Category: loudspeakers

Mixing closed and ported cabinets: Part 1

I made a comment on a forum this week, commenting that, if you mix loudspeakers with closed cabinets with loudspeakers with ported cabinets (or slave drivers), the end result can be a reduction in total output: less sound from more loudspeakers. Of course, the question is “why?” and the short answer is “due to the phase mismatching of the loudspeakers”.

This is the long answer.

Before we begin, we have to get an intuitive understanding of what a ported loudspeaker is. (Note that I’ll keep saying “ported loudspeaker”, but the principle also applies to loudspeakers with slave drivers, as I’ll explain later.) Before we get to a ported loudspeaker, we need to talk about Helmholtz resonators.

Take a block that’s reasonably heavy and hang it using a spring so that it looks like this:

Figure 1.

The spring is a little stretched because the weight of the block (which is the result of its mass and the Earth’s gravity) is pulling downwards. (We’ll ignore the fact that the spring is also holding up its own weight. Let’s keep this simple…) However, it doesn’t fall to the floor because the spring is pulling upwards.

Now pull downwards on the block, so it will look like the example on the right in the figure below.

Figure 2.

The spring is stretched because we’re pulling down on the block. The spring is also pulling upwards more, since it’s pulling against the weight of the block PLUS the force that you’re adding in a downwards direction.

Now you let go of the block. What happens?

The spring is pulling “too hard” on the block, so the block starts rising back to where it started (we’ll call that the “resting position”). However, when it gets there, it has inertia (a body in motion tends to stay in motion… until it hits something big…) so it doesn’t stop. As a result, it moves upwards, higher than the resting position. This squeezes the spring until it gets to some point, at which time the block stops, and then starts going back downwards. When it returns to the resting position, it still has inertia, so it passes that point and goes too far down again. I’ve shown this as a series of positions from left to right in the figure below.

Figure 3

If there were no friction, no air around the block, and no friction within the metal molecules of the spring, then this would bounce up and down forever.

However, there is friction, so some of the movement (“kinetic energy”) is turned into heat and lost. So, each bounce gets smaller and smaller and the maximum velocity of the block (as it passes the resting position) gets lower and lower, until, eventually, it stops moving (at the resting position, where it started).

Notice that I changed the colour of the spring to show when it’s more stretched (lighter blue) and when it’s more compressed (darker blue).

If everything were behaving perfectly, then the RATE at which the bounce repeats wouldn’t change. Only its amplitude (or the excursion of the block, or the height of the bounce) would reduce over time. That bounce rate (let’s say 1 bounce per second, and by “bounce” I mean a full cycle of moment down, up, and back down to where it started again) is the frequency of the repetition (or oscillation).

If you make the weight lighter, then it will bounce faster (because the spring can pull the weight more “easily”). If you make the spring stiffer, then it will bounce faster (because the spring can pull the weight more “easily”). So, we can change the frequency of the oscillation by changing the weight of the block or the stiffness of the spring.

Now take a look at the same weight on a spring next to an up-side down wine bottle that (sadly) has been emptied of wine.

Figure 4.

Notice that I’ve added some colours to the air inside the bottle. The air in the bottle itself is blue, just like the colour of the spring. This is because, if we pull air out of the bottle (downwards), the air inside it will pull back (upwards; just like the metal spring pulling back upwards on the block). I’ve made the small cylinder of air in the neck of the bottle red, just like the block. This is because that air has some mass, and it’s free to move upwards (into the bottle) and downwards (out of the bottle) just like the block.

If I were somehow able to pull the “plug” of air out of the neck of the bottle, the air inside would try to pull it back in. If I then “let go”, the plug would move inwards, go too far (because it also has inertia), squeezing (or compressing) the air inside the bottle, which would then push the plug back out. This is shown in the figure below.

Figure 5.

At the level we’re dealing with, this behaviour is practically identical to the behaviour of the block on the spring. In other words, although the block and the plug are made of different materials, and although the metal spring and the air inside the bottle are different materials, Figures 3 and 5 show the same behaviour of the same kind of system.

How do you pull the plug of air out of the bottle? It’s probably easier to start by pushing it inwards instead, by blowing across the top.

When you do this, a little air leaks into the opening, pushing the plug inwards. The “spring” in the bottle then pushes the plug outwards, and your cycle has started. If you wanted to do the same thing with the block, you’d lift it and let go to start the oscillation.

However, you don’t need to blow across the bottle to make it oscillate. You can just tap it with the palm of your hand, for example. Or, if you put the bottle next to your ear and listen carefully, you’ll hear a note “singing along” with the sound in the room. This is because the air in the bottle resonates; it moves back and forth very easily at the frequency that’s determined by the mass of the air in the neck and the volume of air in of the bottle (the spring).

However, remember that friction can make the oscillation decay (or die away) faster, by turning the movement into heat.

One last thing…

There’s another way to get either the block or the wine bottle oscillating:

You can move the TOP of the spring (for example, if you pull it up, then the spring will pull the block upwards, and it’ll start bouncing). Or, you could tap the bottom of the wine bottle (which is on the top in my drawings).

This method of starting the oscillation will come in handy in part 2.

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.

Excursion vs. Frequency

Last week, I was doing a lecture about the basics of audio and I happened to mention one of the rules of thumb that we use in loudspeaker development:

If you have a single loudspeaker driver and you want to keep the same Sound Pressure Level (or output level) as you change the frequency, then if you go down one octave, you need to increase the excursion of the driver 4 times.

One of the people attending the presentation asked “why?” which is a really good question, and as I was answering it, I realised that it could be that many people don’t know this.

Let’s take this step-by-step and keep things simple. We’ll assume for this posting that a loudspeaker driver is a circular piston that moves in and out of a sealed cabinet. It is perfectly flat, and we’ll pretend that it really acts like a piston (so there’s no rubber or foam surround that’s stretching back and forth to make us argue about changes in the diameter of the circle). Also, we’ll assume that the face of the loudspeaker cabinet is infinite to get rid of diffraction. Finally, we’ll say that the space in front of the driver is infinite and has no reflective surfaces in it, so the waveform just radiates from the front of the driver outwards forever. Simple!

Then, we’ll push and pull the loudspeaker driver in and out using electrical current from a power amplifier that is connected to a sine wave generator. So, the driver moves in and out of the “box” with a sinusoidal motion. This can be graphed like this:

Figure 1: The excursion of a loudspeaker driver playing a 1 kHz sine wave at some output level.

As you can see there, we have one cycle per millisecond, therefore 1000 cycles per second (or 1 kHz), and the driver has a peak excursion of 1 mm. It moves to a maximum of 1 mm out of the box, and 1 mm into the box.

Consider the wave at Time = 0. The driver is passing the 0 mm line, going as fast as it can moving outwards until it gets to 1 mm (at Time = 0.25 ms) by which time it has slowed down and stopped, and then starts moving back in towards the box.

So, the velocity of the driver is the slope of the line in Figure 1, as shown in Figure 2.

Figure 2: The excursion and velocity of the same loudspeaker driver playing the same signal.

As the loudspeaker driver moves in and out of the box, it’s pushing and pulling the air molecules in front of it. Since we’ve over-simplified our system, we can think of the air molecules that are getting pushed and pulled as the cylinder of air that is outlined by the face of the moving piston. In other words, its a “can” of air with the same diameter as the loudspeaker driver, and the same height as the peak-to-peak excursion of the driver (in this case, 2 mm, since it moves 1 mm inwards and 1 mm outwards).

However, sound pressure (which is how loud sounds are) is a measurement of how much the air molecules are compressed and decompressed by the movement of the driver. This is proportional to the acceleration of the driver (neither the velocity nor the excursion, directly…). Luckily, however, we can calculate the driver’s acceleration from the velocity curve. If you look at the bottom plot in Figure 2, you can see that, leading up to Time = 0, the velocity has increased to a maximum (so the acceleration was positive). At Time = 0, the velocity is starting to drop (because the excursion is on its was up to where it will stop at maximum excursion at time = 0.25 ms).

In other words, the acceleration is the slope of the velocity curve, the line in the bottom plot in Figure 2. If we plot this, it looks like Figure 3.

Figure 3: The excursion, velocity and acceleration of the same loudspeaker driver playing the same signal.

Now we have something useful. Since the bottom plot in Figure 3 shows us the acceleration of the driver, then it can be used to compare to a different frequency. For example, if we get the same driver to play a signal that has half of the frequency, and the same excursion, what happens?

Figure 4: Comparing the excursion, velocity and acceleration of the same loudspeaker driver playing two different signals with the same excursion. (The red line is the same in Figure 4 as in Figure 3.)

In Figure 4, two sine waves are shown: the black line is 1/2 of the frequency of the red line, but they both have the same excursion. If you take a look at where both lines cross the Time = 0 point, then you can see that the slopes are different: the red line is steeper than the black. This is why the peak of the red line in the velocity curve is higher, since this is the same thing. Since the maximum slope of the red line in the middle plot is higher than the maximum slope of the black line, then its acceleration must be higher, which is what we see in the bottom plot.

Since the sound pressure level is proportional to the acceleration of the driver, then we can see in the top and bottom plots in Figure 4 that, if we halve the frequency (go down one octave) but maintain the same excursion, then the acceleration drops to 25% of the previous amount, and therefore, so does the sound pressure level (20*log10(0.25) = -12 dB, which is another way to express the drop in level…)

This raises the question: “how much do we have to increase the excursion to maintain the acceleration (and therefore the sound pressure level)?” The answer is in the “25%” in the previous paragraph. Since maintaining the same excursion and multiplying the frequency by 0.5 resulted in multiplying the acceleration by 0.25, we’ll have to increase the excursion by 4 to maintain the same acceleration.

Figure 5: Comparing the excursion, velocity and acceleration of the same loudspeaker driver playing two different signals at two different excursions. Notice that some of the vertical scales in the plots have changed. (The red line is the same in Figure 5 as in Figures 4 and 3.)

Looking at Figure 5: The black line is 1/2 the frequency of the red line. Their accelerations (the bottom plots) have the same peak values (which means that they produce the same sound pressure level). This, means that the slopes of their velocities are the same at their maxima, which, in turn, means that the peak velocity of the black line (the lower frequency) is higher. Since the peak velocity of the black line is higher (by a factor of 2) then the slope of the excursion plot is also twice as steep, which means that the peak of the excursion of the black line is 4x that of the red line. All of that is explained again in Figure 6.

Figure 6. A repeat of Figure 5 with some explanations that (hopefully) help.

Therefore, assuming that we’re using the same loudspeaker driver, we have to increase the excursion by a factor of 4 when we drop the frequency by a factor of 2, in order to maintain a constant sound pressure level.

However, we can play a little trick… what we’re really doing here is increasing the volume of our “cylinder” of air by a factor of 4. Since we don’t change the size of the driver, we have to move it 4 times farther.

However, the volume of a cylinder is

π r2 * height

and we’re just playing with the “height” in that equation. A different way would be to use a different driver with a bigger surface area to play the lower frequency. For example, if we multiply the radius of the driver by 2, and we don’t change the excursion (the “height” of the cylinder) then the total volume increases by a factor of 4 (because the radius is squared in the equation, and 2*2 = 4).

Another way to think of this: if our loudspeaker driver was a square instead of a circle, we could either move it in and out 4 times farther OR we would make the width and the length of the square each twice as big to get the a cube with the same volume. That “r2” in the equation above is basically just the “width * length” of a circle…

This is why woofers are bigger than tweeters. In a hypothetical world, a tweeter can play the same low frequencies as a woofer – but it would have to move REALLY far in and out to do it.

Loud speaker

I live in Denmark where people speak Danish. One interesting word that I use every day is “højtaler” which is the Danish word for “loudspeaker”. I say that this word is “interesting” because, just like “loudspeaker” it is actually two words glued together. “Høj” means “high” or “loud” and “taler” means “talker” or “speaker” (as in “the person who is doing the talking”).

Sometimes, when I have a couple of minutes to spare, instead of looking at cat videos on YouTube, I sift through old audio and electronics magazines for fun. One really good source for these is the collection at worldradiohistory.com. (archive.org is also good!). Today I stumbled across the December, 1921 edition of Practical Electrics Magazine (which changed its name to The Experimenter in November of 1924 and then to Amazing Stories in April 1926*) It had a short description of a stage trick called the “Haunted Violin”, an excerpt from which is shown below.

The trick was that a violin, held by a woman walking around the aisle of the theatre would appear to play itself. In fact, as you can see above, there was an incognito violinist with a “detectophone” that was transmitting through wires connected to metal plates under the carpet in the theatre. The woman was wearing shoes with heels pointy enough to pierce the carpet and make contact with the plates. The heels were then connected with wires running through her dress to a “loud talker” hidden inside the violin.

Seems that, in 1921, it would have been easier to learn at least one word in Danish…

Side note: This is why, when I’m writing about audio systems I try as hard as possible to always use the word “loudspeaker” instead of “speaker”. To me, a “speaker” is a person giving a speech. A “loudspeaker” is a thing I complain about every day at work.

* That April 1926 edition of Amazing Stories had short stories by Jules Verne, H.G. Wells, and Edgar Allan Poe!

Post Script:
My wife reminded me that it’s the same in French: “haut-parleur”. It’s a reminder that the original loudspeakers were never intended for music, I suppose…

Volume controls vs. Output levels

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

One question people often ask about B&O loudspeakers is something like ”Why doesn’t the volume control work above 50%?”.

This is usually asked by someone using a small loudspeaker listening to pop music.

There are two reasons for this, related to the facts that there is such a wide range of capabilities in different Bang & Olufsen loudspeakers AND you can use them together in a surround or multiroom system. In other words for example, a Beolab 90 is capable of playing much, much more loudly than a Beolab 12; but they still have to play together.

Let’s use the example of a Beolab 90 and a Beolab 12, both playing in a surround configuration or a multiroom setup. In both cases, if the volume control is set to a low enough level, then these two types of loudspeakers should play at the same output level. This is true for quiet recordings (shown on the left in the figure below) and louder recordings (shown on the right).

However, if you turn up the volume control, you will reach an output level that exceeds the capability of the Beolab 12 for the loud song (but not for the quiet song), shown in the figure below. At this point, for the loud song, the Beolab 12 has already begun to protect itself.

Once a B&O loudspeaker starts protecting itself, no matter how much more you turn it up, it will turn itself down by the same amount; so it won’t get louder. If it did get louder, it would either distort the sound or stop working – or distort the sound and then stop working.

If you ONLY own Beolab 12s and you ONLY listen to loud songs (e.g. pop and rock) then you might ask “why should I be able to turn up the volume higher than this?”.

The first answer is “because you might also own Beolab 90s” which can go louder, as you can see in the right hand side of the figure above.

The second answer is that you might want to listen to quieter recording (like a violin solo or a podcast). In this case, you haven’t reached the maximum output of even the Beolab 12 yet, as you can see in the left hand side of the figure above. So, you should be able to increase the volume setting to make even the quiet recording reach the limits of the less-capable loudspeaker, as shown below.

Notice, however, that at this high volume setting, both the quiet recording and the loud recording have the same output level on the Beolab 12.

So, the volume allows you to push the output higher; either because you might also own more capable loudspeakers (maybe not today – but some day) OR because you’re playing a quiet recording and you want to hear it over the sound of the exhaust fan above your stove or the noise from your shower.

It’s also good to remember that the volume control isn’t an indicator of how loud the output should be. It’s an indicator of how much quieter or louder you’re making the input signal.

The volume control is more like how far down you’re pushing the accelerator in your car – not the indication of the speedometer. If you push down the accelerator 50% of the way, your actual speed is dependent on many things like what gear you’re in, whether you’re going uphill or downhill, and whether you’re towing a heavy trailer. Similarly Metallica at volume step 70 will be much louder than a solo violin recording at the same volume step, unless you are playing it through a loudspeaker that reached its maximum possible output at volume step 50, in which case the Metallica and the violin might be the same level.

Note 1: For all of the above, I’ve said “quiet song” and “loud song” or “quiet recording” and “loud recording” – but I could just have easily as said “quiet part of the song” and “loud part of the song”. The issue is not just related to mastering levels (the overall level of the recording) but the dynamic range (the “distance” between the quietest and the loudest moment of a recording).

Note 2: I’ve written a longer, more detailed explanation of this in Posting #81: Turn it down half-way.

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: Virtual loudspeakers

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

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. (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…)

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) pinnae (a.k.a. ‘ears’) must be both accurate and precise. (For the same reason that someone else should not try to wear my glasses.)

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.