As part of a listening session today, I put together a playlist to compare piano recordings. I decided that an interesting way to do this was to use the same piece of music, recorded by different artists on different instruments in different rooms by different engineers using different microphone and techniques. The only constant was the notes on the page in front of the performer.
Playing through this, it’s interesting to pay attention to things like:
Overall level of the recording
Notice how much (typically) quieter the Dolby Atmos-encoded recording is than the 2.0 PCM encoded ones. However, there’s a large variation amongst the 2.0 recordings.
Monophonic vs. stereo recordings
Perceived width of the piano
Perceived width of the room
How enveloping the room is (this might be different from the perceived width, but these two attributes can be co-related, possibly even correlated)
Perceived distance to the piano.
On some of the recordings, the piano appears to be close. The attack of each note is quite fast, and there is not much reveberation.
On some of the recordings, the piano appears to be distant – more reveberant, with a soft, slow attack on each note.
On other recordings, it may appear that the piano is both near (because of the fast attack on each hammer-to-string strike) and far (because of the reverberation). (Probably achieved by using a combination of microphones at different distances – or using digital reverb…)
The length of the reverberation time
Whether the piano is presented as one instrument or a collection of strings (e.g. can you hear different directions to (or locations of) individual notes?)
If the piano is presented as a wide source with separation between bass and treble, is the presentation from the pianist’s perspective (bass on the left, treble on the right) or the audience’s perspective (bass on the left, treble on the right… sort of…)
Once upon a time, I did a blog posting about why, when we test digital audio systems, we typically use a 997 Hz sine wave instead of a 1000 Hz tone.
The short version of this is the following:
Let’s say that I digitally create a (not-dithered) 1000 Hz sine wave at 0 dB FS in a 16-bit system running at 48 kHz. This means that every second, there are exactly 1000 cycles of the wave, and since there are 48,000 samples per second, this, in turn means that there is one cycle every 48 samples, so sample #49 is identical to sample #1.
So, we are only testing 48 of the possible 2^16 ( = 65,536) quantisation values, right?
Wrong. It’s worse than you think.
If we zoom in a little more, we can see that Sample #1 = 0 (because it’s a sine wave). Sample #25 is also equal to 0 (because 48,000 / 1,000 is a nice number that is divisible by 2).
Unfortunately, 48,000 / 1,000 is a nice number that is also divisible by 4. So what? This means that when the sine wave goes up from 0 to maximum, it hits exactly the same quantisation values as it does on the way from maximum back down to 0. For example, in the figure below, the values of the two samples shown in red are identical. This is true for all symmetrical points in the positive side and the negative side of the wave.
Jumping ahead, this means that, if we make a “perfect” 1 kHz sine wave at 48 kHz (regardless of how many bits in the system) we only test a total of 25 quantisation steps. 0, 12 positive steps, and 12 negative ones.
Not much of a test – we only hit 25 out of a possible 65,546 values in a 16-bit system (or 25 out of 16,777,216 possible values in a 24-bit system).
What if I wanted to make a signal that tested ALL possible quantisation values in an LPCM system? One way to do this is to simply make a linear ramp that goes from the lowest possible value up to the highest possible value, step by step, sample by sample. (of course, there are other ways, but it doesn’t matter… we’re just trying to hit every possible quantisation value…)
How long would it take to play that test signal?
First we convert the number of bits to the number of quantisation steps. This is done using the equation 2^bits. So, you get the following results
Number of Bits
Number of Quantisation Steps
16
65,536
24
16,777,216
32
4,294,967,296
If the value of each sample has a different quantisation value, and we play the file at the sampling rate then we can calculate the time it will take by dividing the number of quantisation steps by the sampling rate. This results in the following:
Sampling Rate (kHz)
16 Bits
24 Bits
32 Bits
44.1
1.5 seconds
6.4 minutes
27.1 hours
48
1.4 seconds
5.8 minutes
24.9 hours
88.2
0.7 seconds
3.2 minutes
13.5 hours
96
0.7 seconds
2.9 minutes
12.4 hours
176.4
0.4 seconds
1.6 minutes
6.8 hours
192
0.3 seconds
1.5 minutes
6.2 hours
352.8
0.2 seconds
47.6 seconds
3.4 hours
384
0.2 seconds
43.7 seconds
3.1 hours
705.6
0.1 seconds
23.8 seconds
1.7 hours
768
0.1 seconds
21.8 seconds
1.6 hours
So, the moral of the story is, if you’re testing the validity of a quantiser in a 32-bit fixed-point system, and you’re not able to do it off-line (meaning that you’re locked to a clock running at the correct sampling rate) you’d either (1) hope that it’s also a crazy-high sampling rate or (2) that you’re getting paid by the hour.
Why I am thinking about this?
I often get asked for my opinion about audio players; these days, network streamers especially, since they’re in style.
Let’s say, for example, that someone asked me to recommend a network streamer for use with their system. In order to recommend this, I need to measure it to make sure it behaves.
One of the tests I’m going to run is to ensure that every sample value on a file is accurately output from the device. Let’s also make it simple and say that the device has a digital output, and I only need to test 3 LPCM audio file formats (WAV, AIFF and FLAC – since those can be relied to give a bit-for-bit match from file to output). (We’ll also pretend that the digital output can support a 32-bit audio word…)
So, to run this test, I’m going to
create test files that I described above (checking every quantisation value at all three bit depths and all 10 sampling rates)
play them
record them
and then compare whether I have a bit-for-bit match from input (the original file) to the output
If you add up all the values in the table above for the 10 sampling rates and the three bit depths, then you get to a total of 4.2 DAYS of play time (playing audio constantly 24 hours a day) per file format.
So, say I wanted to test three file formats for all of the sampling rates and bit depths, then I’m looking at playing & recording 12.6 days of audio – and then I can start the analysis.
REALLY‽
Of course this is silly… I’m not going to test a 32-bit, 44.1 kHz file… In fact, if I don’t bother with the 32-bit values at all, then my time per file format drops from 4.2 days down to 23.7 minutes of play time, which is a lot more feasible, but less interesting if I’m getting paid by the hour.
However, it was fun to calculate – and it just goes to show how big a number 2^32 is…
This is a radio show by Glenn Gould from 1965 that is the audio version which was expanded by Gould into an article written for High Fidelity magazine’s 15th anniversary edition (which can be downloaded from this site. The article starts on page 46.)
When an analogue audio signal is converted to a digital representation, the value of the level for each sample is rounded to the nearest quantisation step (because a digital audio system does not have an infinite resolution). I’ve talked about this in detail in a past posting.
When a sample value in a digital audio stream is stored or transmitted inside a piece of audio equipment or software, one of the choices the engineer can make is whether the value should be represented using a fixed point or a floating point system. These are related, but fundamentally different, and they have some effects on the audio signal that may be audible if you’re not careful…
Let’s lay down some basic points to start. We’ll say the following:
Audio is a kind of AC signal that has a level that can vary between two values.
For now, we’ll say that the limits on the range of values is -1 and +1, and it can be anything in between.
We’re going to divide up that range into some finite number of steps and round the actual signal value to the closest usable value. (I’ll assume for this posting that you already understand that dither is your friend.)
The value will be stored as a binary number somehow
The question that we’ll look at here is exactly how that binary value represents the number, and a little of what that means to the audio signal.
Fixed Point Representation
The simplest way to represent the value is to divide the total range from the minimum to the maximum number into an equal number of steps, and round the signal’s value to the closest step. This is a really generalised description of a “fixed point” system.
For example, if we have a 3-bit number to play with, we’ll take the first bit and use that one to represent the + or – portion of the value (where 0 means “+” and 1 means “-“). For values from 0 up to (just under) the positive maximum, the other 2 bits are used to just count the steps, from 000 up to 011. The negative values start at the bottom and work their way up to 1 step below 0, from 100 to 111. This can be seen in Figure 1.
Figure 1: A simplified representation of the use of quantisation steps in a 3-bit fixed point system.
If you look carefully at Figure 1, you’ll see that there is one extra negative step, since one of the positive steps is used to represent the value 0 in the middle. This means that, if the signal is symmetrical, then we will wind up using all of the possible quantisation values except for the bottom one (just like I’ve shown in the plot), however, for the rest of this discussion, we’ll be working with numbers that are so big that this one step doesn’t really matter, so I won’t mention it again.
If we are using a 3-bit number to represent the value, then we have a total number of 23 quantisation steps: 8 of them. Each time we add one more bit, we double the number of steps. So, for a 16-bit sample, we have 216, or 65,536 possible quantisation values. For a 24-bit sample, we have 224, or 16,777,216 steps.
By increasing the number of bits in the number, we don’t change the level (it still has a range of -1 to +1), we’re just increasing the resolution that we have to make the measurement. The higher the resolution, the lower the error, and so the lower the level of distortion (if we don’t dither) or noise (if we do) relative to the signal.
If you have a fixed-point system, and you want to calculate the difference in level between the maximum signal level and the noise floor, then you can use a somewhat simplified equation, shown below:
Dynamic Range In dB ≈ 6 * nBits – 3
As I said, this is simplified due to some rounding to keep the numbers nice, but the general idea is that you have a doubling of dynamic range for every extra bit (therefore 6 dB per bit) and you lose 3 dB for the (TPDF) dither (but that’s better than not having the dither and having distortion instead). If you wanted to do it properly, then you can use this math instead:
Dynamic Range In dB ≈ 20*log10(2nBits) – 20*log10(sqrt(2))
So, if you have a 16-bit fixed point system, you have about 93 dB of range from the loudest signal to the noise floor. If you have a 24-bit system, it’s about 141 dB.
Remember that the noise floor is constant (I’m assuming it’s dithered), so as the signal level drops below maximum the current signal to noise ratio will drop by the same amount. Therefore, if your signal is 12 dB below maximum (or -12 dB FS, which means “12 decibels below Full Scale”), then the SNR in a 16-bit system is 93 – 12 = 81 dB.
If that last paragraph didn’t make complete sense, go back and read it again, because it’ll come back later…
Fixed point is a good system for conversion of an audio signal from and to analogue, but if you’re doing some really serious processing, it might not work out so well. This is due to two primary reasons:
If your signal is going to outside the range, it will clip at the maximum positive or the minimum negative value because fixed point is not designed to exceed its range.
If the signal is going to be reduced to a very low level somewhere in your proceeding (say, inside a biquad, for example) then you might need a LOT of bits to keep the noise floor low enough when the signal level is brought back up
Figure 2: The first half of a sine wave (in grey) quantised (without dither) in a simplified 4-bit fixed point system. (I’ve actually cheated a bit and just made 8 equally-spaced steps from 0 to 1 unlike the version shown in Figure 1.) The two plots show identical data, but the bottom plot has a logarithmically-scaled Y-axis.
As can be seen in Figure 2, the equally-spaced steps in a fixed point world mean that the quantisation error is always between -0.5 and 0.5 of a step (a “Least Significant Bit” or LSB), regardless of the level of the signal.
Floating Point Representation
There is another way to use the bits to represent the signal value. This is to divide the binary “word” into two parts and to do a little math involving some subtraction, multiplication, and an exponent to arrive at the value. Just like in the Fixed Point case, we’ll reserve one bit for the +/- indicator.
Let’s say that we have a 32-bit value to work with. We’ll divide this up into the following:
23 bits for the fraction or mantissa, which we’ll abbreviate f
8 bits for the exponent, abbreviated e
1 bit for the +/- sign (just like in Fixed Point)
We’ll then do the following math:
Sample Value = ± (1 – f) * 2e
We need to know a little extra information:
because we’re using 23 bits for f, then it can range from 0 to 223-1. In other words, stated mathematically: 0 ≤ 223*f < 223
because we’re using 8 bits for e, then it has a total range of 28 possible values. In other words it has a range from just over -27 to just under 27. In other words, stated mathematically: -126 ≤ e ≤ 127 (Note that a couple of possible values are reserved for special purposes, but we won’t talk about those)
This is all a little complicated, but there is a “punch line” to which I’m headed:
Unlike Fixed Point representation, the divisions of the values – the number of steps, and therefore the step sizes – are not the same across the entire scale of possible values. It’s divided into sections, where each section has quantisation steps of equal size, but that step size is dependent on what the value is. In other words the step size changes with the value, but on a coarser scale.
That step size can be calculated as follows:
From 2e to 2e+1, the steps all have an equal size of 2e-fBits where fBits is the number of bits used to express f (in the case of a 32-bit floating point word, fBits = 23 bits). In other words, we have 2fBits equally-spaced steps in that range.
Therefore, each time the signal value moves from just below 0.5 to just above (for example) then the resolution changes, and the higher the value, the lower the resolution. This is is how Floating Point representation behaves.
Figure 3: The first half of a sine wave (in grey) quantised (without dither) in a simplified floating point system with 2 bits for the fraction. This means that there are 4 equally-spaced steps from (for example) 0.25 to 0.5 or 0.5 to 1. The two plots show identical data, but the top plot has a linearly-scaled Y-axis, whereas the bottom plot has a logarithmically-scaled Y-axis.
Do I care?
Let’s find out.
In a 32-bit floating point world (therefore, one with a 23-bit fraction), if I have a signal that has a level that has has a maximum positive value of 1 (or 20), then the resolution of the value (which defines the error, which defines the “distance” in dB to the noise floor) is 2-25 (or 1/33,554,432).* This means that the noise floor is about 150 dB below the signal (20 * log10(1 / 2-25). As the signal level drops to 0.5, the noise floor remains the same, so the signal drops by 6 dB, and the SNR reduces to 150 – 6 = 144 dB.
Then, when we drop just below 0.5, the resolution of the value suddenly changes to 2-26 (or 1/67,108,864) , which means that the noise floor is about 150 dB below the signal (20 * log10(0.5 / 2-26). As the signal drops to 0.25 (-6 dB relative to 0.5), the noise floor remains the same, so the signal drops by 6 dB, and the SNR reduces to 150 – 6 = 144 dB.
Then, when we drop just below 0.25, the resolution of the value suddenly changes to 2-27 (or 1/134,217,728), which means that the noise floor is about 150 dB below the signal (20 * log10(0.25 / 2-27). As the signal drops to 0.25 (-6 dB relative to 0.5), the noise floor remains the same, so the signal drops by 6 dB, and the SNR reduces to 150 – 6 = 144 dB.
Hopefully, by now, you’re seeing a pattern here.
Figure 4: Notice that the error of the floating point version is reduced when the signal level (in grey) approaches 0. Figure 6: The errors from the quantisation shown Figure 5. These are just the original signal subtracted from the quantised signals. Notice that, in Floating Point, the general level of the error is dependent on the level of the signal (it’s smaller on the left and right of the plot) whereas in Fixed Point, the overall level of the error is more constant.
The cool thing is that the pattern would have been the same if I had gone above 1 instead of below it. So, the two things to worry about in Fixed Point (inadequate resolution with (temporarily) low-level signals and clipping when the signal goes outside the range) are not problems in floating point.** And, if you have enough bits (32-bit floating point is the standard “single precision” resolution, but 64-bit “double precision” resolution is not uncommon).
Figure 7: The Signal to Distortion+Noise ratio of four different systems, as a function of the signal level in dB FS.**
This is why, in most modern audio systems, you have a fixed-point ADC and a DAC (an Analogue to Digital Converter and a Digital to Analogue converter) at the input and output of your system (because the signal range is reasonably well-defined, and the dynamic range is more than adequate if you do it right) but the processing on the inside is done in 32-bit or 64-bit floating point (or both, in some devices) so that the engineers have the resolution and the range to play with the signals before getting them ready for the output.***
There may be some argument made for a constant noise floor level in a fixed-point system (assuming it’s dithered) over a signal-modulated noise level in a floating-point world (assuming it’s not), however, there are two reasons why this is likely not a real-world issue. The first is that, even in a single-precision floating point system, the worst-case signal to noise ratio is about 144 dB, which is very good. The second is that smart people have already been thinking about dither for floating point systems. If this sounds interesting, you can start reading here…
One last thing
You may be wondering about that sawtooth plot: the red line in Figure 7. It can’t keep going forever, right?
Right.
Eventually, if the signal is quiet enough, then you run out of exponents and the system just behaves as a 23-bit fixed point system (assuming a 32-bit floating point). This will happen when e = -126. Below that, then the SNR just follows a downward slope just like the fixed-point plots. If the signal is loud enough (when e = 127) then you’ll clip, again, just like the fixed-point systems do when the input signal has a level of 0 dB FS.
So, then the question is: “how quiet / loud does the input signal have to be for that to happen?” The answer is very quiet and very loud, as you can see in the plot in Figure 8.
Fig 8. The limits of a 32-bit floating point signal. As you can see, you’ve got plenty of dynamic range to work with before you run out of room on either side. The black line is 16-bit fixed point, the blue line is 24-bit fixed point, and the red line is 32-bit floating-point.
You may be wondering how I calculated those limits:
The first peak in the sawtooth on the left side is at 20*log10(2^-126) = -758.6 dB FS
The last peak in the sawtooth on the right side is at 20*log10(2^127) = 764.6 dB FS
The slope that just below the 0 dB FS Signal level is where e = -1. The slope just above 0 dB FS is where e = 0.
* First small note for the attentive
You may have noticed what appears to be a mistake in my math in there. First I said:
From 2e to 2e+1, the steps all have an equal size of 2e-fBits where fBits is the number of bits used to express f (in our case, fBits = 23 bits). In other words, we have 2fBits equally-spaced steps in that range.
Then I did the math and said
In a 32-bit floating point world (therefore, one with a 23-bit fraction), if I have a signal that has level that has just come up to 1 (or 20), then the resolution of the value (which defines the error, which defines the “distance” in dB to the noise floor) is 2-25 (or 1/133,554,432).
Why did I say 2-25 when maybe I should have said 2-23 (because there are 23 bits in the fraction)? The reason is that the 223 quantisation levels are located between 1 down to 0.5. If I were to continue with the same spacing down to 0, then I would have twice as many quantisation levels, so there would be 224 instead. If I were to continue the spacing all the way down to -1, then there would be twice as many again, or 225.
In other words, a floating point signal ranging from a value of 2-1 to 20 (0.5 to 1) with some number of bits in the fraction that we’re calling fBits will have almost exactly the same signal to noise ratio as an non-dithered fixed point system that is scaled to range from -1 to 1 with fBits+2.
This would be the same from -20 to -2-1 (-1 to -0.5).
At any other signal value, the quantisation behaviours (and therefore the signal-to-noise ratios) of the two systems will be significantly different.
This is visible in Figure 6 where, when the signal is high (in the middle of the plots), the error level is approximately the same in the 4-bit fixed-point system and the floating point system with 2 bits for the fraction.
** Second small note for the attentive
You will notice that the black, blue, and green lines in Figure 7 have a sharp transition when the signal level hits 0 dB FS. This is because, in a fixed point system at signal levels below 0 dB FS, the signal to noise ratio is the difference in level between the dither’s noise floor and the signal. The dither level is constant, so as the signal level increases, it gets “further away” from the noise floor until you reach 0 dB FS (with a sine wave), as which point you reach the maximum possible SNR. However, once the signal goes beyond 0 dB FS (still assuming it’s a sine wave), then it starts to clip and distortion components are generated. It does not take much increase in level to drastically increase the level of the distortion relative to the level of the signal (since the signal level cannot increase – you’re just increasing distortion artefacts). Consequently, the signal to distortion+noise drops dramatically, because the distortion components increase in level dramatically.
This does not happen with the floating point system because, at 0 dB FS, you just change the exponent and keep going up with the signal level until you reach the maximum possible exponent value, which goes far beyond what I’ve plotted here.
Third small note for the attentive
You may be looking at Figure 7 and wondering why the fixed point plots and the floating point plots don’t overlap anywhere. For example, look where the green line (32-bit fixed point) crosses the red line (32-bit floating point). Why don’t they overlap each other there for that little 6 dB-wide range on the X-axis?
The reason is that I’m modelling the fixed point SNRs with TPDF dither, which “costs” 3 dB, but I’m assuming that the floating point signal is not dithered (which would normally be the case). If I were pretending that fixed point didn’t include the dither, then the plots would, indeed, overlap each other for that narrow little window.
***One last comment
You may be saying to yourself “But this is nonsense! Why do I need 150 dB SNR when the signal level is lower than -100 dB FS?” The long answer is in this posting, but the short answer is that the signal can go VERY low and VERY high inside a filter (a biquad), so you need to worry about this if you’re doing any changes to the magnitude response of the signal, for example…
This series has flipped back and forth between talking about high resolution audio files & sources and the processing that happens in the equipment when you play it. For this posting, we’re going to deal exclusively with the playback side – regardless of the source content.
I work for a company that makes loudspeakers (among other things). All of the loudspeakers we make use digital signal processing instead of resistors, capacitors, and inductors because that’s the best way to do things these days…
Point 1: This means that our volume control is a gain (a multiplier) that’s applied to the digital signal.
We also make surround processors (most of our customers call them “televisions”) that take a multichannel audio input (these days, this is under the flag of “spatial audio”, but that’s just a new name on an old idea) and distribute the signals to multiple loudspeakers. Consequently, all of our loudspeakers have the same “sensitivity”. This is a measurement of how loud the output is for a given input.
Let’s take one loudspeaker model, Beolab 90, as an example. The sensitivity of this loudspeaker is set to be the same as all other Bang & Olufsen loudspeakers. Originally, this was based on an analogue signal, but has since been converted to digital.
Point 2: Specifically, if you send a 0 dB FS signal into a Beolab 90 set to maximum volume, then it will produce a little over 122 dB SPL at 1 m in a free field (theoretically).
Let’s combine points 1 and 2, with a consideration of bit depth on the audio signal.
If you have a DSP-based loudspeaker with a maximum output of 122 dB SPL, and you play a 16-bit audio signal with nothing but TPDF dither, then the noise floor caused by that dither will be 122 – 93 = 29 dB SPL which is pretty loud. Certainly loud enough for a customer to complain about the noise coming from their loudspeaker.
Now, you might say “but no one would play a CD at maximum volume on that loudspeaker” to which I say two things:
I do. The “Banditen Galop” track from Telarc’s disc called “Ein Straussfest” has enough dynamic range that this is not dangerous. You just get very loud, but very short spikes when the gunshots happen.
That’s not the point I’m trying to make anyway…
The point I’m trying to make is that, if Beolab 90 (or any other Bang & Olufsen loudspeaker) used 16-bit DACs, then the noise floor would be 29 dB SPL, regardless of the input signal’s bit depth or dynamic range.
So, the only way to ensure that the DAC (or the bit depth of the signal feeding the DAC) isn’t the source of the noise floor from the loudspeaker is to use more than 16 bits at that point in the signal flow. So, we use a 24-bit DAC, which gives us a (theoretical) noise floor of 122 – 141 = -19 dB SPL. Of course, this is just a theoretical number, since there are no DACs with a 141 dB dynamic range (not without doing some very creative cheating, but this wouldn’t be worth it, since we don’t really need 141 dB of dynamic range anyway).
So, there are many cases where a 24-bit DAC is a REALLY good idea, even though you’re only playing 16-bit recordings.
Similarly, you want the processing itself to be running at a higher resolution than your DAC, so that you can control its (the DAC’s) signal (for example, you want to create the dither in the DSP – not hope that the DAC does it for you. This is why you’ll often see digital signal processing running at floating point (typically 32-bit floating point) or fixed point with a wider bit depth than the DAC.
If you read about high resolution audio – or you talk to some proponents of it, occasionally you’ll hear someone talk about “temporal resolution” or “micro details” or some such nonsense… This posting is just my attempt to convince the world that this belief is a load of horse manure – and that anyone using timing resolution as a reason to use higher sampling rates has no idea what they’re talking about.
Now that I’ve gotten that off my chest, let’s look at why these people could be so misguided in their belief systems…
Many people use the analogy of film to explain sampling. Even I do this – it’s how I introduced aliasing in Part 3 of this series. This is a nice analogy because it uses a known concept (converting movement into a series of still “samples”, frame by frame) to explain a new one. It also has some of the same artefacts, like aliasing, so it’s good for this as well.
The problem is that this is just an analogy – digital audio conversion is NOT the same as film. This is because of the details when you zoom in on a time scale.
Film runs at 24 frames per second (let’s say that’s true, because it’s true enough). This means that the time between on frame of film being shot and the next frame being shot is 1/24th of a second. However, the shutter speed – the time the shutter is open to make each individual photograph is less than 1/24th of a second – possibly much less. Let’s say, for the purposes of this discussion, that it’s 1/100th of a second. This means that, at the start of the frame, the shutter opens, then closes 1/100th of a second later. Then, for about 317/10,000ths of a second, the shutter is closed (1/24 – 1/100 ≈ 317/10,000). Then the process starts again.
In film, if something happened while that shutter was closed for those 317 ten-thousandths of a second, whatever it was that happened will never be recorded. As far as the film is concerned, it never happened.
This is not the way that digital audio works. Remember that, in order to convert an analogue signal into a digital representation, you have to band-limit it first. This ensures (at least in theory…) that there is no signal above the Nyquist frequency that will be encoded as an alias (a different frequency) in the digital domain.
When that low-pass filtering happens, it has an effect in the time domain (it must – otherwise it wouldn’t have an effect in the frequency domain). Let’s look at an example of this…
Let’s say that you have an analogue signal that consists of silence and one almost-infinitely short click that is converted to LPCM digital audio. Remember that this click goes through the anti-aliasing low-pass filter, and then gets sampled at some time. Let’s also say that, by some miracle of universal alignment of planets and stars, that click happened at exactly the same time as the sample was measured (we’ll pretend that this is a big deal and I won’t suggest otherwise for the rest of this posting). The result could look like Figure 1.
Figure 1: A click in the analogue domain (red) sampled and encoded as a digital signal (black circles). The analogue output of the digital system will be the black line
If I zoom in on Figure 1 vertically, it looks like the plot in Figure 2.
Figure 2: A vertical zoom of Figure 1.
There are at least three things to notice in these plots.
Since the click happened at the same time as a sample, that sample value is high.
Since the click happened at the same time as a sample, all other sample values are 0.
Once the digital signal is converted back to analogue later (shown as the black line) the maximum point in the signal will happen at exactly the same time as the click
I won’t talk about the fact that the maximum sample value is lower than the original click yet… we’ll deal with that later.
Now, what would happen if the click did not occur at the same time as the sample time? For example, what if the click happened at exactly the half-way point between two samples? This result is shown in Figure 3.
Figure 3: The click happened half-way in time between two samples.
Notice now that almost all samples have some non-zero value, and notice that the two middle samples (8 and 9) are equal. This means that when the signal is converted to analogue (as is shown with the black line) the time of maximum output is half-way between those two samples – at exactly the same time that the click happened.
Let’s try some more:
Figure 4: The click happened sometime between two samplesFigure 5: The click happened some other time between two samples.
I could keep doing this all night, but there’s no point. The message here is, no matter when in time the click happened, the maximum output of the digital signal, after it’s been converted back to analogue, happens at exactly the same time.
But, you ask, what about all that “temporal smearing” – the once-pristine click has been reduced to a long wave that extends in time – both forwards and backwards? Waitaminute… how can the output of the system start a wave before something happened?
Okay, okay…. calm down.
Firstly, I’ve made this example using only one type of anti-aliasing filter, and only one type of reconstruction filter. The waveforms I’ve shown here are valid examples – but so are other examples… This depends on the details of the filters you use. In this case, I’m using “linear phase” filters which are symmetrical in time. I could have used a different kind of filter that would have looked different – but the maximum point of energy would have occurred at the same time as the click. Because of this temporal symmetry, the output appears to be starting to ring before the input – but that’s only because of the way I plotted it. In reality, there is a constant delay that I have removed before doing the plotting. It’s just a filter, not a time machine.
Secondly, the black line is exactly the same signal you would get if you stayed in the analogue domain and just filtered the click using the two filters I just mentioned (because, in this discussion, I’m not including quantisation error or dither – they have already been discussed as a separate topic…) so the fact that the signal was turned into “digital” in between was irrelevant.
Thirdly, you may still be wondering why the level of the black line is so low compared to the red line. This is because the energy is distributed in time – so, in fact, if you were to listen to these two clicks, they’d sound like they’re the same level. Another way to say it is that the black line shows exactly the same as if the red curve was band-limited. The only thing missing is the upper part of the frequency band. (You may notice that I have not said anything about the actual sampling rate in any of this posting, because it doesn’t matter – the overall effect in the time domain is the same.)
Fourthly, hopefully you are able to see now that an auditory event that happens between two samples is not thrown away in the conversion to digital. Its timing information is preserved – only its frequency is band-limited. If you still don’t believe me, go listen to a digital recording (which is almost all recordings today) of a moving source – anything moving more than 7 mm will do*. If you can hear clicks in the sound as the source moves, then I’m wrong, and the arrival time of the sound is quantising to the closest sample. However, you won’t hear clicks (at least not because the source is moving), so I’m not wrong. Similarly, if digital audio quantised audio events to the nearest sample, an interpolated delay wouldn’t work – and since lots of people use “flanger” and “phaser” effects on their guitar solos with their weekend garage band, then I’m still right…
The conclusion
Hopefully, from now on, if you are having an argument about high resolution audio, and the person you’re arguing with says “but what about the timing information!? It’s lost at 44.1 kHz!” The correct response is to state (as calmly as possible) “BullS#!T!!”
Rant over.
* I said “7 mm” because I’m assuming a sampling rate of 48 kHz, and a speed of sound of 344 m/s. This means that the propagation distance in air is 344/48000 = 0.0071666 m per sample. In other words, if you’re running a 48 kHz signal out of a loudspeaker, the amplitude caused by a sample is 7 mm away when the next sample comes out.
Thought another way, if you have a stereo system, and your left loudspeaker is 7 mm further away from you than your right loudspeaker, at 48 kHz, you can delay the right loudspeaker by 1 sample to re-align the times of arrival of the two signals at the listening position.
Almost every audio system has filters or equalisers in it for some reason or another. Originally, equalisers were named that because they were put in on long-distance telephone lines to make the balance of the frequency content more equal. Nowadays, we use equalisers to do things like add bass, or to add more bass.
In the “old days” audio filters were made by building circuits with resistors, capacitors, and inductors: if you choose the relationships between the values of these devices correctly, you can affect the magnitude response as you choose. The problem was production tolerances: if you take two resistors out of the package, and both are supposed to have the same resistance – they’ll be close, but they won’t be identical.
One of the great things about audio filters implemented in a digital system is that you don’t need to worry about variations as a result of production differences. Since digital filters are “just math”, you put the same equation in every device, and you get the same answer for the same input every time. (In the same way that, if I have two calculators on my desk, and I put “2 x 3″ into both of them, and press”=”, I’ll get the same answer on both devices.)
So, to start, let’s talk a little about how a digital filter works. Generally speaking, digital filters work by taking an audio signal, delaying it, changing the level, and adding the result back to the signal itself. Let’s take a simple example, shown in Figure 1.
Figure 1: Simple digital filter
Let’s say that, to start, we make the gain in that signal flow = 1, and set the delay to equal 1 sample. In this case:
At a very low frequency, the output of the delay has almost exactly the same value as its input (because 1 sample is a phase difference of almost 0º at a low frequency). When you add a signal to (nearly) itself, you get twice the output – a gain of 6 dB.
As the frequency of the input goes higher and higher, the delay (of 1 sample) is more and more significant, and therefore its output value gets more and more different from its input value.
When the input signal’s frequency is 1/2 of the sampling rate, then a delay of 1 sample is equal to a 90º phase shift. When you add a sine wave to itself with a “delay” (actually a “phase shift”) of 90º, the result is a magnitude that is 3 dB higher than the original.
When the input signal’s frequency is 1/3 of the sampling rate, then a delay of 1 sample is equal to a 120º phase shift. When you add a sine wave to itself with a “delay” (actually a “phase shift”) of 120º, there’s no change in the magnitude (the level).
When the frequency is 1/2 the sampling rate, then each consecutive sample is 180º out of phase with the previous one, so the sum of the delay and the signal results in complete cancellation, and you get no output at all.
An example of the magnitude response plot of this is shown below in Figure 2.
Figure 2: The magnitude response of the filter shown in Figure 1 when the delay is set to 1 sample, the gain is set to 1 and Fs=48 kHz.
If we reduce the gain to, say 0.5, then the effect of adding the delayed signal is reduced. The overall shape of the magnitude response is the same, it’s just less, as shown in Figure 3.
Figure 3: The magnitude response of the filter shown in Figure 1 when the delay is set to 1 sample, the gain is set to 0.5 and Fs=48 kHz.
Notice in Figure 3 that the boost in the low end is less, and the dip in the high end is also less than in Figure 2. So, by adjusting the gain on the delayed signal that’s added to the original signal, we can adjust how much this filter is affecting the signal.
What happens when we change the delay? If we make it 2 samples instead of 1, then the phase difference between the output and the input of the delay will be bigger for a given frequency. This also means that the delay will be equivalent to a 180º phase shift at 1/4 Fs (instead of 1/2). Also, at 1/2 Fs, the delay will be equivalent to a 360º phase shift, so the signal adds constructively, just like it does in the low frequencies. So, the resulting magnitude response will look like Figure 4.
Figure 4: The magnitude response of the filter shown in Figure 1 when the delay is set to 2 samples, the gain is set to 1 and Fs=48 kHz.
Again, if we reduce the gain, we reduce the effect of the filter, as can be seen in Figure 5.
Figure 5: The magnitude response of the filter shown in Figure 1 when the delay is set to 2 samples, the gain is set to 0.5 and Fs=48 kHz.
Now let’s make things a little more complicated. We can add another delay and another gain to get a little more control of things.
Figure 6: Getting a little more complicated
I’m not going to get very detailed about this – but if each of those delays is just one sample long, and we only play with the gains g1 and g2, we can start getting some nice control over the response of this filter. Below are some examples of the results we can get with just this filter, playing with the gains.
So, as you can see, all I need to do is to play with those two gains to get some nice control over the magnitude response.
Up to now, everything I’ve done is to add a delayed copy of the input to itself. This is what is known as a “feed-forward” design because (as you can see in Figures 1 and 6) I’m feeding the signal forwards in the flow to be added to itself. However, if there’s a “feed-forward”, it must be because we want to distinguish it from a “feed-back” design.
This is a filter where we delay the output (instead of the input) multiply that by a gain, and add it to the signal, as shown in Figure 11.
Figure 11: A simple digital filter using feedback.
This feedback means that, if the gain is not equal to zero, once a signal gets into the input, the output will last forever. This is why this kind of filter is called an infinite impulse response filter (or IIR filter): because if an impulse (a short spike) gets into it, there will be a signal at the output until the end of time (theoretically, at least…).
And, yes… the output of a filter without a feedback loop will eventually stop, which means it’s a finite impulse response filter (or FIR). Stop the input signal, wait for the last delay to send its signal through, and the output stops.
So what?
Most filters in most digital audio devices are built on a combination of these two types of designs. If you take Figure 6 and you combine it with an extended version of Figure 11 (with two delays instead of just one) you get Figure 12:
Figure 12: An FIR filter with two delays and 3 gains combined with an IIR filter with two delays and 2 gains.
This combination of FIR and IIR filters is a powerful little tool that forms the heart of almost every digital audio filter in the world. (yes, there are exceptions, but they’re definitely exceptions…). There are different ways to implement it (for example, you could put the IIR before the FIR, or you could re-draw it to share the delays) but the result is the same.
This little tool is what we call a “biquadratic filter” or “biquad” for short. (The reason it’s called that is that the effect it has on the signal (its “transfer function”) can be mathematically expressed as the ratio of two quadratic equations – but I will not say anything more about that.) Whenever developers are building a new digital audio device like a loudspeaker or a pair of headphones or a car audio system, it’s common in the early meetings to hear someone ask “how many biquads will we need?” which is a way of asking “how much processing power and memory will we need?” (In the same way that I can measure prices in pizzas, or when I was a kid I would ask “how many more Sesame Streets until we’re there?”)
At this point, you may be asking why I’ve gone through all of this, since I haven’t said anything about high resolution audio. The reason is that, in the next posting, we’ll look at what’s going on inside that biquad when you use it to do filtering – and how that changes, not only with the filters you’re building, but how they relate to the sampling rate and the bit depth…
Back in Part 5 of this series, I described an example of a pretty typical / normal signal flow for an audio signal that you’re playing from a streaming service to a “smart-ish” loudspeaker in your house. If you read through that list, you’ll see that I mentioned that the signal might be sampling-rate converted two times (once in your player, and once again in your loudspeaker or headphones).
Let me say something very clearly, before we go any further:
There’s no guarantee that this is happening. For example, many players don’t sampling-rate convert the signal if the device they’re sending the signal is compatible with the sampling rate of the signal. However, many players do sampling-rate convert the signal – and many devices (like DACs, for example) are not compatible with all sampling rates, so the player is forced to do something about it.
Sampling rate conversion is not necessarily a bad thing. There are many good sampling rate converters out there in the world. In fact, you can use a high-quality sampling rate converter to reduce problems with jitter coming in from an “upstream” device or transmission path.
However, sampling rate conversion is not necessarily a good thing either… so the more of them you have in your audio signal path, the better you want them to be. In an optimal case, the artefacts caused by the sampling rate converter will not be the “weakest link” in the audio chain.
However, this last statement is very easy to mis-interpret, as I alluded to in Part 6. The problem is that, if I say “I have a sampling rate converter with a THD+N of -100 dB relative to the signal level” this might look pretty good. However, if the signal and the SRC artefacts are in COMPLETELY different frequency bands, and you’re playing the signal out of a loudspeaker that can’t produce the signal (say, because it’s too low in frequency) then 100 dB might not be nearly good enough. In other words, it’s not a mere numbers-game… you have to know how to interpret the data…
A what?
Maybe we should first back up a little and talk about what a sampling rate converter is. As you saw in Part 1, at its most basic level, LPCM digital audio is just a way of describing a signal by storing a long string of measurements that were made at a regular time interval. Each of those measurements is called a “sample” and the rate at which you measure the samples (per second) is called the “sampling rate”. A CD, for example, uses a standard sampling rate of 44,100 samples per second, or 44.1 kHz. Other systems use other rates.
If you want to listen to a CD on a loudspeaker with built-in digital processing, and the loudspeaker happens to have an internal sampling rate that is NOT 44.1 kHz (let’s say that it’s 48 kHz), then you need to somehow convert the sampling rate from 44.1 kHz to 48 kHz to get things to work properly. (This is a little like having a gearbox in a car – your engine does not turn at the same speed as your wheels – you put gears in-between to convert the rotational speed of the engine to the rotational speed of the wheels.)
One sneaky way to do this is to use an analogue connection – you convert the 44.1 kHz digital signal to an analogue one using a DAC, and then re-sample the analogue signal using an ADC running at 48 kHz. This is simple, and (if you choose your DAC and ADC properly) potentially a really good solution. In the “old days” (up to the 1990s) before digital SRCs became really good, this was the best way to do it (assuming you had access to some decent gear).
There are many ways to make a fully-digital SRC. For example:
Let’s say that you have an audio signal that’s been sampled at some sampling rate that we’ll call “Fs1” (for “Sampling Frequency 1”) , as is shown in Figure 1.
Figure 1: A signal recorded at some sampling rate.
You then want to have the same signal, represented at a different sampling rate, which we’ll call Fs2. The old signal (in black) and the new sampling rate (the red dots and the gridlines) can be seen in Figure 2.
Figure 2: The original signal at Fs1 and the new samples that we want to create (in red) at Fs2.
How do we do this? One way is to draw straight lines between the original samples, and calculate the values at the point on the line that corresponds with the time of the new samples. This is called “linear interpolation” (because it’s based on drawing straight lines between the original samples), and it’s shown in Figure 3.
Figure 3: An example of linear interpolation for converting to the new sampling rate.
A better way to do this is to use some fancy math to calculate where the signal would be after the reconstruction filter smoothed it back to the original (band-limited) input. There are different ways to do this (in other words, different mathematical strategies) that are outside the scope of this posting, however, I’ve shown an example of a piecewise cubic spline interpolation implementation in Figure 4, below.
Figure 4: An example of piecewise cubic spline interpolation for converting to the new sampling rate.
However, let’s say that:
you’ve been given the job of building a sampling rate converter, but
you think that the examples I gave above are way to complicated…
What do you do? One possibility is to look at the sample value that you want to output, find the closest sample (in time) in the original signal, and use that. This is a technique commonly called “nearest neighbour” for obvious reasons – and it’s one of the worst-performing SRC strategies you can use. An example of this is shown in Figure 5, below. Notice that the new values (the red circles) are identical to the closest original value
Figure 5: An example of “nearest neighbour” interpolation for converting to the new sampling rate. Note that each new values in red is a copy of the closest value in black.
If we look at these two signals without the sample values, we’ll see some pretty nasty distortion in the time domain, as shown in Figure 6.
Figure 6: The same signals shown in Figure 5 without the circles.
So what?
The plots above show the results of good and bad SRCs in the time domain, but what does this look like in the frequency domain? Let’s take a couple of specific examples.
Figure 7: 500 Hz sine tone at 0 dB FS, Linear interpolation from 44.1 kHz to 48 kHz.Figure 8: 500 Hz sine tone at 0 dB FS, Piecewise cubic spline interpolation from 44.1 kHz to 48 kHz.
Figures 7 and 8 look almost identical. There are the windowing artefacts of the frequency analysis that I’m doing are larger than most of the artefacts caused by the interpolation implementations. However, you may notice a couple of spikes sticking up between 1 kHz and 10 kHz in Figure 7. These are the most obvious frequency-domain artefacts of the distortion caused by linear interpolation. Notice however, that those artefacts are about 80 dB down from the signal – so that’s pretty good for a cheap implementation.
However, let’s look at the same 500 Hz tone converted using the “nearest neighbour” strategy.
Figure 9: 500 Hz sine tone at 0 dB FS, “nearest neighbour” interpolation from 44.1 kHz to 48 kHz.
Now you can see that things have really fallen apart The artefacts are almost up to 40 dB below the signal level, and they’re quite far away in frequency, so they’ll be easy to hear. Also remember that the artefacts that are generated here are inside the audio band, so they will not be eliminated later in the chain by a reconstruction filter in a DAC, for example. They’re there to stay.
There’s one more interesting thing to consider here. Let’s try the same nearest neighbour algorithm, converting between the same two sampling rates, but I’ll put in signals at different frequencies.
Figure 10: 50 Hz sine tone at 0 dB FS, “nearest neighbour” interpolation from 44.1 kHz to 48 kHz.Figure 11: 5 kHz sine tone at 0 dB FS, “nearest neighbour” interpolation from 44.1 kHz to 48 kHz.
Figure 10 shows the same system, but the input signal is a 50 Hz sine wave (instead of 500 Hz). Notice that the artefacts are now about 60 dB down (instead of 40 dB).
Figure 11 shows the same system again, but the input signal is a 5 kHz sine wave. Notice that the artefacts are now only about 20 dB down.
So, with this poor implementation of an SRC, the distortion-to-signal ratio is not only dependent on the algorithm itself, but the signal’s frequency content. Why is this?
Think back to the way the “nearest neighbour” strategy works. You’re simply copying-and-pasting the value of the nearest sample. However, the lower the frequency, the less change there is in the signal from sample to sample. So, as your signal’s frequency goes down (more accurately, as it gets further away from the sampling rate), the smaller the error that you create with this system. At 0 Hz, there would be no error, because all of the samples would have the same value.
So, (for example) if your job is to build the SRC in the first place, and you measure it with a 50 Hz tone, you’ll see that the artefacts are 60 dB below the signal and you’ll pat yourself on the back and go to lunch. Then, some weeks later, when the customer complaints start coming in about tweeter distortion, you’ll think it must be someone else’s fault… but it isn’t…
Conclusion
What does this have to do with “High Resolution Audio”? Well, the problem is that most audio gear does not run at crazy-high sampling rates (this is not necessarily a bad thing), so if you play a high-res file, you’re probably sampling rate converting (this is not necessarily a bad thing).
play the signal with a different (e.g. not-high-res) sampling rate to find out if it’s better, OR
buy better gear, OR
at least check for a firmware update.
Note that first recommendation of the three: Because the quality of a sampling rate converter is very dependent upon the relationship between the input and the output sampling rates, it can happen that a “normal” resolution audio signal (say, at 44.1 kHz) will sound better on your particular equipment than a “high” resolution audio signal (say, at 192 kHz) because of this. Of course, the opposite could be true (say, because your gear is running at 48 kHz and it’s easier to get to that from 192 kHz (just multiply by 1/4) than it is to get there from 44.1 kHz (just multiply by 480/441…)
This doesn’t mean that “low-res” is better than “high-res” – it just means that your particular equipment deals with it better. (In the same way that purely from the point of view as a fuel, gasoline might have more energy per litre than diesel fuel, but it’s a terrible choice to put in the tank of a car that’s expecting diesel…)
In Part 2 of this series, I talked about the relationship between the noise floor of an LPCM signal and the number of bits used to encode it.
Assuming that the signal is correctly dithered using TPDF dither with a peak-to-peak amplitude of ±1 LSB, then this means that you can easily calculate the dynamic range of your system with a very simple equation:
Dynamic Range in dB = 6.02 * NumberOfBits – 3
(Note that the sampling rate is not part of this equation… That will be useful information later.)
Normally, we’re lazy and we say 6 times the number of bits -3 for the dither – but if you’re really lazy, you leave out the -3 as well.
So, this means that, in a 16-bit system, the noise floor is 93 dB below a sine wave at full scale (6 * 16 -3 = 93) and for a 24-bit system, the noise floor is 141 dB below a sine wave at full scale (you do the math as practice).
Also, we can generalise and say that “adding 1 bit halves the level of the noise floor” (because -6 dB is the same as multiplying by 0.5). However, this is only part of the story.
The noise that’s generated by dither has a “white” characteristic. This means that there is an equal probability of getting some energy per bandwidth (or some say “per Hertz”) over a period of time. This sounds a little complicated, so I’ll explain.
Noise is random. This means that you may or may not get energy at, say 1 kHz, in a given short measurement. However, if you measure white noise for long enough, you’ll eventually see that you got something in every frequency band. Also, you’ll see that, if you look back over the entire length of your measurement of white noise, you got the same amount of energy in the band from 100 Hz to 200 Hz as you did in the band from 1000 Hz to 1200 Hz and the band from 10,000 Hz to 10,200 Hz. (Each of those bandwidths is 200 Hz wide).
There are now two things to discuss:
This distribution of energy is not like the way we hear things. We don’t hear the distance between 100 Hz and 200 Hz as the same distance as going from 1,000 Hz to 1,200 Hz. We hear logarithmically, which means that we hear in multiples of frequency, not additions of bandwidth. So, to use 100 Hz – 200 Hz sounds like the same “distance” as 1,000 Hz to 2,000 Hz. This is why white noise sounds like it is “bright” – or it has emphasis on the high frequencies. If you have a system that has a flat response from 0 to 20,000 Hz, and you play white noise through it, you have the same amount of energy in the top octave (10 kHz to 20 kHz) as you do in all of the octaves below – which is why we hear this as “top-heavy”.
If you had two bands of white noise with equal levels, and let’s say that one ranges from 100 Hz to 200 Hz, and the other is 1000 Hz to 1200 Hz, then the output level of the two of them together will be 3 dB louder than the output level of either of them alone. This is because their powers add together instead of their amplitudes (because the two signals are unrelated to each other).
Let’s put all this (and one or two other things) together:
We know from a previous part in this series that an LPCM digital audio system cannot have signals higher than the Nyquist frequency – 1/2 the sampling rate.
TPDF dither is white noise at a total level that is (6.02 * NumberOfBits – 3) dB below full-scale.
If you add white noise signals with equal levels but different bandwidths, you get a 3 dB increase over the level of just one of them
This means that,
if I have a 16-bit, TPDF dithered LPCM audio signal with a sampling rate of 48 kHz, it has a noise floor that is 93 dB below full scale, and that noise has a white characteristic with a bandwidth of 24 kHz (the Nyquist frequency). There will be no noise above that frequency coming out of the system.
if I have a 16-bit, TPDF dithered LPCM audio signalwith a sampling rate of 192 kHz, it has a noise floor that is 93 dB below full scale, and that noise has a white characteristic with a bandwidth of 96 kHz (the Nyquist frequency). There will be no noise above that frequency coming out of the system.
So, the two systems have the same noise floor level overall, but with very different bandwidths… What does this mean?
Well, let’s start by looking at the level of the noise floor in the 48 kHz system (so the noise “only” extends to 24 kHz).
If I double the sampling rate (to 96 kHz), I double the bandwidth of the noise without changing its level, so this means that the portion of the noise that “lives” in the 0 Hz – 24 kHz region drops by 3 dB (because I’m ignoring the top half of the signal ranging from 24 kHz to 48 kHz in the 96 kHz system.
Figure 1: The noise floor caused by TPDF dither in a 16-bit LPCM system with a sampling rate of 96 kHz. The total power of the noise drawn in red is the same as the total power of the noise drawn in blue. But notice that this is plotted on a linear scale for frequency.Figure 1: Re-plotted on a logarithmic scale for the frequency. To us, the top octave is not so important, but to a measurement device, it’s half the signal…
If I had multiplied the original sampling rate by 4 (to 192 kHz) I multiply the bandwidth of the noise by 4 as well (to 96 kHz). This means that the overall level of the noise from 0 to 24 kHz is now 6 dB down from the original version.
In other words: if I multiply the sampling rate by two, but I don’t increase the bandwidth of the noise floor that I’m interested in (say I only care about 20 Hz – 20 kHz), then its level drops by 3 dB.
So what?
Well, you could jump to the conclusion that this proves that higher sampling rates are better. However, that would be a bit (ha hah) premature. Consider that, if you want to drop the (band-limited) noise floor by 6 dB, you have to quadruple the sampling rate – and therefore quadrupling the data rate (and therefore the disc storage, the bandwidth of the transmission system, the error rate, and so on…) A 400% increase in the data is not insignificant.
OR, you could just add one more bit – going from 16 bits to 17 bits will give you the same result with a data increase of only 6.25% – a much smarter decision, no?
The Real World
This little analysis above makes a basic (and possibly incorrect) assumption. The assumption is that, by quadrupling the sampling rate, all other components in the system will remain predictably identical. This may not be true. For example, many DACs (especially older ones) exhibit an increase in their own noise floor when you use them at a higher sampling rate. So, it could be that the benefit you get theoretically is negated by the detriment that you actually get. This is just one example of a flaw in the theory – but it’s a very typical one – especially if you’re building a product instead of just using one.
P.S.
You may have looked at Figures 1 or 2 and are wondering why, if the noise floor is at -93 dB FS in a 16-bit system, I plotted it around -120 dB FS (give or take). The reason is related to the explanation I just gave above. I said in the captions that it’s from a 96 kHz system. This means that the noise extends to the Nyquist frequency at 48 kHz, and that total level is at -93 dB FS. We also know that, if I keep the noise the same, but half the bandwidth that I’m looking at, the level drops by 3 dB. Therefore I can either do math or I can make the following table:
Bandwidth of noise measurement in Hz
Level in dB FS
48,000
-93
24,000
-96
12,000
-99
6,000
-102
3,000
-105
1,500
-108
750
-111
375
-114
187.5
-117
93.75
-120
If you look carefully at the figures, you’ll see that there’s a point every 100 Hz. (It’s most easily visible in the low-frequency range of Figure 2.) So, the level of the noise that I see on a magnitude response plot like this is not only dependent on the noise level itself, but the bandwidths of the divisions that I’ve used to slice it up. In my case, the bandwidth per “slice” is about 100 Hz, so the noise level of each of those little contributors is at about -120 dB FS. If I had used slices only 50 Hz wide, it would show up at -123 instead…
