How can digital be more spectrum efficient than analog ?

Digital signals needs to represent digits, and for that it pretty much needs quantization in both value *and* time. [And probably some notion of protocol: digital information is not necessarily self-evident or real-time.] Consider the output of a purely analog pulse-width-modulator. The output can't be considered digital, as the information content is continuous, and encoded in the transition times, even though the "values" (voltages) of the output only (nominally) take on two discrete values.

Sometimes official definitions need brushing up as use and techniques change and require finer distinctions.

No, I don't have any experience with FDM or FM microwave.

A given noise floor seems necessary for any analog scheme to merely work. But assuming it works and fills the bandwidth, how does decreasing the noise floor allow FDM to always add more channel capacity?

Thanks.

Except the values are represented by *transition* *times*, not voltages.

The fact that either a voltage or a time (or phase, or whatever) is continuously variable is *not* significant. The physical medium is virtually always continuously variable. The pertinent characteristic is the *value* represented, which is discrete for digital and continuous for analog.

For example, a TTL (binary) digital signal varies from 0 volts to 5 volts. The variation is continuous, and depending on various circuit parameters may change faster or slower (depending on circuit resistance and reactance).

But there is a definition for exactly two values, TRUE and FALSE. Essentially any voltage close to 5 volts is TRUE and anything close to 0 is FALSE.

Here is a typical state chart for TTL digital logic,

TTL input Value 0 voltage < 0.8 Value 1 voltage > 2.0

TTL output Value 0 voltage < 0.4 Value 1 voltage > 2.4

otherwise undefined

The voltage represents three distinct values (two of which are used). When a particular signal line transitions from one value to another, the voltage itself takes time to change and at some point in time during the transition is continuously varied between at least the TTL output limit for the initial logic state and the limit for the final logic state. None of that makes it an analog circuit or an analog signal.

But what needs brushing up is virtually always the way people (mis)understand digital vs. analog, not the definitions.

Stand back for a moment and look at that sentence. There's gotta be something wacky about terminology that says things will work as long as you are above capacity. No wonder people get confused. :-\

Steve

I haven't thought about analog microwave for a *long* time...

The baseband channel will have a given bandwidth and noise floor that depends on the available bandwidth and received signal strength of the microwave system. And the noise floor will go up directly with the amount of signal power applied to the baseband channel.

With no baseband signal at all, the noise floor will be way down. As the baseband power is increased, the noise (which can be measured in any of the FDM channels) will also increase. The application of power in the baseband will cause the RF signal to be modulated of course, and the RF bandwidth transmitted will increase as the power in the baseband is increased.

The transmitter and receiver are bandwidth limited by design. When the actual transmitted bandwidth of the modulated RF signal reaches the bandwidth limit, further increase in baseband power will cause a non-linear change in the demodulator output because the bandwidth cannot be increased (but the spectral distribution within the existing bandwidth will change, which creates noise). The non-linear change results in a very dramatic increase in the noise floor. Instead of being linear, it becomes geometric.

For any given microwave system, the "baseband loading" (power at the baseband input) can be calculated, and measured with a suitable level meter. And it works out just about precisely in practice as the theory says it should, too. All is fine up to the knee of the threshold, and beyond that serious degradation takes place very rapidly with increased power in the baseband.

For example, on a "narrow band" system, say with 12 FDM channels, each channel can have 1/12th of the total power in the baseband, on an average, if they are all loaded equally. For various reasons that level is usually set to be -13 dBm0 (13 dB below the maximum test tone level for a channel). Data tones are set to that level, while signaling tones that are always on are set at -20 dBm0. Active voice channels are assumed to be about the same load as if they were a -13 dBm0 tone. All of which prevents such signals from adding up to enough to overload the entire system, even if every channel is actively being used.

Now consider a wideband FDM system with 1200 channels. The power of a full level test tone is the same, but for any one channel it represents 1/100th as much of the total baseband power as did a test tone on the 12 channel system. That has significance, because it takes 100 times as many full level test tones to overload the baseband on an 1200 channel system as on a

The statement is clearly not correct, so it should not cause confusion. It merely *indicates* confusion...

That's a definition that I can heartily agree with.

I'm afraid that I didn't get that from the supposed governmentally approved definition, which is why I was picking what I perceived to be a nit.

To put the emphasis slightly differently: the pertinent characteristic is the value *represented*: a digit (or digit sequence) or a continuous analog of something.

I'm not convinced that it's necessarily possible to tell by looking at the signal itself, in isolation from its system and purpose. (Although in practice many systems allow a pretty reasonable guess.)

Cheers,

It doesn't confuse comm people who are used to talking about capacity in those terms. e.g., "capacity" for QPSK with R = 1/2 is at ~0.2dB SNR (or Eb/No, it's the same in that case). So if the system is operating at or above ~0.2dB SNR it is possible to achieve error-free transmission in the channel.

So there's no problem with the statement.

Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions.

With the original statement, yes. With a correctly stated case, as you have now provided, no.

You forgot the "C =" in front of that, which will be important in a second.

And with just a teeny bit of algebra one can rearrange that expression to provide the SNR at which a modulation and coding scheme achieving C bps/Hz can operate error-free (i.e., achieve channel capacity). This is commonly done as it is a very practical way to assess the performance of a real scheme.

For a given modulation and coding scheme the spectral efficiency (in bps/Hz) and bandwidth occupancy will be known. So the only unknown, then, is SNR, or Eb/No, however one wishes to compute it. Rearranging the above capacity equation, (well, it's an equation if you take my correction of adding the 'C ='), then provides "capacity" as an SNR for any practical system. It isn't hard at all to normalize out the W, i.e., use W = 1Hz, and then C is still bps/Hz and SNR is per Hz (which is very convenient for computing capacity in terms of Eb/No).

So, contrary to your assertion, "capacity" is very often defined as an SNR and in those cases is a lower limit, below which error-free transmission is not possible.

No need to redefine, since they're the same capacity as I just explained.

So capacity doesn't apply to a twisted pair? That's pretty twisted, alright...

Sure it is. Apply the algebra to the form of the capacity equation you used above and compute capacity as an SNR for the spectral efficiency that the practical system achieves. It's the exact same capacity.

Capacity takes both into account, but when the bandwidth (or the spectral efficiency) is fixed by the modulation and coding then capacity can be easily expressed in terms of an SNR, as I explained above. There's no need to invent new definitions or try to say that they're different, as that's just confusing the issue and obscuring things way more than they need to be. For AWGN channels (which is what we've been discussing, I think), there's just one capacity formula, so there's just one concept of capacity. It can be expressed different ways (e.g., as a curve on a bandwidth-efficiency plane), but it's still the same capacity. That's the only capacity that I've been talking about.

But the coding does, and, as far as I could tell, the table you posted was for uncoded performance (aka, the matched filter bound). You could post the capacity (in SNR) for each modulation and then show how much coding gain one needs to bridge the difference between capacity and the matched filter bound.

Figure one shows up as a red-x for me, but the caption indicates that it's showing the performance of various channel codes for a given system. One could also show a line where "capacity" is (in SNR or Eb/No) for the scheme and show how much gap there is from the various coding schemes to capacity, i.e., how much channel opportunity is left to be exploited. Since none of the codes mentioned are capacity-approaching codes, the gap would be pretty significant.

Uh, yeah, you can. I'll say it again, turning off the FEC wastes channel resources.

If you have enough SNR to make a given modulation scheme work to some desired BER without coding (i.e., with the FEC off), then you have enough SNR to use a higher-order modulation scheme with the FEC turned on, and get more bits through the same channel, probably with better reliability. Again, use the FEC. You're not using the channel efficiently without the FEC turned on.

e.g., your crank the power up until you get your desired BER with uncoded QPSK. With a good code applied (I'm not going to bother to look up the common codes, but this is an easy experiment and a no-brainer with capacity-approaching codes), and without changing the symbol rate (and therefore without changing the bandwidth), you could use 8PSK with R = 2/3 or higher or 16QAM with R = 1/2 or higher. If you use 8PSK with R = 5/6, or 16QAM with R = 3/4, you'll have MORE throughput than you did with the uncoded QPSK and, with good coding, much better reliability.

So if you're ever using a channel with uncoded QPSK, tell me who the customer is and I'll give them more throughput with more reliability by using a higher-order modulation with FEC.

So the reduced information rate due to the FEC is much more than made up for by the increase in power efficiency wth decent codes. If you play the tradeoffs right, and understand them, you'll _always_ be better off with the FEC. This is why we use FEC, and we know that the only way to achieve capacity in a channel is by using powerful FEC.

As I showed earlier, if you understand what you're doing it's all the same capacity formula, so the analysis is consistent.

Hopefully by now you've understood that it is indeed possible, and correct.

Or as an SNR, or as a spectral efficiency. You can rearrange the formula to fit the appropriate unknown. The form that you quoted above, and just below here, defines capacity, C, in terms of spectral efficiency, not BER.

Note the phrase "with arbitrarily small frequency of errors". Pick an "arbitrarily small" error rate, and I'll arbitrarily pick one lower. ergo, error-free transmission.

Already been explained above.

And still waste channel resources. What you're missing is that you should be using a higher-order modulation in that channel. i.e., the channel has a LOT more capacity than you're using if you're transmitting uncoded (i.e., with the FEC turned off). So, yeah, you're wasting the channel resources in a big way.

Or just crank the modulation order up a notch, apply coding, and get even more throughput with better reliability.

Then use a higher-order modulation.

Sigh.

No. You don't seem to understand the concept of channel capacity or how it translates to managing practical channels with practical systems.

Sure, we used to help write the Intelsat standard specs at my old company. I'm not sure how it's relevant to a discussion of capacity, though. All you've shown above is a typical budget for a QPSK, R =

That's like showing the formulas for the speed of sound in certain meteorological conditions and then showing how fast your Cessna 172 flies. So?

So you're not a theory person. I'm guessing you may not even have an engineering degree and so advanced math is something to which you may not have been exposed. This would make your confusion somewhat understandable.

Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions.

Doh! Correction: C is bps unless bandwidth is normalized to W = 1Hz, then it's bps/Hz.

Doh! Correction: As above, C is bps unless bandwidth is normalized to W = 1Hz, then it's bps/Hz. In neither case is the definition BER, rather throughput or spectral efficiency.

Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions.

I've read Shannon, and I've read all of the various "standard" definitions; guess what? They often contradict one another, or are at the very least extremely fuzzy sorts of things for which numerous counter-examples readily come to mind.

If it were ONLY me saying this, you'd have a point. But as to the various standards organizations being wrong - standards organizations don't actually produce the standards they publish, the individuals who make up their various committees do. And like just about any product of a committee, standards are seldom to be considered Holy Writ in any sense other than the "you do it this way, you're in compliance with this standard" one.

As I've already mentioned, I've spent a fair amount of my career rather heavily involved with a number of these same standard organizations, and can say with confidence that I'm reasonably familiar with the processes involved and the quality of the output. In many cases I've come across, I would have to say a very fervent "God help the poor sap who actually tried to learn the basics of this technology from reading the standard!" Like much of what's written by technical specialists, FOR other technical specialists (and therefore not intended as materials intended to train someone new to the field), standards very often contain a high "you already know what we really mean here" factor.

You say toe-may-to, I say toe-mah-to....

But it's still not preferable to a rational argument based on evidence and logic, for the simple fact that we have yet to come up with an "authority" who can realibly be considered to be infallible. If you want to continue to argue that any given group of experts will never be wrong in any pronouncement they make, I can point to any number of counter-examples there as well.

I beg your pardon, but on what basis do you make that judgement? That you disagree with me, and apparently are either incapable or unwilling to understand the arguments I have presented? Or merely because what I have said goes against what you've read in a book that YOU consider to be "Holy Writ"? Can you make an argument in support of your position which does NOT equate to "I'm right because it says so right here in this book!"?

Again, a circular argument. You are arguing for a given definition by saying that any other perspective MUST be wrong because it doesn't agree with that definition!

Bob M.

Not with my original statement, either. If you disagree please state why and make a supportable argument.

Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions.

OH? What about me? :-)

...

Jerry

-- "The rights of the best of men are secured only as the rights of the vilest and most abhorrent are protected." - Chief Justice Charles Evans Hughes, 1927 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Bob you make up a lot of really odd stories. Show me where there is a contradiction between Shannon and the FS-1037C Standard definition that I've quoted.

More bullshit. Show me where someone of any credibility is agreeing with you.

That goes for the rest of your entire article, which was illogical to the point of being nothing but a smoke screen that obfuscates rational discussion of this topic.

If you cannot post rational discussion, or show where someone with some degree of credibility supports something you want to claim, try stuffing it in a trash can, because it doesn't belong here.

We have been over that previously, there is no need to do it again. Your original statement is not the same as your clarification, and if you had merely excused it as an editing blunder or whatever, I think everyone would have accepted that as quite reasonable. But you are *still* insisting that it makes sense... it doesn't.

If you wish to apply for the job, there are forms located over there by the wall. Please simply walk across the pool and pick one up....:-)

Bob M.

I agree it a perfectly *normal* way of expressing it. Its the "no problem" part I can't agree with. :-) Do you think terminology which in plain English says exactly the opposite of what it means is a good idea? There's quite a lot of it as you look around various technologies, especially when mathematicians get involved.

Steve

On the contrary. I've been consistent, but I think your misunderstanding of the concepts makes you think that there's a problem.

If you really think there's a problem, you should be able to point it out and explain why. I've countered all your erroneous "explanations" already, so those don't count.

Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions.