a 60% S/R increase for same
C value."

That is incorrect, because you're confusing SNR with SIGNAL POWER.

We cannot move forward until you realise your mistake.


Let's go further, and unplug the antenna (Why not? It only gives
trouble!). Then, there's something like a drop of 10,000 (let's
say) in S/R. But, according to DAB's method of calculating, this will
reduce channel capacity by approximately 10 bits/s. Not even
measurable, never mind not noticeable! So that's alright, then, since
the reception is entirely unaffected. In fact, we don't really need
antennas at all!


Using identical logic to yours, reduce the bandwidth to zero:

C = B log2 (1 + SNR) = 0

No matter how high the SNR, with a bandwidth of zero you have a capacity
of zero.


--
Steve - www.digitalradiotech.co.uk - Digital Radio News & Info

Find the cheapest Freeview, DAB & MP3 Player Prices:
http://www.digitalradiotech.co.uk/fr..._receivers.htm
http://www.digitalradiotech.co.uk/da...tal_radios.htm
http://www.digitalradiotech.co.uk/mp...rs_1GB-5GB.htm
http://www.digitalradiotech.co.uk/mp...e_capacity.htm

In message
"DAB sounds worse than FM" wrote:

spiney wrote:
However, I have managed to "pin him down" on Information Theory,
where he's shown utter ignorance,


Just for the record, Information Theory was a subject I studied at
university, and IIRC I got about 85% in the exam.

Not 100 per cent, then?

Perhaps it's in the darkness of that missing 15 per cent that your
present disagreement lies....

--
Richard

What, no reply then? Nothing from DAB, Adrian, Etillet? Usually, your
feedback is so quick (and sarcastic)!

spiney wrote:
What, no reply then? Nothing from DAB, Adrian, Etillet? Usually, your
feedback is so quick (and sarcastic)!


You obviously missed this reply to your post yesterday (not surprising
seeing as you sent the same post in reply to 2 different people....):

spiney wrote:
I'm interested in facts, ie, why technical systems behave as they do.
DAB has said lots of things which are completely and very bizarrely
wrong.


FFS, you are one of the most annoying posters I've seen in a long while.


I'm interested in why. Whenever I try to find out, he calls me
rude names, then quickly moves on to something else!


I've I'd have wanted to move on to something else I wouldn't have
carried on replying to your posts.


However, I have managed to "pin him down" on Information Theory,
where he's shown utter ignorance,


Just for the record, Information Theory was a subject I studied at
university, and IIRC I got about 85% in the exam.


and given "examples" which are
complete drivel.


Only yesterday you claimed the the SNR needs to go up 60% when in fact
it was the signal power that needed to go up 60% and the SNR had to go
up 114%.


This is an important point. So, although I detest name dropping,
let's go straight to Shannon's original paper on Information
Theory, to see exactly what he says about "that famous formula".

There's a copy at
http://cm.bell-labs.com/cm/ms/what/s...hannon1948.pdf .

"Theorem 18"

" ......If the noise is itself white, N=N1 and the result reduces to
the formula proved previously: C=Wlog(1+P/N). If the noise is
Gaussian, but with a spectrum which is not necessarily flat, N1 IS
THE GEOMETRIC MEAN OF THE NOISE POWER OVER THE VARIOUS FREQUENCIES IN
THE BAND W. Thus ......."

(and an integral is then given).

I've used capital letters, for emphasis, but the words are entirely
Shannon's (because this is a direct quote!).

Ok, then, there's the famous formula. But, as Shannon makes quite
clear, it's only conditionally true.


http://en.wikipedia.org/wiki/Shannon%27s_theorem

"In information theory, the Shannon-Hartley theorem states the maximum
amount of error-free digital data (that is, information) that can be
transmitted over a communication link with a specified bandwidth in the
presence of noise interference. The law is named after Claude Shannon
and Ralph Hartley. The Shannon limit or Shannon capacity of a
communications channel is the theoretical maximum information transfer
rate of the channel."

Note the last sentence in particular:

"The Shannon limit or Shannon capacity of a communications channel is
the theoretical maximum information transfer rate of the channel"

That's obviously important in digital communication systems design.


There's a further condition,
ie, a separate relationship between W and N, which must also hold
true. Otherwise, the formula becomes invalid, and any "calculations"
done with it will be meaningless gibberish!


No, it never becomes meaningless gibberish. It is a theoretical limit
above which error-free communication is impossible.


So, there's the formula, relating variables C,W, S, N . Initially, we
can give these any values we like!


No, you cannot give any values you like, because for the assumptions
given this is a mathematical proof, so if you provide values of W, S and
N you get a value for C.


But then, if we change W, we must
also accordingly re-calculate the corresponding value of N. If we
don't, then the formula no longer holds.


For God's sake. The formula is an equality, so if one variable in the
equation changes then the others HAVE TO change, OBVIOUSLY.


So, each time we change W, we must also re-calculate N. Consequently,
each time, there's a quite different numerical relationship between
C, W, and S.

(So, there we are! Shannon's exact words, and my explanation of what
they mean for his formula.


Your explanation is so incredibly stupid that it defies belief. What
you're effectively doing is showing surprise that when you change a
value in an equality other values must also change so that both sides of
the equation balance.

Your explanation is about as profound as saying that 2 x 2 = 4. It
really is as ridiculously stupid as suggesting that we should be
surprised that 2 x 2 = 4.

Here's an analogy to what you're saying:

2 x 2 = 4

Now, keep 4 on the right hand side and change the first value to 1:

1 x 2 = 4 - Incorrect

and notice that the equation doesn't balance, so you have to change the
second value to 4 to make the equation balance:

1 x 4 = 4 - Correct

Give spiney a Nobel Prize for mathematics.


Feel entirely free to compare both
versions, very closely, as many times as you like.).


You make it sound like it is difficult to find fault with your logic. Au
contraire. My 8 year old niece is better at maths than you are.


OK, then, what does "DAB" do? He completely ignores this!


I certainly do work on a slightly higher plane than 2 x 2 = 4, I must
admit.


He
changes W, but doesn't re-calculate N. And, the result is gibberish.
Complete rubbish. Not Information Theory at all!


Total bull****, because I use SNR, and don't separate S from N.


Above, I've given an actual example (with "nice" numbers, from a
textbook). A 25% reduction in bandwidth gives a 60% reduction in S/R


You're doing it again.


(requiring a 60% increase in S, if we want to keep C the same!). But,
DAB thinks this is wrong.


It is wrong. Here's your original assertion:

"For example, 4Khz bandwidth and S/R of 7 gives same C value as 3Khz
bandwidth and S/R of 15, but noise power also decreases by a quarter,
which means a 25% bandwidth reduction needs a 60% S/R increase for same
C value."

(i)

C = B log2 (1 + SNR)

C = 4000 log2 (1 + 7) = 12000 bits/s

(ii)

12000 = 3000 log2 (1 + SNR)

SNR = 15

15 / 7 = 2.14

i.e. SNR has increased by 114%, but 60%. SIGNAL POWER HAS INCREASED BY
60%, NOT SNR.



In fact, not only is C very strongly dependent on S/R,


Here's a table of log2 values

SNR - log2 SNR
2 1
4 2
8 3
16 4
32 5
64 6
128 7
256 8
512 9
1024 10

So, increasing the capacity by a factor of 10 for a given B requires the
SNR to be increased by a factor of 512.

To increase the capacity by a factor of 10 for a given SNR requires the
bandwidth only to be increased by a factor of 10 - direct
proportionality.

Conclusion: Capacity is weakly dependent on SNR, but is strongly
dependent on bandwidth.

QED.



but, as W tends
towards infinity, the W term completely disappears,


It doesn't completely disappear at all. Look at the equation:

C = W log2 (1 + SNR)

split SNR into S / N.W (N.W = noise x bandwidth)

C = infinity log2 (1 + S / infinity) = infinity log2 (1) = 0

What you're doing is confusing SNR with SIGNAL POWER. This is what
you've done from the start.


leaving C
dependent only on S/R (a result derived by Shannon, just after the
above quote, and repeated in most text books). It's one of those
"funny"
results, completely counter-intuitive, but quite true!


I don't see why it's counter-intuitive, because if you have a
finite-valued signal power spread across an infinite bandwidth then the
signal will vanish altogether and all that remains is noise.


But, we don't really need Shannon to make this point, just some
experience, and common sense!

Have you got a Sky digibox or Freeview receiver? Look at the "signal
quality" (ie, SNR, or S/R in the formula),


Could you stop calling it S/R? Signal / Ratio? It doesn't make any
sense. Please call it S/N if you're going to put a divide sign in there.


and try slowly rotating
the dish or aerial. What happens? The SNR very rapidly becomes worse.


Shock horror! You have a capacity of zero when you don't receive
anything. My God, a Nobel prize is being loaded into the DHL van as we
speak.


In fact, there's a "safety margin" of about 10dB (20dB for
satellite, to allow for rain!). Once you go beyond that, reception
suddenly stops!

But, according to DAB, S/R has only a very tiny unnoticeable effect on
the channel capacity, C.


No, what I am saying is that within the normal operating region of a
digital communication system, C is far more dependent on bandwidth than
on ***********SNR************. SNR IS NOT NOT NOT NOT NOT NOT SIGNAL
POWER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!

This is what you said originally, and it's what you've maintained all
along:

"For example, 4Khz bandwidth and S/R of 7 gives same C value as 3Khz
bandwidth and S/R of 15, but noise power also decreases by a quarter,
which means a 25% bandwidth reduction needs a 60% S/R increase for same
C value."

That is incorrect, because you're confusing SNR with SIGNAL POWER.

We cannot move forward until you realise your mistake.


Let's go further, and unplug the antenna (Why not? It only gives
trouble!). Then, there's something like a drop of 10,000 (let's
say) in S/R. But, according to DAB's method of calculating, this will
reduce channel capacity by approximately 10 bits/s. Not even
measurable, never mind not noticeable! So that's alright, then, since
the reception is entirely unaffected. In fact, we don't really need
antennas at all!


Using identical logic to yours, reduce the bandwidth to zero:

C = B log2 (1 + SNR) = 0

No matter how high the SNR, with a bandwidth of zero you have a capacity
of zero.


--
Steve - www.digitalradiotech.co.uk - Digital Radio News & Info

Find the cheapest Freeview, DAB & MP3 Player Prices:
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http://www.digitalradiotech.co.uk/mp...rs_1GB-5GB.htm
http://www.digitalradiotech.co.uk/mp...e_capacity.htm

No, DAB, you're completey wrong. About everything.

DAB sounds worse than FM wrote:
spiney wrote:
What, no reply then? Nothing from DAB, Adrian, Etillet? Usually, your
feedback is so quick (and sarcastic)!


You obviously missed this reply to your post yesterday (not surprising
seeing as you sent the same post in reply to 2 different people....):

spiney wrote:
I'm interested in facts, ie, why technical systems behave as they do.
DAB has said lots of things which are completely and very bizarrely
wrong.


FFS, you are one of the most annoying posters I've seen in a long
while.

That's why I killfiled the little twerp long ago. I'm sure he's
deliberately annoying.