Category: perception

Aliasing is Weird: Part 2

In Part 1, we looked at what happens when you try to record a signal whose frequency is higher than 1/2 the sampling rate (which, from now on, I’ll call the Nyquist Frequency, named after Harry Nyquist who was one of the people that first realised that this limit existed). You record a signal, but it winds up having a different frequency at the output than it had at the input. In addition, that frequency is related to the signal’s frequency and the sampling rate itself.

In order to prevent this from happening, digital recording systems use a low-pass filter that hypothetically prevents any signals above the Nyquist frequency from getting into the analogue-to-digital conversion process. This filter is called an anti-aliasing filter because it prevents any signals that would produce an alias frequency from getting into the system. (In practice, these filters aren’t perfect, and so it’s typical that some energy above the Nyquist frequency leaks into the converter.)

So, this means that if you put a signal that contains high frequency components into the analogue input of an analogue-to-digital converter (or ADC), it will be filtered. An example of this is shown in Figure 1, below. The top plot is a square wave before filtering. The bottom plot is the result of low-pass filtering the square wave, thus heavily attenuating its higher harmonics. This results in a reduction in the slope when the wave transitions between low and high states.

Figure 1: A square wave before and after low-pass filtering.

This means that, if I have an analogue square wave and I record it digitally, the signal that I actually record will be something like the bottom plot rather than the top one, depending on many things like the frequency of the square wave, the characteristics of the anti-aliasing filter, the sampling rate, and so on. Don’t go jumping to conclusions here. The plot above uses an aggressively exaggerated filter to make it obvious that we do something to prevent aliasing in the recorded signal. Do NOT use the plots as proof that “analogue is better than digital” because that’s a one-dimensional and therefore very silly thing to claim.

However…

… just because we keep signals with frequency content above the Nyquist frequency out of the input of the system doesn’t mean that they can’t exist inside the system. In other words, it’s possible to create a signal that produces aliasing after the ADC. You can either do this by

  • creating signals from scratch (for example, generating a sine tone with a frequency above Nyquist)
    or
  • by producing artefacts because of some processing applied to the signal (like clipping, for example).

Let’s take a sine wave and clip it after it’s been converted to a digital signal with a 48 kHz sampling rate, as is shown in Figure 2.

Figure 2: The red curve is a clipped version of the black curve.

When we clip a signal, we generate high-frequency harmonics. For example, the signal in Figure 2 is a 1 kHz sine wave that I clipped at ±0.5. If I analyse the magnitude response of that, it will look something like Figure 3:

Figure 3: The magnitude response of Figure 2, showing the upper harmonics that I created by clipping.

The red curve in Figure 2 is not a ‘perfect’ square wave, so the harmonics seen in Figure 3 won’t follow the pattern that you would expect for such a thing. But that’s not the only reason this plot will be weird…

Figure 3 is actually hiding something from you… I clipped a 1 kHz sine wave, which makes it square-ish. This means that I’ve generated harmonics at 3 kHz, 5 kHz, 7 kHz, and so on, up to ∞ Hz..

Notice there that I didn’t say “up to the Nyquist frequency”, which, in this example with a sampling rate of 48 kHz, would be 24 kHz.

Those harmonics above the Nyquist frequency were generated, but then stored as their aliases. So, although there’s a new harmonic at 25 kHz, the system records it as being at 48 kHz – 25 kHz = 23 kHz, which is right on top of the harmonic just below it.

In other words, when you look at all the spikes in the graph in Figure 3, you’re actually seeing at least two spikes sitting on top of each other. One of them is the “real” harmonic, and the other is an alias (there are actually more, but we’ll get to that…). However, since I clipped a 1 kHz sine wave in a 48 kHz world, this lines up all the aliases to be sitting on top of the lower harmonics.

So, what happens if I clip a sine wave with a frequency that isn’t nicely related to the sampling rate, like 900 Hz in a 48 kHz system, for example? Then the result will look more like Figure 4, which is a LOT messier.

Figure 4: The magnitude response of a 900 Hz square wave, plotted with a logarithmic frequency axis in the top axis and a linear axis in the bottom.

A 900 Hz square wave will have harmonics at odd multiples of the fundamental, therefore at 2.7 kHz, 4.5 kHz, and so on up to 22.5 kHz (900 Hz * 25).

The next harmonic is 24.3 kHz (900 Hz * 27), which will show up in the plots at 48 kHz – 24.3 kHz = 23.7 kHz. The next one will be 26.1 kHz (900 Hz * 29) which shows up in the plots at 21.9 kHz. This will continue back DOWN in frequency through the plot until you get to 900 Hz * 53 = 47.7 kHz which will show up as a 300 Hz tone, and now we’re on our way back up again… (Take a look at Figure 7, below for another way to think of this.)

The next harmonic will be 900 Hz * 55 = 49.5 kHz which will show up in the plot as a 1.5 kHz tone (49.5 kHz – 48 kHz).

Depending on the relationship between the square wave’s frequency and the sampling rate, you either get a “pretty” plot, like for the 6 kHz square wave in a 48 kHz system, as shown in Figure 5.

Figure 5: the magnitude response of a 6 kHz square wave in a 48 kHz system

Or, it’s messy, like the 7 kHz square wave in a 48 kHz system in Figure 6.

Figure 6: The magnitude response of a 7 kHz square wave in a 48 kHz system.

The moral of the story

There are three things to remember from this little pair of posts:

  • Some aliased artefacts are negative frequencies, meaning that they appear to be going backwards in time as compared to the original (just like the wheel appearing to rotate backwards in Part 1).
  • Just because you have an antialiasing filter at the input of your ADC does NOT protect you from aliasing, because it can be generated internally, after the signal has been converted to the digital domain.
  • Once this aliasing has happened (e.g. because you clipped the signal in the digital domain), then the aliases are in the signal below the Nyquist frequency and therefore will not be removed by the reconstruction low-pass filter in the DAC. Once they’re mixed in there with the signal, you can’t get them out again.
Figure 7: This is the same as Figure 4, but I’ve removed the first set of mirrored alias artefacts and plotted them on the left side as being mirrored in a “negative frequency” alternate universe.

One additional, but smaller problem with all of this is that, when you look at the output of an FFT analysis of a signal (like the top plot in Figure 7, for example), there’s no way for you to know which components are “normal” harmonics, and which are aliased artefacts that are actually above the Nyquist frequency. It’s another case proving that you need to understand what to expect from the output of the FFT in order to understand what you’re actually getting.

Aliasing is weird: Part 1

One of the best-known things about digital audio is the fact that you cannot record a signal that has a frequency that is higher than 1/2 the sampling rate.

Now, to be fair, that statement is not true. You CAN record a signal that has a frequency that is higher than 1/2 the sampling rate. You just won’t be able to play it back properly, because what comes out of the playback will not be the original frequency, but an alias of it.

If you record a one-spoked wheel with a series of photographs (in the old days, we called this ‘a movie’), the photos (the frames of the movie) might look something like this:

As you can see there, the wheel happens to be turning at a speed that results in it rotating 45º every frame.

The equivalent of this in a digital audio world would be if we were recording a sine wave that rotated (yes…. rotated…) 45º every sample, like this:

Notice that the red lines indicating the sample values are equivalent to the height of the spoke at the wheel rim in the first figure.

If we speed up the wheel’s rotation so that it rotated 90º per frame, it looks like this:

And the audio equivalent would look like this:

Speeding up even more to 135º per frame, we get this:

and this:

Then we get to a magical speed where the wheel rotated 180º per frame. At this speed, it appears when we look at the playback of the film that the wheel has stopped, and it now has two spokes.

In the audio equivalent, it looks like the result is that we have no output, as shown below.

However, this isn’t really true. It’s just an artefact of the fact that I chose to plot a sine wave. If I were to change the phase of this to be a cosine wave (at the same frequency) instead, for example, then it would definitely have an output.

At this point, the frequency of the audio signal is 1/2 the sampling rate.

What happens if the wheel goes even faster (and audio signal’s frequency goes above this)?

Notice that the wheel is now making more than a half-turn per frame. We can still record it. However, when we play it back, it doesn’t look like what happened. It looks like the wheel is going backwards like this:

Similarly, if we record a sine wave that has a frequency that is higher than 1/2 the sampling rate like this:

Then, when we play it back, we get a lower frequency that fits the samples, like this:

Just a little math

There is a simple way to calculate the frequency of the signal that you get out of the system if you know the sampling rate and the frequency of the signal that you tried to record.

Let’s use the following abbreviations to make it easy to state:

  • Fs = Sampling rate
  • F_in = frequency of the input signal
  • F_out = frequency of the output signal

IF
F_in < Fs/2
THEN
F_out = F_in

IF
Fs > F_in > Fs/2
THEN
F_out = Fs/2 – (F_in – Fs/2) = Fs – F_in

Some examples:

If your sampling rate is 48 kHz, and you try to record a 25 kHz sine wave, then the signal that you will play back will be:
48000 – 25000 = 23000 Hz

If your sampling rate is 48 kHz, and you try to record a 42 kHz sine wave, then the signal that you will play back will be:
48000 – 42000 = 6000 Hz

So, as you can see there, as the input signal’s frequency goes up, the alias frequency of the signal (the one you hear at the output) will go down.

There’s one more thing…

Go back and look at that last figure showing the playback signal of the sine wave. It looks like the sine wave has an inverted polarity compared to the signal that came into the system (notice that it starts on a downwards-slope whereas the input signal started on an upwards-slope). However, the polarity of the sine wave is NOT inverted. Nor has the phase shifted. The sine wave that you’re hearing at the output is going backwards in time compared to the signal at the input, just like the wheel appears to be rotating backwards when it’s actually going forwards.

In Part 2, we’ll talk about why you don’t need to worry about this in the real world, except when you REALLY need to worry about it.

One measurement is worse than no measurements

Let’s say that we have to do an audio measurement of a Device Under Test (DUT) that has one input and one output, as shown below.

We don’t know anything about the DUT.

One of the first things we do in the audio world is to measure what most people call the “frequency response” but is more correctly called the “magnitude response”. (It would only be the “frequency response” if you’re also looking at the phase information.)

The standard way to do this is to use an impulse response measurement. This is a method that relies on the fact that an infinitely short, infinitely loud click contains all frequencies at equal magnitude. (Of course, in the real world, it cannot be infinitely short, and if it were infinitely loud, you would have a Big Bang on your hands… literally…)

If we measure the DUT with a single-sample impulse with a value of 1, and use an FFT to convert the impulse response to a frequency-domain magnitude response and we see this:

… then we might conclude that the DUT is as perfect as it can be, within the parameters of a digital audio system. The click comes out just like it went in, therefore the output is identical to the input.

If we measure a different DUT (we’ll call it DUT #2) and we see this:

… then we might conclude that DUT #2 is also perfect. It’s just an attenuator that drops the level by half (or -6.02 dB).

However, we’d be wrong.

I made both of those DUTs myself, and I can tell you that one of those two conclusions is definitely incorrect – but it illustrates the point I’m heading towards.

If I take DUT #1 and send in a sine tone at about 1 kHz and look at the output, I’ll see this:

As you can see there, the output is a sine wave. It looks like one on the top plot, and the bottom plot tells me that there ONLY signal at 1 kHz, which proves it.

If I send the same sine tone through DUT #2 and look at the output, I’ll see this:

As you can see there, DUT #2 clips the input signal so that it cannot exceed ±0.5. This turns the sine wave into the beginnings of a square wave, and generates lots of harmonics that can be seen in the lower half of the plot.

What’s the point?

The point is something that is well-known by people who make audio measurements, but is too easily forgotten:

An Impulse Response measurement only shows you the linear behaviour of an audio device. If the system is non-linear, then your impulse response won’t help you. In a worst case, you’ll think that you measured the system, you’ll think that it’s behaving, and it’s not – because you need to do other measurements to find out more.

The question is “what is ‘non-linear’ behaviour in an audio device?”

This is anything that causes the device to make it impossible to know what the input was by looking at the output. Anything that distorts the signal because of clipping is a simple example (because you don’t know what happened in the input signal when the output is clipped). But other things are also non-linear. For example, dynamic processors like compressors, limiters, expanders and noise gates are all non-linear devices. Modulating delays (like in a chorus or phaser effect), or a transmission system with a drifting clock are other examples. So are psychoaoustic lossy codecs like MP3 and AAC because the signal that gets preserved by the codec changes in time with the signal’s content. Even a “loudness” function can be considered to have a kind of non-linear behaviour (since you get a different filter at different settings of the volume control).

It’s also important to keep in mind that any convolution-based processing is using the impulse response as the filter that is applied to the signal. So, if you have a convolution-based effects unit, it cannot simulate the distortion caused by vacuum tubes using ONLY convolution. This doesn’t mean that there isn’t something else in the processor that’s simulating the distortion. It just means that the distortion cannot be simulated by the convolver.*

P.S.

The reason for the title: “One measurement is worse than no measurements” is that, when you do a measurement (like the impulse response measurement on DUT #2) you gain some certainty about how the device is behaving. In many cases, that single measurement can tell the truth, but only a portion of it – and the remainder of the (hidden) truth might be REALLY bad… So, your one measurement makes you THINK that you’re safe, but you’re really not… It’s not the measurement that’s bad. The problem is the certainty that results in having done it.

* Actually, one of the questions on my comprehensive exams for my Ph.D. was about compressors, with a specific sub-question asking me to explain why you can’t build a digital compressor based on convolution (which was a new-and-sexy way to do processing back then…). The simple answer is that you can’t use a linear time-invariant processor to do non-linear, time-variant processing. It would be like trying to carry water in a net: it’s simply the wrong tool for the job.

Dynamic Styli Correlator Pt. 3

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.

SNR vs DNR

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:

  1. 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.
  2. 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.

What is a “virtual” loudspeaker? Part 1

#91.1 in a series of articles about the technology behind Bang & Olufsen

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…

on to Part 2

Filters and Ringing: Part 10

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.

Fixed point vs. Floating Point

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…

Further Reading

Floating Point Numbers posted by Cleve Moler at Mathworks

Floating Point Denormals, Insignificant But Controversial posted by Cleve Moler at Mathworks