People talk of cd's have a 96db noise floor and high rez sources as having a 120db noise floor. What does this mean, in layman's terms?

Thanks for reading.

Just like analogue media (e.g. tape) digital representations of audio signals are also noisy. The noise floor refers to the theoretical lowest noise level possible that you'll get when recording silence. There are many sources of noise (e.g. microphone) but the noise level in question when we compare different digital formats is the noise that comes from digitizing an analogue signal alone (regardless how noisy your analogue signal already was). Usually the digital data is in the "linear PCM" format which means that the digitizing noise floor is always the same regardless of your input audio signal. Linear PCM is characterized by a sampling frequency (in case of CDs: 44100 samples/sec) and a sample resolution (in this case: 16 bits/sample). The choice of sampling rate affects the frequency bandwidth you can capture and the choice of sample resolution affects the level of the noise floor.

96 dB is a typical number you see for CD-DA (44kHz/16bit). 120 dB can be reached by using 20bits/sample. Theoretically it means that due to the difference of 24 dB the perceived loudness of the digitizing noise of the 96dB version is roughly five times higher than the 120dB version (2^(24dB/10)). However, I think most HA members would agree with me when I say: A noise floor that's "96 dB below full scale" is already inaudible on calibrated playback systems.

In the light of formats like DSD (alternative to PCM, used on Super-Audio-CDs) that heavily rely on a technique called "noise shaping" it's important to note that it's possible to trade high sample resolution for a high sampling rate while keeping the noise floor in the audible band at the same level. For example: In theory you could use 44100*2 Hz and 16/2 bits/sample to get nearly the same noise floor in the audible frequency band (0-20 kHz) like a 44100/16 version without noise shaping. Now, the DSD format carries this to its extreme end: The smapling frequency is 2.8 Megahertz and the sample resolution is only 1 bit/sample. At sampling frequencies above 44 kHz What really counts in terms of a possible noise floor is the data rate (bits/sec) because the noise shaping technique can be used.

HTH,
SG

The noise floor is the signal level of the self noise of an electronic device or a system (multiple devices working together). For a soundcard + computer you obtain the noise floor by recording with nothing plugged into the inputs -- you record just the noise that the system generates, noise that is always there whether any other signal is present or not. What you are referring to is actually the dynamic range and it is the range of the format, not any particular offering.

Sample value are what a CD contains, amplitude measurements made on the original audio (then generally modified by mixing and mastering before writing to CD).

With one bit it is possible to express two values, 0 & 1. With two bits it is possible to express four values 00, 01, 10, & 11. These are in binary rather than decimal, since a bit is binary, only either on or off.

With 16 bits samples it is possible to express 65,536 values or amplitude levels, 2 to the 16th power. The dynamic range that can be expressed with these 16 bits is thus 96.33dB which is 20 times the log of the ratio of the maximum possible value (65536) to the minium possible value (0), which is the definition of dynamic range. This is in decimal rather than binary, since we are used to using only decimal values. In an audio file, the minimum is actually the maximum negative swing of the waveform, rather than zero, which is in the center, but the numbers come out the same.

If any particular part of a CD contained absolute silence, it would be ~96dB down from the maximum possible signal level. In fact, a music CD would not contain zero values, so the actual dynamic range is always less. Dither alone rarely reaches -90dB. With much modern pop and rock music the dynamic range may involve only the upper (maximum value) few bits, making it very shallow.

The dynamic range for 24 bits is 144dB, according to the numbers, and to what can be created digitally, such as on a computer. However, nothing recorded can approach that unless one goes to cryogenic cooling because there is an absolute noise floor well above -120dB from thermal noise is the best electronics possible. I much doubt any music recordings are made with liquified gas cooling, although this is done for deep space research and military equipment.

How can you have quantisation noise when there is no signal being quantised ?

Quantizing triangular dithered perfect silence results in noise. Maybe that's what Andy had in mind.

Cheers!
SG

No, the post was written too late at night. It is simple enough to write digital zeros, silence, to a CD, although it is probably not done except for test tracks. Without dither, anything more than 96dB down can only involve the least significant bit. It will either be on, at -96dB, or off, not there at all. When a music track is dithered, its “silent” sections will probably be above -90dB. The lowest signal level dither I find useful measures -78dB peak, -87dB average RMS. Since it is noise shaped, it is just about impossible for me to hear but someone with young enough ears might find it a little different.

Thanks guys.

One thing I don't understand though, how does the sound floor of SACD compare to dvda and ccda? IIRC, SACD uses 1 bit sample resolution and 2.64 MHz sample frequency. Does DVDa win out here?

The noise floor of a DSD signal within the audible band (0-20 kHz) is supposedly at -120 dB. It really depends on what noise shaping filter has been used. As a rule of thumb you'll have 6 dB per bit. Without noise shaping the noise floor would be 6 dB below full scale. But DSD runs at a very high sampling rate. So, there's a lot of room in the ultrasonic range where you can push most of the noise into using noise shaping (the average log response of the noise shaping filter is always 0 dB). So, you could use a filter that has a 114 dB rejection below 20 kHz (114+6=120) and a gain of 3 dB for 52-1411 kHz (because the average must be 0 dB). That's what the DSD people are trying to do. The whole process is really difficult, though. There are some unsolved problems and also a loss of efficiency but the SACD creators wouldn't admit it.

However we should be careful when comparing DVDA to SACD. What really counts in the end is how they sound to us. On this level you can hardly distinguish between these two (let alone distinguish between them and plain CDs if it's not multi-channel audio we're dealing with). But we can compare the formats on a technical & theoretical level. For example you can imitate the performance of DSD in terms of noise floor and frequency response below 80 kHz with a specific LPCM setting that only requires half the data rate of DSD. So, DSD is clearly inefficient plus it's a lot easier to do things right in the LPCM domain.


HTH,
Sebastian

There's some tutorial material on this at http://www.aes.org/sections/pnw/ppt/adc.ppt

The usual 16 bits quantizer's noise corresponds to -101.1 dB on the spectrum plot and I think it should be consider a "Noise Floor" for the 16 bits audio.

I'd put it differently if 'n' is the number of bits, 'm' is the number of octaves of oversampling, and p is the number of poles in the noise shaper, the assymtotic (best case) low frequency SNR is approximately

6.02*n+ (6.02*p + 3.01) * m dB.





Um, hold it. The total energy should be about 93dB or so down from full scale.

How you evaluate this in terms of "noise floor" per bin depends entirely on how wide your bins are in frequency. -101.1 dB is just for one particular frequency analysis.

-101.1 dB is the standard deviation of the random variable used to model the quantization noise of a quantizer (- I'll check its type). The white noise which is used to model the quantization noise of the sound digitalized by such quantilizer has the same spectral density. Or in other words - a white noise with a spectral density of -101.1 dB is added to the 16 bits sound representation.
So, I think it is appropriate to use this value as a "Noise floor"

Uh, we're talking about a TPD with what total extent in terms of an LSB now?

Sorry, I do not understand some acronymes.
What I am talking about is:
A linear quantizer like this:



introduses a noise which is consider random variable with sigma squared = (quantizer step)squared/12.
For 16 bits quantizer the step is 1/2^15 and the spectral density of the white noise, which is used to model the error vs time, is -101.1 dB.

Uh, for what input signal would that be?

It is realistic to think that would be true for music digitalized by 16 bits quantizer.

This only shows that you didn't bother to at least take a look at the presentation slides Woodinville linked to.
I'm not going to repeat the essence of dithering here.

- SG

Maybe it is better for me to use the full reference from:
Digital Signal Processing
Alan V. Oppenheim , Ronald W. Schafer
Maybe I'll do it.

Where the noise floor "lies" in a spectrum plot depends entirely on the length of the FFT, you may very well have a spectrum where the noise floor lies at -120dB on the y-axis, but still only have 96dB SNR.

The SNR is defined as the signal power divided by the total noise power. To find the total noise power, you must integrate the noise's power spectral density (PSD) across the entire frequency range of interest.

Imagine you have taken a 1024-point FFT of the 0-20kHz frequency range. Then each FFT bin has a width of approx. 20Hz. What you see in the power spectrum plot then is the power (the integrated PSD) within each bin. The total in-band noise will be the combined noise power in all bins containing noise. The longer the FFT, the less noise power is contained in each bin and the lower the noise-floor appears to be.

Thus the FFT out of a CD-player may well look like this:

Then you argue that all music digitized by a 16 bit quantizer is in fact full-amplitude square waves?

No, don't think that's quite reasonable, to say the least. Perhaps you need to consider what to compare to what to determine your dynamic range, please.

So you lose 3dB. Then you lose another 4.78dB for dithering.

Now you're cookin'

You seem determined to argue that Al and Ron don't suggest you use dither, when you argue for a raw (2^-30)/12. I'm quite sure that neither party would agree with failing to dither your quantizer.

Obviously I have to take the discission seriously.
I began by saying I consider -101.1 dB as a suitable "noise floor" for a CD (16 bits) sound.
Maybe I have to proove it?
But I still think it is the "noise floor" - I still think you cannot be under it if you stay at 16 bits.
And I'll show this.
P.S. "be under it" - have the noise below -101.1 dB in the whole frequency range and preserving the whole frequency range of the original sound.

Have you read all posts carefully?

What you're saying is mathematically true, but today no one uses a non dithered quantizer for digital audio.

It's easy to show a counterexample to your assertion.

Take a sine wave at 1kHz. Multiply it by 1/65537 (i.e. well under 1 lsb).

Add the proper dithering.

Quantize it.

Take the fft.

Do a 65536 long transform.

Pooft, there's your sine wave.

Now, if you do not add the dithering, you don't see it. This shows graphically that not adding dithering MEANS THAT YOU LOSE INFORMATION.

Dithering is essential, mathematically required for quantization. This is all covered in the slide deck that was cited here, and what's more, the author of that slide deck contributes to a lot of modern DSP books, so the snide cite of O&S was simply gratuitous, as well as wrong.

Not true.
FFT is a method to calculate (fastly) a DFT which is sampled version of the DTFT. The lenght of the FFT determines its resolution vs the frequency parameter.
So, within a bin you have an interpolation of the DTFT an the longer FFT the more precise the interpolation is.
At the sampling points (in your example at 20 Hz) you have exactly the power spectrum.

And "where the noise floor lies" does not depends on the length of the FFT.

"The longer the FFT, the less noise power is contained in each bin and the lower the noise-floor appears to be." again is not true for the same reason. The power spectrum is not a power vs the frequency parameter but power dencity vs the frequency parameter.
(If you want a derivative - it has a non-zero value when the frequency interval goes to zero - a frequency point on the axis.)

P.S. I have to correct myself - actually a power spectrum is a power vs the frequency parameter - it is a power dencity.

It pretty much depends on what scaling people use. The common FFT employs an implicit sqrt(n) scaling (*). So, if you're interested in the PSD you compute abs(fft(x,n)).^2/n. Note the division by the blocksize 'n' to preserve energy. Some like to compute fft(x,n)/n to get an idea of the sinusoids' amplitudes. This isn't energy preserving anymore and probably the case ilo had in mind.

* -> fft([1 0 0 0...]) = [1 1 1 1...]

HTH,
SG

You'd be lucky to find many university courses of that quality!

Each of the "throw away" comments could be a course in its own right. Personally, I'd like to hear more about the audibility of sharp low-pass filters with ultrasonic cut-off frequencies. I know why they shouldn't be audible (the ringing is ultrasonic!) and why they may be (energy doesn't stay where it should in imperfect systems), but I'd love to see hard research or even interesting theories.

Cheers,
David.

Re SpasV:

-101.1dB:

In Bennett's classic paper he proved that for a "busy" input signal with large range compared to the quantization step, one could approximate the quantization error as a noise source with power 1/12 (normalized to the quantization step). A 16-bit quantizer can have up to 65536 levels.

Thus a sinewave, which has the power App^2/8 (pp=peak-peak), can have a maximum power of 65536^2/8 without overloading the quantizer.

10*log((1/12)/(65536^2/8)) gives -98.08dB.

If the input is random (uniform PDF => power 65536^2/12) the result is -96.32dB.

Generalizing to B bits, you get an SNR of 6.02*B for random input and 6.02*B+1.76dB for sinewave input. This is the "6dB per bit"-rule which is pretty ubiquitously used in data converter design.

For RPDF dither subtract 3dB and for TPDF dither subtract 4.77dB.


The DFT:

The DFT is a sampled DTFT yes. I'm not sure what you mean by "within a bin you have an interpolation of the DTFT" since the DTFT is a function of a continuous frequency domain; but if the DFT is longer than the sequence available for DTFT (N>L in typical "Fourier notation") the sequence is interpolated. You only need to do so to facilitate an FFT though (FFT's work best for DFT's where N is a power of 2), but for simulations you typically choose to apply an input sequence where the length L is a power of 2 and use L=N.

The total noise power equals the integrated PSD across the frequency band. You can use the autocorrelation function and Wiener–Khinchin to get there. The total noise power will equal the sum of all bins in the DFT containing noise so the noise floor will appear "lower" if the DFT is longer. It also follows from Parseval's theorem on the DFT (the 1/N factor). Try it in Matlab to see. Run a short and a long DFT/FFT of a quantized sinusoid or something. That "the power spectrum plot then is the the integrated PSD within each bin" may not have been mathematically correct, a little too late for me to look into that right now.

Dude, the FFT is equal mathematically to a multiplication by a tight frame, i.e. an orthonormal transform. Ergo, it bleeping well better ought to obey Parsival's theorem, which it does, and which says that energy is conserved. The person making a point about SQRT(n) has a point in that scaling is necessary due to the FFT algorithm, but the necessary numbers are easily calcuated by putting in a single '1' and doing the forward FFT, and seeing what comes out

Actually this scaling is intentional and not just an implementation issue. You usually want the unit pulse ("convolutional neutral" signal) to be mapped on a vector with all ones which is then the "multiplicative neutral" signal. This is the way the DFT is defined in many textbooks. See Wolfram MathWorld for example.

Cheers,
SG

I would like to mention a couple of things.
First, I accept the idea to do some research on noise model.

But to say it right now, not in the form that has been offered. This model has nothing in common with the real quantization sound noise. And here is why I think so.
First, here is a reference from the book Discrete Time Signal Processing by Alan V. Oppenheim, Ronald W. Schafer, I have promised.
In their Analysis of Quantization Errors chapter they introduce a model and state the assumptions their results are based on.
(The essence of the model and assumptions are: the noise is additive white noise and its random variables have a uniform distribution over the range of the quantization error.)
Further, they say: "...when the signal is a complicated signal, such a speech or music, where the signal fluctuates rapidly in somewhat unpredictable manner, the assumptions are more realistic. Experiments have shown that, as the signal becomes more complicated, the measured correlation between the signal and the quantization error decreases, and the error also becomes uncorrelated. In a heuristic sense, the assumptions of the statistical model appear to be valid if the signal is sufficiently complex and the quantization steps are sufficiently small so that the amplitude of the signal is likely to traverse many quantization steps from sample to sample."
Next, there are two points to mention (in my interpretation):
--"sufficiently complex" means all frequencies (or at least 20 Hz - 20 kHz) are present in the signal spectrum,
--"quantization steps are sufficiently small" means that the most samples are sufficiently greater than the quantization step.
It is obvious the simple sinusoid is not "sufficiently complex". As to its small amplitude here is a statistic I have extracted from a CD rip showing the distribution of the sound sample values.



Here bits mean how many bits were used for a specific sample value or a sample value was equal or grater than 2^bits. For example: there are 5.7% of the samples represented in 10 bits, 30% - represented in 13 bits.
As it is seen samples less than 2^10 are less than 15% and correspondingly samples greater than or equal to 2^10 are 85% of whole number of samples, equal or greater than 2^13 are 50% of the samples and so on. So, the “simple small amplitude” model is not suitable but I could extract a real noise from a real sound.

Second, about the dynamic range on the power spectrum plot (I have seen many free interpretations of it).
What follows is a reference from the book Mathematics of the Discrete Fourier Transform (DFT) with Audio Applications by Julius O. Smith III.
(begin of the reference) A decibel (dB) is defined as one tenth of a bel. The bel is an amplitude unit defined for a sound as the logarithm (base 10) of the intensity relative to some reference intensity. (Intensity is physically power per unit area. Bels may also be defined in terms energy or power.) The choice of reference intensity or power defines the particular choice of dB scale.
Since there are 10 decibels to a bel:


(end of the reference)

For 16 bits sound the Reference Amplitude is the maximum |Amplitude| and it is 32767 or 0 dB.
The minimum |Amplitude| is 1 and it is -90.309 dB.
The dynamic range of a sound is maximum dB level minus the minimum sound level and it is 90.309 dB.
The dynamic range of a signal is the maximum dB level minus the "Noise floor" level (But I prefer to bring it 20 dB up.) which is the quantization noise level. If the noise is generated by a linear quantizer then, I still think, the "Noise floor" is -101.1 dB and the signal dynamic range is 101.1 dB.

Third, about the quantization noise itself.

There is a method of noise changing, mentioned earlier in the discussion, dithering.
As long as I understand it, the method is based on adding a noise. The new noise random variable Y is chosen so as the resulting noise random variable X + Y, where X is the random variable corresponding to the quantization noise, has a proper probability distribution function (PDF). This can be done easily because the joint PDF of the sum of independent random variables is a product of their PDFs - p(x,y) = p(x)p(y).
The method adds noise and the variance of the sum is sum of variances. So, the "Noise floor" can only be up than this of the quantization error. For example if a noise with the same variance is added the "Noise floor" would be -94.6 dB. So, I do not want to guess about its scope of applications.
If my understanding is not true or a dithering is another method of rounding the numbers then there would be only one random variable Y and if I knew its PDF I could calculate the corresponding "Noise floor".

There exist methods to decrease the quantization noise explained in the book "Discrete Time Signal Processing" (Alan V. Oppenheim, Ronald W. Schafer):
--oversampling, filtering, and downsampling,
--oversampled A/D conversion with noise shaping (sampled-data Data-Sigma modulator), which includes filtering also
The first method decreases the noise variance by the upsampling factor (for example with sampling frequency of 88,200 Hz - upsampling factor of 2 - the "Noise floor" would be -104.1 dB), while the second additionally to decreasing the noise variance shapes the spectral density so as its maximums are at the ends of the frequency range.
There is something I do not understand and unless I have understood it I would doubt the methods work for the purpose of CD recordings. What I do not understand is how the filters work and more specifically: to me, a filter cannot be implemented as integer number coefficient filter. Then, after everything has been done, there should be a quantization (converting the new calculated samples into an integer form) again and a new quantization noise added again. Actually, there is an analysis of the filter's coefficient quantization in the book, which can be applied to study this but it seems to me this is the same quantization noise.

It all depends on if you want to be able to measure energy in the same units or not. Personally, I prefer to be able to do that. It makes validating some kinds of filters a whole lot easier.



Actually, my example is exactly what real quantization noise is like.

If I thought it was worth it, I'd point your "analysis" at Ron in a bar next month at ICASSP and then buy him a beer.

Have a nice day.

N.B. I'm still waiting here for you to compare apples to apples, and to compare your calculated values for something other than a full-amplitude square wave, which is actually not a possible signal in the real world. Try it vs. a sine wave that reaches +- full scale, why don't you, and see what happens to your number. Then add the obligatory dithering, and find out what happens to that number. You've simply ignored how to measure the overload point now a couple of times, now, and you seem to be determined to insist that dither is not necessary, when, of course, dither is necessary because low-level reberberation or instrument decay in music is exactly the kind of low-level signal that your harvested quotes do not describe.