Excruciating minutiae: Part 4

I mentioned in this posting that lately I’ve been doing some measurements of a DUT that:

  • required a frequency analysis with a very big dynamic range
  • … which meant that I was testing it using a sine tone with a frequency that was exactly the same as an FFT bin’s frequency centre
  • … and the sine tone had to be sent through the device by playing a standard audio file (wav and/or FLAC)

So, I did this, but I saw some weirdness that I didn’t expect down in the noise floor of the FFT output. Whenever I’m testing something and I see something weird, I start working my way back through the audio chain to verify that the weirdness is coming from the thing that I’m testing, and not from my test system itself.

So, the first step was to do of an FFT of both the .wav and the .flac files that I was sending through the DUT. The results of this test looked something like Figure 1.

Figure 1.

Before I go further, let’s clarify exactly what I did to generate those three plots.

  • Using Matlab, I made a sine wave with a frequency identical to an FFT bin that was a close as possible to 997 Hz as I could get with a 65,536-point FFT at 48 kHz. (See this posting for more information about this.)
  • I exported the signal using Matlab’s “audiowrite” function, both as a 24-bit wav and a 24-bit FLAC.
  • I imported the two files back into Matlab
  • I ran an FFT on the original, and the two imported files

I would not expect the bottom two plots to be as “good” as the top plot, since they’ve been reduced to a 24-bit fixed point version of the original floating-point signal. However, there are two things to notice in Figure 1.

  1. The most important thing is that the FLAC and WAVE imports produce different results. This is weird.
  2. The less-important (but more interesting, later…) thing is that, for the FLAC import, every odd-numbered FFT bin is -∞ dB, which means that there is absolutely NO energy at those frequencies.

First things first

Let’s address that first issue first. The FFTs show us that the signals coming back from the .wav and .flac import are different. But I’m interested in (1) how they’re different and (2) why they’re different.

Let’s try to answer the first question first. I made a linear ramp that had the same number of samples as the number of quantisation values and had a range of -1 to 1 (just like my sine wave…). So, to test a 16-bit export, I made a ramp that was 216 = 65,536 samples long (shown in the top plot in Figure 2). To test a 24-bit export, the ramp was 224 samples long.

In theory, if I export this ramp to a file type with the matching number of bits, then each sample should quantise to the next quantisation level from the bottom to the top. I then exported this ramp out to .wav and .flac, imported it again, and looked at the result, which is shown in Figure 2.

Figure 2.

If I subtract the results of the imported files from the original, I get the result shown in the middle plot in Figure 2. I would NOT expect either the .wav or the .flac to be identical to the original, since information is lost in the export to a 16- or 24-bit fixed point LPCM format. However, I WOULD expect the .wav and .flac to be the same, which they obviously aren’t.

As can be seen in the bottom plot in Figure 2, there is a 1-quantisation level difference between the .wav and .flac files for signal values higher than 0.

Now the question is whether this difference is inherent in the file format, or if something else is going on. To test this, I did the same test on the 997-ish Hz sine wave (again) without dither, but with my own quantisation (using the code shown in this post). The result of this test is shown in Figure 3.

Figure 3.

As you can see there, the imported .wav and .flac files behave identically. But, if you look carefully and compare to the .flac version in Figure 1, you’ll see that they’re different from THAT version.

The fact that the red and blue plots in Figure 3 are identical tell me that .wav and .flac are identical.

The fact that my quantisation results in identical results in .wav and .flac, but are different from Matlab’s “audiowrite” results (which produces .wav and .flac files that are different from each other) tells me that Matlab’s quantisation is different for .wav and .flac – and different from what I’m doing.

So, I go back to the ramp shown in Figure 2 and dig into the details again, zooming in on the samples near a value of -1, 0, and 1. These are shown below in Figure 4.

Figure 4.

It’s a bit cryptic to see the results in Figure 4, but let’s walk through it.

  • The top plot shows the ramp signal that I encoded as a 16-bit audio file in 4 different ways: as a .wav and a .flac, using audiowrite’s quantiser and mine.
  • The second plot shows the differences in the imported files relative to the original for the first 20 samples, which correspond to the bottom 20 quantisation levels. As can be seen there, the audiowrite quantiser’s result appears to be identical to the original (they’re not, as we saw in the middle plot of Figure 1, but they’re close…). My quantiser is one level higher. This is because I’m scaling my original signal so that it can’t reach the bottom, as I talked about in Part 2.
  • The third plot shows the behaviour of the three quantisers (2 audiowrites and mine) around the 0 value ±10 quantisation levels. Note that there’s no sample with a value of 0 (Because two’s complement is not symmetrical around the 0 value.). It’s not immediately obvious there, but all three quantisers have an “error” of 1/2 a quantisation level step around 0.
    • Below 0, both of audiowrite’s quantisers have a negative offset, and mine has a positive offset.
    • Above 0, audiowrite’s .flac quantiser has a positive offset whereas both audiowrite’s .wav quantiser and mine have a negative offset

If the signal were a sine wave, then we’d see the same thing, it would just be harder to interpret, as shown in Figure 5. (There’s nothing useful shown in the third plot there because when you zoom in so closely , the slope of the sine wave as it passes 0 is really steep…)

Figure 5.

I titled this series of posts “Excruciating minutiae” for a reason. The “error” (let’s call it a “difference” instead) is VERY small. It’s a difference of 1 quantisation level on a portion of the signal, which raises the very pragmatic question: “So what?”

Unless you’re REALLY digging into the bottom of the noise floor of a device, you probably never need to care about this. (In fact, even if you ARE digging into the bottom of the noise floor, you might not need to care.)

You CERTAINLY don’t have to worry about it if you’re just writing audio files to listen to, since you should be dithering those with TPDF dither, which will create a noise floor that is FAR above the “errors” caused by the differences I described above. This can be seen in Figures 6 and 7 below.

Figure 6
Figure 7

In other words, I’ve been using Matlab to export test files both in .wav and .flac for at least 20 years, and it’s only now that I’ve noticed this issue, which is another way of saying “don’t worry about it…”

Nota Bene

If you’re still awake, you might notice that there is one loose end… At the top of this posting I said

The less-important (but more interesting, later…) thing is that, for the FLAC import, every odd-numbered FFT bin is -∞ dB, which means that there is absolutely NO energy at those frequencies.

That will be the topic of another posting, since it’s more or less unrelated to this one – it was just an artefact of the test I described above.

Excruciating minutiae: Part 3

In Part 2 of this series, I wrote the following sentence:

The easiest (and possibly best) way to do this is to create white noise with a triangular probability distribution function and a peak-to-peak amplitude of ± 1 quantisation level.

That’s a very busy sentence, so let’s unpack it a little.

Rolling the dice

If you roll one die, you have an equal probability of rolling any number between 1 and 6 (inclusive). Let’s roll one die 100 times counting the number of times we get a 1, or a 2, or a 3, and so on up to 6.

Number rolledNumber of times the number was rolledPercentage of times the number was rolled
11717
21414
31515
41515
52121
61818

(Note that the percentage of times each number was rolled is the same as the number of times each number was rolled only because I rolled the die 100 times.)

If I plot those results, it looks like Figure 1.

Figure 1. The results of rolling 1 die 100 times.

It may be weird, but I’ve plotted the number of times I rolled -5 or 13 (for example). These are 0 times because it’s impossible to get those numbers by rolling one die. But the reason I put those results in there will make more sense later.

Let’s keep rolling the die. If I do it 1,000,000 times instead of 100, I get these results:

edNumber of times the number was rolledPercentage of times the number was rolled
116622516.6225
216640016.6400
316693016.6930
416705516.7055
516650116.6501
616688916.6889

Now, since I rolled many, many, more times, it’s more obvious that the six results have an equal probability. The more I roll the die, the more those numbers get closer and closer to each other.

Figure 2.

Take a look at the shape of the plot above. The area under the line from 1 to 6 (inclusive) is almost a rectangle because the six numbers are all almost the same.

The shape of that plot shows us the probability of rolling the six numbers on the die, so we call it a probability density function or PDF. In this case, we see a rectangular PDF.

But what happens if we roll two dice instead? Now things get a little more complicated, since there is more than one way to get a total result, as shown in the table below.

Total
21+1
31+22+1
41+32+23+1
51+42+33+24+1
61+52+43+34+25+1
71+62+53+44+35+26+1
82+63+54+45+36+2
93+64+55+46+3
104+65+56+4
115+66+5
126+6

As can be (hopefully) seen in the table, there is only one way to roll a 2, and there’s only one way to roll a 12. But there are 6 different ways to roll a 7. Therefore, if you’re rolling two dice, it’s 6 times more likely that you’ll roll a 7 than a 12, for example.

If I were to roll two dice 1,000,000 times, I would get a PDF like the one shown in Figure 3.

Figure 3.

I won’t explain why this would be considered to be a triangular PDF.

Whether you roll one die or two dice, the number you get is random. In other words, you can’t use the past results to predict what the next number will be. However, if you are rolling one die, and you bet that you’ll roll a 6 every time, you’ll be right about 16.7% of the time. If you’re rolling two dice and you bet that you’ll roll a 12 every time, you’ll only be right about 2.8% of the time.

Let’s take two dice of different colours, say, one red die and one blue die. We’ll roll both dice again, but instead of adding the two values, we’ll subtract the blue value from the red one. If we do this 1,000,000 times, we’ll get something like the results shown below in Figure 4.

Figure 4.

Notice that the probability density function keeps the same shape, it’s just moved down to a range of ±5 instead of 2 to 12.

Generating noise

In audio, noise is a sound that is completely random. In other words, just like the example with the dice, in a digital audio signal, you can’t predict what the next sample value will be based on the past sample values. However, there are many different ways of generating that random number and manipulating its characteristics.

Let’s start with a computer algorithm that can generate a random number between 0 and 1 (inclusive) with a rectangular PDF. We’ll then ask the algorithm to spit out 1,000,000 values. If the numbers really are random, and the computer has infinite precision, then we’ll probably get 1,000,000 different numbers. However, we’re not really interested in the numbers themselves – we’re interested in how they’re distributed between 0.00 and 1.00. Let’s say we divide up that range into 100 steps (or “buckets”) that are 0.01 wide and count how many of our random numbers fall into each group. So, we’ll count how many are between 0.0 and 0.01, between 0.01 and 0.02, and so on up to 0.99 to 1.00. We’ll get something like Figure 5.

Figure 5.

I’ve only plotted the probabilities of the possible values: 0 to 1, which winds up showing only the top of the rectangle in the rectangular PDF.

If I generate 1,000,000 random numbers with that algorithm, and then subtract 1,000,000 other random numbers, one by one, and find the probabilities of the result, the answer will be familiar.

Figure 6.

So, this is how we make the noise that’s added to the signal. If, for each sample, you generate two random numbers (making sure that your algorithm has a rectangular PDF) and subtract one from the other, you have the dither signal that will have a maximum level of ±1 quantisation level.

  • The signal (with a maximum range of ±1) is scaled up by multiplying it by 2(NumberOfBits-1)-2
  • then you add the result of the dither generator
  • then the total is rounded to the nearest integer value
  • and then the result is scaled back down by a factor of 2(NumberOfBits-1) to bring its back down to a range of ±1 to get it ready for exporting to a standard audio file format like .wav or .flac.

In other words, assuming that you have an audio signal called “Signal” that has a range of ±1 and consists of floating point values:

ScaleUp = 2^(Bitdepth-1)-2
ScaleDown = 2^(Bitdepth-1)

TpdfDither = rand(LengthOfSignal) - rand(LengthOfSignal) 

QuantisedDitheredSignal = round(Signal * ScaleUp + TpdfDither) / ScaleDown;

Internal vs. External Volume Control

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

A question came to my desk this week from a customer who would like to connect a third-party streaming device to his Beolab 50s. He plans to use a USB-Audio connection and his question was “Should I control the volume of the audio signal in the streamer or in the Beolab 50s?” There are three different ways to configure these two options:

  1. Control the volume in the streamer using its interface, and send a signal that has been volume-regulated to the Beolab 50s, which should then be set to have a start up default volume such that the maximum volume on the streamer results in a level that is as loud as the customer will ever want it to be. In order to do this, the Beolab 50s need to be set to ignore the volume information that is received on the USB-Audio connection.
  2. Set the streamer to output an unregulated signal, and set the Beolab 50s to obey the volume information that is received on the USB-Audio connection, then use the streamer’s interface for the volume control (which would actually be happening inside the Beolab 50s).
  3. Set the streamer to output an unregulated signal, and set the Beolab 50s to disobey the volume information that is received on the USB-Audio connection, then use the Beolab 50’s interface for the volume control (which would actually be happening inside the Beolab 50s).

Of course, one way to answer the question is “where do you want to control the volume?” For example, if it’s with a remote control for the Beolab 50s, then the answer is “use option #3”. If you’d prefer to use the streamer’s app, for example, then the answer is “use option #1 or #2”.

However, the question came to my desk because it was specifically about the technical performance of the audio signal. Which of these three options results in the highest audio “quality”? (I put the word “quality” in quotation marks because it is a loaded term, and might mean different things to different persons…)

The simplest answer without getting into any details is “it probably doesn’t matter“. However, that answer is based on a couple of assumptions that may or may not be wrong.

Hypothetically, the Beolab 50 can output an audio signal that peaks at about 122 dB SPL measured at 1 m in a free field, albeit not at all frequencies present at its output. (This is because there are some physical limitations of how far the woofers can move, which means that you can’t get 122 dB SPL at 20 Hz, for example.) The noise floor of the Beolab 50s is about 0 dB SPL measured in the same place (again, this is frequency-dependent). So, it has a total dynamic range at its output of about 122 dB.

The maximum output level is a result of a combination of the loudspeaker drivers, the amplifiers, and the power supply, however, these have all been chosen to reach their maximum outputs approximately simultaneously, so changing one of the three won’t make a big difference.

The noise floor is a result of the combination of the loudspeaker drivers’ sensitivities, the amplifiers’ noise floors, and the signal that feeds the amplifiers: the DAC outputs’ noise floors. For the purposes of this discussion, I’m sticking with a digital input, so we don’t need to worry about the noise floor of the ADC at the loudspeaker’s input.

If you have an audio signal at one of the digital inputs of the Beolab 50, and that signal is at its loudest possible level (for a sine wave, that’s 0 dB FS; or 0 dB relative to Full Scale). At Beolab 50’s maximum volume setting, this will produce a peak output level of 122 dB SPL (depending on the frequency as I mentioned above).

All digital inputs of the Beolab 50 accept at least a 24 bit word length. This means that the dynamic range of the digital input signal itself is about 6 * 24 – 3 = 141 dB. This in turn means that the hypothetical noise floor of a correctly-dithered 24-bit signal is 19 dB below the noise floor of the loudspeakers even at their maximum volume setting. (because 122 – 141 = -19)

In other words, if we assume that the streamer has a correctly-implemented gain function for its volume control, using TPDF dither implemented at the 24-bit level, then its noise floor will be 19 dB below the “natural” noise floor of the Beolab 50. Therefore, if the volume is controlled in the streamer, any artefacts will be masked by the 50s themselves.

On the other hand, the Beolab 50s volume control is done using a gain function that is performed in a 32-bit floating point calculation, which means that it has a dynamic range of 144 to 150 dB. (See this posting for an explanation and comparison of fixed point and floating point systems.) So the noise generated by the internal volume control will be somewhere between 22 and 26 dB below the “natural” noise floor of the Beolab 50.

So, (assuming my assumptions are correct) the noise floor that is produced by controlling the volume control in either the streamer or the Beolab 50s is FAR below the constant noise floor of the DAC / amplifiers.

In addition, the noise floors have roughly the same spectra (in other words, you don’t have pink noise in one case but white noise in the other; they’re all producing white noise). And since both are so far below, it really doesn’t matter. Arguing about whether the noise is 19 dB lower or 22 dB lower is a waste of good argument time, unless you paid for the four-and-a-half-hour argument instead of the five-minute one…

Important Notes

If the customer was asking about using the analogue input, then the answer MIGHT have been different.

Also, if my assumption about a 24-bit signal coming from the streamer, or that it has a correctly-implemented gain function for its volume control are incorrect, the this answer MIGHT be incorrect as well.

What is a “virtual” loudspeaker? Part 3

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

In Part 1 of this series, I talked about how a binaural audio signal can (hypothetically, with HRTFs that match your personal ones) be used to simulate the sound of a source (like a loudspeaker, for example) in space. However, to work, you have to make sure that the left and right ears get completely isolated signals (using earphones, for example).

In Part 2, I showed how, with enough processing power, a large amount of luck (using HRTFs that match your personal ones PLUS the promise that you’re in exactly the correct location), and a room that has no walls, floor or ceiling, you can get a pair of loudspeakers to behave like a pair of headphones using crosstalk cancellation.

There’s not much left to do to create a virtual loudspeaker. All we need to do is to:

  • Take the signal that should be sent to a right surround loudspeaker (for example) and filter it using the HRTFs that correspond to a sound source in the location that this loudspeaker would be in. REMEMBER that this signal has to get to your two ears since you would have used your two ears to hear an actual loudspeaker in that location.
  • Send those two signals through a crosstalk cancellation processing system that causes your two loudspeakers to behave more like a pair of headphones.
Figure 1: A block diagram of the system described above.

One nice thing about this system is that the crosstalk cancellation is only there to ensure that the actual loudspeakers behave more like headphones. So, if you want to create more virtual channels, you don’t need to duplicate the crosstalk cancellation processor. You only need to create the binaurally-processed versions of each input signal and mix those together before sending the total result to the crosstalk cancellation processor, as shown below.

Figure 2: You only need one crosstalk cancellation system for any number of virtual channels.

This is good because it saves on processing power.

So, there are some important things to realise after having read this series:

  • All “virtual” loudspeakers’ signals are actually produced by the left and right loudspeakers in the system. In the case of the Beosound Theatre, these are the Left and Right Front-firing outputs.
  • Any single virtual loudspeaker (for example, the Left Surround) requires BOTH output channels to produce sound.
  • If the delays (aka Speaker Distance) and gains (aka Speaker Levels) of the REAL outputs are incorrect at the listening position, then the crosstalk cancellation will not work and the virtual loudspeaker simulation system won’t work. How badly is doesn’t work depends on how wrong the delays and gains are.
  • The virtual loudspeaker effect will be experienced differently by different persons because it’s depending on how closely your actual personal HRTFs match those predicted in the processor. So, don’t get into fights with your friends on the sofa about where you hear the helicopter…
  • The listening room’s acoustical behaviour will also have an effect on the crosstalk cancellation. For example, strong early reflections will “infect” the signals at the listening position and may/will cause the cancellation to not work as well. So, the results will vary not only with changes in rooms but also speaker locations.

Finally, it’s worth nothing that, in the specific case of the Beosound Theatre, by setting the Speaker Distances and Speaker Levels for the Left and Right Front-firing outputs for your listening position, then you have automatically calibrated the virtual outputs. This is because the Speaker Distances and Speaker Levels are compensations for the ACTUAL outputs of the system, which are the ones producing the signal that simulate the virtual loudspeakers. This is the reason why the four virtual loudspeakers do not have individual Speaker Distances and Speaker Levels. If they did, they would have to be identical to the Left and Right Front-firing outputs’ values.

What is a “virtual” loudspeaker? Part 2

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

In Part 1, I talked at how a binaural recording is made, and I also mentioned that the spatial effects may or may not work well for you for a number of different reasons.

Let’s go back to the free field with a single “perfect” microphone to measure what’s happening, but this time, we’ll send sound out of two identical “perfect” loudspeakers. The distances from the loudspeakers to the microphone are identical. The only difference in this hypothetical world is that the two loudspeakers are in different positions (measuring as a rotational angle) as shown in Figure 1.

Figure 1: Two identical, “perfect” loudspeakers in a free field with a single “perfect” microphone.

In this example, because everything is perfect, and the space is a free field, then output of the microphone will be the sum of the outputs of the two loudspeakers. (In the same way that if your dog and your cat are both asking for dinner simultaneously, you’ll hear dog+cat and have to decide which is more annoying and therefore gets fed first…)

Figure 2: The output from the microphone is the sum of the outputs from the two loudspeakers. At any moment in time, the value of the top plot + the value of the middle plot = the value of the bottom plot.

IF the system is perfect as I described above, then we can play some tricks that could be useful. For example, since the output of the microphone is the sum of the outputs of the two loudspeakers, what happens if the output of one loudspeaker is identical to the other loudspeaker, but reversed in polarity?

Figure 3: If the output of Loudspeaker 1 is exactly the same as the output of Loudspeaker 2 except for polarity, then the sum (the output of the microphone) is always 0.

In this example, we’re manipulating the signals so that, when they add together, you nothing at the output. This is because, at any moment in time, the value of Loudspeaker 2’s output is the value of Loudspeaker 1’s output * -1. So, in other words, we’re just subtracting the signal from itself at the microphone and we get something called “perfect cancellation” because the two signals cancel each other at all times.

Of course, if anything changes, then this perfect cancellation won’t work. For example, if one of the loudspeakers moves a little farther away than the other, then the system is broken, as shown below.

Figure 4: A small shift in time in the output of Loudspeaker 2 cases the cancellation to stop working so well.

Again, everything that I’ve said above only works when everything is perfect, and the loudspeakers and the microphone are in a free field; so there are no reflections coming in and ruining everything.

We can now combine these two concepts:

  1. using binaural signals to simulate a sound source in a location (although this would normally be done using playback over earphones to keep it simple) and
  2. using signals from loudspeakers to cancel each other at some location in space as a

to create a system for making virtual loudspeakers.

Let’s suspend our adherence to reality and continue with this hypothetical world where everything works as we want… We’ll replace the microphone with a person and consider what happens. To start, let’s just think about the output of the left loudspeaker.

Figure 5: The output of the left loudspeaker reaches both ears with different time/frequency characteristics caused by the HRTF associated with that sound source location.

If we plot the impulse responses at the two ears (the “click” sound from the loudspeaker after it’s been modified by the HRTFs for that loudspeaker location), they’ll look like this:

Figure 6: The impulse responses of the HRTFs for a sound source at 30º left of centre.

What if were were able to send a signal out of the right loudspeaker so that it cancels the signal from the left loudspeaker at the location of the right eardrum?

Figure 7: What if we could cancel the signal from the left loudspeaker at the right ear using the right loudspeaker?

Unfortunately, this is not quite as easy as it sounds, since the HRTF of the right loudspeaker at the right ear is also in the picture, so we have to be a bit clever about this.

So, in order for this to work we:

  • Send a signal out of the left loudspeaker.
    We know that this will get to the right eardrum after it’s been messed up by the HRTF. This is what we want to cancel…
  • …so we take that same signal, and
    • filter it with the inverse of the HRTF of the right loudspeaker
      (to undo the effects of the HRTF of the right loudspeaker’s signal at the right ear)
    • filter that with the HRTF of the left loudspeaker at the right ear
      (to match the filtering that’s done by your head and pinna)
    • multiply by -1
      (so that it will cancel when everything comes together at your right eardrum)
    • and send it out the right loudspeaker.

Hypothetically, that signal (from the right loudspeaker) will reach your right eardrum at the same time as the unprocessed signal from the left loudspeaker and the two will cancel each other, just like the simple example shown in Figure 3. This effect is called crosstalk cancellation, because we use the signal from one loudspeaker to cancel the sound from the other loudspeaker that crosses to the wrong side of your head.

This then means that we have started to build a system where the output of the left loudspeaker is heard ONLY in your left ear. Of course, it’s not perfect because that cancellation signal that I sent out of the right loudspeaker gets to the left ear a little later, so we have to cancel the cancellation signal using the left loudspeaker, and back and forth forever.

If, at the same time, we’re doing the same thing for the other channel, then we’ve built a system where you have the left loudspeaker’s signal in the left ear and the right loudspeaker’s signal in the right ear; just like a pair of headphones!

However, if you get any of these elements wrong, the system will start to under-perform. For example, if the HRTFs that I use to predict your HRTFs are incorrect, then it won’t work as well. Or, if things aren’t time-aligned correctly (because you moved) then the cancellation won’t work.

on to Part 3

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

Beosound Theatre: Outputs

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

A “naked” Beosound Theatre

Beosound Theatre has a total of 11 possible outputs, seven of which are “real” or “internal” outputs and four of which are “virtual” loudspeakers. As with all current Beovision televisions, any input channel can be directed to any output by setting the Speaker Roles in the menus.

Internal outputs

On first glance of the line drawing above it is easy to jump to the conclusion that the seven real outputs are easy to find, however this would be incorrect. The Beosound Theatre has 12 loudspeaker drivers that are all used in some combination of level and phase at different frequencies to all contribute to the total result of each of the seven output channels.

So, for example, if you are playing a sound from the Left front-firing output, you will find that you do not only get sound from the left tweeter, midrange, and woofer drivers as you might in a normal soundbar. There will also be some contribution from other drivers at different frequencies to help control the spatial behaviour of the output signal. This Beam Width control is similar to the system that was first introduced by Bang & Olufsen in the Beolab 90. However, unlike the Beolab 90, the Width of the various beams cannot be changed in the Beosound Theatre.

The seven internal loudspeaker outputs are

  • Front-firing: Left, Centre, and Right
  • Side-firing: Left and Right
  • Up-firing: Left and Right

Looking online, you may find graphic explanations of side-firing and up-firing drivers in other loudspeakers. Often, these are shown as directing sound towards a reflecting wall or ceiling, with the implication that the listener therefore hears the sound in the location of the reflection instead. Although this is a convenient explanation, it does not necessarily match real-life experience due to the specific configuration of your system and the acoustical properties of the listening room.

The truth is both better and worse than this reductionist view. The bad news is that the illusion of a sound coming from a reflective wall instead of the loudspeaker can occur, but only in specific, optimised circumstances. The good news is that a reflecting surface is not strictly necessary; therefore (for example) side-firing drivers can enhance the perceived width of the loudspeaker, even without reflecting walls nearby.

However, it can be generally said that the overall benefit of side- and up-firing loudspeaker drivers is an enhanced impression of the overall width and height of the sound stage, even for listeners that are not seated in the so-called “sweet spot” (see Footnote 1) when there is appropriate content mixed for those output channels.

Virtual outputs

Devices such as the “stereoscope” for representing photographs (and films) in three-dimensions have been around since the 1850s. These work by presenting two different photographs with slightly different perspectives two the two eyes. If the differences in the photographs are the same as the differences your eyes would have seen had you “been there”, then your brain interprets into a 3D image.

A similar trick can be done with sound sources. If two different sounds that exactly match the signals that you would have heard had you “been there” are presented at your two ears (using a binaural recording) , then your brain will interpret the signals and give you the auditory impression of a sound source in some position in space. The easiest way to do this is to ensure that the signals arriving at your ears are completely independent using headphones.

The problem with attempting this with loudspeaker reproduction is that there is “crosstalk” or “bleeding of the signals to the opposite ears”. For example, the sound from a correctly-positioned Left Front loudspeaker can be heard by your left ear and your right ear (slightly later, and with a different response). This interference destroys the spatial illusion that is encoded in the two audio channels of a binaural recording.

However, it might be possible to overcome this issue with some careful processing and assumptions. For example, if the exact locations of the left and right loudspeakers and your left and right ears are known by the system, then it’s (hypothetically) possible to produce a signal from the right loudspeaker that cancels the sound of the left loudspeaker in the right ear, and therefore you only hear the left channel in the left ear. (see Footnote 2)

Using this “crosstalk cancellation” processing, it becomes (hypothetically) possible to make a pair of loudspeakers behave more like a pair of headphones, with only the left channel in the left ear and the right in the right. Therefore, if this system is combined with the binaural recording / reproduction system, then it becomes (hypothetically) possible to give a listener the impression of a sound source placed at any location in space, regardless of the actual location of the loudspeakers.

Theory vs. Reality

It’s been said that the difference between theory and practice is that, in theory, there is no difference between theory and practice, whereas in practice, there is. This is certainly true both of binaural recordings (or processing) and crosstalk cancellation.

In the case of binaural processing, in order to produce a convincing simulation of a sound source in a position around the listener, the simulation of the acoustical characteristics of a particular listener’s head, torso, and (most importantly) pinnæ (a.k.a. “ears”) must be both accurate and precise. (see Footnote 3)

Similarly, a crosstalk cancellation system must also have accurate and precise “knowledge” of the listener’s physical characteristics in order to cancel the signals correctly; but this information also crucially includes the exact locations of the loudspeakers and the listener (we’ll conveniently pretend that the room you’re sitting in does not exist).

In the end, this means that a system with adequate processing power can use two loudspeakers to simulate a “virtual” loudspeaker in another location. However, the details of that spatial effect will be slightly different from person to person (because we’re all shaped differently). Also, more importantly, the effect will only be experienced by a listener who is positioned correctly in front of the loudspeakers. Slight movements (especially from side-to-side, which destroys the symmetrical time-of-arrival matching of the two incoming signals) will cause the illusion to collapse.

Beosound Theatre gives you the option to choose Virtual Loudspeakers that appear to be located in four different positions: Left and Right Wide, and Left and Right Elevated. These signals are actually produced using the Left and Right front-firing outputs of the device using this combination of binaural processing and crosstalk cancellation in the Dolby Atmos processing system. If you are a single listener in the correct position (with the Speaker Distances and Speaker Levels adjusted correctly) then the Virtual outputs come very close to producing the illusion of correctly-located Surround and Front Height loudspeakers.

However, in cases where there is more than one listener, or where a single listener may be incorrectly located, it may be preferable to use the “side-firing” and “up-firing” outputs instead.

Wrapping up

As I mentioned at the start, Beosound Theatre on its own has 11 outputs:

  • Front-firing: Left, Centre, and Right
  • Side-firing: Left and Right
  • Up-firing: Left and Right
  • Virtual Wide: Left and Right
  • Virtual Elevated: Left and Right

In addition to these, there are 8 wired Power Link outputs and 8 Wireless Power Link outputs for connection to external loudspeakers, resulting in a total of 27 possible output paths. And, as is the case with all Beovision televisions since Beoplay V1, any input channel (or output channel from the True Image processor) can be directed to any output, giving you an enormous range of flexibility in configuring your system to your use cases and preferences.


1. In the case of many audio playback systems, the “sweet spot” is directly in front of the loudspeaker pair or at the centre of the surround configuration. In the case of a Bang & Olufsen system, the “sweet spot” is defined by the user with the help of the Speaker Distance and Speaker Level adjustments.

2. Of course, the cancelling signal of the right loudspeaker also bleeds to the left ear, so the left loudspeaker has to be used to cancel the cancellation signal of the right loudspeaker in the left ear, and so on…

3. For the same reason that someone else should not try to wear my glasses.

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

Quantisation of Poles in the Z-plane

Back in this posting, I talked about biquads and their use in digital signal processing for making linear filters (what most of us call “equalisers”). Part of that explanation showed that a biquad is formed of a feed-forward section and a feed-back section, and you could swap the order of these two and get the same results. In a “Direct Form 1” implementation, the feed-forward comes first, as shown in Figure 1.

Figure 1: A Direct Form 1 implementation of a biquad

in the “Direct Form 2” implementation, the feed-back comes first, as shown in Figure 2.

Figure2: A Direct Form 2 implementation of a biquad.

There are advantages and disadvantages to each of these implementations, depending on things like how the rest of your system is implemented, and what, exactly you’re expecting the biquad to do.

However, for this posting, we’re going to “zoom in” a bit to the feed-back portion of the above diagrams. This portion of the biquad provides the “poles” in the Z-plane, as I described in the “Intuitive Z-plane” series of postings last week.

If I separate the feed-back portion of the above figures, it would look like Figure 3:

Figure 3: A Direct Form implementation of the feed-back portion of a biquad.

The locations of the poles (and therefore the magnitude response) of this portion of the filter are dependent on the gains at the outputs of the two 1-sample delays. What happens when these gains do not have an infinite resolution (which they can’t, because everything in a digitally-represented world is quantised to a finite number of steps)?

Before we go any further with this, I have to put in a reminder that quantisation (or “rounding”) in a DSP world is done in binary – not decimal. So, the quantisation that I’m about to do isn’t the same as stopping a couple of digits after the decimal…

In a world with infinite resolution, I can set the values of those two gains to be anything I want, and therefore the poles in my system can be anywhere within the circle. (We’ll assume that we don’t want a pole on the circle because that would result in a gain of ∞ dB, which is very loud.) However, let’s say that we quantise the gain values to a 4-bit binary value, where the first bit is reserved for the +/- indicator. This means that we only have +/- 2^3 possibilities, or 8 values, one of which is +0, and another is -0 (which is the same thing…). So, in other words, we have 7 possible negative values, 7 possible positive values, and 0.

The end result of this is that the poles have a limited number of locations where they can be placed. For the 4-bit quantisation described above, the resulting locations look like Figure 4.

Figure 4: The circles indicate the possible locations of the poles as a result of quantisation of the gain coefficients in a 4-bit Direct Form implementation.

Remember that we’re not talking about quantisation of the audio signal – it’s quantisation of the gain coefficients inside the biquad that will have an impact on the response of the filter.

Of course, it would be crazy to implement a biquad using only 4-bit quantisation for the gain coefficients. However, the point of this posting is not to show that biquads suck. It’s only to show one possibly important aspect of them if you’re a DSP engineer – but we’re getting there.

Just for fun, let’s increase the resolution of the system to 6 bits:

Figure 5: The dots indicate the possible locations of the poles as a result of quantisation of the gain coefficients in a 6-bit Direct Form implementation.

… or to 8 bits:

Figure 6: The dots indicate the possible locations of the poles as a result of quantisation of the gain coefficients in a 4-bit Direct Form implementation.

Any higher than this and the pattern will just get so dense that it’ll turn black, so I’ll stop.

So what?

At this point, you may be asking “so what?” The answer to this very important question lies on the far right side of all of those graphs. Notice that there aren’t any locations near the point on the graph where X=1 and Y=0. You may remember from Figure 7 in this posting that I did last week, that that’s the point on the Z-plane that corresponds to very low frequencies.

This means that, if you want to create a digital filter, and you need a pole near 0 Hz, you’re going to run into some trouble with a Direct Form implementation of the biquad. Yes, the higher the bit depth of the gain coefficients, the closer you can get, but this might not be the best way to do things.

There is another option for implementing your feed-back portion of the biquad. You can use a design called a Coupled Form instead, which is shown in Figure 7 (compare it to Figure 3).

Figure 7: A Coupled Form implementation of the feed-back portion of a biquad.

Notice that you still have two 1-sample delays, however there are now 4 gains instead of 2. How can this be better? Well, in a system with infinite resolution on the gain coefficients, it’s not. Given the appropriate choice of gain values, this implementation will do exactly the same thing as the Direct Form implementation if your resolution is infinite.

However, if you have a limited resolution, then the available locations of the poles on the Z-plane are very different. Let’s use the 4-bit, 6-bit, and 8-bit quantisations of the gain values again: these are shown in Figures 8 to 10.

Figure 8: The circles indicate the possible locations of the poles as a result of quantisation of the gain coefficients in a 4-bit Coupled Form implementation.
Figure 9: The dots indicate the possible locations of the poles as a result of quantisation of the gain coefficients in a 6-bit Coupled Form implementation.
Figure 10: The dots indicate the possible locations of the poles as a result of quantisation of the gain coefficients in a 8-bit Direct Form implementation.

As you can see in Figures 8 through 10, the Coupled Form implementation, given the same resolution on the gain coefficients, will give you much better placement opportunities for the poles in the low frequency region than the Direct Form implementation.

Of course, in all of these examples, I’m only showing up to an 8-bit word, and a typical DSP runs uses a lot more than 8 bits for the gain coefficients. So, it’s possible that, in the real world, the actual resolution is high enough that this is of no concern whatsoever.

However, if you’re building a very-low-frequency filter an if the magnitude response isn’t exactly what you’re looking for, this might help you get a little closer to your goal.

It’s important to point out here that the quantisation of the locations of the zeros is not the same as this. Someday, I’ll come back to this and plot the expected vs. actual magnitude responses of some biquads where I’ve quantised the zeros and poles, just to see how badly things go wrong when they do…

Further Reading

This posting is a simple summary of the discussion of a section called “Poles of Quantized Second-Order Sections” in “Discrete-Time Signal Processing” by Alan V. Oppenheim and Ronald W. Schafer.

The Coupled Form implementation was introduced by C.M. Rader and B. Gold in their paper called “Digital Filter Design Techniques in the Frequency Domain” from the Proceesings of the IEEE, Vol. 55, pp. 149 – 171 , from February, 1967.

Intuitive Z-plane: Part 5 – Conclusion

If you’ve read through the first four parts of this series, then you’re already at a point where you can intuitively understand what’s going on. We just have a couple of details to take care of before finishing off.

Firstly, the plots showing the zeros and poles in the figures you’ve been looking at plots of the “Z-plane” or “Complex-plane“. As I said at the start, we’re only trying to get to an intuitive understanding of these plots – so I’m not going to get into complex numbers, or even much math (apart from what you’ll see below… which isn’t very complicated, and avoids complex numbers).

When I’m developing a new DSP algorithm, I use an application called Max from cycling74.com. Figure 1 shows a screenshot from Max, where I’m using an object to calculate the biquad coefficients to make a low pass filter, as you can see. I’ve then connected the output of that object (it looks like a magnitude response) to a Z-plan representation that shows me the same thing in a different way.

Figure 1: The top plot shows the magnitude response of the filter. The bottom plot shows the Z-plane representation of the same filter.

You may notice that this plot has two poles, one at (0, 0.408) and the other at (0, -0.408). In fact there are two zeros there as well, but they’re situated in the same place, on “on top” of the other, at (-1, 0). This is always true for a biquad – there are always two zeros and two poles. Sometimes, they’re located in the same place, sometimes not, sometimes they’re placed symmetrically, sometimes not, depending on the filter, as we’ll see below.

Let’s look at that Z-plane representation in 3-dimensions:

Figure 2: A 3D view of the Z-plane representation of the filter shown in Figure 1.
Figure 3: The same plot as shown in Figure 2, rotated to show the back of the plot.

So, as you would now expect, the poles pull up the edge of the circle, and the zeros (both in the same place) pull down, giving the red line the height that it has.

Now, think back to this Figure from earlier in the series:

Figure 4: Think of the edge of the circle as the frequency

If you therefore look at Figure 3, which is like looking at Figure 4 from the top, you’ll notice that the height of the red line (the edge of the circle is high on the left (in the low frequencies) and drops as you go to the right (the high frequencies). This is the magnitude response that’s shown on the top of Figure 1. The only difference is that it’s on a linear scale instead of a logarithmic scale, so the shape looks a little weird.

Let’s do another one:

Figure 5: A reciprocal peak/dip filter.
Figure 6: A 3D view of the Z-plane representation of the filter shown in Figure 5.
Figure 7: The same plot as shown in Figure 6, rotated to show the back of the plot.

Hopefully, now you are able to look at a Z-plane representation of a filter and think about the effect of the poles and zeros on the edge of the circle, and therefore get a rough idea of the magnitude response of the filter…

If not, I apologize for wasting your time. On the other hand, if you’re in a life-threatening situation, this knowledge probably wouldn’t help you anyway… Very few people have gotten a critical injury in a biquad accident.

How I did it

If you want to make these plots for yourself, the math is pretty simple.

Figure 8: How to do the math.

Start by choosing the frequency, which will be a point on the circle. You then find the four distances from the zeros and poles to that point (I’ve indicated those distances in Figure 8 with the variables z1, z2, p1, and p2.) This can be done using the Pythagorean theorem.

To find the gain of the filter at the frequency, you divide the sum of the zeros’ distances by the sum of the poles’ distances. In other words:

(z1 + z2) / (p1 + p2)

That will give you the result as a linear value. If you then want to convert it to decibels, like I’ve done, you do a little extra math like this:

20 * log10 ( (z1 + z2) / (p1 + p2) )

That’s it! You just need to do repeat that math for each frequency that you’re interested in, and you’re done!