I thought that I was finished talking about (and even thinking about) the RCA Dynagroove Dynamic Styli Correlator as well as tracking and tracing distortion… and then I got an email about the last two postings pointing out that I didn’t mention two-channel stereo vinyl, and whether there was something to think about there.
My first reaction was: “There’s nothing interesting about that. It’s just two channels with the same problem, and since (at least in a hypothetical world) the two axes of movement of the needle are orthogonal, then it doesn’t matter. It’ll be the same problem in both channels. End of discussion.”
Then I took the dog out for a walk, and, as often happens when I’m walking the dog, I re-think thoughts and come home with the opposite opinion.
So, by the time I got home, I realised that there actually is something interesting about that after all.
Starting with Emil Berliner, record discs (original lacquer, then vinyl) have been cut so that the “mono” signal (when the two channels are identical) causes the needle to move laterally instead of vertically. This was originally (ostensibly) to isolate the needle’s movement from vibrations caused by footsteps (the reality is that it was probably a clever manoeuvring around Edison’s patent).
This meant that, when records started supporting two audio channels, a lateral movement was necessary to keep things backwards-compatible.
What does THIS mean? It means that, when the two channels have the same signal (say, on the lead vocal of a pop tune, for example) when the groove of the left wall goes up, the groove of the right wall goes down by the same amount. That causes the needle to move sideways, as shown below in Figure 1.
Figure 1. A two-channel groove with identical information in the two channels.
What are the implications of this on tracing distortion? Remember from the previous posting that the error in the movement of the needle is different on a positive slope (where the needle is moving upwards) than a negative slope (downwards). This can be seen in a one-channel representation in Figure 2.
Figure 2. The grey line is the groove wall. The blue line shows the actual movement of the needle and the red line shows the difference between the two – the error contained in the output signal.
Since the two groove walls have an opposite polarity when the audio signals are the same, then the resulting movement of the two channels with the same magnitude of error will look like Figure 3.
Figure 3. The physical movement of the two channels, and their independent errors.
Notice that, because the two groove walls are moving in opposite polarity (in other words, one is going up while the other is going down) this causes the two error signals to shift by 1/2 of a period.
However, Figure 3 doesn’t show the audio’s electrical signals. It shows the physical movement of the needle. In order to show the audio signals, we have to flip the polarity of one of the two channels (which, in a real pickup would be done electrically). That means that the audio signals will look like Figure 4.
Figure 4. The electrical outputs of the two audio channels and their error components.
Notice in Figure 4 that the original signals are identical (that’s why it looks like there’s only one sine wave) but their actual outputs are different because their error components are different.
But here’s the cool thing:
One way to think of the actual output signals is to consider each one as the sum of the original signal and the error signal. Since (for a mono signal like a lead vocal) their original signals are identical, then, if you sit in the right place with a properly configured pair of loudspeakers (or a decent pair of headphones) then you’ll hear that part of the signal as a phantom image in the middle. However, since the error signals are NOT correlated, they will not be localised in the middle with the voice. They’ll move to the sides. They’re not negatively correlated, so they won’t sound “phase-y” but they’re not correlated either, so they won’t be in the same place as the original signal.
So, although the distortion exists (albeit not NEARLY on the scale that I’ve drawn here…) it could be argued that the problem is attenuated by the fact that you’ll localise it in a different place than the signal.
Of course, if the signal is only in one channel (like Aretha Franklin’s backup singers in “Chain of Fools” for example) then this localisation difference will not help. Sorry.
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.
Without connecting external loudspeakers, Bang & Olufsen’s Beosound Theatre has a total of 11 independent outputs, each of which can be assigned any Speaker Role (or input channel). Four of these are called “virtual” loudspeakers – but what does this mean? There’s a brief explanation of this concept in the Technical Sound Guide for the Theatre (you’ll find the link at the bottom of this page), which I’ve duplicated in a previous posting. However, let’s dig into this concept a little more deeply.
To begin, let’s put a “perfect” loudspeaker in a free field. This means that it’s in a space that has no surfaces to reflect the sound – so it’s an acoustic field where the sound wave is free to travel outwards forever without hitting anything (or at least appear as this is the case). We’ll also put a “perfect” microphone in the same space.
Figure 1: A loudspeaker and a microphone (the circle) in a free field: an infinite space completely free of reflective surfaces.
We then send an impulse; a very short, very loud “click” to the loudspeaker. (Actually a perfect impulse is infinitely short and infinitely loud, but this is not only inadvisable but impossible, and probably illegal.)
Figure 2: The “click” signal that’s sent to the input of the loudspeaker.
That sound radiates outwards through the free field and reaches the microphone which converts the acoustic signal back to an electrical one so we can look at it.
Figure 3: The “click” signal that is received at the microphone’s location and sent out as an electrical signal.
There are three things to notice when you compare Figure 3 to Figure 2:
The signal’s level is lower. This is because the microphone is some distance from the loudspeaker.
The signal is later. This is because the microphone is some distance from the loudspeaker and sound waves travel pretty slowly.
The general shape of the signals are identical. This is because I said that the loudspeaker and the microphone were both “perfect” and we’re in a space that is completely free of reflections.
What happens if we take away the microphone and put you in the same place instead?
Figure 4: The microphone has been replaced by something more familiar.
If we now send the same click to the loudspeaker and look at the “outputs” of your two eardrums (the signals that are sent to your brain), these will look something like this:
Figure 5: The outputs of your two eardrums with the same “click” signal from the loudspeaker.
These two signals are obviously very different from the one that the microphone “hears” which should not be a surprise: ears aren’t microphones. However, there are some specific things of which we should take note:
The output of the left eardrum is lower than that of the right eardrum. This is largely because of an effect called “head shadowing” which is exactly what it sounds like. The sound is quieter in your left ear because your head is in the way.
The signal at the right eardrum is earlier than at the left eardrum. This is because the left eardrum is not only farther away, but the sound has to go around your head to get there.
The signal at the right eardrum is earlier than the output of the microphone output (in Figure 3) because it’s closer to the loudspeaker. (I put the microphone at the location of the centre of the simulated head.) Similarly the left ear output is later because it’s farther away.
The signal at the right eardrum is full of spikes. This is mostly caused by reflections off the pinna (the flappy thing on the side of your head that you call your “ear”) that arrive at slightly different times, and all add together to make a mess.
The signal at the left eardrum is “smoother”. This is because the head itself acts as a filter reducing the levels of the high frequency content, which tends to make things less “spiky”.
Both signals last longer in time. This is the effect of the ear canal (the “hole” in the side of your head that you should NOT stick a pencil in) resonating like a little organ pipe.
The difference between the signals in Figures 2 and 4 is a measurement of the effect that your head (including your shoulders, ears/pinnae) has on the transfer of the sound from the loudspeaker to your eardrums. Consequently, we geeks call it a “head-related transfer function” or HRTF. I’ve plotted this HRTF as a measurement of an impulse in time – but I could have converted it to a frequency response instead (which would include the changes in magnitude and phase for different frequencies).
Here’s the cool thing: If I put a pair of headphones on you and played those two signals in Figure 5 to your two ears, you might be able to convince yourself that you hear the click coming from the same place as where that loudspeaker is located.
Although this sounds magical, don’t get too excited right away. Unfortunately, as with most things in life, reality tends to get in the way for a number of reasons:
Your head and ears aren’t the same shape as anyone else’s. Your brain has lived with your head and your ears for a long time, and it’s learned to correlate your HRTFs with the locations of sound sources. If I suddenly feed you a signal that uses my HRTFs, then this trick may or may not work, depending on how similar we are. This is just like borrowing someone else’s glasses. If you have roughly the same prescription, then you can see. However, if the prescriptions are very different, you’ll get a headache very quickly.
In reality, you’re always moving. So, even if the sound source is not moving, the specific details of the HRTFs are always changing (because the relative positions and angles to your ears are changing) but my system doesn’t know about this – so I’m simulating a system where the loudspeaker moves around you as you rotate your head. Since this never happens in real life, it tends to break the simulation.
The stuff I showed above doesn’t include reflections, which is how you determine distance to sources. If I wanted to include reflections, each reflection would have to have its own HRTF processing, depending on its angle relative to your head.
However, hypothetically, this can work, and lots of people have tried. The easiest way to do this is to not bother measuring anything. You just take a “dummy head” -a thing that is the same size as an average human head (maybe with an average torso) and average pinnae* – but with microphones where the eardrums are – and you plunk it down in a seat in a concert hall and record the outputs of the two “ears”. You then listen to this over earphones (we don’t use headphones because we want to remove your pinnae from the equation) and you get a “you are there” experience (assuming that the dummy head’s dimensions and shape are about the same as yours). This is what’s known as a binaural recording because it’s a recording that’s done with two ears (instead of two or more “simple” microphones).
If you want to experience this for yourself, plug a pair of headphones into your computer and do a search for the “Virtual Barber Shop” video. However, if you find that it doesn’t work for you, don’t be upset. It just means that you’re different: just like everyone else.* Typically, recordings like this have a strange effect of things sounding very close in the front, and farther away as sources go to the sides. (Personally, I typically don’t hear anything in the front. All of the sources sound like they’re sitting on the back of my neck and shoulders. This might be because I have a fat head (yes, yes… I know…) and small pinnae (yes, yes…. I know…) – or it might indicate some inherent paranoia of which I am not conscious.)
* Of course, depressingly typically, it goes without saying that the sizes and shapes of commercially-available dummy heads are based on averages of measurements of men only. Neither women nor children are interested in binaural recordings or have any relevance to such things, apparently…
There’s one last thing that I alluded to in a previous part of this series that now needs discussing before I wrap up the topic. Up to now, we’ve looked at how a filter behaves, both in time and magnitude vs. frequency. What we haven’t really dealt with is the question “why are you using a filter in the first place?”
Originally, equalisers were called that because they were used to equalise the high frequency levels that were lost on long-distance telephone transmissions. The kilometres of wire acted as a low-pass filter, and so a circuit had to be used to make the levels of the frequency bands equal again.
Nowadays we use filters and equalisers for all sorts of things – you can use them to add bass or treble because you like it. A loudspeaker developer can use them to correct linear response problems caused by the construction or visual design of the device. They can be used to compensate for the acoustical behaviour of a listening room. Or they can be used to compensate for things like hearing loss. These are just a few examples, but you’ll notice that three of the four of them are used as compensation – just like the original telephone equalisers.
Let’s focus on this application. You have an issue, and you want to fix it with a filter.
IF the problem that you’re trying to fix has a minimum phase characteristic, then a minimum phase filter (implemented either as an analogue circuit or in a DSP) can be used to “fix” the problem not only in the frequency domain – but also in the time domain. IF, however, you use a linear phase filter to fix a minimum phase problem, you might be able to take care of things on a magnitude vs. frequency analysis, but you will NOT fix the problem in the time domain.
This is why you need to know the time-domain behaviour of the problem to choose the correct filter to fix it.
For example, if you’re building a room compensation algorithm, you probably start by doing a measurement of the loudspeaker in a “reference” room / location / environment. This is your target.
You then take the loudspeaker to a different room and measure it again, and you can see the difference between the two.
In order to “undo” this difference with a filter (assuming that this is possible) one strategy is to start by analysing the difference in the two measurements by decomposing it into minimum phase and non-minimum phase components. You can then choose different filters for different tasks. A minimum phase filter can be used to compensate a resonance at a single frequency caused by a room mode. However, the cancellation at a frequency caused by a reflection is not minimum phase, so you can’t just use a filter to boost at that frequency. An octave-smoothed or 1/3-octave smoothed measurement done with pink noise might look like you fixed the problem – but you’ve probably screwed up the time domain.
Another, less intuitive example is when you’re building a loudspeaker, and you want to use a filter to fix a resonance that you can hear. It’s quite possible that the resonance (ringing in the time domain) is actually associated with a dip in the magnitude response (as we saw earlier). This means that, although intuition says “I can hear the resonant frequency sticking out, so I’ll put a dip there with a filter” – in order to correct it properly, you might need to boost it instead. The reason you can hear it is that it’s ringing in the time domain – not because it’s louder. So, a dip makes the problem less audible, but actually worse. In this case, you’re actually just attenuating the symptom, not fixing the problem – like taking an Asprin because you have a broken leg. Your leg is still broken, you just can’t feel it.
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…
If you you get an audiometry test done, you’ll be shown into a small room, about the size of a public bathroom stall. Someone will put a pair of headphones on you, and pass you a small handle with a button. Your instructions are to press the button if you hear a tone. Then the audiometrist will leave the room, closing the door, and you’ll suddenly realise that if there’s any noise in this room, it’s because you’re making it.
Then you hear a beep in your left ear. You press the button. You hear a quieter beep. Press. Quieter beep. Press…. …. …. Beep, press… …. …. …. Beep, press…. New frequency beep, loud again. Press… and so on.
What’s happening here is that you’re presented with a sine tone at some frequency, probably loud enough for you to hear. You press. The tone gets quieter, and you press again. Eventually, the tone is so quiet that you cannot hear it (this is normal) so you don’t press. So, the tone gets louder, and you press. Then it gets quieter again, until you can’t hear it again.
By crossing over that threshold of “can hear” and “can’t hear” a couple of times, the audiometrist finds out whether or not you got lucky… If you bottom out at the same level a couple of times in a row, then that’s your threshold of hearing at that frequency in that ear.
The frequency changes (usually by 1 octave, but sometimes less), and the whole process is repeated.
If you get a full test done, then this is probably done at 9 frequencies (250, 500, 1k, 1.5k, 2k, 3k, 4k, 6k, and 8kHz) in both ears individually – 18 tests in all.
You’ll then be given a sheet of paper, or at least shown a plot of your hearing threshold. Typically, if you have “normal” hearing (whatever that means) your thresholds will all be sitting on a horizontal line marked 0 dB. If you’re “better than normal” then you get a negative score, if you’re “worse than normal” you get a positive score.
What does this mean?
Let’s start over.
If a lot of people do this test, and we only test at 1 kHz, we’ll find out that, after the results are averaged, the group can hear the 1 kHz sine tone when the change in air pressure at the ear entrance is 20 µPa. We’re not going to talk about what this means other than to say that “sound is a change in air pressure over time, and that pressure is measured in pascals, abbreviated Pa”. Needless to say, 20 µPa is pretty quiet, since it’s the quietest sound a group of people can hear at 1 kHz when you take their average.
If you did that test at a much lower frequency, you would find out that people aren’t as good at hearing quiet sounds. In other words, at 100 Hz, the sine tone has to be louder than 20 µPa for people to hear it.
The same is true if you repeated the test at a much higher frequency – say, 10,000 Hz.
If you did this test at a lot of frequencies, then you’d find out that, on average, the threshold of hearing for a human follows the bottom red line of the plot in Figure 1, borrowed from Wikipedia.
Figure 1: The bottom red curve is the average threshold of hearing for a human being.
That bottom plot shows the threshold of hearing for different frequencies, plotted in dB SPL. Notice that, at 1 kHz, the line is at 0 dB SPL. This is because 0 dB SPL is defined to be the average threshold of hearing of a human at 1 kHz, which is 20 µPa. So, it’s not an accident…
Looking at that plot, you can see that, in order to hear a sine tone at 20 Hz, the tone has got to be more than 70 dB louder (that’s a LOT louder). So, a microphone “sees” a 73 dB SPL, 20 Hz sine tone as being louder than a 0 dB SPL, 1 kHz sine tone – but as far as you’re concerned, they’re both “the quietest sound you can hear” – therefore, they’re the same level.
If we take that threshold of hearing curve, and we play tones at those levels for those frequencies, then you should “just be able to” hear them. So, we’ll call those levels “0 dB” – since it’s the same as what is expected of you.
In other words, the piece of paper you got from the audiometrist tells you how much above or below that red threshold of hearing YOU sit.
Now, let’s back up a bit.
I said that, in your test, you only went up to 8 kHz. This is because, above that (and possibly even before that) the headphones might not be trust-worthy, and even a tiny movement (say a couple of millimetres) in the position of the headphones will have a (relatively) big effect on the level at your eardrum. So, rather than get people worried about losing their hearing at 20,000 Hz (when, in fact, they were actually just wearing the headphones 1 mm too far forward), you won’t get tested.
Notice how variable that threshold of hearing line is. There are big changes in level over the “audible” frequency range.
Remember that the threshold of hearing curve is an AVERAGE of a lot of people. Just like no one has 2.6 children, no one has this exact response. And, if you are some freak of nature and you DO have exactly that response, you don’t for long… we all get old…
Notice how that threshold of hearing curve only goes up to about 16 kHz, and above that it says “estimated”. See point #1.
Now, you should know that your ability to hear a sine tone at some frequency is defined as how your ability compares to an expectation based on an average, within a relatively small frequency band: 250 to 8 kHz.
Then you look at a textbook or you read a website that says “humans can hear from 20 Hz to 20 kHz”, which is not enough information to be either true or false… It’s like saying “humans are usually between 0 and 10 m tall” which is also sort of true, but also adequately vague to be potentially worse-than-useless information.
The truth is, unfortunately, much more complicated… However, it’s fair to say that, in order for you to just hear a sine tone at 20 kHz, it would have to be much, much louder than one at 1 kHz. In fact, if I played a 20 kHz sine tone loud enough for you to hear, measured that level, and then played a 1 kHz sine tone for you at the same level, you’d probably punch me – after you had passed out due to the pain, woken up, hunted me down, and found me… (I’d already have run away by then….)
So what?
We humans like nice, tidy, answers. “It will rain tomorrow” is preferable to “there is a 70 – 80% chance of scattered showers in the afternoon tomorrow”. We even get mad when the information is correct, but we interpret it tidily… For example, we’ll complain about getting rained on in the middle of our hike, when there was only a 10% chance of rain. On the other hand, if there was a 10% chance of winning 1 Million dollars in the lottery, we’d all buy a ticket.
Anyways, once-upon-a-time, when the committee for inventing the compact disc was holding meetings, they said “what should the sampling rate be?” and someone said “at least 40 kHz, because we can hear up to 20 kHz”. (The reason it’s 44100 is related to the fact that the bits were stored as black and white stripes on video tape, and NTSC and PAL come close to meeting each other close to that number, when you look at the numbers of lines per field and frames per second.)
Of course, like any first-generation thing, digital recording equipment wasn’t very good at the start (back around 1980 or so) – so the first DDD recordings that were released on CD sounded… well…. weird. There was quantisation distortion because they hadn’t figured out dither yet, only 12 or 13 of the bit values were working properly on the ADC’s, the anti-aliasing filters were implemented as analogue circuits, so they let some stuff through that aliased, and they rang (“sang along”) with the signal at a high frequency… All of that added up to “weird” – possibly even “bad”. Then, people who had good equipment (high-end turntables or, even better, 1/4″ tape running at 30 ips) listened to this new format, decided it was bad, and that was that.
Some of them asked “why is is bad?” and one answer they came up with was the band limiting… If the system can’t capture or store or play materials above 20 kHz, then it’s useless… Right? Maybe…
Then, instruments were put in front of measurement microphones and spectra were measured – and the proof was in. Trumpets with harmon (wah-wah) mutes, when pointing directly at the microphone, contain harmonics as high as 50 kHz! This must explain why CDs sound bad! Right? Maybe…
Then Rupert Neve did a demo at an AES (Audio Engineering Society) convention where he played people two tones. Both were at 7 kHz, but one was a sine wave and the other was a square wave (at some level). The question was: have a listen and tell me which is which. The results were the same as if everyone was just guessing. (Remember that, in order to make a square wave, you need to add odd harmonics – so the lowest-frequency content difference between a 7 kHz sine wave and a 7 kHz square wave is at 21 kHz.) Proof that we don’t need to go above 20 kHz, right? Maybe…
Some years ago, I took some “high resolution” audio files and measured their spectral content. One particularly interesting result is shown in Figures 2, below.
Figure 2: The spectral content of a 96/24 “high resolution” audio file I bought.
Look at that spike in the top end – around 20 kHz. What musical instrument makes that sound? The answer is “no musical instrument makes that sound – at least none of the baroque instruments in that recording make that sound. As I wrote back in 2014:
If you’re wondering what it might be, I asked a bunch of smart friends, and the best explanation we can come up with is that it’s noise from a switched-mode power supply that is somehow bleeding into the recording. HOW it’s bleeding into the recording is a potentially interesting question for recording engineers. One possibility is that one of the musicians was charging up a phone in the room where the microphones were – and the mic’s just picked up the noise. Another possibility is that the power supply noise is bleeding electrically into the recording chain – maybe it’s a computer power supply or the sound card and the manufacturer hasn’t thought about isolating this high frequency noise from the audio path. Or, maybe it’s something else.
Interestingly, this is a conflict of two engineers. The designer of the power supply (assuming that’s what it is…) said “I’ll put the switching frequency above 20 kHz so that no one will hear it” and the recording engineer said “I’ll record this at 96 kHz so that people can get the content they’re missing…” The problem is that the content you’re missing is something you don’t want…
Similarly, if you listen to Eric Clapton’s “Unplugged” album with headphones or loudspeakers that have a low-enough low-frequency range, you’ll hear a loud thump, thump, thump going along with the music. This is the sound of someone tapping their foot on a temporary stage floor, shaking a vocal microphone. In my not-very-humble opinion, that should never have made it out to the public release. However, my guess is that the speakers it was mastered on didn’t go low enough… (OR, it was an artistic decision, and I would have done it differently.) Assuming that I’m right, then this is a second example where a “better” system sounds “worse”.
Of course, through all of this, I have assumed that your loudspeakers or headphones can produce the signals that we’re talking about in the direction that you’re sitting in, and that those signals are not being masked by other sounds in the room (like phone chargers singing…) However, to complicate things with reality would just be too far to go today…
Conclusions?
I don’t have any, but I have some questions and (as usual) some opinions…
Does a harmon mute on a trumpet produce energy at 50 kHz, if you’re sitting right in front of it? Yes.
Do you want to sit right in front of a trumpet with a harmon mute? Debatable.
Can a high-res audio recording include the sound of a phone charger? Yes.
Do you want to have an expensive recording of a baroque ensemble with obligato phone charger? Probably not – the charger is not in Buxtehude’s original score as far as I can see.
Can you hear the difference between a 7 kHz sine and a 7 kHz square wave? Depends on the speaker / headphone, the listening position, the background noise level, and whether or not you were out clubbing last night. Heads or tails?
Will you feel better by knowing that your file contains “audio” content above 20 kHz? Probably. Placebos have been known to work bigger miracles than this. (But don’t forget the stuff I said about sampling rate converters earlier…)
If you’ve read the three introductory parts of this series, linked above; and if you’re still awake, then we are ready to start putting things together and jumping to incorrect conclusions…
Let’s say that you’ve been hired to specify a digital audio system for some reason (we’ll assume that it’s an LPCM system – nothing exotic). Using the information I’ve told you so far, you can make two decisions in your specification:
You select a bit depth to be enough to give you the dynamic range you desire. In this case, “dynamic range” means the “distance” in level between the loudest sound you can record / store / transmit (I isn’t say what the “digital audio system” was going to be used for) and the inherent noise floor of the system. If you’re recording the background noise on an airplane while it’s in flight, you don’t need a big dynamic range, because it’s always loud, and never changes. However, if you’re recording a Japanese Taiko Drummer group, you’ll need a huge usable dynamic range because the loud parts of the performance are a LOT louder than the quietest parts.
As we saw in Part 3, an LPCM digital audio system cannot record any audio that has a frequency higher than 1/2 the sampling rate. So, you select a sampling rate that is at least 2x the highest frequency you’re interested in. For example, if you believe the books that say you can hear from 20 Hz to 20,000 Hz, then you might decide that your sampling rate has to be at least 40,000 Hz. On the other hand, if you’re making a subwoofer that you know will never be fed a signal above 120 Hz, then you don’t need a sampling rate higher than 240 Hz.
Don’t get angry yet. I’m just keeping these numbers simple to make the math easy. Later on, I’ll explain why what I just said might not be correct.
Mistake #1
I just jumped to at least three conclusions (probably more) that are going to haunt me.
The first was that my “digital audio system” was something like the following:
Figure 1
As you can see there, I took an analogue audio signal, converted it to digital, and then converted it back to analogue. Maybe I transmitted it or stored it in the part that says “digital audio”.
However, the important, and very probably incorrect assumption here is that I did nothing to the signal. No volume control, no bass and treble adjustments… nothing.
Mistake #2
We assumed above that we can define the system’s dynamic range based on the dynamic range of the audio signal itself. However, this makes the assumption that the noise floor of the digital system and the noise floor of your audio signal are identical, which is probably not true. As we saw in Part 2, the noise generated by TPDF dither is white – it has the same probability of having a given amount of energy per Hertz. Since we hear sound logarithmically (meaning that, to us, octaves are equal widths. Equal spacings in Hz are not.) This means that the noise sound “bright” to us – because there’s just as much energy in the top octave (say, 10 kHz to 20 kHz, if you believe the books) as there is in all other frequencies combined from 0 Hz up to 10 kHz.
If, however, the noise floor in your concert hall where the taiko drummers are playing is caused by the air conditioning system, then this noise will be a lot louder in the low frequencies than the the highs – which is not the same.
Therefore it’s too simplistic to say “the noise floor of the digital system” and the “noise floor of the signal” – since these two noise floors are different. (As Steven Wright said: “It doesn’t matter what temperature the room is, it’s always room temperature.”)
Mistake #3
As we’ll see later, if you’re going to do anything to the signal while it’s in the “digital domain”, then you need to take that into consideration when you’re deciding on your sampling rate. It’s not enough to say “useful audio bandwidth times 2” because there are some side effects that need to be remembered…
However, counter-intuitively, it could be that, in order to improve your system, you’ll want to make the sampling rate LOWER instead of HIGHER – so this is not a simple case of “more is better”.
We’ll get to that topic later. For now, I’ll leave you in suspense.
Some details
One thing we saw in Part 3 was that, if we have an audio signal with energy at a frequency higher than 1/2 the sampling rate, and if that signal gets into the analogue-to-digital converter (ADC), then the output of the ADC will contain an error. We’ll get out energy at frequencies that were not in the original, due to the effect called “aliasing“.
Once that’s in the digital audio signal, there’s no removing it, so we need to make sure that the too-high-frequency signals don’t get into the ADC’s input in the first place. This is done using a low-pass filter that (in theory) removes all energy in the signal above the Nyquist frequency (which is equal to 1/2 the sampling rate). Since that low-pass filter prevents aliasing, we call it an anti-aliasing filter. Normally, these days, that antialiasing filter is built into the ADC itself.
As we also saw in Part 3, the digital-to-analogue converter (DAC) has to smooth out the digital signal to convert it from a “staircase” wave to a smoother one. That’s also done with a low-pass filter that eliminates all the harmonics that would be required to make the staircase have sharp corners. Since this is done to re-construct the analogue signal, it’s called a “reconstruction filter“.
This means that, if we pull apart some of the components in the signal chain I showed in Figure 1, it really looks more like this:
Reminder: This is still just the lead-up to the real topic of this series. However, we have to get some basics out of the way first…
Just like the last posting, this is a copy-and-paste from an article that I wrote for another series. However, this one is important, and rather than just link you to a different page, I’ve reproduced it (with some minor editing to make it fit) here.
In the first posting in this series, I talked about digital audio (more accurately, Linear Pulse Code Modulation or LPCM digital audio) is basically just a string of stored measurements of the electrical voltage that is analogous to the audio signal, which is a change in pressure over time… In the second posting in the series, we looked at a “trick” for dealing with the issue of quantisation (the fact that we have a limited resolution for measuring the amplitude of the audio signal). This trick is to add dither (a fancy word for “noise”) to the signal before we quantise it in order to randomise the error and turn it into noise instead of distortion.
In this posting, we’ll look at some of the problems incurred by the way we carve up time into discrete moments when we grab those samples.
Let’s make a wheel that has one spoke. We’ll rotate it at some speed, and make a film of it turning. We can define the rotational speed in RPM – rotations per minute, but this is not very useful. In this case, what’s more useful is to measure the wheel rotation speed in degrees per frame of the film.
Fig 1. The position of a clockwise-rotating wheel (with only one spoke) for 9 frames of a film. Each column shows a different rotational speed of the wheel. The far left column is the slowest rate of rotation. The far right column is the fastest rate of rotation. Red wheels show the frame in which the sequence starts repeating.
Take a look at the left-most column in Figure 1. This shows the wheel rotating 45º each frame. If we play back these frames, the wheel will look like it’s rotating 45º per frame. So, the playback of the wheel rotating looks the same as it does in real life.
This is more or less the same for the next two columns, showing rotational speeds of 90º and 135º per frame.
However, things change dramatically when we look at the next column – the wheel rotating at 180º per frame. Think about what this would look like if we played this movie (assuming that the frame rate is pretty fast – fast enough that we don’t see things blinking…) Instead of seeing a rotating wheel with only one spoke, we would see a wheel that’s not rotating – and with two spokes.
This is important, so let’s think about this some more. This means that, because we are cutting time into discrete moments (each frame is a “slice” of time) and at a regular rate (I’m assuming here that the frame rate of the film does not vary), then the movement of the wheel is recorded (since our 1 spoke turns into 2) but the direction of movement does not. (We don’t know whether the wheel is rotating clockwise or counter-clockwise. Both directions of rotation would result in the same film…)
Now, let’s move over one more column – where the wheel is rotating at 225º per frame. In this case, if we look at the film, it appears that the wheel is back to having only one spoke again – but it will appear to be rotating backwards at a rate of 135º per frame. So, although the wheel is rotating clockwise, the film shows it rotating counter-clockwise at a different (slower) speed. This is an effect that you’ve probably seen many times in films and on TV. What may come as a surprise is that this never happens in “real life” unless you’re in a place where the lights are flickering at a constant rate (as in the case of fluorescent or some LED lights, for example).
Again, we have to consider the fact that if the wheel actually were rotating counter-clockwise at 135º per frame, we would get exactly the same thing on the frames of the film as when the wheel if rotating clockwise at 225º per frame. These two events in real life will result in identical photos in the film. This is important – so if it didn’t make sense, read it again.
This means that, if all you know is what’s on the film, you cannot determine whether the wheel was going clockwise at 225º per frame, or counter-clockwise at 135º per frame. Both of these conclusions are valid interpretations of the “data” (the film). (Of course, there are more – the wheel could have rotated clockwise by 360º+225º = 585º or counter-clockwise by 360º+135º = 495º, for example…)
Since these two interpretations of reality are equally valid, we call the one we know is wrong an alias of the correct answer. If I say “The Big Apple”, most people will know that this is the same as saying “New York City” – it’s an alias that can be interpreted to mean the same thing.
Wheels and Slinkies
We people in audio commit many sins. One of them is that, every time we draw a plot of anything called “audio” we start out by drawing a sine wave. (A similar sin is committed by musicians who, at the first opportunity to play a grand piano, will play a middle-C, as if there were no other notes in the world.) The question is: what, exactly, is a sine wave?
Get a Slinky – or if you don’t want to spend money on a brand name, get a spring. Look at it from one end, and you’ll see that it’s a circle, as can be (sort of) seen in Figure 2.
Fig 2. A Slinky, seen from one end. If I had really lined things up, this would just look like a shiny circle.
Since this is a circle, we can put marks on the Slinky at various amounts of rotation, as in Figure 3.
Fig 3. The same Slinky, marked in increasing angles of 45º.
Of course, I could have put the 0º mark anywhere. I could have also rotated counter-clockwise instead of clockwise. But since both of these are arbitrary choices, I’m not going to debate either one.
Now, let’s rotate the Slinky so that we’re looking at from the side. We’ll stretch it out a little too…
Fig 4. The same Slinky, stretched a little, and viewed from the side.
Let’s do that some more…
Fig 5. The same Slinky, stretched more, and viewed from the “side” (in a direction perpendicular to the axis of the rotation).
When you do this, and you look at the Slinky directly from one side, you are able to see the vertical change of the spring from the centre as a result of the change in rotation. For example, we can see in Figure 6 that, if you mark the 45º rotation point in this view, the distance from the centre of the spring is 71% of the maximum height of the spring (at 90º).
Fig 6. The same markings shown in Figure 3, when looking at the Slinky from the side. Note that, if we didn’t have the advantage of a little perspective (and a spring made of flat metal), we would not know whether the 0º point was closer or further away from us than the 180º point. In other words, we wouldn’t know if the Slinky was rotating clockwise or counter-clockwise.
So what? Well, basically, the “punch line” here is that a sine wave is actually a “side view” of a rotation. So, Figure 7, shows a measurement – a capture – of the amplitude of the signal every 45º.
Fig 7. Each measurement (a black “lollipop”) is a measurement of the vertical change of the signal as a result of rotating 45º.
Since we can now think of a sine wave as a rotation of a circle viewed from the side, it should be just a small leap to see that Figure 7 and the left-most column of Figure 1 are basically identical.
Let’s make audio equivalents of the different columns in Figure 1.
Fig 8. A sampled cosine wave where the frequency of the signal is equivalent to 90º per sample period. This is identical to the “90º per frame” column in Figure 1.Fig 9. A sampled cosine wave where the frequency of the signal is equivalent to 135º per sample period. This is identical to the “135º per frame” column in Figure 1.Fig 10. A sampled cosine wave where the frequency of the signal is equivalent to 180º per sample period. This is identical to the “180º per frame” column in Figure 1.
Figure 10 is an important one. Notice that we have a case here where there are exactly 2 samples per period of the cosine wave. This means that our sampling frequency (the number of samples we make per second) is exactly one-half of the frequency of the signal. If the signal gets any higher in frequency than this, then we will be making fewer than 2 samples per period. And, as we saw in Figure 1, this is where things start to go haywire.
Fig 11. A sampled cosine wave where the frequency of the signal is equivalent to 225º per sample period. This is identical to the “225º per frame” column in Figure 1.
Figure 11 shows the equivalent audio case to the “225º per frame” column in Figure 1. When we were talking about rotating wheels, we saw that this resulted in a film that looked like the wheel was rotating backwards at the wrong speed. The audio equivalent of this “wrong speed” is “a different frequency” – the alias of the actual frequency. However, we have to remember that both the correct frequency and the alias are valid answers – so, in fact, both frequencies (or, more accurately, all of the frequencies) exist in the signal.
So, we could take Fig 11, look at the samples (the black lollipops) and figure out what other frequency fits these. That’s shown in Figure 12.
Fig 12. The red signal and the black samples of it are the same as was shown in Figure 11. However, another frequency (the blue signal) also fits those samples. So, both the red signal and the blue signal exist in our system.
Moving up in frequency one more step, we get to the right-hand column in Figure 1, whose equivalent, including the aliased signal, are shown in Figure 13.
Fig 13. A signal (the red curve) that has a frequency equivalent to 280º of rotation per sample, its samples (the black lollipops) and the aliased additional signal that results (the blue curve).
Do I need to worry yet?
Hopefully, now, you can see that an LPCM system has a limit with respect to the maximum frequency that it can deal with appropriately. Specifically, the signal that you are trying to capture CANNOT exceed one-half of the sampling rate. So, if you are recording a CD, which has a sampling rate of 44,100 samples per second (or 44.1 kHz) then you CANNOT have any audio signals in that system that are higher than 22,050 Hz.
That limit is commonly known as the “Nyquist frequency“, named after Harry Nyquist – one of the persons who figured out that this limit exists.
In theory, this is always true. So, when someone did the recording destined for the CD, they made sure that the signal went through a low-pass filter that eliminated all signals above the Nyquist frequency.
In practice, however, there are many cases where aliasing occurs in digital audio systems because someone wasn’t paying enough attention to what was happening “under the hood” in the signal processing of an audio device. This will come up later.
Two more details to remember…
There’s an easy way to predict the output of a system that’s suffering from aliasing if your input is sinusoidal (and therefore contains only one frequency). The frequency of the output signal will be the same distance from the Nyquist frequency as the frequency if the input signal. In other words, the Nyquist frequency is like a “mirror” that “reflects” the frequency of the input signal to another frequency below Nyquist.
This can be easily seen in the upper plot of Figure 14. The distance from the Input signal and the Nyquist is the same as the distance between the output signal and the Nyquist.
Also, since that Nyquist frequency acts as a mirror, then the Input and output signal’s frequencies will move in opposite directions (this point will help later).
Fig 14. Two plots showing the same information about an Input Signal above the Nyquist frequency and the output alias signal. Notice that, in the linear plot on top, it’s easier to see that the Nyquist frequency is the mirror point at the centre of the frequencies of the Input and Output signals.
Usually, frequency-domain plots are done on a logarithmic scale, because this is more intuitive for we humans who hear logarithmically. (For example, we hear two consecutive octaves on a piano as having the same “interval” or “width”. We don’t hear the width of the upper octave as being twice as wide, like a measurement system does. that’s why music notation does not get wider on the top, with a really tall treble clef.) This means that it’s not as obvious that the Nyquist frequency is in the centre of the frequencies of the input signal and its alias below Nyquist.
Reminder: This is still just the lead-up to the real topic of this series. However, we have to get some basics out of the way first…
Just like the last posting, this is a copy-and-paste from an article that I wrote for another series. However, this one is important, and rather than just link you to a different page, I’ve reproduced it (with some minor editing to make it fit) here.
In the last posting, I talked about digital audio (more accurately, Linear Pulse Code Modulation or LPCM digital audio) is basically just a string of stored measurements of the electrical voltage that is analogous to the audio signal, which is a change in pressure over time…
For now, we’ll say that each measurement is rounded off to the nearest possible “tick” on the ruler that we’re using to measure the voltage. That rounding results in an error. However, (assuming that everything is working correctly) that error can never be bigger than 1/2 of a “step”. Therefore, in order to reduce the amount of error, we need to increase the number of ticks on the ruler.
Now we have to introduce a new word. If we really had a ruler, we could talk about whether the ticks are 1 mm apart – or 1/16″ – or whatever. We talk about the resolution of the ruler in terms of distance between ticks. However, if we are going to be more general, we can talk about the distance between two ticks being one “quantum” – a fancy word for the smallest step size on the ruler.
So, when you’re “rounding off to the nearest value” you are “quantising” the measurement (or “quantizing” it, if you live in Noah Webster’s country and therefore you harbor the belief that wordz should be spelled like they sound – and therefore the world needz more zees). This also means that the amount of error that you get as a result of that “rounding off” is called “quantisation error“.
In some explanations of this problem, you may read that this error is called “quantisation noise”. However, this isn’t always correct. This is because if something is “noise” then is is random, and therefore impossible to predict. However, that’s not strictly the case for quantisation error. If you know the signal, and you know the quantisation values, then you’ll be able to predict exactly what the error will be. So, although that error might sound like noise, technically speaking, it’s not. This can easily be seen in Figures 1 through 3 which demonstrate that the quantisation error causes a periodic, predictable error (and therefore harmonic distortion), not a random error (and therefore noise).
Sidebar: The reason people call it quantisation noise is that, if the signal is complicated (unlike a sine wave) and high in level relative to the quantisation levels – say a recording of Britney Spears, for example – then the distortion that is generated sounds “random-ish”, which causes people to jump to the conclusion that it’s noise.
Fig 1: The first cycle of a periodic signal (in this case, a sinusoidal waveform) that we are going to quantise using a 4-bit system (notice the 4 bits in the scale on the left).
Fig 2: The same waveform shown in Figure 1 after quantisation (rounding off) in a 4-bit world.
Fig 3: The difference between Figure 2 and Figure 1. I made this by subtracting the original signal from the quantised version. This is the error in the quantised waveform – the quantisation error. Notice that it is not noise… it’s completely predictable and it will repeat with repetitions of the signal. Therefore the result of this is distortion, not noise…
Now, let’s talk about perception for a while… We humans are really good at detecting patterns – signals – in an otherwise noisy world. This is just as true with hearing as it is with vision. So, if you have a sound that exists in a truly random background noise, then you can focus on listening to the sound and ignore the noise. For example, if you (like me) are old enough to have used cassette tapes, then you can remember listening to songs with a high background noise (the “tape hiss”) – but it wasn’t too annoying because the hiss was independent of the music, and constant. However, if you, like me, have listened to Bob Marley’s live version of “No Woman No Cry” from the “Legend” album, then you, like me, would miss the the feedback in the PA system at that point in the song when the FoH engineer wasn’t paying enough attention… That noise (the howl of the feedback) is not noise – it’s a signal… Which makes it just as important as the song itself. (I could get into a long boring talk about John Cage at this point, but I’ll try to not get too distracted…)
The problem with the signal in Figure 2 is that the error (shown in Figure 3) is periodic – it’s a signal that demands attention. If the signal that I was sending into the quantisation system (in Figure 1) was a little more complicated than a sine wave – say a sine wave with an amplitude modulation – then the error would be easily “trackable” by anyone who was listening.
So, what we want to do is to quantise the signal (because we’re assuming that we can’t make a better “ruler”) but to make the error random – so it is changed from distortion to noise. We do this by adding noise to the signal before we quantise it. The result of this is that the error will be randomised, and will become independent of the original signal… So, instead of a modulating signal with modulated distortion, we get a modulated signal with constant noise – which is easier for us to ignore. (It has the added benefit of spreading the frequency content of the error over a wide frequency band, rather than being stuck on the harmonics of the original signal… but let’s not talk about that…)
For example…
Let’s take a look at an example of this from an equivalent world – digital photography.
The photo in Figure 4 is a black and white photo – which actually means that it’s comprised of shades of gray ranging from black all the way to white. The photo has 272,640 individual pixels (because it’s 640 pixels wide and 426 pixels high). Each of those pixels is some shade of gray, but that shading does not have an infinite resolution. There are “only” 256 possible shades of gray available for each pixel.
So, each pixel has a number that can range from 0 (black) up to 255 (white).
Fig 4: A photo of a building in Paris. Each pixel in this photo has one of 256 possible levels of gray – from white (255) down to black (0).
If we were to zoom in to the top left corner of the photo and look at the values of the 64 pixels there (an 8×8 pixel square), you’d see that they are:
What if we were to reduce the available resolution so that there were fewer shades of gray between white and black? We can take the photo in Figure 1 and round the value in each pixel to the new value. For example, Figure 5 shows an example of the same photo reduced to only 6 levels of gray.
Fig 5: The same photo of the same building. Each pixel in this photo has one of 6 possible levels of gray. Notice that some details are lost – like the smooth transitions in the clouds, or the stripes in the marble in the pillars.
Now, if we look at those same pixels in the upper left corner, we’d see that their values are
They’ve all been quantised to the nearest available level, which is 102. (Our possible values are restricted to 0, 51, 102, 154, 205, and 255).
So, we can see that, by quantising the gray levels from 256 possible values down to only 6, we lose details in the photo. This should not be a surprise… That loss of detail means that, for example, the gentle transition from lighter to darker gray in the sky in the original is “flattened” to a light spot in a darker background, with a jagged edge at the transition between the two. Also, the details of the wall pillars between the windows are lost.
If we take our original photo and add noise to it – so were adding a random value to the value of each pixel in the original photo (I won’t talk about the range of those random values…) it will look like Figure 6. This photo has all 256 possible values of gray – the same as in Figure 1.
Fig 6: A photo of noise with the same width and height as the original photo, with random values (ranging from 0 to 255) in each pixel.
If we then quantise Figure 6 using our 6 possible values of gray, we get Figure 7. Notice that, although we do not have more grays than in Figure 5, we can see things like the gradual shading in the sky and some details in the walls between the tall windows.
Fig 7: The same photo of the same building in Figure 4. Each pixel in this photo ALSO only has one of 6 possible levels of gray – just like in Figure 5. However, this version is the result of quantising the original photo with the noise added before quantisation. The result is admittedly noisy – but we are able to see pattens in the noise that preserve some of the details that we lost in Figure 5.
That noise that we add to the original signal is called dither – because it is forcing the quantiser to be indecisive about which level to quantise to choose.
I should be clear here and say that dither does not eliminate quantisation error. The purpose of dither is to randomise the error, turning the quantisation error into noise instead of distortion. This makes it (among other things) independent of the signal that you’re listening to, so it’s easier for your brain to separate it from the music, and ignore it.
Addendum: Binary basics and SNR
We normally write down our numbers using a “base 10” notation. So, when I write down 9374 – I mean 9 x 1000 + 3 x 100 + 7 x 10 + 4 x 1 or 9 x 103 + 3 x 102 + 7 x 101 + 4 x 100
We use base 10 notation – a system based on 10 digits (0 through 9) because we have 10 fingers.
If we only had 2 fingers, we would do things differently… We would only have 2 digits (0 and 1) and we would write down numbers like this: 11101
which would be the same as saying 1 x 16 + 1 x 8 + 1 x 4 + 0 x 2 + 1 x 1 or 1 x 24 + 1 x 23 + 1 x 22 + 0 x 21 + 1 x 20
The details of this are not important – but one small point is. If we’re using a base-10 system and we increase the number by one more digit – say, going from a 3-digit number to a 4-digit number, then we increase the possible number of values we can represent by a factor of 10. (in other words, there are 10 times as many possible values in the number XXXX than in XXX.)
If we’re using a base-2 system and we increase by one extra digit, we increase the number of possible values by a factor of 2. So XXXX has 2 times as many possible values as XXX.
Now, remember that the error that we generate when we quantise is no bigger than 1/2 of a quantisation step, regardless of the number of steps. So, if we double the number of steps (by adding an extra binary digit or bit to the value that we’re storing), then the signal can be twice as “far away” from the quantisation error.
This means that, by adding an extra bit to the stored value, we increase the potential signal-to-error ratio of our LPCM system by a factor of 2 – or 6.02 dB.
So, if we have a 16-bit LPCM signal, then a sine wave at the maximum level that it can be without clipping is about 6 dB/bit * 16 bits – 3 dB = 93 dB louder than the error. The reason we subtract the 3 dB from the value is that the error is +/- 0.5 of a quantisation step (normally called an “LSB” or “Least Significant Bit”).
Note as well that this calculation is just a rule of thumb. It is neither precise nor accurate, since the details of exactly what kind of error we have will have a minor effect on the actual number. However, it will be close enough.
I’ve been debating writing a series of postings about “high resolution” audio for a long time – years. Lately, (probably because of some hype generated by some recent press releases) I’ve been getting lots of question (no, that’s not a typo) about it, so it appears the time has come…
To start: the question that I get (a lot) is “If I can’t hear above 20 kHz, then what’s the use of high-res?” As I’ll explain as we go through, this is only one, rather small aspect to consider in this topic. In fact, it might be the least important issue to consider.
However, before I write too much, I’ll say that I’m not going to argue for or against higher resolutions in digital audio systems. I’m only going to go through a bunch of issues that can be used to argue either for or against them. So, there’s not going to be a big reveal at the end of this series telling you that high-res is either better, worse, or no different than whatever you’re using now. It’s merely going to be a discussion of a number of issues that need to be weighed. The problem is that this entire topic is complicated – and there’s no single “right” answer, as I’ll argue as we go along.
To start, let’s get down to basics and look (once again, from the perspectives of this website) at what sound is, and how it’s converted from an analogue electrical signal into a digital representation. The good thing is that I’ve written this introduction before in a different series of postings. So, I’m going to be extremely lazy and just copy-and-paste that information here. I’m not just referring you to another page because I’m intentionally leaving some things out because we’re headed into having a different discussion this time.
A quick introduction to sound
At the simplest level, sound can be described as a small change in air pressure (or barometric pressure) over short periods of time. If you’d like to have a better and more edu-tain-y version of this statement with animations and pretty colours, you could take 10 minutes to watch this video, for example.
That change in pressure can be “captured” by using a microphone, that is (at the simplest level) a device that has a change in air pressure at its input and a change in electrical voltage at its output. Ignoring a lot of details, we could say that if you were to plot a measurement of the air pressure (at the input of the microphone) over time, and you were to compare it to a plot of the measurement of the voltage (at the output of the microphone) over time, you would see the same curve on the two graphs. This means that the change in voltage is analogous to the change in air pressure.
Fig 1. Notice that (in theory, and ignoring a lot of things…) the change in air pressure over time at the input of the microphone is identical to the change in voltage over time at its output. Of course, this is not true in real life – microphones lie like a cheap rug…
At this point in the conversation, I’ll make a point to say that, in theory, we could “zoom in” on either of those two curves shown in Figure 1 and see more and more details. This is like looking at a map of Canada – it has lots of crinkly, jagged lines. If you zoom in and look at the map of Newfoundland and Labrador, you’ll see that it has finer, crinkly, jagged lines. If you zoom in further, and stand where the water meets the shore in Trepassey and take a photo of your feet, you could copy it to draw a map of the line of where the water comes in around the rocks – and your toes – and you would wind up with even finer, crinkly, jagged lines… You could take this even further and get down to a microscopic or molecular level – but you get the idea… The point is that, in theory, both of the plots in Figure 1 have infinite resolution, both in time and in air pressure or voltage.
Now, let’s say that you wanted to take that microphone’s output and transmit it through a bunch of devices and wires that, in theory, all do nothing to the signal. Let’s say, for example, that you take the mic’s output, send it through a wire to a box that makes the signal twice as loud. Then take the output of that box and send it through a wire to another box that makes it half as loud. You take the output of that box and send it through a wire to a measuring device. What will you see? Unfortunately, none of the wires or boxes in the chain can be perfect, so you’ll probably see the signal plus something else which we’ll call the “error” in the system’s output. We can call it the error because, if we measure the input voltage and the output voltage at any one instant, we’ll probably see that they’re not identical. Since they should be identical, then the system must be making a mistake in transmitting the signal – so it makes errors…
Fig 2. If you send an audio signal through some wires and devices that (in theory) do nothing to the signal, you’ll find out that they add some extra stuff that you don’t want.
Pedantic Sidebar: Some people will call that error that the system adds to the signal “noise” – but I’m not going to call it that. This is because “noise” is a specific thing – noise is random – so if it’s not random, it’s not noise. Also, although the signal has been distorted (in that the output of the system is not identical to the input) I won’t call it “distortion” either, since distortion is a name that’s given to something that happens to the signal because the signal is there. (We would probably get at least some of the error out of our system even if we didn’t send any audio into it.) So, we could be slightly geeky and adequately vague and call the extra stuff “Distortion plus noise” but not “THD+N” – which stands for “Total Harmonic Distortion Plus Noise” – because not all kinds of distortion will produce a harmonic of the signal… but I’m getting ahead of myself…
So, we want to transmit (or store) the audio signal – but we want to reduce the noise caused by the transmission (or storage) system. One way to do this is to spend more money on your system. Use wires with better shielding, amplifiers with lower noise floors, bigger power supplies so that you don’t come close to their limits, run your magnetic tape twice as fast, and so on and so on. Or, you could convert the analogue signal (remember that it’s analogous to the change in air pressure over time) to one that is represented (and therefore transmitted or stored) digitally instead.
What does this mean?
Conversion from analogue to digital and back (but skipping important details)
IMPORTANT: If you read this section, then please read the following postings as well. This is because, in order to keep things simple to start, I’m about to leave out some important details that I’ll add afterwards. However, if you don’t add the details, you could (understandably) jump to some incorrect conclusions (that many others before you have concluded…) So, if you don’t have time to read both sections, please don’t read either of them.
In the example above, we made a varying voltage that was analogous to the varying air pressure. If we wanted to store this, we could do it by varying the amount of magnetism on a wire or a coating on a tape, for example. Or we could cut a wiggly groove in a bit of vinyl that has a similar shape to the curve in the plots in Figure 1. Or, we could do something else: we could get a metronome (or a clock) and make a measurement of the voltage every time the metronome clicks, and write down the measurements.
For example, let’s zoom in on the first little bit of the signal in the plots in Figure 1
Fig. 3 The same curve as was shown in Figure 1 – but zoomed in to the very beginning.
We’ll then put on a metronome and make a measurement of the voltage every time we hear the metronome click…
Fig 4. The same curve (in red) measured at regular intervals (in black)
We can then keep the measurements (remembering how often we made them…) and write them down like this:
We can store this series of numbers on a computer’s hard disk, for example. We can then come back tomorrow, and convert the measurements to voltages. First we read the measurements, and create the appropriate voltage…
Fig. 5. The voltages that we stored as measurements
We then make a “staircase” waveform by “holding” those voltages until the next value comes in.
Fig 6. We make a “staircase” curve using the voltages.
All we need to do then is to use a low-pass filter to smooth out the hard edges of the staircase.
Fig 7. When we smooth out the staircase, we get back the original signal (in red).
So, in this example, we’ve gone from an analogue signal (the red curve in Figure 3) to a digital signal (the series of numbers), and back to an analogue signal (the red curve in Figure 7).
In some ways, this is a bit like the way a movie works. When you watch a movie, you see a series of still photographs, probably taken at a rate of 24 pictures (or frames) per second. If you play those photos back at the same rate (24 fps or frames per second), you think you see movement. However, this is because your eyes and brain aren’t fast enough to see 24 individual photos per second – so you are fooled into thinking that things on the screen are moving.
However, digital audio is slightly different from film in two ways:
The sound (equivalent to the movement in the film) is actually happening. It’s not a trick that relies on your ears and brain being too slow.
If, when you were filming the movie, something were to happen between frames (say, the flash of a gunshot, for example) then it would never be caught on film. This is because the photos are discrete moments in time – and what happens between them is lost. However, if something were to make a very, very short sound between two samples (two measurements) in the digital audio signal – it would not be lost. This is because of something that happens at the beginning of the chain that I haven’t described… yet…
However, there are some “artefacts” (a fancy term for “weird errors”) that are present both in film and in digital audio that we should talk about.
The first is an error that happens when you mess around with the rate at which you take the measurements (called the “sampling rate”) or the photos (called the “frame rate”) – and, more importantly, when you need to worry about this. Let’s say that you make a film at 24 fps. If you play this back at a higher frame rate, then things will move very quickly (like old-fashioned baseball movies…). If you play them back at a lower frame rate, then things move in slow motion. So, for things to look “normal” you have to play the movie at the same rate that it was filmed. However, as long as no one is looking, you can transfer the movie as fast as you like. For example, if you wanted to copy the film, you could set up a movie camera so it was pointing at a movie screen and film the film. As long as the movie on the screen is running in sync with the camera, you can do this at any frame rate you like. But you’ll have to watch the copy at the same frame rate as the original film… (Note that this issue is not something that will come up in this series of postings about high resolution audio)
The second is an easy artefact to recognise. If you see a car accelerating from 0 to something fast on film, you’ll see the wheels of the car start to get faster and faster, then, as the car gets faster, the wheels slow down, stop, and then start going backwards… This does not happen in real life (unless you’re in a place lit with flashing lights like fluorescent bulbs or LED’s). I’ll do a posting explaining why this happens – but the thing to remember here is that the speed of the wheel rotation that you see on the film (the one that’s actually captured by the filming…) is not the real rotational speed of the wheel. However, those two rotational speeds are related to each other (and to the frame rate of the film). If you change the real rotational rate or the frame rate, you’ll change the rotational rate in the film. So, we call this effect “aliasing” because it’s a false version (an alias) of the real thing – but it’s always the same alias (assuming you repeat the conditions…) Digital audio can also suffer from aliasing, but in this case, you put in one frequency (which is actually the same as a rotational speed) and you get out another one. This is not the same as harmonic distortion, since the frequency that you get out is due to a relationship between the original frequency and the sampling rate, so the result is almost never a multiple of the input frequency. (We’re going to dig into this a lot deeper through this series of postings about high resolution audio, so if it doesn’t immediately make sense, don’t worry…)
Some important details that I left out…
One of the things I said above was something like “we measure the voltage and store the results” and the example I gave was a nice series of numbers that only had 4 digits after the decimal point. This statement has some implications that we need to discuss.
Let’s say that I have a thing that I need to measure. For example, Figure 8 shows a piece of metal, and I want to measure its width.
Fig 8. A piece of metal with a width of “approximately 57 mm”.
Using my ruler, I can see that this piece of metal is about 57 mm wide. However, if I were geeky (and I am) I would say that this is not precise enough – and therefore it’s not accurate. The problem is that my ruler is only graduated in millimetres. So, if I try to measure anything that is not exactly an integer number of mm long, I’ll either have to guess (and be wrong) or round the measurement to the nearest millimetre (and be wrong).
So, if I wanted you to make a piece of metal the same width as my piece of metal, and I used the ruler in Figure 8, we would probably wind up with metal pieces of two different widths. In order to make this better, we need a better ruler – like the one in Figure 9.
Fig 9. The same piece of metal being measured with a vernier caliper. This gives us additional precision (down to 0.05 mm) so we can make a more accurate measurement.
Figure 9 shows a vernier caliper (a fancy type of ruler) being used to measure the same piece of metal. The caliper has a resolution of 0.05 mm instead of the 1 mm available on the ruler in Figure 8. So, we can make a much more accurate measurement of the metal because we have a measuring device with a higher precision.
The conversion of a digital audio signal is the same. As I said above, we measure the voltage of the electrical signal, and transmit (or store) the measurement. The question is: how accurate and precise is your measurement? As we saw above, this is (partly) determined by how many digits are in the number that you use when you “write down” the measurement.
Since the voltage measurements in digital audio are recorded in binary rather than decimal (we use 0 and 1 to write down the number instead of 0 up to 9) then we use Binary digITS – or “bits” instead of decimal digits (which are not called “dits”). The number of bits we have in the number that we write down (partly) determines the precision of the measurement of the voltage – and therefore (possibly), our accuracy…
Just like the example of the ruler in Figure 8, above, we have a limited resolution in our measurement. For example, if we had only 4 bits to work with then the waveform in 4 – the one we have to measure – would be measured with the “ruler” shown on the left side of Figure 10, below.
Fig 10: The waveform from Figure 4 as a voltage (notice the Y-axis on the right). We have to measure these values using the ruler with the resolution shown on the Y-axis on the left.
When we do this, we have to round off the value to the nearest “tick” on our ruler, as shown in Figure 11.
Fig 11: The values from figure 10 (shown as the circles) rounded off to the nearest value on our 4-bit ruler (the red staircase).
Using this “ruler” which gives a write-down-able “quantity” to the measurement, we get the following values for the red staircase:
When we “play these back” we get the staircase again, shown in Figure 12.
Fig 12: The output of the measurements. Notice that all values sit exactly on one of the values for the “ruler” on the left Y-axis of the plot.
Of course, this means that, by rounding off the values, we have introduced an error in the system (just like the measurement in Figure 8 has a bigger error than the one in Figure 9). We can calculate this error if we just subtract the original signal from the output signal (in other words, Figure 12 minus Figure 10) to get Figure 13.
Fig 13: The error that we produced due to the rounding off of the signal when we did the measurements. Notice that the error is always less than 0.5 of a “tick” of the ruler on the left Y-axis.
In order to improve our accuracy of the measurement, we have to increase the precision of the values. We can do this by adding an extra digit (or bit) to the number that we use to record the value.
If we were using decimal numbers (0-9) then adding an extra digit to the number would give us 10 times as many possibilities. (For example, if we were using 4 digits after the decimal in the example at the start of this posting, we have a total of 10,000 possible values – 0.0000 to 0.9999. If we add one more digit, we increase the resolution to 100,000 possible values – 0.00000 to 0.99999 ).
In binary, adding one extra digit gives us twice as many “ticks” on the ruler. So, using 4 bits gives us 16 possible values. Increasing to 5 bits gives us 32 possible values.
If you’re listening to a CD, then the individual measurements of each voltage – the “sample values” – are stored with 16 bits, which means that we have 65,536 possible values to pick from.
Remember that this means that we have more “ticks” on our ruler – but we don’t necessarily increase its range. So, for example, we’re still measuring a voltage from -1 V to 1 V – we just have more and more resolution with which we can do that measurement.
