saf@clyde.UUCP (Steve Falco) (01/15/85)
> It seems to me that the disc itself was created with a given sampling rate. > How can a player change this? > > Bob The disc has a sampling rate of 44.1k. However, you can feed that into a digital filter along with additional "made up" samples (a value of 0 is used). The new sampling rate depends on how many zeros are added. If you add 3 zeros, you now have 4x the rate or 176.4k. According to the article in the Phillips Technical Journal (1982), the net effect is to suppress the aliases of the original spectrum. They achieve about 50db of suppression which makes the analog filter that much easier to build. Note that there are still aliased components at 24.1k (44.1k - 20k) but they are much lower, hence easier to suppress. This stuff looks like out-of-band noise in some sense. By the way, the output of the digital filter is 28 bits, 16 from the original sample, plus 12 that are added by the multiplier coeficients in the filter. This gets averaged down to 14 and then fed to the D/A. Now for the part that bothers me: The Nyquist theorem always seems to be based on continuous signals. I.e, the reason you can sample a 20k signal at just slightly more than 40k is that EVENTUALLY you will get some samples around the peak amplitude - you will also get some around zero. But music isn't like that - notes start and end. In effect, you are multiplying (or modulating) the sine wave by an envelope which also contributes to the spectrum. HOW MUCH ADDED SAMPLING DOES THIS REQUIRE? I don't want to listen to a 20k sine wave, I want transients (anybody remember TIM distortion?). How about it digital types? - any guesses? Steve Falco AT&T Bell Laboratories Whippany, NJ
karn@petrus.UUCP (01/16/85)
> Now for the part that bothers me: The Nyquist theorem always seems to > be based on continuous signals. I.e, the reason you can sample a 20k > signal at just slightly more than 40k is that EVENTUALLY you will get > some samples around the peak amplitude - you will also get some around > zero. But music isn't like that - notes start and end. In effect, you > are multiplying (or modulating) the sine wave by an envelope which also > contributes to the spectrum. HOW MUCH ADDED SAMPLING DOES THIS REQUIRE? > I don't want to listen to a 20k sine wave, I want transients (anybody > remember TIM distortion?). How about it digital types? - any guesses? > The Nyquist theorem is valid for ANY bandlimited signal, it doesn't matter whether it is periodic or not. One way to visualize this intuitively is that any signal which has been bandlimited (whether it be a transient or a periodic waveform) is limited in how fast the signal voltage can change with respect to time. This places a bound on how fast the signal must be sampled in order to capture *all* of the information within it. If the transient contains high frequency signals which cannot be sampled fast enough, they must be removed by the pre-sampling low pass filter or else aliasing distortion will result. But as long as the sampling rate is at least twice as fast as the cutoff of the low pass filter, then the sampling operation will retain ALL of the information coming OUT (not necessarily going IN) to the low pass filter. Check out your communications theory textbook on this one. In terms of multiplying the signal by a modulating waveform, yes, this affects the spectrum of the result. The theory of sampling always begins by using a train of Dirac delta functions as the sampling waveform; Dirac delta functions are infintesimally short, impulses which do not exist in the real world but have nice flat spectral properties. Real world sampling waveforms are generally retangular because of the sample-and-hold circuits used in real A/D converters. Since the fourier transform of a rectangular pulse is a sinc (sin(x)/x) function, this has the effect of giving a sinc shape to the resulting spectrum. This is corrected in the player with an inverse sinc function which is added either to the CD low pass reconstruction filter or to the pre-sampling anti-aliasing filter, I'm not sure which one. Does anybody know? Phil
newton2@ucbtopaz.CC.Berkeley.ARPA (01/17/85)
Here's a way to think about how the sampling theorem works, without needing to believe that "eventually" you need to sample a continuous repetitive waveform "everywhere" to be sure you get its peak amplitude. Actually, two ways: First, disabuse yourself of the notion that because music is transient-ridden or is characterized by an envelope that modulates (multiplies) a steady-state excitation, that therefore this implies the need for infinite bandwidth in a digital audio system. Yes, an impulse or discontinuous sinewave has a continuous spectrum that is not bandlimited, *but* the *premise* of the sampling theorem is that the signal is bandlimited to <1-2X the sampling rate. Thus, even if the acoustic signal is a spectral smear with non-zero magnitude in the nuclear phonon realm, *assume* that no significant components are present at more than <1-2X the sample rate. In practice, we use a big bad low-pass filter to attenuate such out-of-band signals sufficiently so the aliases resulting from their presence are less than an arbitrarily-decided bound. Second, even assuming continuous sinewaves, the system doesn't have to wait until it's tasted every chunk of a sinewave to know what frequency and magnitude it represents- this is the role of the complementary anti-imaging filter, which outputs, for example, a damped sinewave when excited by a pulse, or a continuous sinewave when excited by a pulse train, and selects only the first of the repeated ensemple of imaged spectra implied by discrete or stepwise output of the D/A. The real answer to nagging uncertainties about the propriety of hacking up the silken-smooth continuity of music is to settle on a bandwidth that is acceptable and then hack away; everything will bne provably copacetic *within those agreed-upon constraints*. It's not fair to keep coming back and claiming that things fall through cracks that never troubled anyone when we depended on even narrower-band media before digital. Regards, Doug Maisel
ed@mtxinu.UUCP (Ed Gould) (01/18/85)
> The Nyquist theorem is valid for ANY bandlimited signal, it doesn't matter > whether it is periodic or not. ... > ... Check out your communications > theory textbook on this one. > > Phil When I studied signal theory briefly about 12 years ago, there was a theorem stating that it was *impossible* to push more information through a signal than the bandwidth of the signal, e.g., one can't send more than k bits per second through a k Hz bandlimnited channel. Telephone voice-grade channels are 2700 Hz limited, filtering to allow signals only from 300 Hz to 3000 Hz. So how do 4800 and 9600 bps modems work over dialup circuits? (The telco carriers, by the way, are strict about bandlimiting their signals, since they frequency- multiplex them onto higher-bandwidth channels.) The answer seems to be that the theory that generated that theorem wasn't completely correct. Maybe the Nyquist theorem shouldn't be regarded as gospel, either. -- Ed Gould mt Xinu, 739 Allston Way, Berkeley, CA 94710 USA {ucbvax,decvax}!mtxinu!ed +1 415 644 0146
jans@mako.UUCP (Jan Steinman) (01/21/85)
In article <272@mtxinu.UUCP> ed@mtxinu.UUCP (Ed Gould) writes: >When I studied signal theory briefly about 12 years ago, there was a theorem >stating that it was *impossible* to push more information through a signal >than the bandwidth of the signal, e.g., one can't send more than k bits per >second through a k Hz bandlimnited channel. It must have been very brief study -- you're confusing signalling rate (commonly known as "baud" rate) with the maximum frequency rate at which a bandwidth limited channel will pass a sine wave. Quadrature phase encoding (the kind used in 1200 and up modems) isn't magic, and doesn't betray Nyquist, it simply takes advantage of the relatively greater time-domain \binary/ bandwidth contained in the linear, but frequency-domain limited channel. The relationships between bits-per-second and Hz is a very complex thing -- seldom a 1:1 relationship. If the bandwidth is perfectly "square" with no phase distortion, the number of bits you could pack to a Hz is limited by the noise floor, not the frequency of the carrier. -- :::::: Jan Steinman Box 1000, MS 61-161 (w)503/685-2843 :::::: :::::: tektronix!tekecs!jans Wilsonville, OR 97070 (h)503/657-7703 ::::::
herbie@watdcsu.UUCP (Herb Chong [DCS]) (01/21/85)
In article <272@mtxinu.UUCP> ed@mtxinu.UUCP (Ed Gould) writes: >When I studied signal theory briefly about 12 years ago, there was >a theorem stating that it was *impossible* to push more information >through a signal than the bandwidth of the signal, e.g., one can't >send more than k bits per second through a k Hz bandlimnited channel. This is almost true; your memory isn't quite right. There is also the different signal levels to consider. Suppose I were to agree decide that the S/N of the channel was sufficiently high to use a 4-level encoding of the data being transmitted (2 bits in one pulse). The pulses are distinguished by which one of the 4 levels is received. Thus, I'm transmitting 2*n bits/s in a channel bandlimited to n bits/s. This is equivalent to saying I'm transmitting 2*n baud. 1 baud != 1 bit/s in general. >Telephone voice-grade channels are 2700 Hz limited, filtering to allow >signals only from 300 Hz to 3000 Hz. So how do 4800 and 9600 bps >modems work over dialup circuits? (The telco carriers, by the way, >are strict about bandlimiting their signals, since they frequency- >multiplex them onto higher-bandwidth channels.) They don't transmit at 9600 bps, they transmit at 9600 baud. See above. >The answer seems to be that the theory that generated that theorem >wasn't completely correct. Maybe the Nyquist theorem shouldn't be >regarded as gospel, either. I'm afraid Nyquist knew what he was talking about. Theoretically, in a completely noise-free channel, it is possible to transmit all the information known to Man in a single pulse of a specified amplitude within any non-zero bandwidth. When there is noise, there is a limit to the information transfer rate. >-- >Ed Gould mt Xinu, 739 Allston Way, Berkeley, CA 94710 USA >{ucbvax,decvax}!mtxinu!ed +1 415 644 0146 Your's for enlightenment, Herb Chong, BASc Computer Consultant I'm user-friendly -- I don't byte, I nybble.... UUCP: {decvax|utzoo|ihnp4|allegra|clyde}!watmath!water!watdcsu!herbie CSNET: herbie%watdcsu@waterloo.csnet ARPA: herbie%watdcsu%waterloo.csnet@csnet-relay.arpa NETNORTH, BITNET, EARN: herbie@watdcs, herbie@watdcsu POST: Department of Computing Services University of Waterloo N2L 3G1 (519)885-1211 x3524
karn@petrus.UUCP (01/21/85)
> > The Nyquist theorem is valid for ANY bandlimited signal, it doesn't matter > > whether it is periodic or not. ... > > ... Check out your communications > > theory textbook on this one. > > > > Phil > > When I studied signal theory briefly about 12 years ago, there was > a theorem stating that it was *impossible* to push more information > through a signal than the bandwidth of the signal, e.g., one can't > send more than k bits per second through a k Hz bandlimnited channel. > > Telephone voice-grade channels are 2700 Hz limited, filtering to allow > signals only from 300 Hz to 3000 Hz. So how do 4800 and 9600 bps > modems work over dialup circuits? (The telco carriers, by the way, > are strict about bandlimiting their signals, since they frequency- > multiplex them onto higher-bandwidth channels.) > > The answer seems to be that the theory that generated that theorem > wasn't completely correct. Maybe the Nyquist theorem shouldn't be > regarded as gospel, either. > > -- > Ed Gould mt Xinu, 739 Allston Way, Berkeley, CA 94710 USA > {ucbvax,decvax}!mtxinu!ed +1 415 644 0146 Wrongo! The answer seems to be that you don't remember the theory correctly. The Nyquist limit still holds. High speed modems work by sending more than one bit per transition on the line, e.g., the Bell 212 sends two bits per signal transition, which can therefore take one of 2^2 or 4 possible states. The theoretical limit on the bit rate through a given channel also depends on the signal-to-noise ratio as well as the bandwidth, and is given by C = B * log2(1 + S/(N0 * B)) where C = channel capacity in bits/sec B = bandwidth, hertz S = signal power, watts N0 = noise spectral density, watts/hertz This is the famous Shannon limit.
jlg@lanl.ARPA (01/21/85)
> Telephone voice-grade channels are 2700 Hz limited, filtering to allow > signals only from 300 Hz to 3000 Hz. So how do 4800 and 9600 bps > modems work over dialup circuits? They don't! In order to get the higher bandwidth modems to work over the phone you have to lease a special line from your local telephone operating company (you pay by the mile). If you want 9600 baud long distance, then you'd better get a satellite channel leased. Standard phone lines carry a maximum standard baud rate of 2400, which explains the popularity of 1200/2400 baud modems these days for 'hackers' and other amateurs. J. Giles
cmoore@amdimage.UUCP (chris moore) (01/23/85)
> > Telephone voice-grade channels are 2700 Hz limited, filtering to allow > > signals only from 300 Hz to 3000 Hz. So how do 4800 and 9600 bps > > modems work over dialup circuits? > > Ed Gould mt Xinu, 739 Allston Way, Berkeley, CA 94710 USA > ... High speed modems work by sending > more than one bit per transition on the line, e.g., the Bell 212 sends > two bits per signal transition, which can therefore take one of 2^2 or > 4 possible states.... This brings up an often misundertood point - there _is_ a difference between 1200 bps and 1200 baud. The baud rate is the number of signal transitions on the line. By recording 2 bits per baud interval, the Bell 212 modem sends 1200 _bps_ at 600 _baud_. The V.24bis 2400 bps modem sends 4 bits per transition, thereby givin 2400 bps at 600 baud. 600 baud is considered to be the practical limit for full duplex communications, and 1200 baud is the limit for half duplex. The bell 202 uses 1200 baud, but only one modem can transmit at a time. -- "You can't get out backwards, you have to go forwards to go back" Chris Moore (408) 749-4692 UUCP: {ucbvax,decwrl,ihnp4,allegra}!amdcad!amdimage!cmoore ARPA: amdcad!amdimage!cmoore@decwrl.ARPA
phil@amdcad.UUCP (Phil Ngai) (01/23/85)
> phone you have to lease a special line from your local telephone operating > company (you pay by the mile). If you want 9600 baud long distance, then > you'd better get a satellite channel leased. This is simply not true. You can rent long distance lines on which you can operate modems at 9600 baud. Even 14.4K or 16Kbaud. There's also 56Kb DDS, 256Kb DDS, 1.544Mb T1, etc. None of these services are satellite based (thank god!). > Standard phone lines carry > a maximum standard baud rate of 2400, which explains the popularity of > 1200/2400 baud modems these days for 'hackers' and other amateurs. > > J. Giles 4800 baud full duplex over DDD is coming out, it's not cheap but just wait til we get it into a single chip! -- This is my opinion, I guess. Phil Ngai (408) 749-5720 UUCP: {ucbvax,decwrl,ihnp4,allegra}!amdcad!phil ARPA: amdcad!phil@decwrl.ARPA
mikey@trsvax.UUCP (01/23/85)
Does the theory say "bits/sec" or does it say "data/sec" The higher speed modems, for example a 1200 baud, send dibits. Each data is decoded into 2 bits, that is 4 possible values. That's why a 1200 baud modem can work full duplex over a phone line and be roughly the same bandwidth as a 300 baud modem. Any phase discrepencies become critical as these modems use the phase angle for detection of which of the 4 quadrants the signal is in. I haven't looked close at the higher speed modems, but I believe they use phase angle and amplitude to divide each datum into more and more bits. mikey at trsvax
herbie@watdcsu.UUCP (Herb Chong [DCS]) (01/26/85)
In article <55100071@trsvax.UUCP> mikey@trsvax.UUCP writes: > > >Does the theory say "bits/sec" or does it say "data/sec" The higher >speed modems, for example a 1200 baud, send dibits. Each data is >decoded into 2 bits, that is 4 possible values. That's why a 1200 baud >modem can work full duplex over a phone line and be roughly the same >bandwidth as a 300 baud modem. Any phase discrepencies become critical >as these modems use the phase angle for detection of which of the 4 >quadrants the signal is in. I haven't looked close at the higher speed >modems, but I believe they use phase angle and amplitude to divide >each datum into more and more bits. > >mikey at trsvax The theory says bits because that is the smallest possible unit of information. A good book on information theory for the types who don't deal with it every day is Pierce, John R., An Introduction to Information Theory, Dover Publications, Inc., New York, 1980. It's a relatively nonmathematical treatment of the subject based more on intuitive arguments than lots of equations. It won't help you design communications channels much, but the theory is all there. Herb Chong... I'm user-friendly -- I don't byte, I nybble.... UUCP: {decvax|utzoo|ihnp4|allegra|clyde}!watmath!water!watdcsu!herbie CSNET: herbie%watdcsu@waterloo.csnet ARPA: herbie%watdcsu%waterloo.csnet@csnet-relay.arpa NETNORTH, BITNET, EARN: herbie@watdcs, herbie@watdcsu
tweed@apollo.uucp () (01/29/85)
ced at
intervals of the sampling frequency above the baseband carrier (0 Hz). Any
aliasing you hear in a digital reproduction occured in the manufacturer's lab,
not in the CD player -- and oversampling can't help it at all.
Oversampling *will* help in the following situation:
Suppose there is a (sine-wave) signal at 20000 Hz. (Bells and other metallic
instruments can produce strong signals in this range.) In the reconstructed
waveform, there will also be a signal at 44100 - 20000 = 24100 Hz, which is
not harmonically related to the original signal (the ear is much less sensitive
to harmonic distortion than to other forms). If your system can reproduce this
signal, and you are capable of hearing it, you will find it objectionable.
Furthermore, if your system is subject to intermodulation distortion (and most
power amplifiers' distortion figures go way up in this frequency range), a signal
will be produced at 24100 - 20000 = 4100 Hz, which is *certainly* hearable, and
again, not harmonically related. As someone mentioned, oversampling can help
suppress that 24100 Hz image by as much as 50 db (before filtering), which will
make it much less likely to show up the flaws in the rest of your reproduction
system.
-- Dave Tweed
...!apollo!tweeded@mtxinu.UUCP (Ed Gould) (01/29/85)
> "The answer seems to be that the theory that generated that > theorem wasn't completely correct. Maybe the Nyquist theorem > shouldn't be regarded as gospel, either." > > The answer *in fact* is that when you "studied" information theory you > didn;t learn anything, and you've apparently spent the last dozen years > polishing your ignorance while wholesaling your groundless opinions. As various people have pointed out, some necely, others not, my recollection of the Shannon theorem was faulty. However, what I was really remembering, and this recollection is *not* faulty, was that there were practicing professionals in the communications business who *believed* that one couldn't stuff more than k bits per second through a k Hz channel (yes, I mean bits, not baud). My point with the Nyquist theorem is that even though it is correct (I have no reason to disbelieve the result itself), peoples' understanding of what it *means* shouldn't be taken as absolute. If I can hear the difference between a 44 kHz sample and a 50 kHz sample, then Nyquist or no, the fact is that the music comes out different. Since it's the music I care about, not the theory that makes for "good enough" designs, I'll give up this discussion with a saying I saw some years back: "If 'good enough' were really good enough, there wouldn't be any need to convince anyone." (I couldn't resist a bit of vilification. Someone pointed out in this voluminous discussion that the Nyquist theorem holds for *perfect* samples. Since 16 bits [or 14 as some systems use] isn't perfect [it might be "close enough"], maybe we shouldn't assume that the theorem is gospel, after all!) -- Ed Gould mt Xinu, 739 Allston Way, Berkeley, CA 94710 USA {ucbvax,decvax}!mtxinu!ed +1 415 644 0146