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#441
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
"glen herrmannsfeldt" wrote in
message Jerry Avins wrote: (snip) So Information Theory tells us that a quantized signal is digital? Consider the output of the limiters in an FM IF driving a Foster-Seely discriminator. It has two states -- saturated and zero -- before the tank that smooths the edges. I guess Information theory says that FM radio is digital (maybe unless you use an Avins-Seely ratio detector, but even those work better with at least one limiter). This sounds like what I previously tried to describe as quantized but not sampled. The signal has two states, but the transition can happen at any time. The signal might be thought of as being quantized in the aplitude domain, but it is clearly not quantized in the time domain. For a signal to be quantized, it has to be fully quantized, that is quantized in both the time domain and the amplitude domain. |
#443
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Floyd L. Davidson wrote: Actually, they are good definitions, but they are *not* "valid standard definitions" for this discussion. They are valid for any discussion in which I care to use them. But thanks for your input. jk |
#444
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
"Arny Krueger" wrote:
"Jerry Avins" wrote in message Floyd L. Davidson wrote: ... Nyquist rate: The reciprocal of the Nyquist interval, i.e., the minimum theoretical sampling rate that fully describes a given signal, i.e., enables its faithful reconstruction from the samples. Note: The actual sampling rate required to reconstruct the original signal will be somewhat higher than the Nyquist rate, because of quantization errors introduced by the sampling process. It does not say what you claimed it does. Do you buy the "because clause? I don't. You do well to disagree with it. It is false. The errors that are introduced are aliasing, not quantization errors. You could change the size of the quantization steps any which way you want, and the aliasing would still be there. That is another technically *absurd* statement. Can you explain how one gets aliasing under the circumstances stated? "The actual sampling rate required to reconstruct the original signal will be somewhat higher than the Nyquist rate, because of quantization errors introduced by the sampling process." All qualified practitioners will recognize that as wrong. Agreed. It's an incorrect statement for the reason I stated above. The reason you state is incorrect, and that should be obvious. Incidentally, I took a look at "Telecommunications System Engineering, Third Edition" by Roger L. Freeman to see what definitions he uses. He cited another reference, and quoted this as "the Nyquist sampling theorem", "If a band-limited signal is sampled at regular intervals of time at a rate equal to or higher than twice the highest significant signal frequency, then the sample contains all the information of the original signal. The original signal may then be reconstructed by use of a low-pass filter." His source is "/Reference/ /Data/ /for/ /Radio/ /Engineers/, 6th Ed., ITT/Howard W. Sams, Indianapolis, 1976. That's two more absolutely credible sources who say you need to learn more about this. -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
#445
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Jim Kelley wrote:
Floyd L. Davidson wrote: Actually, they are good definitions, but they are *not* "valid standard definitions" for this discussion. They are valid for any discussion in which I care to use them. But thanks for your input. That is true. You can use any definition for any word you like, in Alice's Wonderland. And you won't be understood by anyone else, which seems to be the point of many who post this sort of drivel. -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
#446
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Floyd L. Davidson wrote: And you won't be understood by anyone else, which seems to be the point of many who post this sort of drivel. Perhaps true insofar as the word might not be understood by someone who also does not know how to use a dictionary. Admitedly, I had not accounted for that possibility. 8-| jk |
#447
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Jim Kelley wrote:
Floyd L. Davidson wrote: And you won't be understood by anyone else, which seems to be the point of many who post this sort of drivel. Perhaps true insofar as the word might not be understood by someone who also does not know how to use a dictionary. You, for example. Admitedly, I had not accounted for that possibility. 8-| That was fairly obvious, and still is. You are implying that any dictionary definition (hence not your unique Alice in Wonderland definition) is correct in any context. That is not the way a dictionary is properly used. "Digital", for example, has at least three different definitions. You want to be able to pull any one of them out of a hat, and say that it means what it means... But that is back to Alice in Wonderland. The word "Digital" is a Term of Art, and your common language dictionary definition is not valid in a technical discussion. -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
#448
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
"Floyd L. Davidson" wrote in message
"Arny Krueger" wrote: "Jerry Avins" wrote in message Floyd L. Davidson wrote: ... Nyquist rate: The reciprocal of the Nyquist interval, i.e., the minimum theoretical sampling rate that fully describes a given signal, i.e., enables its faithful reconstruction from the samples. Note: The actual sampling rate required to reconstruct the original signal will be somewhat higher than the Nyquist rate, because of quantization errors introduced by the sampling process. It does not say what you claimed it does. Do you buy the "because clause? I don't. You do well to disagree with it. It is false. The errors that are introduced are aliasing, not quantization errors. You could change the size of the quantization steps any which way you want, and the aliasing would still be there. That is another technically *absurd* statement. Can you explain how one gets aliasing under the circumstances stated? Real world signals have finite, not zero bandwidth. That means that they have sidebands. As soon as any of the sidebands are at or exceed the nyquist rate, there is distortion of the wave because of aliasing. Therefore the sample rate has to be somewhat higher than the sample rate of the signal, which is characterized by its carrier frequency. "The actual sampling rate required to reconstruct the original signal will be somewhat higher than the Nyquist rate, because of quantization errors introduced by the sampling process." All qualified practitioners will recognize that as wrong. Agreed. It's an incorrect statement for the reason I stated above. The reason you state is incorrect, and that should be obvious. Incidentally, I took a look at "Telecommunications System Engineering, Third Edition" by Roger L. Freeman to see what definitions he uses. He cited another reference, and quoted this as "the Nyquist sampling theorem", "If a band-limited signal is sampled at regular intervals of time at a rate equal to or higher than twice the highest significant signal frequency, then the sample contains all the information of the original signal. The original signal may then be reconstructed by use of a low-pass filter." Congrats to Freeman for avoiding making a different statement that avoids the error made in the previous statement. |
#449
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Floyd L. Davidson wrote: You are implying that any dictionary definition (hence not your unique Alice in Wonderland definition) is correct in any context. I'm not implying that at all. I'm simply saying that the terms and definitions I used are correct in the context in which I used them. jk |
#450
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Arny Krueger wrote: Incidentally, I took a look at "Telecommunications System Engineering, Third Edition" by Roger L. Freeman to see what definitions he uses. He cited another reference, and quoted this as "the Nyquist sampling theorem", "If a band-limited signal is sampled at regular intervals of time at a rate equal to or higher than twice the highest significant signal frequency, then the sample contains all the information of the original signal. The original signal may then be reconstructed by use of a low-pass filter." Congrats to Freeman for avoiding making a different statement that avoids the error made in the previous statement. Although, unless the original signal is a squarewave, saying "the sample contains all the information of the original signal" may be overly optimistic. jk |
#451
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Randy Yates wrote:
Jerry Avins writes: Randy Yates wrote: "Bob Myers" writes: "Floyd L. Davidson" wrote: Lets be clear... The definitions I cited are standard. I posted 5 or 6 varied references to the same definitions. And everyone knows, ""standards" are holy. Not that I necessarily agree or disagree with Floyd's original point, but citing a written reference holds more water than a post from an individual on a usenet newsgroup, in my opinion. Floyd maintains that any signal whose values are restricted to a finite set -- IOW, "quantized" -- is digital. I cited a two-level analog signal and I can demonstrate a digital signal with a relatively large continuous range of values. His definitions are simply too restrictive to accommodate those, and he seems to be having a fit. I've decided that it's not fruitful to continue this discussion since the knowledge I work with admits anough understanding to get a lot of real work done. These sorts of discussions take too much time and produce little or no fruit. My ability to do work does not depend on others' judgement of the correctness of my definitions. So you figure that posting invalid definitions to Usenet or proving that you cannot understand standard definitions won't be a problem for you? It's amazing though, just who finds what with web searches. Anyone who reads what you've had to say here... well it could easily affect your work! Whatever, an article with out of hand invalid statements quoted from the two people who clearly post from an IEEE host is a really good place to throw out something that I've asked them for repeatedly, and they have weaseled around the question in odd ways: IEEE definitions of "digital" and "analog". I think it was Randy Yates who claimed they had been posted but did admit that nobody had ever cited IEEE as a source. Well, it appears that the person who posted it was me. IEEE apparently uses the standard definitions which I have posted from other sources. However, here is a very interesting discussion from an IEEE dictionary: An analog signal implies /continuity/, as contrasted to a digital signal that is concerned with /discrete/ states. Often the means of carrying information is the distinguishing feature between analog and digital. The information content of an analog signal is conveyed by the value or magnitude of some characteristics of the signal such as phase, amplitude, frequency of the voltage, the amplitude or duration of a pulse, and so on. To extract the information, it is necessary to compare the value or magnitude of the signal to a standard. The information content of a digital signal is concerned with discrete states of the signal, such as the presence or absence of a voltage, a contrast in the open or closed position, or a hole or no hole in certain positions on a card. The digital signal is given meaning by assigning numerical values or other information to the various possible combinations of the discrete states of the signal." "The New IEEE Standard Dictionary of Electrical and Electronic Terms", 5th ed., IEEE Std. 100-1992, IEEE Press, New York, 1992. I'm quoting it from Roger L. Freeman's "Telecommunications System Engineering", 3rd ed., 1996. For one, it clearly shows the FM-signal-through-a-limiter example given by Jerry very clearly to be exactly as my analysis indicated, and not at all what Jerry said. Also they clearly state that the values assigned to a digital symbol need not be "numerical" as someone argued repeatedly in earlier posts. Another example of credible references that support each and every point that I've made. And it again highlights that none of those saying it isn't so can find *anything* credible to support their statements. (And that of course is why it is not "fruitful" to argue with me. I don't buy rotten fruit.) -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
#452
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Jim Kelley wrote:
Floyd L. Davidson wrote: You are implying that any dictionary definition (hence not your unique Alice in Wonderland definition) is correct in any context. I'm not implying that at all. I'm simply saying that the terms and definitions I used are correct in the context in which I used them. The were easily demonstrated as *not* correct in the context used, and your whole discussion since has been to argue the implication as stated above. Do you read what you write? Lets review it: Here are some valid standard defintions: "quantize - to subdivide into small but measurable increments." (Merriam Webster's Collegiate Dictionary, Tenth Edition) Your source is for common English, it is not defining the Terms of Art used in technical discussions. This *is* a discussion about those terms of art, not about common English word usages. To whatever degree your cited definition differs from the standard definitions I've cited, you are wrong because of the context. Here is the standardized definition for "quantization" quantization: A process in which the continuous range of values of an analog signal is sampled and divided into nonoverlapping (but not necessarily equal) subranges, and a discrete, unique value is assigned to each subrange. And you looked at a invalid definition for this context and stated: Note that in the definition, there appears no mention of assigning a value. Assigning a value would then be considered a part of a separate and distinct process of converting to digital form, as in Obviously the Term of Art used in technical discussion does indeed mean there *must* be a value assigned. In fact it makes no sense at all unless that step is included. Then you go on to produce other, equally invalid in this context, definitions for other terms of art: "digital - of, or relating to data in the form of numerical digits", and as opposed to "analog - of, relating to, or being a mechanism in which data is represented by continuously variable physical quantities." The first definition states that the form must be "of numerical digits", and that is simply unnecessary. It must be assigned a "value". The value can be numerical digits, but it can be otherwise too. The point is that it must be from a finite set of discrete values. (I used the example of flags, where the value might be a square flag, a round flag, or a triangular flag. No numerical digits at all.) As you can see, knowing how to use a dictionary is vitally important... -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
#453
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Floyd L. Davidson wrote: Jim Kelley wrote: Floyd L. Davidson wrote: You are implying that any dictionary definition (hence not your unique Alice in Wonderland definition) is correct in any context. I'm not implying that at all. I'm simply saying that the terms and definitions I used are correct in the context in which I used them. The were easily demonstrated as *not* correct in the context used, and your whole discussion since has been to argue the implication as stated above. Do you read what you write? Lets review it: Here are some valid standard defintions: "quantize - to subdivide into small but measurable increments." (Merriam Webster's Collegiate Dictionary, Tenth Edition) Your source is for common English, it is not defining the Terms of Art used in technical discussions. It perfectly defines the term as I use it. This *is* a discussion about those terms of art, not about common English word usages. To whatever degree your cited definition differs from the standard definitions I've cited, you are wrong because of the context. I don't use the term the way your reference defines it. Here is the standardized definition for "quantization" quantization: A process in which the continuous range of values of an analog signal is sampled and divided into nonoverlapping (but not necessarily equal) subranges, and a discrete, unique value is assigned to each subrange. And you looked at a invalid definition for this context and stated: Note that in the definition, there appears no mention of assigning a value. Assigning a value would then be considered a part of a separate and distinct process of converting to digital form, as in Obviously the Term of Art used in technical discussion does indeed mean there *must* be a value assigned. In fact it makes no sense at all unless that step is included. It seems there are a great many things that don't make sense to you. If a "term of art" were to be defined in such a way that it contradicts the definition for the exact same term as it is published in Websters dictionary, I would be inclined to disregard it. Then you go on to produce other, equally invalid in this context, definitions for other terms of art: "digital - of, or relating to data in the form of numerical digits", and as opposed to "analog - of, relating to, or being a mechanism in which data is represented by continuously variable physical quantities." The first definition states that the form must be "of numerical digits", and that is simply unnecessary. It must be assigned a "value". The value can be numerical digits, but it can be otherwise too. The point is that it must be from a finite set of discrete values. (I used the example of flags, where the value might be a square flag, a round flag, or a triangular flag. No numerical digits at all.) As you can see, knowing how to use a dictionary is vitally important... Perhaps almost as important as realizing at which side of the analog to digital convertor you're looking. jk |
#454
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
"Arny Krueger" wrote:
"Floyd L. Davidson" wrote in message "Arny Krueger" wrote: "Jerry Avins" wrote in message Floyd L. Davidson wrote: ... Nyquist rate: The reciprocal of the Nyquist interval, i.e., the minimum theoretical sampling rate that fully describes a given signal, i.e., enables its faithful reconstruction from the samples. Note: The actual sampling rate required to reconstruct the original signal will be somewhat higher than the Nyquist rate, because of quantization errors introduced by the sampling process. It does not say what you claimed it does. Do you buy the "because clause? I don't. You do well to disagree with it. It is false. The errors that are introduced are aliasing, not quantization errors. You could change the size of the quantization steps any which way you want, and the aliasing would still be there. That is another technically *absurd* statement. Can you explain how one gets aliasing under the circumstances stated? Real world signals have finite, not zero bandwidth. That means that they have sidebands. As soon as any of the sidebands are at or exceed the nyquist You meant 1/2 the Nyquist Rate. But the point is that the statement above excludes all such signals. It specifically says that the actual sampling rate will be *above* the Nyquist Rate. There will not be any aliasing at samplying rate which is higher than the Nyquist Rate. Therefore there *cannot* *be* any problem with aliasing. There will, however, be quantization distortion. rate, there is distortion of the wave because of aliasing. Therefore the sample rate has to be somewhat higher than the sample rate of the signal, which is characterized by its carrier frequency. What???? The rate of the signal is characterized by its carrier frequency? 1) What if there is no carrier? 2) What if there is a carrier and there is an upper sideband? Clearly the carrier is not what characterizes the signal frequency or the sample rate. "The actual sampling rate required to reconstruct the original signal will be somewhat higher than the Nyquist rate, because of quantization errors introduced by the sampling process." All qualified practitioners will recognize that as wrong. Agreed. It's an incorrect statement for the reason I stated above. The reason you state is incorrect, and that should be obvious. Incidentally, I took a look at "Telecommunications System Engineering, Third Edition" by Roger L. Freeman to see what definitions he uses. He cited another reference, and quoted this as "the Nyquist sampling theorem", "If a band-limited signal is sampled at regular intervals of time at a rate equal to or higher than twice the highest significant signal frequency, then the sample contains all the information of the original signal. The original signal may then be reconstructed by use of a low-pass filter." Congrats to Freeman for avoiding making a different statement that avoids the error made in the previous statement. He said "at a rate equal to or higher than twice the highest significant signal frequency". That is precisely the same as the one you claim is wrong: Nyquist's theorem: A theorem, developed by H. Nyquist, which states that an analog signal waveform may be uniquely reconstructed, without error, from samples taken at equal time intervals. The sampling rate must be equal to, or greater than, twice the highest frequency component in the analog signal. Synonym sampling theorem. Did you forget what you were arguing previously? -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
#455
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Will you two just get a room. Please.
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#456
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Jim Kelley wrote:
Floyd L. Davidson wrote: Jim Kelley wrote: Floyd L. Davidson wrote: You are implying that any dictionary definition (hence not your unique Alice in Wonderland definition) is correct in any context. I'm not implying that at all. I'm simply saying that the terms and definitions I used are correct in the context in which I used them. The were easily demonstrated as *not* correct in the context used, and your whole discussion since has been to argue the implication as stated above. Do you read what you write? Lets review it: Here are some valid standard defintions: "quantize - to subdivide into small but measurable increments." (Merriam Webster's Collegiate Dictionary, Tenth Edition) Your source is for common English, it is not defining the Terms of Art used in technical discussions. It perfectly defines the term as I use it. Yes. And in a technical discussion, you are off in Alice's Wonderland. This *is* a discussion about those terms of art, not about common English word usages. To whatever degree your cited definition differs from the standard definitions I've cited, you are wrong because of the context. I don't use the term the way your reference defines it. But nobody else, unless they are also in Wonderland, knows what you mean if you don't use standard definitions. And in a *technical* discussion, that means the technical term of art, not the common English term. .... Obviously the Term of Art used in technical discussion does indeed mean there *must* be a value assigned. In fact it makes no sense at all unless that step is included. It seems there are a great many things that don't make sense to you. Why people want to make absurd claims about terminology is one of them. Maybe you can explain that... If a "term of art" were to be defined in such a way that it contradicts the definition for the exact same term as it is published in Websters dictionary, I would be inclined to disregard it. It isn't a contradiction. The term of art is usually a stricter definition. It is certainly going to be more rigorous. Regardless, to ignore it is abject foolishness. Can you imagine a lawyer in court ignoring all of the term of art definitions in favor of the standard common English dictionary definitions? The other side would be ecstatic... and the judge would probably throw the lawyer out the door! If you don't want to use the technical terminology, why not just step out the door and avoid all technical discussions? You won't make sense, it it does become annoying. As you can see, knowing how to use a dictionary is vitally important... Perhaps almost as important as realizing at which side of the analog to digital convertor you're looking. Exactly. Now, if you don't want to use terms in a what that will be understood, how can you even tell which side is which? You'll end up making grossly trivial errors the way Jerry Arvins and a couple of others here do... -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
#457
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
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#458
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Floyd L. Davidson wrote:
Jerry Avins wrote: Floyd L. Davidson wrote: ... Nyquist rate: The reciprocal of the Nyquist interval, i.e., the minimum theoretical sampling rate that fully describes a given signal, i.e., enables its faithful reconstruction from the samples. Note: The actual sampling rate required to reconstruct the original signal will be somewhat higher than the Nyquist rate, because of quantization errors introduced by the sampling process. It does not say what you claimed it does. Do you buy the "because clause? I don't. "The actual sampling rate required to reconstruct the original signal will be somewhat higher than the Nyquist rate, because of quantization errors introduced by the sampling process." All qualified practitioners will recognize that as wrong. Are you qualified? All qualified practitioners will recognize that you are wrong, and obviously unqualified. It is in fact a correct statement. Do you know what quantization distortion is? Yes. How is it related to sampling rate? Why would sampling exactly at the Nyquist rate be adequate if the quantization were fine enough? Don't you read the crap you cut and paste? Don't answer. I'm through. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ |
#459
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
"Jim Kelley" wrote in message ... "If a band-limited signal is sampled at regular intervals of time at a rate equal to or higher than twice the highest significant signal frequency, then the sample contains all the information of the original signal. The original signal may then be reconstructed by use of a low-pass filter." Although, unless the original signal is a squarewave, saying "the sample contains all the information of the original signal" may be overly optimistic. Did you not understand the words "band-limited signal", or the words "sampled at a rate equal to or higher than twice the *highest significant signal frequency* " ? And I guess you have no idea how Fourier analysis applies to square waves either? MrT. |
#460
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Floyd L. Davidson wrote:
Jerry Avins wrote: Floyd maintains that any signal whose values are restricted to a finite set -- IOW, "quantized" -- is digital. I cited a two-level analog signal and I can demonstrate a digital signal with a relatively large continuous range of values. His definitions are simply too restrictive to accommodate those, and he seems to be having a fit. I'll admit to a really great fit of laughter at that one! You are so thoroughly confused that it is hilarious. The recognized standard definitions say that a quantized signal is digital. You can indeed have a two-level analog signal, but the fact is that the *possible* values are infinite (all values between your two listed ones, for example). You cannot possibly have a digital signal with a continuous range of values (large or small, relative or otherwise). I've cited multiple credible sources that agree with what I say. You can't cite even one. There are none. There are only two possible values for the output of a hard limiter. Make something of it. OTOH, bu using CMOS logic and varying the power supply voltage randomly between the limits of 3 volts and 18 volts, I can hive a digital signal an infinite number of values. Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ |
#461
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
Jerry Avins wrote:
Floyd L. Davidson wrote: Jerry Avins wrote: Floyd maintains that any signal whose values are restricted to a finite set -- IOW, "quantized" -- is digital. I cited a two-level analog signal and I can demonstrate a digital signal with a relatively large continuous range of values. His definitions are simply too restrictive to accommodate those, and he seems to be having a fit. I'll admit to a really great fit of laughter at that one! You are so thoroughly confused that it is hilarious. The recognized standard definitions say that a quantized signal is digital. You can indeed have a two-level analog signal, but the fact is that the *possible* values are infinite (all values between your two listed ones, for example). You cannot possibly have a digital signal with a continuous range of values (large or small, relative or otherwise). I've cited multiple credible sources that agree with what I say. You can't cite even one. There are none. There are only two possible values for the output of a hard limiter. There are an infinite number of errors in your statement. An FM signal, such as you specified earlier, does not encode information as voltage levels. The phase or frequency contains the information, and a hard limiter does not restrain the values to only two. There might in fact be only two, but we don't know that and the limiter does not assure that. For a typical FM signal modulated by audio we might well have an infinite number of possible values after the signal is passed through a hard limiter. It is also true that we could modulate it with something else that has only two values, and then it would be digital. Make something of it. OTOH, bu using CMOS logic and varying the power supply voltage randomly between the limits of 3 volts and 18 volts, I can hive a digital signal an infinite number of values. That would be funny if you were joking, but Jerry I realize that you are dead serious. Lets use, just for this example, an RS-232C interface definition. We could use some other interface, and it would always be the same result. But since your voltage range fits within the range of voltages that are correct for RS-232 (and few others do) we'll use that. If your signal varies between +3 and +18 volts, it has exactly 1 value. That is a space signal. If you want it to be mark, you can vary it between -3 and -18 volts... Yes, you do have an infinite number of *voltages* between 3 and 18, but only one signal value. Tell me Jerry, do you actually get paid to do engineering? -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
#462
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech
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Questions about equivalents of audio/video and digital/analog.
On Mon, 27 Aug 2007 20:52:29 -0800, (Floyd L.
Davidson) wrote: Jerry Avins wrote: Floyd L. Davidson wrote: Jerry Avins wrote: Floyd maintains that any signal whose values are restricted to a finite set -- IOW, "quantized" -- is digital. I cited a two-level analog signal and I can demonstrate a digital signal with a relatively large continuous range of values. His definitions are simply too restrictive to accommodate those, and he seems to be having a fit. I'll admit to a really great fit of laughter at that one! You are so thoroughly confused that it is hilarious. The recognized standard definitions say that a quantized signal is digital. You can indeed have a two-level analog signal, but the fact is that the *possible* values are infinite (all values between your two listed ones, for example). You cannot possibly have a digital signal with a continuous range of values (large or small, relative or otherwise). I've cited multiple credible sources that agree with what I say. You can't cite even one. There are none. There are only two possible values for the output of a hard limiter. There are an infinite number of errors in your statement. An FM signal, such as you specified earlier, does not encode information as voltage levels. We all know that. You seem to be trying to catch up, though. The phase or frequency contains the information, and a hard limiter does not restrain the values to only two. The example is of the sort of hard limiting achieved by something like a Class-C amplifier where the limiter does limit the output values to two voltages. This is a specific function in an important stage of a common FM demodulator architecture. Jerry spelled this out pretty clearly, but you're having a real hard time understanding the folks here who know what they're talking about. Pay attention. There might in fact be only two, but we don't know that and the limiter does not assure that. A limiter in this sense absolutely does assure that. The idea is to remove any AM modulation, which is fine in an FM signal since there's no information in the amplitude. Limiting it to two voltages helps clean up a lot of noise and distortion with a relatively simple implementation. For a typical FM signal modulated by audio we might well have an infinite number of possible values after the signal is passed through a hard limiter. You're not following; the limiter limits it to two voltages. That's what a limiter does in this case. It's a pretty common function, with a very common symbol used frequently in block diagrams and schematics. Jerry's question, which you've been dancing around without any sort of coherent response that I can detect, is whether the signal at the output of the limiter is digital or analog according to you? The situations that you've been describing and explanations for definitions that you claim to be absolute and universal don't seem to work well for some of these cases. It is also true that we could modulate it with something else that has only two values, and then it would be digital. LOL! Okay, so we can modulate it FM with audio, and limit it, and then it's two voltages, but it's not digital. But if we modulate with only two values, then it's two voltages and it is digital! Right? Floyd, you must have a mind for detail to be able to keep track of all the goofy, byzantine distinctions you try to make up. Make something of it. OTOH, bu using CMOS logic and varying the power supply voltage randomly between the limits of 3 volts and 18 volts, I can hive a digital signal an infinite number of values. That would be funny if you were joking, but Jerry I realize that you are dead serious. Lets use, just for this example, an RS-232C interface definition. Why not just stick with the example Jerry gave? You have a digital CMOS circuit, the high voltage is one, the low voltage is zero (or, just to keep you comfortable, mark and space). Vary the supply voltage, over some range, so that the voltages for one and zero (oh, wait, mark and space) change. You could even stretch the example a little bit so that the receiving CMOS gate supply voltage stays at the same level, because there's a lot of slop built into these things. So does it stop being a digital circuit because there is an infinite range of voltages being used? We could use some other interface, and it would always be the same result. But since your voltage range fits within the range of voltages that are correct for RS-232 (and few others do) we'll use that. If your signal varies between +3 and +18 volts, it has exactly 1 value. That is a space signal. If you want it to be mark, you can vary it between -3 and -18 volts... Yes, you do have an infinite number of *voltages* between 3 and 18, but only one signal value. So, in your mind, is it digital or analog? Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org |
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Questions about equivalents of audio/video and digital/analog.
"Floyd L. Davidson" wrote in message ... Yes, you do have an infinite number of *voltages* between 3 and 18, Only in theory, in practice nothing is infinite in this universe. Noise will set the resolution limit. MrT. |
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Questions about equivalents of audio/video and digital/analog.
"Mr.T" MrT@home wrote:
"Floyd L. Davidson" wrote in message ... Yes, you do have an infinite number of *voltages* between 3 and 18, Only in theory, in practice nothing is infinite in this universe. Noise will set the resolution limit. The noise itself has an infinite number of possible values, and therefore even if the signal itself is supposed to be just 1 value, add the noise and there are an infinite number of values. -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
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Questions about equivalents of audio/video and digital/analog.
Floyd L. Davidson wrote:
Jerry Avins wrote: (snip on analog vs. digital signals) There are only two possible values for the output of a hard limiter. There are an infinite number of errors in your statement. An FM signal, such as you specified earlier, does not encode information as voltage levels. The phase or frequency contains the information, and a hard limiter does not restrain the values to only two. Previously when I was in these discussions instead of digital vs. analog it was modulated (and so in need of a modem) vs. not modulated. Passing a digital signal through an analog channel is said to require a modulated carrier. In that case, it is an analog representation of a digital signal. There might in fact be only two, but we don't know that and the limiter does not assure that. For a typical FM signal modulated by audio we might well have an infinite number of possible values after the signal is passed through a hard limiter. It is also true that we could modulate it with something else that has only two values, and then it would be digital. (snip) OK, so when is it a direct representation of a digital signal instead of an analog channel with a modulated signal? I might believe it for NRZ, but just about anything else I don't. Lets use, just for this example, an RS-232C interface definition. We could use some other interface, and it would always be the same result. But since your voltage range fits within the range of voltages that are correct for RS-232 (and few others do) we'll use that. If your signal varies between +3 and +18 volts, it has exactly 1 value. That is a space signal. If you want it to be mark, you can vary it between -3 and -18 volts... Say you have a space alien trying to decode the signal. In this case, as with other NRZ signals, there are two choices: The higher voltage represents '1', or the lower voltage. There also needs to be a way to know when the value is there, usually an external clock signal. How about this for a digital signal: I take an analog telephone and say into it either "one" or "zero". That is, the english words. The information content is digital, but the representation is pretty definitely analog. -- glen |
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Questions about equivalents of audio/video and digital/analog.
glen herrmannsfeldt wrote:
Floyd L. Davidson wrote: Previously when I was in these discussions instead of digital vs. analog it was modulated (and so in need of a modem) vs. not modulated. Okay... Passing a digital signal through an analog channel is said to require a modulated carrier. That is not necessarily true. The problem has nothing to with analog vs digital. It instead a question of whether DC can be amplified or only AC, and over what bandwidth, on that analog channel. In that case, it is an analog representation of a digital signal. I'm not sure that makes sense for an electrical signal. A digital representation of an analog signal does (the result when restored is called quasi-analog though). However, I do think that your last example, discussed below, appears to be a valid example of an analog representation of digital signal, except it is not an electronic representation... There might in fact be only two, but we don't know that and the limiter does not assure that. For a typical FM signal modulated by audio we might well have an infinite number of possible values after the signal is passed through a hard limiter. It is also true that we could modulate it with something else that has only two values, and then it would be digital. (snip) OK, so when is it a direct representation of a digital Any time you encode discrete values from a finite set, that is digital. Period. Whether it can be passed over an inherently analog channel or not is fairly meaningless. The voltage, phase, or whatever that is encoded with the information may have only a discrete set of values for the information, but they obviously take on an infinite number of possible values for the characteristic itself. Hence the voltage on a binary system carries only two values, on and off, but that is the value of the information. The voltage that is encoded might be +3 to +15 for an on and -3 to -15 for an off. And when the state switch from on or off to the opposite value, those voltages do not change instantly, and they do cover an infinite number of voltages during that change. That is an infinite number of possible voltages, but they have a value of either on or off. The information values are what makes it a digital signal. But indeed, you can pass that signal through an analog amplifier. Depending on the characteristics, it may or may not destroy the information. Obviously the amplifier would need to pass DC voltages unless we encode the information in a way that guarantees some set minimum time interval between state changes (T1 digital carrier systems typically do that, for example). signal instead of an analog channel with a modulated signal? I might believe it for NRZ, but just about anything else I don't. If the digital signal has DC components it can be modulated onto an analog carrier to pass through an AC coupled analog channel. It could also be re-encoded in a manner that will pass through an AC only channel, and then be transmitted over the same AC coupled analog channel. Most physical channels are inherently analog! Wire cables and fiber optic cables are two examples. Digital signals are commonly sent via either of them. Lets use, just for this example, an RS-232C interface definition. We could use some other interface, and it would always be the same result. But since your voltage range fits within the range of voltages that are correct for RS-232 (and few others do) we'll use that. If your signal varies between +3 and +18 volts, it has exactly 1 value. That is a space signal. If you want it to be mark, you can vary it between -3 and -18 volts... Say you have a space alien trying to decode the signal. In this case, as with other NRZ signals, there are two choices: The higher voltage represents '1', or the lower voltage. Yes, but you aren't saying anything of significane to the rest of the discussion. There also needs to be a way to know when the value is there, usually an external clock signal. This is even farther off track. (Not that it isn't interesting! clock synchronization is a really fascinating topic in my opinion.) How about this for a digital signal: I take an analog telephone and say into it either "one" or "zero". That is, the english words. The information content is digital, but the representation is pretty definitely analog. The information content encoded and sent electronically is not digital. There is nothing digital about the sound of your voice. And while "one" and "two" might be digital, those are not the values that are encoded and set over the telephone wires. It is the difference between "one" and "uno" that is encoded, and that is an infinite number of variations. But, I do think you've stated a good case for the analog representation of a digital signal! The "one/two" code (like Paul Revere's message) is in fact digital, and it is being sent over an analog channel, which is the channel from your mouth to your ear! -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
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Questions about equivalents of audio/video and digital/analog.
Floyd L. Davidson wrote:
(I wrote) Floyd L. Davidson wrote: Previously when I was in these discussions instead of digital vs. analog it was modulated (and so in need of a modem) vs. not modulated. Okay... Passing a digital signal through an analog channel is said to require a modulated carrier. That is not necessarily true. The problem has nothing to with analog vs digital. It instead a question of whether DC can be amplified or only AC, and over what bandwidth, on that analog channel. (snip) When DSL started to become popular, there was discussion that a DSL modem wasn't a modem because the DSL signal was digital. Everyone (just about) knows that v.90 needs a modem because it goes through the voice telephone system. But DSL does use a modulated carrier, and the box is a modem. However, I do think that your last example, discussed below, appears to be a valid example of an analog representation of digital signal, except it is not an electronic representation... (snip) Any time you encode discrete values from a finite set, that is digital. Period. Whether it can be passed over an inherently analog channel or not is fairly meaningless. The voltage, phase, or whatever that is encoded with the information may have only a discrete set of values for the information, but they obviously take on an infinite number of possible values for the characteristic itself. But why phase modulation, for example? Two reasons that I see. One is that the channel does not have the appropriate frequency response, and the other is the need for a clock. Consider 10baseT ethernet. Phase modulation allows for transformer coupling that is needed to avoid ground loops and ensure a balanced signal to avoid EMI problems. Using synchronous phase modulation allows for the recovery of the clock from the signal, which is also important. If, for example, the signal was not modulated and one decided to send 1000000 zeros in a row, there would be no way to recover the clock to know how many zeros were sent. If you can't separate the bits, you are losing an important part of a digital signal. Hence the voltage on a binary system carries only two values, on and off, but that is the value of the information. The voltage that is encoded might be +3 to +15 for an on and -3 to -15 for an off. And when the state switch from on or off to the opposite value, those voltages do not change instantly, and they do cover an infinite number of voltages during that change. That is an infinite number of possible voltages, but they have a value of either on or off. The information values are what makes it a digital signal. But indeed, you can pass that signal through an analog amplifier. Depending on the characteristics, it may or may not destroy the information. If the information is destroyed, then the signal didn't get through. Obviously the amplifier would need to pass DC voltages unless we encode the information in a way that guarantees some set minimum time interval between state changes (T1 digital carrier systems typically do that, for example). signal instead of an analog channel with a modulated signal? I might believe it for NRZ, but just about anything else I don't. If the digital signal has DC components it can be modulated onto an analog carrier to pass through an AC coupled analog channel. It could also be re-encoded in a manner that will pass through an AC only channel, and then be transmitted over the same AC coupled analog channel. Most physical channels are inherently analog! Wire cables and fiber optic cables are two examples. Digital signals are commonly sent via either of them. Some can pass a digital signal without modulation, others can't. (snip) -- glen |
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Questions about equivalents of audio/video and digital/analog.
On Mon, 27 Aug 2007 23:54:40 -0800, (Floyd L.
Davidson) wrote: Any time you encode discrete values from a finite set, that is digital. Period. In that case, the quantum nature of matter means that *everything* *everywhere* is digital. So no need to talk about this analog stuff anymore. Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org |
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Questions about equivalents of audio/video and digital/analog.
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Questions about equivalents of audio/video and digital/analog.
"isw" wrote in message ... In article , (Floyd L. Davidson) wrote: Most physical channels are inherently analog! Wire cables and fiber optic cables are two examples. Digital signals are commonly sent via either of them. I'll probably regret jumping in here, but: The *message* may be digital, but the *signals* are most definitely analog. The SIGNALS are electrical or optical. Bob M. |
#471
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Questions about equivalents of audio/video and digital/analog.
glen herrmannsfeldt wrote:
Floyd L. Davidson wrote: glen herrmannsfeldt wrote: Floyd L. Davidson wrote: Previously when I was in these discussions instead of digital vs. analog it was modulated (and so in need of a modem) vs. not modulated. Okay... Passing a digital signal through an analog channel is said to require a modulated carrier. That is not necessarily true. The problem has nothing to with analog vs digital. It instead a question of whether DC can be amplified or only AC, and over what bandwidth, on that analog channel. (snip) When DSL started to become popular, there was discussion that a DSL modem wasn't a modem because the DSL signal was digital. Everyone (just about) knows that v.90 needs a modem because it goes through the voice telephone system. Everyone who thinks that is wrong. A v.90 modem is digital on both sides, and will not pass through a "voice telephone system". It requires a *digital* switching system to work. V.90 is not a D/A-A/D protocol, it is a digital level encoding scheme. (Indeed, a lot equipment originally used for 56Kbps digital services was often called a "modem" by not just customers, but also by telecommunications people. None of them were technically modems, they were all level changers, with digital signals in one side and out the other in a different, but equally digital, format.) But DSL does use a modulated carrier, and the box is a modem. But is that because it has a digital side, or is that because the bandwidth restrictions of the channel cannot pass the input signal due to the low frequency components? It is a bandwidth problem, and it has nothing at all to do with digital or analog. In fact, a v.90 modem sends an entirely digital signal down that very same line. Of course in the process it necessarily uses the same bandwidth that on a DSL is allocated for a normal voice channel, and therefore while DSL can co-exist separately with POTS the v.90 modem cannot. However, I do think that your last example, discussed below, appears to be a valid example of an analog representation of digital signal, except it is not an electronic representation... (snip) Any time you encode discrete values from a finite set, that is digital. Period. Whether it can be passed over an inherently analog channel or not is fairly meaningless. The voltage, phase, or whatever that is encoded with the information may have only a discrete set of values for the information, but they obviously take on an infinite number of possible values for the characteristic itself. But why phase modulation, for example? It has *nothing* do to with whether it is digital or analog... Two reasons that I see. One is that the channel does not have the appropriate frequency response, and the other is the need for a clock. Phase modulation does not uniquely conserve bandwidth (Manchester encoding uses twice the bandwidth of NRZ, for example) nor is it unique in the ability to recover a clocking rate from the data. Consider 10baseT ethernet. Phase modulation allows for transformer coupling that is needed to avoid ground loops and ensure a balanced signal to avoid EMI problems. Alternate mark inversion (AMI) provides the same characteristics. But while all of this is indeed very interesting stuff... it has *nothing* to do with analog vs. digital or the definitions of either. I don't see any point to your discussion. Using synchronous phase modulation allows for the recovery of the clock from the signal, which is also important. That is one way to do it, but there are others. If, for example, the signal was not modulated and one decided to send 1000000 zeros in a row, there would be no way to recover the clock to know how many zeros There would be no way *only* if you select an encoding scheme such as NRZ. Manchester encoding provides for easy clock recover, but so do other encoding schemes. were sent. If you can't separate the bits, you are losing an important part of a digital signal. That is not true. Consider AMI using B8ZS encoding... Hence the voltage on a binary system carries only two values, on and off, but that is the value of the information. The voltage that is encoded might be +3 to +15 for an on and -3 to -15 for an off. And when the state switch from on or off to the opposite value, those voltages do not change instantly, and they do cover an infinite number of voltages during that change. That is an infinite number of possible voltages, but they have a value of either on or off. The information values are what makes it a digital signal. But indeed, you can pass that signal through an analog amplifier. Depending on the characteristics, it may or may not destroy the information. If the information is destroyed, then the signal didn't get through. Of course. But it has *nothing* to do with the amplifier being analog. There *are* analog amplifiers that will not destroy it. (And that is no different for analog data either, which will also be destroyed if the amplifier does not have suitable characteristics to pass it.) Obviously the amplifier would need to pass DC voltages unless we encode the information in a way that guarantees some set minimum time interval between state changes (T1 digital carrier systems typically do that, for example). Don't ignore what has already been made clear! signal instead of an analog channel with a modulated signal? I might believe it for NRZ, but just about anything else I don't. If the digital signal has DC components it can be modulated onto an analog carrier to pass through an AC coupled analog channel. It could also be re-encoded in a manner that will pass through an AC only channel, and then be transmitted over the same AC coupled analog channel. Most physical channels are inherently analog! Wire cables and fiber optic cables are two examples. Digital signals are commonly sent via either of them. Some can pass a digital signal without modulation, others can't. But is has nothing to do with analog vs. digital. If the amplifier cannot handle the signal's bandwidth, it makes no difference if the signal is analog or digital, it will not "pass" the data from input to output. You are trying to impute something to digital that is actually common to analog as well. -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
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Questions about equivalents of audio/video and digital/analog.
isw wrote:
In article , (Floyd L. Davidson) wrote: Most physical channels are inherently analog! Wire cables and fiber optic cables are two examples. Digital signals are commonly sent via either of them. I'll probably regret jumping in here, but: The *message* may be digital, but the *signals* are most definitely analog. That is not correct. Whether a message is or not digital is entirely unrelated to whether the signal used to transmit it is analog or digital (and it can indeed be either, without regard to the message). "Message" specifically means a complete set of data formatted for transmission, and is not related to analog/digital data signals. For the term "signal", you can choose from several definitions (FS-1037C): signal: 1. Detectable transmitted energy that can be used to carry information. 2. A time-dependent variation of a characteristic of a physical phenomenon, used to convey information. 3. As applied to electronics, any transmitted electrical impulse. 4. Operationally, a type of message, the text of which consists of one or more letters, words, characters, signal flags, visual displays, or special sounds, with prearranged meaning and which is conveyed or transmitted by visual, acoustical, or electrical means. Hence you can see that using "message" and "signal" in the same sentence is bound to cause confusion in the context of this particular discussion. It simply does not mean what you were thinking of. When used properly the terms "signal" and "message" would be something like this, "Our actions are intended to send Congress a message, and we wish to signal our intense displeasure with corruption." But we've been discussing signals that meet either the 1st or 2nd definition above, and specifically not numbers 3 or 4. In context, the signals are either digital or analog, and which they are depends mostly on whether the data, or individual parts of the information (message) that the signal carries, is digital or analog. -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
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Questions about equivalents of audio/video and digital/analog.
"Bob Myers" wrote:
"isw" wrote in message ... In article , (Floyd L. Davidson) wrote: Most physical channels are inherently analog! Wire cables and fiber optic cables are two examples. Digital signals are commonly sent via either of them. I'll probably regret jumping in here, but: The *message* may be digital, but the *signals* are most definitely analog. The SIGNALS are electrical or optical. Really? Nothing could be acoustical? Are you deaf? Regardless, that does not address the incorrectness of stating that all signals are analog. Morse code is not analog. -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
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Questions about equivalents of audio/video and digital/analog.
"Floyd L. Davidson" wrote in message ... Yes, you do have an infinite number of *voltages* between 3 and 18, Only in theory, in practice nothing is infinite in this universe. Noise will set the resolution limit. The noise itself has an infinite number of possible values, OK, prove it. and therefore even if the signal itself is supposed to be just 1 value, add the noise and there are an infinite number of values. Only for those who failed mathematics at high school. MrT. |
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Questions about equivalents of audio/video and digital/analog.
In article ,
(Floyd L. Davidson) wrote: isw wrote: In article , (Floyd L. Davidson) wrote: Most physical channels are inherently analog! Wire cables and fiber optic cables are two examples. Digital signals are commonly sent via either of them. I'll probably regret jumping in here, but: The *message* may be digital, but the *signals* are most definitely analog. That is not correct. Whether a message is or not digital is entirely unrelated to whether the signal used to transmit it is analog or digital (and it can indeed be either, without regard to the message). You specifically mentioned "wire cables and fiber optic cables", so lets talk about those and ignore other possible transmission media. In both of those (as they are actually used in the real world), communication is accomplished by the propagation along them of electromagnetic fields; never anything else. Doesn't matter one whit whether you turn the field on and off, or vary the amplitude or any other characteristic of it continuously, as a means of sending a message from one end to the other, those fields can take on *any value* from the maximum level injected into the cable by the transmitter down to far below the ambient noise level, depending (for example) on the length of cable being used. IOW, those signals are, without exception, *analog*. Isaac |
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Questions about equivalents of audio/video and digital/analog.
In article ,
(Floyd L. Davidson) wrote: "Bob Myers" wrote: "isw" wrote in message ... In article , (Floyd L. Davidson) wrote: Most physical channels are inherently analog! Wire cables and fiber optic cables are two examples. Digital signals are commonly sent via either of them. I'll probably regret jumping in here, but: The *message* may be digital, but the *signals* are most definitely analog. The SIGNALS are electrical or optical. Really? Nothing could be acoustical? Are you deaf? Regardless, that does not address the incorrectness of stating that all signals are analog. Morse code is not analog. The carrier or tone that is keyed on and off to send the code starts out at a certain strength at the transmitter and grows weaker in a continuous fashion as the receiver moves further and further away, until at some point it becomes impossible to understand the *message* it is carrying. That is, the signal is analog. How can it be digital if it can take on *any* value? Isaac |
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Questions about equivalents of audio/video and digital/analog.
"Mr.T" MrT@home wrote:
"Floyd L. Davidson" wrote in message ... Yes, you do have an infinite number of *voltages* between 3 and 18, Only in theory, in practice nothing is infinite in this universe. Noise will set the resolution limit. The noise itself has an infinite number of possible values, OK, prove it. Eh? You seem to have misunderstood what was said. It has *nothing* to do with resolution. It has to do with the fact that no matter what level the noise is, it could be either a little bit more or a little bit less. That means whatever value you think you have resolved, could in fact actually have had two other possible values. Which of course means that the number of voltages between 3 and 18 is indeed infinite, with or without noise. Your ability to resolve those values is an entirely different topic. and therefore even if the signal itself is supposed to be just 1 value, add the noise and there are an infinite number of values. Only for those who failed mathematics at high school. Add a random number with an infinite range of possible values to *anything*, and you have a result with an infinite range of possible values. Pretty simple math. I'm sorry to hear that you didn't do well with math in highschool. -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
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Questions about equivalents of audio/video and digital/analog.
isw wrote:
(Floyd L. Davidson) wrote: In both of those (as they are actually used in the real world), communication is accomplished by the propagation along them of electromagnetic fields; never anything else. Doesn't matter one whit whether you turn the field on and off, or vary the amplitude or any other characteristic of it continuously, as a means of sending a message from one end to the other, those fields can take on *any value* from the maximum level injected into the cable by the Yes, the electrical fields can take any value. It is inherently an analog medium. But that has no relationship to the signal which is used to send a message. The *signal* does or does not have the ability to take on various values. If the signal uses discrete symbols, it is a digital signal. If the symbols have a continuous range of values, it is an analog signal. This is not an insignificant distinction. It precisely the reason that at Bell Labs Claude Shannon studied the theory of how the two differ. As a result of his Theory of Information the telecommunications industry began to develop the technology required to implement digital system to replace the existing analog systems. They did that based on what Shannon had shown to be theoretically the most effective for telecommunications. Digital systems typically trade SNR (which can be very low) for bandwidth (which will be very high) compared to using analog signaling. The inherent noise immunity of a digital system is great, and because the analog noise in the medium is *not* directly proportional to the signal value, a digital signal can be transmitted with zero errors (due to noise) if the SNR on the analog medium is above a minimal level. It happens that that SNR is so low that a system using analog signals would be unusable at the same SNR. (Fiber optic cables are an example, where they are virtually useless with analog signaling for the typical long circuit lengths that are provided when digital signaling is used.) The actual minimum SNR depends on the type of digital encoding used. But some typical values for various communications purposes are interesting to look at. A dialup telephone connection is supposed to have at least a 24 dB SNR. That is relatively useful for voice communications, but a typical dialup modem won't work very well unless the SNR of the connection is above 30 dB above random noise (because it has been converted to a bandwidth limited analog signal). On the other hand a binary polar signal (such as the RS-232C digital interface to that dialup modem) will have an error rate of less than 1 in 10^5 with an SNR of only 9.5 dB. But that isn't even the most significant benefit! Noise is additive on an analog system, but not on a digital system, which is specifically the difference between digital and analog that has revolutionized all telecommunications in the years since Shannon showed that digital was superior. What that means is if we use 5 analog channels tandem linked to get our message from one location to another, the end to end noise must be added to determine the SNR, and that total SNR must meet the above criteria for a higher SNR than is needed by a digital system. But if 5 digital channels are tandem linked, only the errors are additive and not the noise. That is, with analog both the noise and the errors in the first link are sent to all succeeding links, and that noise causes errors in each link on analog system. On a digital system only the errors are inputted to the succeeding links but not the noise, so noise in the first link does not cause errors in the succeeding links as it does with an analog system. transmitter down to far below the ambient noise level, depending (for example) on the length of cable being used. IOW, those signals are, without exception, *analog*. No, those signals are digital if the symbols they use are discrete. The fact that the voltage, for example can range from 0 to 1 volt has no significance in terms of the signal *if* that signal uses exactly two symbols, one of which is represented by any voltage less than 0.4 volts and one of which is represented by any voltage greater than 0.6 volts. That describes a digital signal (which indeed is being sent through an inherently analog channel). encoder medium decoder +------+ +------+ | | | | input ----+ +---------------+ +---- output | | | | +------+ +------+ | | |---Analog ---| | Channel | | | |--------------- Digital --------------| | Channel | Typical examples of the above are where the input to the "encoder" is a DS1 and the "medium" is a twisted pair cable, or where that input is an OC3 and the "medium" is a fiber optic cable, or where the encoder is a satellite modem and the medium is a "bent-pipe" geosynchronous satellite. -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
#479
Posted to sci.electronics.basics,rec.video.desktop,comp.dsp,rec.audio.tech,rec.photo.digital
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Questions about equivalents of audio/video and digital/analog.
isw wrote:
(Floyd L. Davidson) wrote: The SIGNALS are electrical or optical. Really? Nothing could be acoustical? Are you deaf? I should have put a smiley on that, sorry for missing it. Regardless, that does not address the incorrectness of stating that all signals are analog. Morse code is not analog. The carrier or tone that is keyed on and off to send the code starts out Morse code does not necesarily have either a carrier or a tone involved, but we can ignore that for this discussion without changing the validity of our conclusions. at a certain strength at the transmitter and grows weaker in a continuous fashion as the receiver moves further and further away, until at some point it becomes impossible to understand the *message* it is carrying. Well, lets take exactly that as an example, because it is a good one. We could use a tone as the carrier if you like, and send it down a regular twisted pair cable. I'm going to describe this for Morse Code signaling, but I'd like to point out that virtually any FSK modem does exactly the same thing with exactly the dynamic range I'm describing here. Instead of on/off though, it uses two tones. Everything else is the same, except the modem is many times faster than a human can decode Morse Code. If we put it on the cable at 0 dBm, we'll likely have an SNR of roughly 50 dB or so, plus or minus a few. The message is sent using on/off keying of a tone, so at the cable head we have a 50 dB range which is used to determine on vs. off. If we head down the road several miles and get to a point where the signal level has dropped 10 dB (about the maximum that can be used by a POTS line), we now have a 40 dB SNR range to deal with. We could go twice that distance again (losing 10 dB of signal each time) and get to a point where our signal is -30 dB and we have only a 20 dB SNR. At 20 dB SNR there is no reason at all that you won't get perfect copy, with no errors. Clearly the *signal* has not changed, even though it has dropped 30 dB in power. That is because the symbols used are discrete. From perhaps -40 dBm to 0 dBm there is *no* *change* *in* *the* *value* *of* *the* *symbols*! That is, the signal is analog. How can it be digital if it can take on *any* value? Obviously it does *not* take on any value. The value for Morse code is either on or off. There is no "on at -22.4 dB" value, just on. -- Floyd L. Davidson http://www.apaflo.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
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