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.
What caught my eye was the discussion of gramophone needles made of “hard wood”, and also the prediction that “the growth of electrical recording steps … to grapple with that problem of wear and tear.”
The fact that electrical (instead of mechanical) recording and playback was seen as a solution to “wear and tear” reminded me of my first textbook in Sound Recording where “Digital Audio” was introduced only within the chapter on Noise Reduction.
Later in that same issue, there is a little explanation of the “Electrocolor” and “Burmese” needles.
The March 1935 issue raises the point of wear vs. fidelity in the Editorial (which starts by comparing players with over-sized horns).
I like the comment about having to be in the “right mood” for Ravel. Some things never change.
What’s funny is that, now that I’ve seen this, I can’t NOT see it. There are advertisements for fibre, thorn, and wood needles all over the place in 1930s audio magazines.
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.
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:
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
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.
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…
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.
In a perfect sound system, all loudspeakers are identical, and they are all able to play a full frequency range at any listening level. However, most often, this is not a feasible option, either due to space or cost considerations (or both…). Luckily, it is possible to play some tricks to avoid having to install a large-scale sound system to listen to music or watch movies.
Humans have an amazing ability to localise sound sources. With your eyes closed, you are able to point towards the direction sounds are coming from with an incredible accuracy. However, this ability gets increasingly worse as we go lower in frequency, particularly in closed rooms.
In a sound system, we can use this inability to our advantage. Since you are unable to localise the point of origin of very low frequencies indoors, it should not matter where the loudspeaker that’s producing them is positioned in your listening room. Consequently, many simple systems remove the bass from the “main” loudspeakers and send them to a single large loudspeaker whose role it is to reproduce the bass for the entire system. This loudspeaker is called a “subwoofer”, since it is used to produce frequency bands below those played by the woofers in the main loudspeakers.
The process of removing the bass from the main channels and re-routing them to a subwoofer is called bass management.
It’s important to remember that, although many bass management systems assume the presence of at least one subwoofer, that output should not be confused with an (Low-Frequency Effects) LFE (Low-Frequency Effects) or a “.1” input channel. However, in most cases, the LFE channel from your media (for example, a Blu-ray disc or video streaming device) will be combined with the low-frequency output of the bass management system and the total result routed to the subwoofer. A simple example of this for a 5.1-channel system is shown below in Figure 1.
Figure 1: A simple example of a bass management system for a 5.1 channel system.
Of course, there are many other ways to do this. One simplification that’s usually used is to put a single Low Pass Filter (LPF) on the output of the summing buss after the signals are added together. That way, you only need to have the processing for one LPF instead of 5 or 6. On the other hand, you might not want to apply a LPF to an LFE input, so you may want to add the main channels, apply the LPF, and then add the LFE, for example. Other systems such as Bang & Olufsen televisions use a 2-channel bass management system so that you can have two subwoofers (or two larger main loudspeakers) and still maintain left/right differences in the low frequency content all the way out to the loudspeakers.
However, the one thing that older bass management systems have in common is that they typically route the low frequency content to a subset of the total number of loudspeakers. For example, a single subwoofer, or the two main front loudspeakers in a larger multichannel system.
In Bang & Olufsen televisions starting with the Beoplay V1 and going through to the Beovision Harmony, it is possible to change this behaviour in the setup menus, and to use the “Re-direction Level” to route the low frequency output to any of the loudspeakers in the current Speaker Group. So, for example, you could choose to send the bass to all loudspeakers instead of just one subwoofer.
There are advantages and disadvantages to doing this.
The first advantage is that, by sending the low frequency content to all loudspeakers, they all work together as a single “subwoofer”, and thus you might be able to get more total output from your entire system.
The second advantage is that, since the loudspeakers are (probably) placed in different locations around your listening room, then they can work together to better control the room’s resonances (a.k.a. room modes).
One possible disadvantage is that, if you have different loudspeakers in your system (say, for example, Beolab 3s, which have slave drivers, and Beolab 17s, which use a sealed cabinet design) then your loudspeakers may have different frequency-dependent phase responses. This can mean in some situations that, by sending the same signal to more loudspeakers, you get a lower total acoustic output in the room because the loudspeakers will cancel each other rather than adding together.
Another disadvantage is that different loudspeakers have different maximum output levels. So, although they may all have the same output level at a lower listening level, as you turn the volume up, that balance will change depending on the signal level (which is also dependent on frequency content). For example, if you own Beolab 17s (which are small-ish) and Beolab 50s (which are not) and if you’re listening to a battle scene with lots of explosions, at lower volume levels, the 17s can play as much bass as the 50s, but as you turn up the volume, the 17s reach their maximum limit and start protecting themselves long before the 50s do – so the balance of bass in the room changes.
Beosound Theatre
Beosound Theatre uses a new Bass Management system that is an optimised version of the one described above, with safeguards built-in to address the disadvantages. To start, the two low-frequency output channels from the bass management system are distributed to all loudspeakers in the system that are currently being used.
However, in order to ensure that the loudspeakers don’t cancel each other, the Beosound Theatre has Phase Compensation filters that are applied to each individual output channel (up to a maximum of 7 internal outputs and 16 external loudspeakers) to ensure that they work together instead against each other when reproducing the bass content. This is possible because we have measured the frequency-dependent phase responses of all B&O loudspeakers going as far back in time as the Beolab Penta, and created a custom filter for each model. The appropriate filters are chosen and applied to each individual outputs accordingly.
Secondly, we also know the maximum bass capability of each loudspeaker. Consequently, when you choose the loudspeakers in the setup menus of the Beosound Theatre, the appropriate Bass Management Re-direction levels are calculated to ensure that, for bass-heavy signals at high listening levels, all loudspeakers reach their maximum possible output simultaneously. This means that the overall balance of the entire system, both spatially and timbrally, does not change.
The total result is that, when you have external loudspeakers connected to the Beosound Theatre, you are ensured the maximum possible performance from your system, not only in terms of total output level, but also temporal control of your listening room.
There’s one last thing that I alluded to in a previous part of this series that now needs discussing before I wrap up the topic. Up to now, we’ve looked at how a filter behaves, both in time and magnitude vs. frequency. What we haven’t really dealt with is the question “why are you using a filter in the first place?”
Originally, equalisers were called that because they were used to equalise the high frequency levels that were lost on long-distance telephone transmissions. The kilometres of wire acted as a low-pass filter, and so a circuit had to be used to make the levels of the frequency bands equal again.
Nowadays we use filters and equalisers for all sorts of things – you can use them to add bass or treble because you like it. A loudspeaker developer can use them to correct linear response problems caused by the construction or visual design of the device. They can be used to compensate for the acoustical behaviour of a listening room. Or they can be used to compensate for things like hearing loss. These are just a few examples, but you’ll notice that three of the four of them are used as compensation – just like the original telephone equalisers.
Let’s focus on this application. You have an issue, and you want to fix it with a filter.
IF the problem that you’re trying to fix has a minimum phase characteristic, then a minimum phase filter (implemented either as an analogue circuit or in a DSP) can be used to “fix” the problem not only in the frequency domain – but also in the time domain. IF, however, you use a linear phase filter to fix a minimum phase problem, you might be able to take care of things on a magnitude vs. frequency analysis, but you will NOT fix the problem in the time domain.
This is why you need to know the time-domain behaviour of the problem to choose the correct filter to fix it.
For example, if you’re building a room compensation algorithm, you probably start by doing a measurement of the loudspeaker in a “reference” room / location / environment. This is your target.
You then take the loudspeaker to a different room and measure it again, and you can see the difference between the two.
In order to “undo” this difference with a filter (assuming that this is possible) one strategy is to start by analysing the difference in the two measurements by decomposing it into minimum phase and non-minimum phase components. You can then choose different filters for different tasks. A minimum phase filter can be used to compensate a resonance at a single frequency caused by a room mode. However, the cancellation at a frequency caused by a reflection is not minimum phase, so you can’t just use a filter to boost at that frequency. An octave-smoothed or 1/3-octave smoothed measurement done with pink noise might look like you fixed the problem – but you’ve probably screwed up the time domain.
Another, less intuitive example is when you’re building a loudspeaker, and you want to use a filter to fix a resonance that you can hear. It’s quite possible that the resonance (ringing in the time domain) is actually associated with a dip in the magnitude response (as we saw earlier). This means that, although intuition says “I can hear the resonant frequency sticking out, so I’ll put a dip there with a filter” – in order to correct it properly, you might need to boost it instead. The reason you can hear it is that it’s ringing in the time domain – not because it’s louder. So, a dip makes the problem less audible, but actually worse. In this case, you’re actually just attenuating the symptom, not fixing the problem – like taking an Asprin because you have a broken leg. Your leg is still broken, you just can’t feel it.
So far, we have looked at minimum phase and linear phase examples of one basic kind of filter, but everything I’ve shown you are just examples. There are other filters (I could be showing you shelving filters instead of peaking filters – which would be a through-put plus a low- or high-pass instead of a bandpass) and other implementations (for example, there are other ways to make a linear phase filter).
We won’t go through these other versions because we’re not here to learn how to make filters, we’re here to learn why, when someone asks you “which is better, minimum-phase or linear phase?” or “aren’t you worried about the unnatural effects of pre-ringing?” you can answer “it depends” (which is almost always the correct answer for any question related to audio).
At this point you should know that
a filter that changes the frequency response also changes the time response. This is unavoidable.
some filters will also ‘pre-ring’ ahead of the sound
a filter may or may not have an effect on the phase of the signal
If you’re designing (or choosing) a filter, nothing comes for free. (For example, if you want linear phase, the price is latency. There are other prices attached to other design decisions.)
The question that we have not yet addressed is “so what?”
This is the point in the discussion where things are going to get a little fuzzy… Hang on!
You probably read somewhere that human hearing extends from 20 Hz to 20 kHz, therefore anything outside that range is not audible. This is not true. You might have read a little farther where they added an extra detail saying something about your age: the older you are, the lower that top number gets. That might be true.
You can also read that the quietest sound you can hear has a sound pressure level of 20 µPa, which is equivalent to 0 dB SPL, and that the loudest sound you can “hear” (over the sound of your screams of pain) is about 120 dB SPL or so. You might have also read a little farther where they added an extra detail that points out that this is only at 1 kHz. But this is probably also not true.
There are many reasons why all of those numbers are basically meaningless – but the main one is that they’re numbers based on averages. Imagine if an optometrist only carried one strength of glasses, which happened to be the average of the prescriptions required by all of his or her patients. We’d all be tripping over out own feet, and getting blinding headaches caused by wearing the wrong glasses.
Actually, a similar point was once proven by the American Air Force. They wanted to design the perfect airplane cockpit, so they measured all their pilots, averaged all the numbers, and built a seat for the result. Of course, the result was that the seat fit no one, since there was no single person that matched the average.
There are other examples that prove that averages are useless information. For example, I have more than the average number of legs. Also, since one in every three mammals on the earth is a bat, if I have meeting with two other people, one of us must be Batman…
Of course, the message is that we’re all different. For example. I don’t taste things very well. I don’t understand people who talk about the various elements in the taste of wine. All I know is that it’s drinkable, or it’s good for putting on fish & chips. When I eat food, I tend to put on pepper and chill so that I can at least taste something…
Hearing is the same: so all of the stuff I’m about to say is based on averages, which may or may not apply to you. You might be more or less sensitive than the average. Don’t email me to argue about the numbers I’m using here. You’re different. I know… We all are…
Psychoacoustic Masking
Let’s go to an AC/DC concert together. Halfway through You Shook Me (All Night Long), I’ll whisper a secret code word into your left ear. Then, you whisper it back to me.
This exercise will not work. You won’t hear me, which is strange because the changes in air pressure that I’m making by whispering are exactly the same as when AC/DC is not playing – and normally, you’d be able to hear that. Also, your eardrum is wiggling in exactly the same way as a result of that whispering – it also happens to be wiggling to the AC/DC more. So, why can’t you hear me?
The answer lies in your brain. It decides that I’m not as important as AC/DC, so the signal is thrown out as being irrelevant, and so the sound my my whispering is ‘psychoacoustically masked’ by AC/DC. Notice the ‘psycho-‘ part of that, which is the indication that the acoustic masking is happening in your brain, not as a result of a mechanical or physical issue.
This probably doesn’t come as a surprise – at some point in your life, you’ve probably been somewhere where you’ve asked someone to speak up because you can’t hear them over the noise. You’ve been the victim of the limitations of your own brain.
Temporal Masking
What might come as a surprise is that this effect also works when the loud sound (AC/DC) and the quiet sound (me whispering) don’t happen simultaneously.
For example, let’s say that you are getting your photo taken by someone with a flash camera. You’re looking right at the camera, and the flash goes off. For a short while after that, you can’t see anything but the leftover spot in the middle of your vision. If, right after the flash, someone were to hold up some number of fingers and ask “how many fingers?”, you’d have to guess. (This is only an analogous example; the spot in your eyes is not happening in your brain, but the effect is similar.)
Similarly, if I were to fire a gun (not at you… don’t worry), and quickly whisper a word immediately afterwards, you wouldn’t hear the whisper. The gunshot and the whisper didn’t happen simultaneously, but there is temporal masking that causes you to not hear the quieter sound for a little which after the loud sound has happened.
Even weirder, there is an effect called pre-masking. If I were REALLY quick and VERY well-timed, I could whisper the word right BEFORE the gunshot and you also wouldn’t hear it. The loud sound (the gun shot) not only masks quiet sounds that come after it, but also sounds that come before it.
Fig 1. Generalised representation of temporal masking. The gray block is a loud noise. Anything in the pink area won’t be heard by most people most of the time. This plot is intentionally vague. The point is the effect, not the actual values.
So, if I were to play a loud noise (a gunshot, a blast of pink noise, a portion of an AC/DC song) and play a quiet sound before, during, or afterwards, and ask what the quiet sound was (to test if you heard it) your behaviour will match something like the plot in Figure 1. The gray block represents the loud sound. Anything in the pink shape is “stuff you can’t hear”. If the quiet sound occurs much earlier or much later than the loud sound, then it will have to be really quiet for you to not hear it. The closer in time the quiet sound occurs to the loud sound, the louder it has to be for you to hear it.
Of course, this is very general plot, so don’t use it for arguments while you’re drinking beer with your friends. For example, one element that I have not mentioned is frequency content. If the loud sound is the low frequency effects of a recording of distant thunder, and the quiet sound is a kitten mewing, then these two sounds are too far apart in frequency to have any influence on each other, and the graph is just plain wrong.
Signal Envelope
Unless you, like me, spend a lot of time listening to sinusoidal tones, you’ll notice that everything you listen to varies in level over time. This happens on different time scales. The sound pressure level (SPL) in the car while you’re driving to work or at the daycare picking up the kids is much louder than the SPL in your bedroom while you’re dealing with the free-floating existential anxiety that comes to visit at 2:00 in the morning. This is the ‘slow’ time scale. On the ‘fast’ end of the time scale, you have the extreme, and short SPL cause by closing a car door, or the very short, but not very loud sound of a high heel impacting a hardwood or tiled floor. Speech is somewhere in between these two. There are short, spikes caused by “t-” and “k-” sounds, and long-ish portions produced by vowels.
Note that we’re not really talking about how loud or how quiet things are. We’re talking about the change in level from quieter to louder and back again.
If we plot that change over time, we have a view of the signal’s envelope. For example, a single note on a piano has a fast attack (from quieter to louder) and a slow decay (from louder to quieter). A car horn honked in an open area (not a city intersection, where there are buildings to reflect the sound) has almost identical attack and decay envelopes.
Fig 1. Three different representations of an audio signal in time.
Figure 1 shows an 8-second slice of a recording of the Alleluia Chorus from Handel’s “Messiah”, chosen because it’s easy for me to load that into Matlab using the “load handel” command, and I’m very lazy.
The top plot shows the signal in the way we’re used to looking at sound files. The x-axis is time, in ms. The Y-axis shows the linear value of each of the samples. Oddly, this is the way we normally look at sound files, but it represents how we hear sound very poorly, because we don’t hear amplitude linearly.
So, in the middle plot, I’ve taken the same data and, sample-by-sample, plotted each value on a decibel scale (using the equation DisplayOutput = 20*log10(abs(signal)). The absolute value is there because calculating the log of a negative number gives you strange results)
The third plot is the one we’re really interested in. That’s created by connecting the peaks in the middle plot, which results in a running plot of the signal’s level over time. This is its envelope. As you can see in that particular musical example, the attacks (the changes from quieter to louder) are steeper (and therefore faster) than the decays. This is not surprising. It’s hard to get an orchestra and choir to all stop instantaneously…
Fig 2. The same treatment to a different audio signal. In this case, it’s female speech recorded in an anechoic chamber.
Figure 2 shows the same three ways of plotting an audio signal, but in this case, the signal is female speech recorded in an anechoic chamber. Notice that, partly because it’s only one sound source and partly because there is no reverberation in the signal, the decays are almost as fast as the attacks. However, this is a very strange recording. Most people don’t listen to anything in an anechoic chamber, and we don’t typically go to the middle of a football field or a frozen lake to have a conversation.
Back to the “so what?”
Let’s assemble the three collections of information that we’ve been throwing around.
Firstly, focusing on filter response:
We know that filters can ring.
We also know that some filters can pre-ring.
We also know that, unless the Q is really high, that ringing decays pretty quickly.
Finally, we know that, if the frequency that’s ringing in the filter is not present in the signal, it won’t ring because there’s nothing there to ring. (Similarly if you turn up the low bass while listening to a solo piccolo recording, you won’t hear a difference because there’s no bass to boost.)
Secondly, we looked at our own response to quiet and loud signals in time
A loud sound will simultaneously mask a quiet sound that occurs at the same time (fancy-talk for “drown it out so I can’t hear it)
The loud sound will also post-mask a quiet sound that occurs after within a short time window (on the order of 100 – 200 ms)
The loud sound will also pre-mask a quiet sound that occurs before it within a short time window (on the order of 10 – 20 ms)
Finally, we looked at signal envelopes – the change in level of an audio signal over time
Sounds never start instantaneously. In order to do so, they would have to have an infinite frequency range measured at the receiver (e.g. your eardrum, which is most certainly band-limited as well). Even very fast attacks take milliseconds to ramp up.
Sounds in real life have longer decay times
I know, I know, you can name sounds that have very fast attacks (e.g. the pluck of a harpsichord plectrum on a high string, or a single xylophone bar struck in an anechoic environment) and very fast decays (ummm…. good luck finding something that decays more than 100 dB in less than 100 ms…)
The question is: if you have a filter that rings or pre-rings, and you apply it to a sound,
Is that ringing the thing that defines the envelope of the resulting sound? In other words, does the signal’s attack and/or decay envelope change significantly as a result of the time response of the filter?
Is that change in the envelope outside the limits of your ability to detect it due to pre-masking and post-masking?
There is no single answer to this question. If the audio signal is someone hitting a rim shot, and the filter is a peak filter with 12 dB of gain at 500 Hz with a Q of 100, then you’ll hear it ringing. In fact, it will sound like a rim shot with a sine wave generator. If the audio signal is a single bowed note on a ‘cello, and the filter is a dip filter with -3 dB of gain at 100 Hz with a Q of 0.707, then you won’t.
However, (for example) you CANNOT automatically jump to a conclusion that “pre-ringing sounds unnatural” because that starts with the assumption that you can hear it, and therefore it sounds “like” anything.
Whether or not you can hear the effects of a filter applied to an audio signal is dependent not only on the very specific characteristics of the filter, but its interaction with the signal. Change the filter OR change the signal, and the result will change.
This means that you CANNOT say things like “linear phase filters are better (or worse) than minimum phase filters” or “the pre-ringing of a linear phase filter sounds unnatural” because both of those statements start with the (possibly incorrect) assumption that you can hear the effect of the filter.
Now, don’t mis-interpret what I’ve said to mean “you can’t hear a filter ringing” – I didn’t say that. What I said was “just because a filter can be measured to be ringing doesn’t necessarily mean that you can hear it with a given audio signal”.
In this part, we’re going to do something a little weird.
We’ll start by making a minimum phase peaking filter that has Q of 2 and a gain of +6 dB. The response of that filter is shown in Figure 1.
Fig 1. A minimum phase implementation of a peaking filter with a centre frequency of 1 kHz, a gain of 6 dB and a Q of 2.
Almost everything in the plots above should look familiar. The one weird thing is that I’ve left out the data in the time response before T = 0 ms. That’s because the future doesn’t matter, since we’re talking about a non-causal filter, right?
What if I (for some reason) wanted to make a filter that had a desired magnitude response, but a flat phase response? Let’s say that you’re allergic to phase shifts or you belong to an ancient religious cult that thinks that phase shifts are against the order of nature. How would we create that filter?
One way to do it is to create a filter that has the same magnitude response as the one above, but with an opposite phase response so that the two cancel each other. But, how do we create a filter with the opposite phase response?
One trick for doing this is to ignore what I said earlier. Remember when I was being pedantic about how filters advance signals in phase BUT NOT TIME? What if you were to ignore that for a brief moment? Could we use that ignorance to an advantage?
For example, what would happen if we took the time response shown above and reversed it? All of the component frequencies are still there, at all the same relative amplitudes. So the Magnitude Response won’t change. But what happens to the phase response? The answer is this:
Fig 2. The same impulse response as shown in Figure 1, reversed in time.
Now we have a weird filter that can only have an output based on future inputs (therefore it’s non-causal, and therefore not minimum-phase), with the same magnitude response as the one shown in Figure 1. But check out that phase response. It’s the inverse of the first filter.
So reversing time has the effect of flipping the polarity of the phase (not the polarity of the signal!) without modifying the magnitude response. What would have been a 45º phase shift (earlier) becomes a -45º phase shift (later).
So, if I were (conceptually – not really…) to put these two filters in series, feeding the output of one into the input of the other then their total magnitude responses would add (so I’d get a 12 dB boost instead of a 6 dB boost at 1 kHz) and their phase responses would cancel each other out. The end result would look like this:
Fig 3. The combination of the filters in Figure 1 and Figure 2. Notice that it has a gain of 12 dB at 1 kHz (because 6 + 6 = 12)
By now, you’ve probably figured out that what we’re looking at here is a linear phase filter, since can be used to change the magnitude response of a signal without mucking up its phase response.
The catch with this way of implementing a linear phase filter is that it has to see into the future. Of course, in real life this is difficult, but there is a trick you can use to fake it.
If you look at the impulse response plot in Figure 1, you can see that the peak in the response is at Time = 0 ms. As you get later, moving away from that moment in time, the signal is quieter and quieter until it dies away to (almost) nothing. The problem is that ‘almost nothing’ is not the same as ‘nothing’. In fact, if we’re being pedantic, the ringing keeps going forever, which is why it’s called a filter with an Infinite Impulse Response – an IIR filter.
This also means that the same is true for the filter in Figure 2, but the ringing extends infinitely backwards in time.
However, our resolution in measuring and storing the amplitude is not infinite – when the signal is quiet enough, we run out of resolution (‘ticks’ on the ruler) – and when the signal gets that quiet, it effectively becomes the same as nothing.
So, if we decide on the level where ‘almost nothing’ is the same as ‘nothing’ (this is more-or-less up to us if we’re the ones designing the filter), then we can look at the filter’s response and decide when that signal level is reached (both forwards and backwards in time).
For example, with the extremely limited resolution of the plots that I’ve made above, we can decide that ‘nothing’ is what’s left when you get ±10 ms from the impulse peak. Of course, the resolution of the pixels on that plot is not even close to the resolution of the audio signal, so ‘nothing’ on our plot and ‘nothing’ for the audio signal are two different things – but the concept is the same.
Back to the Future
Let’s pretend for a moment that, for the impulse response in Figure 3, we decide that ±10 ms is the window of time we need to get to ‘nothing’. This means that we could implement this filter by sending audio into it, letting it pre-ring with an increasing amount until it hits the maximum peak 10 ms after the sound comes into it, then ringing for another 10 ms until it’s out.
This means that the latency (the delay time between the input and the output of the filter) would appear to be 10 ms. Yes, when you feed in a signal, you immediately start getting something out of the filter, but the output is loudest after the signal has been feeding into the filter for 10 ms.
In other words, if you want to have a linear-phase filter that’s implemented using this method, you are going to have to accept that the cost will be latency. In our case (a filter with a Q of 2, with a centre frequency of 1 kHz, and the decisions we’ve made here) this results in a 10 ms latency. Change a parameter, and you change the latency. For example, as we’ve already seen, the lower the centre frequency, the longer in time the filter will ring. So, if we change the centre frequency of this filter to 100 Hz, then it will have 10 times the latency (because 100 Hz is 1/10 of 1 kHz – therefore the periodicity of the ringing is 10x longer). If we increase the Q, then the ringing lasts longer – both forwards and backwards and time.
Generally speaking, this means that, if you’re building a linear phase filter, you need to decide what characteristics the filter has, and then you need to start making decisions about the quality of the filter (e.g. is the response exactly what you want, or is just almost what you want?) and its latency.
Although I’m not going to talk about implementation much, the latency has two primary considerations. The first is obvious: are you willing to wait for the output? If you’re building a filter for a PA system, then you don’t want the output delayed so much that it sounds like an echo. However, the second issue is one for the person building the filter: latency means memory. This was a bigger problem in the ‘old days’ when memory was expensive, but it’s still an issue.
Why? The problem is that the latency is defined by the frequency and Q of the filter, which define the total time the filter takes to get through the entire impulse response. However, the filter’s processing doesn’t think of time in milliseconds, it thinks in samples. If you double the sampling rate, then you double the amount of memory you need to implement the filter.
So, going from 48 kHz to 192 kHz requires 4x the memory.
A little perspective
One thing to notice in the three figures above is the vertical scale of the time response plots. You can see there that the range is ±0.1, which is a zoomed-in view of the amplitude. This may give you a distorted impression of the level of the ringing relative to the signal (the impulse). Figure 4 should clear this up.
Fig 4. Three plots showing exactly the same data in three different ways.
The top plot in Figure 4 is identical to the top plot in Figure 3. The middle plot is also on a linear amplitude scale, but the range of the plot shows the entire range of the signal. You can see there that the impulse at T = 0 hits a value of 1, and so the pre- and post-ringing isn’t as loud as you might have thought based on the first three figures.
The bottom plot shows the same information, plotted on a dB scale instead. The impulse at T = 0 ms hits 0 dB. Relative to that, the pre- and post-ringing has a maximum value of about -24 dB. You can also see there that the ringing is at least 100 dB down by the time we’re 6 ms away from the main impulse, looking both backwards and forwards in time.
Of course, all of these characteristics would be different for filters with different parameters. The point here is to understand the general characteristics, not the specifics.
Additional Comment
You may read on various fora someone who claims that there’s no such thing as “pre-ringing”. “It’s ‘ripple'” They’ll claim.
This is incorrect.
Pre-ringing and ringing are behaviours that occur in the time domain.
Ripple is a wiggle in the response in the frequency domain. (Say, for example, you zoom into the magnitude response, it won’t be completely flat in some cases – and if it’s not, you have ripple.)
