# Mixing closed and ported cabinets: Part 6

As I showed in Part 5, the phase response of a loudspeaker driver in a closed cabinet is different from one in a ported cabinet in the low frequency region because, the low frequency output of the ported system is actually coming from the port, not the driver.

If we take the phase response plots from the two systems shown in Part 5 and put them on the same graph, the result is Figure 1.

If we calculate the difference in these two plots by subtracting the blue curve from the red curve at each frequency then we can see that a ported cabinet is increasingly out of phase relative to a sealed cabinet as you go lower and lower in frequency. This difference is shown in Figure 2.

Now, don’t look at that graph and say “but you never get to 180º so what’s the problem?” All of the plots I’ve shown in this series are for one specific driver in one specific enclosure, with and without a port of one specific diameter and length. I could have been more careful and designed two different enclosures (with and without a port) that does get to 180º (or something else up to 180º).

In other words: “results may vary”. Every loudspeaker in every cabinet has some magnitude response and some phase response (these are directly related to each other), and they’ll all be different by different amounts. (This is also the reason why I’m neglecting to talk about the fact that, as you go lower in frequency, the ported loudspeaker also drops faster in output level, so even if it were a full 180º out of phase, it would cancel less and less when combined with the sealed cabinet loudspeaker.)

The point of all of this was to show that, if you take two different loudspeakers with two different enclosure types, you get two different phase responses, particularly in the low frequency region.

This means that if you take those two loudspeaker types (the original question that inspired this series was specifically about mixing Beolab 9, Beolab 20, and Beolab 2 in a system where all of those loudspeakers are “helping” to produce the bass) and play identical signals from them in the same room, it’s not only possible, but highly likely that they will wind up cancelling each other. This results in LESS bass instead of MORE, ignoring all other effects like loudspeaker placement, room modes, and so on.

## But Beolab 2 has slave drivers, not ports…

Take a look at Figure 3. I’ve shown a conceptual drawing of a ported loudspeaker (showing the mass of the air in the port as a red rectangle) on the left and a loudspeaker with a slave driver (on the bottom – notice it’s missing a former and voice coil, and the diaphragm is thicker to make it heavy) on the right.

This should make it intuitively obvious that a ported loudspeaker and an enclosure with a slave driver are effectively identical. This raises the question of why you would do one rather than the other.

The advantages of using a port instead of a slave driver is that a port will be more “stable” on a production line (since all of the ports on all the loudspeakers you make will be identical in size) and they’ve very cheap to make. The disadvantage of a port is that if the velocity of the air moving in and out of it is too high, then you hear it “chuffing”, which is a noise caused by turbulence around the edges of the port. (If you blow across the top of a wine bottle, you don’t hear a perfect sine wave, you hear a very noisy “breathy” one. The noise is the chuffing.)

The advantage of a slave driver is that you don’t get any turbulence, and therefore no chuffing. A slave driver can also be heavier than the air in a port in a smaller space, so you can get the response of a large port in a smaller loudspeaker. There is a small disadvantage in the fact that there will be production line tolerance variations (but this is not really a big worry), and then there’s the price, which is much higher than a hole in a box.

This means that if you take anything I’ve said above about ported loudspeakers, and replace the word “port” with “slave driver” then it’s still true.

## P.S.

If you do have a surround system that not only has a bass management system, but is also capable of re-directing the bass to more loudspeakers than just your subwoofer (as is the case with all current Bang & Olfusen surround processors in the televisions), then all of this is important to remember. You can’t just send the bass to more loudspeakers and expect to get more output. You might get less.

This is true unless you have a Beosound Theatre. This is because the Theatre has an extra bit of processing in the signal path called “Phase Compensation” which applies an allpass filter to the outputs, compensating for the phase differences between loudspeakers in the low frequency region. So, in this one particular case, you should expect to get more output from more loudspeakers.

# Mixing closed and ported cabinets: Part 5

Let’s build a ported box and put a woofer in it. If we measure the magnitude responses of the individual outputs of the driver and the port as well as the total output of the entire loudspeaker, they might look like the three curves shown in Figure 1.

If you take a look at the curves at 1 kHz, you can see that the total output (the blue curve) is the same as the woofer’s output (the red curve) because the port’s output (the yellow curve) is so low that it’s not contributing anything.

As we come down in frequency, we see the output of the port coming up and the output of the driver coming down. At around 20 Hz, the port reaches its maximum output and the woofer reaches its minimum as a result. In fact that woofer’s output is about 15 dB lower than the port’s at that frequency.

As we go farther down in frequency, we can see that the woofer comes up and then starts to drop again, but the port just drops in level the lower we go.

Now look at the total output (the blue curve) from 20 Hz and down. Notice that the total output of the system from 20 Hz down to about 15 Hz is LOWER than the output of the port alone. As you go below about 15 Hz, you can see that the total output is lower than either the woofer or the port.

This means that the port and the woofer are cancelling each other, just like I described in the previous part in this series. This can be seen when we look at their respective phase responses, shown in the middle plot in Figure 2. I’ve also plotted the difference in the woofer and the port phase responses in the bottom plot.

Notice that, below 20 Hz, the woofer and the port are about 180º apart. So, as the woofer moves out of the enclosure, the air in the port moves inwards, and the total sum is less than either of the two individual outputs.

What happens when you put a woofer in a sealed enclosure instead of one with a port? The responses from this kind of system are shown below in Figure 3.

The first thing that you’ll notice in the plots in Figure 3 is that there is only one curve in each graph. This is because the total output is the driver output.

You’ll also notice in the top plot that a woofer in a cabinet acts as a second-order high-pass filter because the cabinet is not too small for the driver. If the cabinet were smaller, then you’d see a peak in the response, but let’s say that I’m not that dumb…

Because it’s a second-order high-pass filter, it has a phase response that approaches 180º as you go down in frequency.

Now, compare that phase response in the low end of Figure 3 to the phase response of the low end in Figure 2. This is where we’re headed, since the purpose of all of this discussion is to talk about what happens when you have a system that combines sealed enclosures with ported ones. That brings us to Part 6.

# Mixing closed and ported cabinets: Part 4

In Part 1, I showed how a wine bottle behaves exactly like a mass on a spring where the mass is the cylinder of air in the bottle’s neck and the spring is the air inside the bottle itself.

I also showed how a loudspeaker driver (like a woofer) in a closed box is the same thing, where the spring is the combination of the surround, the spider and the air in the box.

But what happens if the speaker enclosure is not sealed, but instead is open to the outside world through a “port” which is another way of saying “a tube”. Then, conceptually, you are combining the loudspeaker driver with the wine bottle like I’ve shown in Figure 3.

If I were to show this with all the masses in red and all the springs in blue, it would look like Figure 4.

Now things are getting a little complicated, so let’s take things slowly… literally.

If the loudspeaker driver in Figure 4 moves into the cabinet very slowly (say, you push it with your fingers or you play a very low frequency with an electrical signal), then the air that it displaces in of the bottle (the enclosure) will just push the plug of air out the bottle’s neck (the port). The opposite will happen if you pull the driver out of the enclosure: you’ll suck air into the port.

If, instead you move the driver back and forth very quickly (by playing a very high frequency) then the inertia of the air inside the cabinet (shown as the big blue spring in the middle) prevents it from moving down near the port. In fact, if the frequency is high enough, then the air at the entrance of the port doesn’t move at all. This means that, for very high frequencies, the system will behave exactly the same as if the enclosure were sealed.

But somewhere between the very low frequencies and the very high frequencies, there is a “magic” frequency where the air in the port resonates, and there, things don’t behave intuitively. At that frequency, whenever the driver is trying to move into the enclosure, the air in the port is also moving into the enclosure. And, although the air has less mass than the driver, it’s free to move more. The end result is that, at the port’s resonant frequency, the driver (in theory) doesn’t move at all*, and the air in the port is moving a lot.**

In other words, you can think of a single driver in a ported cabinet as being basically the same as a two-way loudspeaker, where the woofer (for example) is one driver and the port is the other “driver”.

• At high frequencies, the sound is only coming out of the woofer (for example).
• As you come down in frequency and get closer to the port’s resonance, you get less and less from the woofer and more and more from the port.
• At the port’s resonant frequency, all* of the sound is coming from the movement of the air in and out of the port
• As you go lower than the port’s resonant frequency, the woofer starts working again, but now as the woofer moves out of the enclosure (making a positive pressure) it sucks air into the port (making a negative pressure). So, at very low frequencies, the woofer is working very hard, but you get very little sound output because the port cancels it out.

If you look at this as a magnitude response (the correct term for “frequency response” for this discussion), you can think of the woofer having one response, the port having a different response, and the two adding together somehow to produce a total response for the entire loudspeaker.

However, as you can see from the short 4-point list above, something happens with the phase of the signal at different frequencies. This is most obvious in the “very low frequency” part, where the woofer’s and the port’s outputs are 180º out of phase with each other.

In Part 5 we’ll look at these different components of the total output separately, both in terms of magnitude and phase responses (which, combined are the frequency response).

* Okay okay…. I say “the driver (in theory) doesn’t move at all” and “all of the sound is coming from the movement of the air in and out of the port” which is a bit of an exaggeration. But it’s not MUCH of an exaggeration…

** This is an oversimplified explanation. The slightly less simplified version is that the air inside the cabinet is acting like a spring that’s getting squeezed from two sides: the driver and the air in the port. The driver “sees” the “spring” (the air in the box) as pushing and pulling on it just as much as its pulling and pushing, so it can’t move (very much…).

# Mixing closed and ported cabinets: Part 3

Before starting on this portion of the series, I’ll ask you to think about how little energy (or movement) it takes to get a resonant system oscillating. For example, if you have a child on a swing, a series of very gentle pushes at the right times can result in them swinging very high. Also, once the child is swinging back and forth, it takes a lot of effort to stop them quickly.

## Moving onwards…

So far, we’ve seen that a loudspeaker driver in a closed cabinet can be thought of as just a mass on a spring, and, as a result, it has some natural resonance where it will oscillate at some frequency.

The driver is normally moved by sending an electrical signal into its voice coil. This causes the coil to produce a magnetic field and, since it’s already sitting in the magnetic field of a permanent magnet, it moves. The surround and spider prevent it from moving sideways, so it can only move outwards (if we send electrical current in one direction) or inwards (if we send current in the other direction).

When you try to move the driver, you’re working against a number of things:

• the inertia of the mass of the moving parts
Pick up a heavy book, for example, and try to push and pull it back and forth. It’s hard work!
• the inertia of the air directly in front of and behind the driver
Pick up a big sheet of stiff plastic (like the thing you put on the floor under an office chair) and try to push it back and forth. It’s also hard work!
• the compliance (springiness) of the surround, spider, and air trapped in the cabinet behind the driver
Blow up a ballon, and use your two hands to squeeze it repeatedly. It’s also hard work!

These three things can be considered separately from each other as a static effect. In other words:

• It’s hard work to pick up a book or push a car that’s broken down (forget about pushing-and-pulling – just push OR pull)
• It’s hard work to run into a headwind with that big piece of stiff plastic
• It’s hard work to squeeze a balloon and keep it compressed

But, if you’re pushing AND pulling the loudspeaker driver there is another effect that’s dynamic.

When you’re moving the driver at a VERY low frequency, you’re mostly working against the “spring” which is probably quite easy to do. So, at a low frequency, the driver is pretty easy to move, and it’s moving so slowly that it doesn’t push back electrically. So, it does not impede the flow of current through the voice coil.

When you’re moving the driver at a VERY high frequency, you’re mostly working against the inertia of the moving parts and the adjacent air molecules. The higher the frequency, the harder it is to move the driver.

However, when you’re trying to moving the driver at exactly the resonant frequency of the driver, you don’t need much energy at all because it “wants” to move at that same rate. However, at that frequency, the voice coil is moving in the magnetic field of the permanent magnet, and it generates electricity that is trying to move current in the opposite direction of what your amp is going. In other words, at the driver’s resonant frequency, when you’re trying to push current into the voice coil, it generates a current that pushes back. When you try to pull current out of the voice coil, it generates a current that pulls back.

In other words, at the driver’s resonant frequency, your amplifier “sees” the driver as as a thing that is trying to impede the flow of electrical current. This means that you get a lot of movement with only a little electrical current; just like the child on the swing gets to go high with only a little effort – but only at one frequency.

This is a nice, simple case where you have a moving mass (the moving parts of the driver) and a spring (the surround, spider, and air in the sealed box). But what happens when the speaker has a port?

On to Part 4…

# Mixing closed and ported cabinets: Part 2

Let’s look at a typical moving coil loudspeaker driver like the woofer shown in Figure 1.

If I were to draw a cross-section of this and display it upside-down, it would look like Figure 2.

Typically, if we send a positive voltage/current signal to a driver (say, the attack of a kick drum to a woofer) then it moves “forwards” or “outwards” (from the cabinet, for example). It then returns to the rest position. If we send it a negative signal, then it moves “backwards” or “inwards”. This movement is shown in Figure 3.

Notice in Figure 3 that I left out all of the parts that don’t move: the basket, the magnet and the pole piece. That’s because those aren’t important for this discussion.

Also notice that I used only two colours: red for the moving parts that don’t move relative to each other (because they’re all glued together) and blue for the stretchy parts that act as a spring. These colours relate directly to the colours I used in Part 1, because they’re doing exactly the same thing. In other words, if you hold a woofer by the basket or magnet, and tap it, it will “bounce” up and down because it’s just a mass suspended by a spring. And, just like I talked about in Part 1, this means that it will oscillate at some frequency that’s determined by the relationship of the mass to the spring’s compliance (a fancy word for “springiness” or “stiffness” of a spring. The more compliant it is, the less stiff.) In other words, I’m trying to make it obvious that Figure 3, above is exactly the same as Figures 3 and 5 in Part 1.

However, it’s very rare to see a loudspeaker where the driver is suspended without an enclosure. Yes, there are some companies that do this, but that’s outside the limits of this discussion. So, what happens when we put a loudspeaker driver in a sealed cabinet? For the purposes of this discussion, all it means is that we add an extra spring attached to the moving parts.

I’ve shown the “spring” that the air provides as a blue coil attached to the back of the dust cap. Of course, this is not true; the air is pushing against all surfaces inside the loudspeaker. However, from the outside, if you were actually pushing on the front of the driver with your fingers, you would not be able to tell the difference.

This means that the spring that pushes or pulls the loudspeaker diaphragm back into position is some combination of the surround (typically made of rubber nowadays), the spider (which might be made of different things…) and the air in the sealed cabinet. Those three springs are in parallel, so if you make one REALLY stiff (or lower its compliance) then it becomes the important spring, and the other two make less of a difference.

So, if you make the cabinet too small, then you have less air inside it, and it becomes the predominant spring, making the surround and spider irrelevant. The bigger the cabinet, the more significant a role the surround and spider play in the oscillation of the system.

Sidebar: If you are planning on making a lot of loudspeakers on a production line, then you can use this to your advantage. Since there is some variation in the compliance of the surround and spider from driver-to-driver, then your loudspeakers will behave differently. However, if you make the cabinet small, then it becomes the most important spring in the system, and you get loudspeakers that are more like each other because their volumes are all the same.

Remember from part 1 that if you increase the stiffness of the spring, then the resonant frequency of the oscillation will increase. It will also ring for longer in time. In practical terms, if you put a woofer in a big sealed cabinet and tap it, it will sound like a short “thump”. But if the cabinet is too small, then it will sound like a higher-pitched and longer-ringing “bonnnnnnnggggg”.

So far, we’ve only been talking about physical things: masses and springs. In the next part, we’ll connect the loudspeaker driver to an amplifier and try to push and pull it with electrical signals.

# 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”.

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:

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.

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.

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.

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.

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.

# Burmese Colour Needles

Once-upon-a-time I posted a little thing about Burmese Colour Needles for grammophones. Then, today, I received an email from the owner of www.burmesecolourneedles.com a company owned by Andy Briggs, where you can buy brand new ones!

Andy also has a YouTube channel that is well worth visiting.

# Dynamic Styli Correlator Pt. 5

In the last posting, I showed a scale drawing of a 15 µm radius needle on a 1 kHz sine tone with a modulation velocity of 50 mm/s (peak) on the inside groove of a record. Looking at this, we could see that the maximum angular rotation of the contact point was about 13º away from vertical, so the total range of angular rotation of that point would be about 27º.

I also mentioned that, because vinyl is mastered so that the signal on the groove wall has a constant velocity from about 1 kHz and upwards, then that range will not change for that frequency band. Below 1 kHz, because the mastering is typically ensuring a constant amplitude on the groove wall, then the range decreases with frequency.

We can do the math to find out exactly what the angular rotation the contact point is for a given modulation velocity and groove speed.

Looking at Figure 1, the rotation is ±13.4º away from vertical on the maximum; so the total range is 26.8º. We convert this to a time modulation by converting that angular range to a distance, and dividing by the groove speed at the location of the needle on the record.

If we repeat that procedure for a range of needle radii from 0 µm to 75 µm for the best-case (the outside groove) and the worst-case (the inside groove), we get the results shown in Figure 2.

# Dynamic Styli Correlator Pt. 4

Back in Part II of what is turning out to be a series of postings on this topic, I wrote

If this were a digital system instead of an analogue one, we would be describing this as ‘signal-dependent jitter’, since it is a time modulation that is dependent on the slope of the signal. So, when someone complains about jitter as being one of the problems with digital audio, you can remind them that vinyl also suffers from the same basic problem…

As I was walking the dog on another night, I got to thinking whether it would be possible to compare this time distortion to the jitter specifications of a digital audio device. In other words, is it possible to use the same numbers to express both time distortions? That question led me here…

Remember that the effect we’re talking about is caused by the fact that the point of contact between the playback needle and the surface of the vinyl is moving, depending on the radius of the needle’s curvature and the slope of the groove wall modulation. Unless you buy a contact line needle, then you’ll see that the radius of its curvature is specified in µm – typically something between about 5 µm and 15 µm, depending on the pickup.

Now let’s do some math. The information and equations for these calculations can be found here.

We’ll start with a record that is spinning at 33 1/3 RPM. This means that it makes 0.556 revolutions per second.

The Groove Speed relative to the needle is dependent on the rotation speed and the radius – the distance from the centre of the record to the position of the needle. On a 12″ LP, the groove speed at the outside groove where the record starts is 509.8 mm/sec. At the inside groove at the end of the record, it’s 210.6 mm/sec.

Let’s assume that the angular rotation of the contact point (shown in Figure 1) is 90º. This is not based on any sense of scale – I just picked a nice number.

We can convert that angular shift into a shift in distance on the surface of the vinyl by finding the distance between the two points on the surface, as shown below in Figure 2. Since you might want to choose an angular rotation that is not 90º, you can do this with the following equation:

2 * sin(AngularRotation / 2) * radius

So, for example, for a needle with a radius of 10 µm and a total angular rotation of 90º, the distance will be:

2 * sin(90/2) * 10 = 14.1 µm

We can then convert the “jitter” as a distance to a jitter in time by dividing it by the distance travelled by the needle each second – the groove speed in µm per second. Since that groove speed is dependent on where the needle is on the record, we’ll calculate it as best-case and a worst-case values: at the outside and the inside of the record.

Jitter Distance / Groove Speed = Jitter in time

For example, at the inside of the record where the jitter is worst (because the wavelength is shortest and therefore the maximum slope is highest), the groove speed is about 210.6 mm/sec or 210600 µm/sec.

Then the question is “what kind of jitter distance should we really expect?”

Looking at Figure 3 which shows a scale drawing of a 15 µm radius needle on a 1 kHz tone with a modulation velocity of 50 mm/s (peak) on the inside groove of a record, we can see that the angular rotation at the highest (negative) slope is about 13.4º. This makes the total range about 27º, and therefore the jitter distance is about 7.0 µm.

If we have a 27º angular rotation on a 15 µm radius needle, then the jitter will be

7.0 / 210600 = 0.0000332 or 33.2 µsec peak-to-peak

Of course, as the radius of the needle decreases, the angular rotation also decreases, and therefore the amount of “jitter” drops. When the radius = 0, then the jitter = 0.

It’s also important to note that the jitter will be less at the outside groove of the record, since the wavelength is longer, and therefore the slope of the groove is lower, which also reduces the angular rotation of the contact point.

Since the groove on records are typically equalised to ensure that you have a (roughly) constant velocity above 1 kHz and a constant amplitude below, then this means that the maximum slope of the signal and therefore the range of angular rotation of the contact point will be (roughly) the same from 1 kHz to 20 kHz. As the frequency of the signal descended from 1 kHz and downwards, the amplitude remains (roughly) the same, so the velocity decreases, and therefore the range of the angular rotation of the contact point does as well.

In other words, the amount of jitter is 0 at 0 Hz, and increases with frequency until about 1 kHz, then it remains the same up to 20 kHz.

As one final thing: as I was drawing Figure 3, I also did a scale drawing of a 20 kHz signal with the same 50 mm/s modulation velocity and the same 15 µm radius needle. It’s shown in Figure 4.

As you can see there, the needle’s 15 µm radius means that it can’t drop into the trough of the signal. So, that needle is far too big to play a CD-4 quad record (which can go all the way up to 45 kHz).

# Dynamic Styli Correlator Pt. 3

I thought that I was finished talking about (and even thinking about) the RCA Dynagroove Dynamic Styli Correlator as well as tracking and tracing distortion… and then I got an email about the last two postings pointing out that I didn’t mention two-channel stereo vinyl, and whether there was something to think about there.

My first reaction was: “There’s nothing interesting about that. It’s just two channels with the same problem, and since (at least in a hypothetical world) the two axes of movement of the needle are orthogonal, then it doesn’t matter. It’ll be the same problem in both channels. End of discussion.”

Then I took the dog out for a walk, and, as often happens when I’m walking the dog, I re-think thoughts and come home with the opposite opinion.

So, by the time I got home, I realised that there actually is something interesting about that after all.

Starting with Emil Berliner, record discs (original lacquer, then vinyl) have been cut so that the “mono” signal (when the two channels are identical) causes the needle to move laterally instead of vertically. This was originally (ostensibly) to isolate the needle’s movement from vibrations caused by footsteps (the reality is that it was probably a clever manoeuvring around Edison’s patent).

This meant that, when records started supporting two audio channels, a lateral movement was necessary to keep things backwards-compatible.

What does THIS mean? It means that, when the two channels have the same signal (say, on the lead vocal of a pop tune, for example) when the groove of the left wall goes up, the groove of the right wall goes down by the same amount. That causes the needle to move sideways, as shown below in Figure 1.

What are the implications of this on tracing distortion? Remember from the previous posting that the error in the movement of the needle is different on a positive slope (where the needle is moving upwards) than a negative slope (downwards). This can be seen in a one-channel representation in Figure 2.

Since the two groove walls have an opposite polarity when the audio signals are the same, then the resulting movement of the two channels with the same magnitude of error will look like Figure 3.

Notice that, because the two groove walls are moving in opposite polarity (in other words, one is going up while the other is going down) this causes the two error signals to shift by 1/2 of a period.

However, Figure 3 doesn’t show the audio’s electrical signals. It shows the physical movement of the needle. In order to show the audio signals, we have to flip the polarity of one of the two channels (which, in a real pickup would be done electrically). That means that the audio signals will look like Figure 4.

Notice in Figure 4 that the original signals are identical (that’s why it looks like there’s only one sine wave) but their actual outputs are different because their error components are different.

But here’s the cool thing:

One way to think of the actual output signals is to consider each one as the sum of the original signal and the error signal. Since (for a mono signal like a lead vocal) their original signals are identical, then, if you sit in the right place with a properly configured pair of loudspeakers (or a decent pair of headphones) then you’ll hear that part of the signal as a phantom image in the middle. However, since the error signals are NOT correlated, they will not be localised in the middle with the voice. They’ll move to the sides. They’re not negatively correlated, so they won’t sound “phase-y” but they’re not correlated either, so they won’t be in the same place as the original signal.

So, although the distortion exists (albeit not NEARLY on the scale that I’ve drawn here…) it could be argued that the problem is attenuated by the fact that you’ll localise it in a different place than the signal.

Of course, if the signal is only in one channel (like Aretha Franklin’s backup singers in “Chain of Fools” for example) then this localisation difference will not help. Sorry.