Once-upon-a-time I posted a little thing about Burmese Colour Needles for grammophones. Then, today, I received an email from the owner of www.burmesecolourneedles.com a company owned by Andy Briggs, where you can buy brand new ones!

Andy also has a YouTube channel that is well worth visiting.

When you look at the datasheet of an audio device, you may see a specification that states its “signal to noise ratio” or “SNR”. Or, you may see the “dynamic range” or “DNR” (or “DR”) lists as well, or instead.

These days, even in the world of “professional audio” (whatever that means), these two things are similar enough to be confused or at least confusing, but that’s because modern audio devices don’t behave like their ancestors. So, if we look back 30 years ago and earlier, then these two terms were obviously different, and therefore independently usable. So, in order to sort this out, let’s take a look at the difference in old audio gear and the new stuff.

Let’s start with two of basic concepts:

All audio devices (or storage media or transmission systems) make noise. If you hold a resistor up in the air and look at the electrical difference across its two terminals and you’ll see noise. There’s no way around this. So, an amplifier, a DAC, magnetic tape, a digital recording stored on a hard drive… everything has some noise floor at the bottom that’s there all the time.

All audio devices have some maximum limit that cannot be exceeded. A woofer can move in and out until it goes so far that it “bottoms out” on the magnet or rips the surround. A power amplifier can deliver some amount of current, but no higher. The headphone output on your iPhone cannot exceed some voltage level.

So, the goal of any recording or device that plays a recording is to try and make sure that the audio signal is loud enough relative to that noise that you don’t notice it, but not so loud that the limit is hit.

Now we have to look a little more closely at the details of this…

If we take the example of a piece of modern audio equipment (which probably means that it’s made of transistors doing the work in the analogue domain, and there’s lots of stuff going on in the digital domain) then you have a device that has some level of constant noise (called the “noise floor”) and maximum limit that is at a very specific level. If the level of your audio signal is just a weeee bit (say, 0.1 dB) lower than this limit, then everything is as it should be. But once you hit that limit, you hit it hard – like a brick wall. If you throw your fist at a brick wall and stop your hand 1 mm before hitting it, then you don’t hit it at all. If you don’t stop your hand, the wall will stop it for you.

In older gear, this “brick wall” didn’t exist in lots of gear. Let’s take the sample of analogue magnetic tape. It also has a noise floor, but the maximum limit is “softer”. As the signal gets louder and louder, it starts to reach a point where the top and bottom of the audio waveform get increasingly “squished” or “compressed” instead of chopping off the top and bottom.

I made a 997 Hz sine wave that starts at a very, very low level and increases to a very high level over a period of 10 seconds. Then, I put it through two simulated devices.

Device “A” is a simulation of a modern device (say, an analogue-to-digital converter). It clips the top and bottom of the signal when some level is exceeded.

Device “B” is a simulation of something like the signal that would be recorded to analogue magnetic tape and then played back. Notice that it slowly “eases in” to a clipped signal; but also notice that this starts happening before Device “A” hits its maximum. So, the signal is being changed before it “has to”.

Let’s zoom in on those two plots at two different times in the ramp in level.

Device “A” is the two plots on the top at around 8.2 seconds and about 9.5 seconds from the previous figure. Device “B” is the bottom two plots, zooming in on the same two moments in time (and therefore input levels).

Notice that when the signal is low enough, both devices have (roughly) the same behaviour. They both output a sine wave. However, when the signal is higher, one device just chops off the top and bottom of the sine wave whereas the other device merely changes its shape.

Now let’s think of this in terms of the signals’ levels in relationship to the levels of the noise floors of the devices and the distortion artefacts that are generated by the change in the signals when they get too loud.

If we measure the output level of a device when the signal level is very, very low, all we’ll see is the level of the inherent noise floor of the device itself. Then, as the signal level increases, it comes up above the noise floor, and the output level is the same as the level of the signal. Then, as the signal’s level gets too high, it will start to distort and we’ll see an increase in the level of the distortion artefacts.

If we plot this as a ratio of the signal’s level (which is increasing over time) to the combined level of the distortion and noise artefacts for the two devices, it will look like this:

On the left side of this plot, the two lines (the black door Device “A” and the red for Device “B”) are horizontal. This is because we’re just seeing the noise floor of the devices. No matter how much lower in level the signals were, the output level would always be the same. (If this were a real, correct Signal-to-THD+N ratio, then it would actually show negative values, because the signal would be quieter than the noise. It would really only be 0 dB when the level of the noise was the same as the signal’s level.)

Then, moving to the right, the levels of the signals come above the noise floor, and we see the two lines increasing in level.

Then, just under a signal level of about -20 dB, we see that the level of the signal relative to the artefacts starts in Device “B” reaches a peak, and then starts heading downwards. This is because as the signal level gets higher and higher, the distortion artefacts increase in level even more.

However, Device “A” keeps increasing until it hits a level 0 dB, at which point a very small increase in level causes a very big jump in the amount of distortion, so the relative level of the signal drops dramatically (not because the signal gets quieter, but because the distortion artefacts get so loud so quickly).

Now let’s think about how best to use those two devices.

For Device “A” (in red) we want to keep the signal as loud as possible without distorting. So, we try to make sure that we stay as close to that 0 dB level on the X-axis as we can most of the time. (Remember that I’m talking about a technical quality of audio – not necessarily something that sounds good if you’re listening to music.) HOWEVER: we must make sure that we NEVER exceed that level.

However, for Device “B”, we want to keep the signal as close to that peak around -20 dB as much as possible – but if we go over that level, it’s no big deal. We can get away with levels above that – it’s just that the higher we go, the worse it might sound because the distortion is increasing.

Notice that the red line and the black line cross each other just above the 0 dB line on the X-axis. This is where the two devices will have the same level of distortion – but the distortion characteristics will be different, so they won’t necessarily sound the same. But let’s pretend that the the only measure of quality is that Y-axis – so they’re the same at about +2 dB on the X-axis.

Now the question is “What are the dynamic ranges of the two systems?” Another way to ask this question is “How much louder is the loudest signal relative to the quietest possible signal for the two devices?” The answer to this is “a little over 100 dB” for both of them, since the two lines have the same behaviour for low signals and they cross each other when the signal is about 100 dB above this (looking at the X-axis, this is the distance between where the two lines are horizontal on the left, and where they cross each other on the right). Of course, I’m over-simplifying, but for the purposes of this discussion, it’s good enough.

The second question is “What are the signal-to-noise ratios of the two systems?” Another way to ask THIS question is “How much louder is the average signal relative to the quietest possible signal for the two devices?” The answer to this question is two different numbers.

Device “A” has a signal-to-noise ratio of about 100 dB , because we’re going to use that device, trying to keep the signal as close to clipping as possible without hitting that brick wall. In other words, for Device “A”, the dynamic range and the signal-to-noise ratio are the same because of the way we use it.

Device “B” has a signal-to-noise ratio of about 80 dB because we’re going to try to keep the signal level around that peak on the black curve (around -20 dB on the X-axis). So, its signal-to-noise ratio is about 20 dB lower than its dynamic range, again, because of the way we use it.

The problem is, these days, a lot of engineers aren’t old enough to remember the days when things behaved like Device “B”, so they interchange Signal to Noise and Dynamic Range all willy-nilly. Given the way we use audio devices today, that’s okay, except when it isn’t.

For example, if you’re trying to connect a turntable (which plays vinyl records that are mastered to behave more like Device “B”) to a digital audio system, then the makers of those two systems and the recordings you play might not agree on how loud things should be. However, in theory, that’s the problem of the manufacturers, not the customers. In reality, it becomes the problem of the customers when they switch from playing a record to playing a digital audio stream, since these two worlds treat levels differently, and there’s no right answer to the problem. As a result, you might need to adjust your volume when you switch sources.

This episode of 99 Percent Invisible tells the story of the Recording Ban of 1942, the impact on the rise of modern jazz music, and the parallels with the debates between artists and today’s streaming services. It’s worth the 50 minutes and 58 seconds it takes to listen to this!

At the end of that episode, the ban on record manufacture is mentioned, almost as an epilogue. This page from the January, 1949 issue of RCA’s “Radio Age” magazine discusses the end of that ban.

Interestingly, that same issue of the magazine has an article that introduces a new recording format: 7-inch records operating at 45 revolutions per minute! The article claims that the new format is “distortion free” and “noise-free”, stating that this “new record and record player climax more than 10 years of research and refinement in this field by RCA.”

What caught my eye was the discussion of gramophone needles made of “hard wood”, and also the prediction that “the growth of electrical recording steps … to grapple with that problem of wear and tear.”

The fact that electrical (instead of mechanical) recording and playback was seen as a solution to “wear and tear” reminded me of my first textbook in Sound Recording where “Digital Audio” was introduced only within the chapter on Noise Reduction.

Later in that same issue, there is a little explanation of the “Electrocolor” and “Burmese” needles.

The March 1935 issue raises the point of wear vs. fidelity in the Editorial (which starts by comparing players with over-sized horns).

I like the comment about having to be in the “right mood” for Ravel. Some things never change.

What’s funny is that, now that I’ve seen this, I can’t NOT see it. There are advertisements for fibre, thorn, and wood needles all over the place in 1930s audio magazines.

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.

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 2^{3} 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 2^{16}, or 65,536 possible quantisation values. For a 24-bit sample, we have 2^{24}, 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(2^{nBits}) – 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

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) * 2^{e}

We need to know a little extra information:

because we’re using 23 bits for f, then it can range from 0 to 2^{23}-1. In other words, stated mathematically: 0 ≤ 2^{23}*f < 2^{23}

because we’re using 8 bits for e, then it has a total range of 2^{8} possible values. In other words it has a range from just over -2^{7} to just under 2^{7}. 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 2^{e} to 2^{e+1}, the steps all have an equal size of 2^{e-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 2^{fBits} 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.

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 2^{0}), 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.

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).

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.

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 2^{e} to 2^{e+1}, the steps all have an equal size of 2^{e-fBits} where fBits is the number of bits used to express f (in our case, fBits = 23 bits). In other words, we have 2^{fBits} 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 2^{0}), 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 2^{23} 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 2^{24} instead. If I were to continue the spacing all the way down to -1, then there would be twice as many again, or 2^{25}.

In other words, a floating point signal ranging from a value of 2^{-1} to 2^{0} (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 -2^{0} 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…