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This thread was posted on com.dcom.modems, however I thought perhaps it may
be more relevant to this group... Thanks for reading/replying, Sam

Thanks Floyd, brilliant, however I'm still a 'bit' unclear about a few
things...
I'm a software developer with a Maths and Computing degree, so this physcial
stuff tends to fry my brain a bit!
I did high school physics but that was 20+ years ago!
Comments (questions) inline...

"Floyd L. Davidson" <> wrote in message
news:...
> (SamD) wrote:
> >I'm writing a high school textbook. The current chapter includes
> >descriptions of bandwidth, baud, etc...
> >
> >Bandwidth is the frequency range of the channel. Baud is the number of
> >signal events per second, or symbols transferred per second. Most
> >modulation schemes alter the amplitude and phase of the signal to
> >create each symbol, but not the frequency.
>
> Yep.

Great, I got something right!!

>
> >Furthermore each symbols
> >represents a pattern of bits.
>
> That is true, but you don't want to look at it that way because
> it will confuse people.

I understand your point, however it's an IT textbook, hence the starting
point for this part of the book is the transmission of binary digital data.

>
> Each symbol represents a unique /value/ assigned to each
> different state in a finite set of discrete states. (BTW, that
> is the definition of "digital data"). Whatever is varied by the
> modulation (voltage, frequency, phase, etc.) is divided into
> that many distinct divisions. It does not necessarily relate
> directly to bits as such. There could, for example, be 47
> discrete states. A bit is a binary, or 2 state, symbol.
>
> While a binary system has 2 states, there are other "m-ary"
> systems that use multiple states. Another common encoding is
> PCM where there are 255 states. ISDN uses 4 states.
>
> Generally it is true that these states in practical systems can
> indeed be translated directly to a binary encoding, mostly
> because we use the data for digital computers that use binary
> data. But it isn't necessarily so and viewing it only as so
> many bits per symbol limits ones perspective on what digital
> data actually is.
>
> Modems typically use 2 states only for FSK coding. Bell 212A
> modems at 1200 bps use 4 states (two bits are sent per symbol).
> With v.32 and v.34 there are several variations possible.
>
> The point is that if the encoding scheme can send 2 states,
> there are exactly two value that a symbol can have, and that is
> a binary system that sends 1 bit per symbol. If it has a set of
> 4 states, it can send two bits with each symbol. Every time the
> number of states is doubled, another bit is added to the number
> that each symbol represents.
>
> The interesting part, which relates to your next question, is
> what happens as a result of increasing the number of discrete
> values a symbol can have. The symbol rate essentially
> determines the bandwidth required for the signal. But if you
> recall the Shannon-Hartley equations for channel capacity, in
> addition to bandwidth, signal to noise ratio is also factor.
> Think about that terms of the number of discrete values that a
> symbol can have...
>

Yep, I get that in terms of Shannon-Hartley Law. Maybe my problem is to
understand why the Shannon-Hartley law holds.

> If everything else remains constant, whatever type of modulation
> we use to change the signal is divided into smaller increments
> for each value when we increase the number of values each symbol
> can have. If we divide it into only two, it takes relatively
> twice the amount of "noise" to mask the difference between fully
> on and fully off than it does to mask a change if we have
> divided it into 4 different values. Hence the signal to noise
> ratio required for 4 values is 3 dB greater than that required
> for only 2 values.
>
> Obviously dividing a symbol into 64 or 256 or some even higher
> number requires even lower levels of noise. This is true no
> matter what the modulation is. If it is amplitude modulation,
> the "noise" is any undesired variation in amplitude. If it is
> phase, then "noise" is any undesired change in phase.
>
> Hence, multi-level encoding schemes trade bandwidth for SNR.
>
> That is why a bandwidth limited channel, such as a telephone
> connection, can make good use of an m-ary encoding scheme. A
> simple binary scheme uses up the bandwidth, but makes poor use
> of the available SNR. Increasing the number of values for each
> symbol makes use of that SNR without increasing the bandwidth
> necessary.
>

Yes, intuitively I understand this as something like:
Some level of noise is present on any channel (we assume this noise to be
constant). As the number of symbols represented per second increases, then
the noise present makes it more and more difficult to distinguish between
each symbol. Now, if the bandwidth is increased then the differences between
symbols can be greater, hence we are able to better distinguish individual
symbols.
Is this correct?

My problem is grasping how increased bandwidth physically permits larger
differences between symbols. I think my fundamental understanding of
electro-magnetic waves may well be the problem. In my muddled brain I'm
thinking: "OK, only the the amplitude and phase of the wave are being
altered (using say QAM), so how the hell does the provision of extra
frequencies or a wider range of frequencies (ie. via increased bandwidth)
help? Frequency is not even being used to represent the symbols."

> >From what I've read, increasing the bandwidth allows a larger baud
> >rate.
>
> More bandwidth means a higher symbols rate can be used. But
> "baud" *is* a rate, so saying "baud rate" is technically
> referring to how fast the symbols rate is changing! (Of course
> it is commonly misused to mean symbols rate, but if you are
> writing a text book you want to stay away from "baud rate". And
> since "baud" is correct, but confusing, don't use it either!)
>

Unfortunately baud is a term used in the syllabus for the course, so I must
define and use it. My understanding is (or was) that a baud is an individual
signal event, hence baud rate refers to the number of signal events per
second (or sym/s). So 1200baud is really shorthand for a baud rate of 1200,
meaning 1200 signal events occur each second.

> >In fact it seems bandwidth and baud have roughly a 1:1
> >relationship. e.g. a bandwidth of 6MHz accommodates just over 5Msym/s,
> >similarly a bandwidth of 200KHz accommodates 160Ksym/s, etc...
>
> For any given type of modulation, at any given SNR, there will
> be a 1:1 relationship between increasing the symbols rate and
> the bandwidth used.
>

Yes, but why? (see above)

> >Is this all correct? And if so then what is the reason for this near
> >1:1 relationship?
>
> Consider a very simple example using a binary system with
> amplitude modulation of a voltage. +1 volt is 1, and -1 volt is
> 0. The lowest frequency is sent when any string of symbol is
> all 1's or all 0's. That has a frequency of 0!
>

It would be a straight line? Is this a frequency of 0 or is it infinite?

> The highest frequency is sent when symbols alternate between 1
> and 0 continuously. The symbol rate is then exactly twice the
> fundamental frequency of the signal. (And in fact, everything
> above that can be filtered out without introducing errors.) So
> for a binary system, the minimum bandwidth is twice the symbols
> rate, and that is a constant ratio. Increasing the symbols rate
> will necessarily increase the highest fundamental frequency to
> maintain the ratio.
>

I understand that alternating 1s and 0s result in the highest frequency.
However, I'm not sure what you mean by the "fundamental frequency of the
signal"? Hence I'm lost... And I think this may well be precisely what I
need to grasp!

> Note that with multi-level encoding the ratio might be
> different, depending on the type of modulation. But once a
> modulation scheme is selected, whatever the ratio is it will be
> constant and increasing the symbols rate will increase the
> bandwidth by that ratio.
>

Are modulation schemes developed and taylored so that the symbol rate to
bandwidth ratio is close to 1:1? Is this an aim?

> Not all modulation schemes are created equal, and trying to find
> one that balances bandwidth to SNR to match whatever the link
> that carries the data is important. It varies dramatically
> between such things as a typical PSTN telephone connection to
> what would be expected on over a fiber optic cable, for example.
>
> --
> FloydL. Davidson <http://web.newsguy.com/floyd_davidson>
> Ukpeagvik (Barrow, Alaska)