We all know that music CD’s can generally contain a lot more
music, in terms of playing time, than LP’s. But I’ve always been puzzled by the fact that if you look through the little window on some players, you can see that the CD is spinning quite rapidly, far faster than an LP. Assuming that a particular segment of music is arranged along an arc or circle on the playing surface, why does the disc have to spin so fast? I would have expected the disc to rotate extremely slowly while the optical scanner gleans every bit of digital data from this highly compact medium.
Or is the rapid spinning of the disc essential to the original concept, like those little black-and-white carboard disks you can spin and see other colors “magically” appear?
CD’s don’t contain audio recordings the same way LP’s do… LP’s have grooves that a needle runs in this needle picks up the wave vibrations that are in the groove, and the vibration of the needle dupilcates the sound on the LP.
CD’s on the other hand have a dull plastic core covered with a highly reflective surface… when a cd is “burned” the reflective coating is burned down to the non reflecitve layer. When played the laser of your cd player reflects off the reflective surfcae, but doesn’t reflect off the burned down non reflecive surface. working in an almost morse code type manner.
In short the data on a CD is encoded and needs to be decoded to be played, the high amount of data that makes up a song means more data needs to be tranferred before it can be fully decoded. An LP is just a solid casting of a sound wave.
Specifically about the spinning- the “optical scanner” doesn’t move around the CD (nor does the laser beam get “aimed” around the CD). It does move back and forth in a straight line from the outside of the CD to the inside. The spinning of the CD brings the bits past the optics.
BurnMeUp kind of of waffled when it came to how the music was encoded. What is actually recorded is a digital representation of the sound waves. Little ones and zeroes physically etched into the disk as the pits described above. The CD player turns a certain amount of those pits into a certain amount of sound output, so it turns digital input into analog output. Because of microprocessors hardwired with the algorithims used, this happens very fast. Because of the large amount of information used to describe a set amount of sound (Try ripping a CD into an uncompressed format. Wow.), a lot of pits and rises are needed. So the disk has to spin very fast in front of the laser.
finally a topic to which I can speak as a (sort of) expert, as I deal with digital audio formats for a living. Derleth, you are a little off the mark in your references to CD’s being “compressed”. A little crash course. There are two sets of parameters that describe sound frequency and amplitude (waves) when those are changed from acoustic sound or eletrical current into data. The first and easier to describe is Sample rate. which works similarly to how motion pictures, the computer takes a picture of the amplitude of the wave at regular intervals and with that creates a series of steps that aproximate an acoustic wave. Where a movie takes 29 or so pictures per second, a standard CD quality digital recording takes 44,100 pictures (cross sections) per second. The second half or the digital audio conversion is bit depth, which is a little harder to visualize. Bit depth refers to how many digits (bits) the recorder uses to desribe the amplitude of a given sample. So a 4 bit recording would have 2 to the fourth power different available positoons for the amplitude. Obviously, the more different amplitudes (bits0 used to desrible the wave, the more accurate the wave is particularly in describing music (classical in particular) which has a great deal of dynamic range (loud/quiet contrast). the current CD standard is 16 bit wich translates to 2 to the 16th power different amplitudes possible. What all this adds up to, (and further more how it relates to the original post) is that this means that every second you play a CD it has to process 44,100x2to the sixteenth power bits of info ot create CD quality audio. That is why it has to read spin so quickly, lots to read. Interesting side note, CDs actually do not spin at a constant rate either They spin faster at the outside than at the inside annd gradually slow down as they work inwards.
Derleth, the compression issue has to do more with lower end consumer formats like MP3’s and Minidiscs, which use “compression” to reduce the amount of data needed to describe audio data. they do this by intuitively guessing wich frequencies and compound harmonic structurers are actually an audible part of the signal and which are noise. MP3s in particular reduce “compress” the audio by as much as 90% and while the schemes are getting more convincing, there is still a noticeable difference between CD pure uncompressed digital audio, and MP3s, which tend to do strange things to higher frequencies and often as a result of ignoring some harmonics, they make low frequencies less rich and detailed sounding.
Hope this was helpful and didn’t cause anyone to die of boredom…
CJ
Bad Hat gave a good simple explanation as far as rotation is concerned. He’s right that the angular velocity changes, this is so the laser can read the data at a constant rate. CDs are CLV (Constant Linear Velocity). So it slows down as it reaches the edge in order to keep the velocity constant. Yes, CDs are read from the center out to the edge. As far as the data’s concerned, a standard pressed CD consists of millions of pits. When the laser is reading the CD, it reflects back to the sensor on both pits and lands, and the light scatters when it hits a pit boundary. These boundaries are the 1s. Everything else is a 0. Also, the data is not sequential…a single sample can be spread across an area roughly 3cm in length…it does this interleaving so that spots and scratches only erase 1 bit or so out of many samples, rather than 60 consecutive bits. This ensures that the (very) robust error correction algorithm can reconstruct the intended data. Basically, the data is just the amplitude, and the D/A converter simply outputs that value at the sampling rate in order to reproduce the sound.
The sampling rate of 44.1kHz allows for reproduction of all frequencies up to 22.05kHz, which is plenty since humans top out at hearing 20kHz. There’s a little bit more to it, such as oversampling and such, but that’s too much to get into now.
I never said CDs were compressed. I just said that turning CD audio into a sound file you do not then compress takes up massive amounts of disk space.
Nitpick: Machines (computers included) are not intuitive. The types that compress files use algorithms, explicit sets of instructions, to do their work.
Now that I’m done picking out flaws, good work on describing how CDs work.
Having scanned through the replies & not seeing the answer to your question:
In order to read the information from the disc at a sufficiently fast rate needed to reproduce the original audio, the bitstream coming from the disc must be somewhere in the neighborhood of 1.41 Mbps. Somebody may come by & post the exact number (I taught D/A conversion circuitry but that was almost ten years ago).
No one seems to have mentioned yet that the 44,100 samples per second used to describe an anolog wave form by CD’s is not enough to give an accurate representation of sounds above 8820 Hz. A complete cycle of such a sound is represented by only 5 sampled voltages which is too few to distinguish between a sine and a triangle wave. Even a signal at 4410 Hz is only represented by 10 data points, which cannot give an accurate representation of many complex wave forms. Perhaps as dvd starts getting more popular we’ll start seeing music sampled at higher data rates ?
You’ve been reading too many audiophile magazines. Actually, a 44100 Hz sample rate is enough to exactly reproduce any sound containing frequencies less than half the sample rate (22050 Hz). So as long as we can be sure that the sampled signal is limited in its frequency content, having no signal components above 22050 Hz, then that sample rate is enough to allow us to exactly reproduce the original wave.
Did I say “exactly” enough times? It’s not an approximation, but a rigorous mathematical equivalence, known as the Nyquist Criterion.
I really need to start reading my posts before i submit them. I am so embarrassed by all of the unfinished thoughts and sentences in my first post. I assure you, I and not deficient, I just have a mild ADD buzz going on. Sorry, Derleth, if I misinterpreted your use of the term “compression”. There are so many differnt processes in audio and in computers that use the term, that it gets hard to follow. Also embarrassed by my misassretion that CD’s read outside to inside. duhh. I need to go turn in my card now.
adressing the arguement about whether 44.1 k 16 bit is to human ears acoustical identical to analogue sound, blind A/B tests with so called "golden ears" engineers have proven more than once that 44.1/16 is distinguishable from analogue. The acousticians explanation for this is that, while we can't actually hear frequecies above the Nyquest frequecy of 44.1 (22.05, half of 44.1: according to the nyquest principle the greatest frequency accurately represented by a digital recording is half the sample rate), the harmonics that occur above 22.05 affect the way we hear lower frequencies and give them a richness that when 22.05+ freqs are lost or ignored, result in a duller, less complex sound.
There is however hope for digital music in this regard. A currently proposed standard for DVD Audio and the other potential new format, CD24, is digital recording with a sample rate of 94Khz and a bit depth of 24. it is my understanding that even to the most sensitive ears, A/B tests have demonstrated that 24/96 is undistinguishable from analogue.
CJ
But it is an approximation. Common sense dictates that the Nyquist Criterion only applies to situations where exact samples are taken. The exact samples aren’t recorded; digital approximations are recorded.
You are using some dubious language there Ryan, so much so that I am not sure what you are saying. digital recording, as distinct from analogue recording, is always an approximation. Because it involves turning curves into right angles. Nyquest does refer speciffically to digital recording (which is also called"digital sampling")and the language asociated with nyquest should always include accurately “represent” analogue waves.
so the assertion that "The exact samples aren't recorded; digital approximations are recorded" is nonsensical, and this has me confused. samples are digital aproximations and this is what nyquest refers to. In order to make reasonably accurate representations of a sound wave or portion of a sound wave, the sample rate must be atleast twice the frequency or said sound wave. In fact the distortion that occurs during A/D conversion of unsupportable frequencies is fairly nasty, which is why almost all if not all digital recording systems have a form of pre-conversion filtering that deadens unsupported frequencies before digital encoding begins. Nyquist is in fact fairly mathematically rigorous, and its really not all that difficult to figure out why, if you draw it on paper or conceptualize it a little in your head. think about it...
Actually, The Ryan’s comment made sense to me. The classic Nyquist criterion only applies to the sampling of a continuous function into a discrete-time sequence. A discrete-time sequence is a quantized-time, but continuous-amplitude sequence. An implementation in the real world would require an infinite-bit A/D, which doesn’t exist. Introducing a real A/D turns the discrete-time sequence into a digital sequence. As you mentioned, there is pre-A/D and post-D/A filtering to help things along, as well as dithering techniques in the digital domain, but it is still an approximation, technically.
I think that’s the point The Ryan was making- the exact analog value wasn’t record, but a quantized value (for which the Nyquist theorem isn’t applicable). Of course, on a practical level, 16 bits is plenty for high-quality CD recording, and in most cases can approximate the true analog value.
I think the real case for 24-bit sampling is that it lets you be much less rigourous with recording levels. A 16-bit A/D system is plenty if the peak of the signal is carefully matched with the peak of the A/D, but it’s easy to screw that up and either clip or reduce your dynamic range. Having a 24-bit A/D should make things much easier, especially when combining music from different sources. It also helps when applying certain wacky digital filters popular in some audio equipment.
Something’s not quite right here. If you do the math you describe you come up with a REALLY big number. On the order of 800 quintillion bits of information per second. Computers are fast but not that fast (not even remotely close).
If my math is right a CD holds on the order of 5.2 billion bits of information total (650 million bytes * 8 [for conversion into bits]).
As to the OP realize that CD’s often spin faster than necessary. The player reads the data off of the CD and places the info into a buffer (memory) and plays the music from that. The bigger the buffer the fewer skips you get from bumps and read errors. Your portable CD players made for a car do nothing more than sample the data at a higher rate than a normal CD player and store the info in a bigger buffer. If you hit a bump the drive head has several seconds to find its original place before you hear a skip in the music.
The CD player can also use this technique to ‘oversample’ the data. If it encounters a scratch it can read that spot several times and try to figure what data is really there.
This about 745 megabytes, and more than the 650 MB of data you can get on a CD-ROM (due the addition of more error correction bits for CD-ROM data).
Jeff_42’s exactly right about the disc spinning faster than necessary to provide a buffer for skips. I’ve never heard of “oversampling” being used to describe a CD player reading the same bits on the disc more than once, though, although I’m certainly not a CD expert.
When recording to CD, the audio waveform (as represented by a continuously-varying voltage) is sampled at a rate of 44,100 samples per second. For each channel. Each sample is comprised of exactly 16 bits. For example, 5 samples of the left channel may look like this:
I know what you’re talking about, but the traditional term “oversampling” has nothing to do with disc speed.
Oversampling occurs in the digital stage of the CD player. After sampling the data off the CD, the CD player calculates and injects “interpolated” samples between each sample. For example, let’s say the following two words were read off the disc:
1001001110101001
1001001110101101
Using 2X oversampling, the CD player will calculate the midpoint (i.e. average) of these two samples and inject it between them as so:
1001001110101001 - original
1001001110101011 - interpolated
1001001110101101 - original
So why is this beneficial?
Let’s back up… Let’s assume we have a normal (non-oversampling) CD player. After reading digital data off a disc, the data from a must be sent to a 44.1 kHz digital-to-analog converter (DAC), which does exactly what it says. The problem with the DAC, however, is that it introduces a bunch of crap above f/2 = 22.05 kHz. (We like to call them “artifacts”.) Therefore, you must send the analog signal through an analog filter to get rid of the high frequency crap. Ideally you would want to use an analog filter that is perfectly flat between 20 Hz and 22.05 kHz and stops anything above 22.05 kHz. Such a brickwall analog filter is very difficult to design. (Even if you did design one, it would probably have a crappy phase response.) Anyway, by oversampling the digital data we can use (in fact, must use) a DAC operating at a higher frequency. These are readily available, fortunately. For example, a 2X CD player would have a DAC running at 88.2 kHz. This CD player would need an analog filter to get rid of everything above 44.1 kHz. The good news is that we do not need a fancy analog filter with “brickwall” performance; a plain-Jane Butterworth filter will do fine. This is because your speakers can’t reproduce anything up that high. Even if they could, your ear wouldn’t be able to.