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.
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.
Figure 2
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.
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.
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.
Figure 1
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.
Figure 2
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.
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.
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…).
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?
Let’s look at a typical moving coil loudspeaker driver like the woofer shown in Figure 1.
Figure 1.
If I were to draw a cross-section of this and display it upside-down, it would look like Figure 2.
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.
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.
Figure 4
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.
I made a comment on a forum this week, commenting that, if you mix loudspeakers with closed cabinets with loudspeakers with ported cabinets (or slave drivers), the end result can be a reduction in total output: less sound from more loudspeakers. Of course, the question is “why?” and the short answer is “due to the phase mismatching of the loudspeakers”.
This is the long answer.
Before we begin, we have to get an intuitive understanding of what a ported loudspeaker is. (Note that I’ll keep saying “ported loudspeaker”, but the principle also applies to loudspeakers with slave drivers, as I’ll explain later.) Before we get to a ported loudspeaker, we need to talk about Helmholtz resonators.
Take a block that’s reasonably heavy and hang it using a spring so that it looks like this:
Figure 1.
The spring is a little stretched because the weight of the block (which is the result of its mass and the Earth’s gravity) is pulling downwards. (We’ll ignore the fact that the spring is also holding up its own weight. Let’s keep this simple…) However, it doesn’t fall to the floor because the spring is pulling upwards.
Now pull downwards on the block, so it will look like the example on the right in the figure below.
Figure 2.
The spring is stretched because we’re pulling down on the block. The spring is also pulling upwards more, since it’s pulling against the weight of the block PLUS the force that you’re adding in a downwards direction.
Now you let go of the block. What happens?
The spring is pulling “too hard” on the block, so the block starts rising back to where it started (we’ll call that the “resting position”). However, when it gets there, it has inertia (a body in motion tends to stay in motion… until it hits something big…) so it doesn’t stop. As a result, it moves upwards, higher than the resting position. This squeezes the spring until it gets to some point, at which time the block stops, and then starts going back downwards. When it returns to the resting position, it still has inertia, so it passes that point and goes too far down again. I’ve shown this as a series of positions from left to right in the figure below.
Figure 3
If there were no friction, no air around the block, and no friction within the metal molecules of the spring, then this would bounce up and down forever.
However, there is friction, so some of the movement (“kinetic energy”) is turned into heat and lost. So, each bounce gets smaller and smaller and the maximum velocity of the block (as it passes the resting position) gets lower and lower, until, eventually, it stops moving (at the resting position, where it started).
Notice that I changed the colour of the spring to show when it’s more stretched (lighter blue) and when it’s more compressed (darker blue).
If everything were behaving perfectly, then the RATE at which the bounce repeats wouldn’t change. Only its amplitude (or the excursion of the block, or the height of the bounce) would reduce over time. That bounce rate (let’s say 1 bounce per second, and by “bounce” I mean a full cycle of moment down, up, and back down to where it started again) is the frequency of the repetition (or oscillation).
If you make the weight lighter, then it will bounce faster (because the spring can pull the weight more “easily”). If you make the spring stiffer, then it will bounce faster (because the spring can pull the weight more “easily”). So, we can change the frequency of the oscillation by changing the weight of the block or the stiffness of the spring.
Now take a look at the same weight on a spring next to an up-side down wine bottle that (sadly) has been emptied of wine.
Figure 4.
Notice that I’ve added some colours to the air inside the bottle. The air in the bottle itself is blue, just like the colour of the spring. This is because, if we pull air out of the bottle (downwards), the air inside it will pull back (upwards; just like the metal spring pulling back upwards on the block). I’ve made the small cylinder of air in the neck of the bottle red, just like the block. This is because that air has some mass, and it’s free to move upwards (into the bottle) and downwards (out of the bottle) just like the block.
If I were somehow able to pull the “plug” of air out of the neck of the bottle, the air inside would try to pull it back in. If I then “let go”, the plug would move inwards, go too far (because it also has inertia), squeezing (or compressing) the air inside the bottle, which would then push the plug back out. This is shown in the figure below.
Figure 5.
At the level we’re dealing with, this behaviour is practically identical to the behaviour of the block on the spring. In other words, although the block and the plug are made of different materials, and although the metal spring and the air inside the bottle are different materials, Figures 3 and 5 show the same behaviour of the same kind of system.
How do you pull the plug of air out of the bottle? It’s probably easier to start by pushing it inwards instead, by blowing across the top.
When you do this, a little air leaks into the opening, pushing the plug inwards. The “spring” in the bottle then pushes the plug outwards, and your cycle has started. If you wanted to do the same thing with the block, you’d lift it and let go to start the oscillation.
However, you don’t need to blow across the bottle to make it oscillate. You can just tap it with the palm of your hand, for example. Or, if you put the bottle next to your ear and listen carefully, you’ll hear a note “singing along” with the sound in the room. This is because the air in the bottle resonates; it moves back and forth very easily at the frequency that’s determined by the mass of the air in the neck and the volume of air in of the bottle (the spring).
However, remember that friction can make the oscillation decay (or die away) faster, by turning the movement into heat.
One last thing…
There’s another way to get either the block or the wine bottle oscillating:
You can move the TOP of the spring (for example, if you pull it up, then the spring will pull the block upwards, and it’ll start bouncing). Or, you could tap the bottom of the wine bottle (which is on the top in my drawings).
This method of starting the oscillation will come in handy in part 2.
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.
Figure 1: 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. These two points are the two extremes of the angular rotation of the contact point.
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.
Figure 2. The peak-to-peak equivalent “jitter” values of the inside and outside grooves for a range of needle radii.
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.
Figure 1. Artists rendition of the range of the point of contact between the surface of the vinyl and the pickup needle.
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
Figure 2. The angular range from Figure 1 converted to a linear distance on the vinyl’s surface.
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?”
Figure 3. Scale drawing of a needle on a record.
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.
Figure 4. Scale drawing of a needle on a record.
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).
After writing the previous posting, I couldn’t stop thinking about it. Mostly, I wanted to get a better idea of the shape of the waveform that results from the difference in a groove cut with a stylus and a spherically-tipped needle on a turntable pickup. To be perfectly honest, I’m not even interested in a ‘real’ simulation. I just wanted to get an intuitive idea of what’s happening down at that nearly-microscopic level. So, I used Matlab to draw some pictures.
Let’s take one period of a sine wave cut into the vinyl master with a chisel-shaped stylus:
Figure 1: The black line shows the wave cut into the vinyl surface. The grey shapes are “artist’s renditions” of the chisel-shaped stylus that cut it.
In theory, the pickup needle tracks this vertical movement exactly, as shown in Figure 2.
Figure 2. The black line is the original signal. The Red line is the signal tracked by a needle that has the same shape as the cutting stylus.
However, we already know that the pickup needle is NOT the same shape as the cutting stylus. In 1964, the needle would have had a spherical tip, which I’ve shown in Figure 3 as a series of semicircles (certainly NOT to scale…).
Figure 3: The black line is the original signal. The grey semicircles are the outline of a spherically-tipped pickup needle. The small grey circles are the centres of those semicircles. When you connect those circles, you get the red line.
In Figure 3, I’ve connected the centres of the semicircles to make the red line. However, you may notice that this line is not directly above the black line because of the interaction between the slope of the original signal and the radius of the ‘sphere’ that I’m showing. This might be easier to see in Figure 4 which is the same as Figure 3, but I’ve ‘connected the dots’.
Figure 4. This is the same as Figure 3, but I’ve shown the radii of the ‘spheres’ connecting the centre to the surface where it’s touching the vinyl.
(1) One interesting thing about the figure above is that it shows that the point where the needle is resting on the vinyl surface isn’t always vertical – it’s 90º from the tangent of the groove wall (assuming a spherically-ground needle). This means that the output of the needle (which, we assume is determined only by its vertical movement) is actually sliding forwards and backwards in time on the recording, depending on whether the slope is positive or negative.
For example, if you look at the far left of Figure 4, you can see that the centre of the needle is to the left of the point where it’s touching the vinyl. If this is drawn so that the vinyl is moving from right to left (or the needle is moving from left to right – so it’s drawn from the perspective of someone looking in from the edge of the record) then this means that the output of the system is looking ahead in time.
When the needle drops back downwards, it’s delaying the signal, looking back in time.
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…
(2) Another interesting thing is that, if we subtract the original signal on the vinyl’s surface from the actual path traced by the needle, we can see the tracing error itself. This is shown below as the red curve in Figure 5.
Figure 5. The original signal is in grey, the movement of the needle is in blue, and the difference (the tracing error) is in red.
Notice that, although the original signal is symmetrical, the blue curve (the actual signal) is not. This means that it has a DC offset, which is easily seen in the error curve in red, which never drops below the vertical 0 line; the mean of the original signal.
(Remember, I’m exaggerating everything here just to get an intuitive understanding of what’s going on.)
Although I’ve done all of this analysis numerically using Matlab, I’ve also found a paper that describes this error analytically. It’s “Integrated Treatment of Tracing and Tracking Error” by Duane H. Cooper in the Journal of the Audio Engineering Society from January, 1964. In that paper, he shows the following drawing shown below in Figure 6. Compare the dotted line to the blue one above, for example. (It seems that I wasted my time doing math when I should have been reading old papers instead…)
Figure 6.
The horizontal distance in Figure 6 between the bold capital ‘X’ and the small ‘x’ is an angular rotation from the centre of the needle’s spherical tip and therefore a time shift in the playback of the recording. Later in the same paper, Cooper proposes an analogue computer that can predict this distortion by modulating a delay applied to the audio signal as a function of the signal itself. A representation of this from the paper is shown below in Figure 7. This prediction can then be used to generate the pre-emphasis distortion of RCA’s “Dynamic Styli Correlator”.
Figure 7.
(3) The last thing that I’ve found is an extreme case that should never happen in real life, but it might. This is when the trough that the needle is dropping into is narrower than the diameter of the stylus. When this happens, the point where the stylus is touching jumps instantaneously from one side of the trough to the other. This is shown in Figure 8.
Figure 8. When the stylus is too big to fit into the trough, parts of the waveform are skipped.
This is the same thing that happens when a tire of your car drops into a bad pothole. You roll off one edge of the hole, and hit the edge on the opposite side, but the part of the tire that is actually IN the pothole never actually touches the bottom.
This problem is the same as I described above; but instead of the output signal merely sliding in time, it’s jumping. One example I can think where this would happen in real life is when you play a CD-4 quad LP with a needle that isn’t made for it. However, in this case you won’t notice the problem, since the high-frequency FM modulated surround channels result in a more-or-less constant “ripple” on the groove wall. This means that your needle is just surfing along the tops of the ripples and never drops into a trough at all.
Many audio recording systems are based on a concept known as “pre-emphasis” and “de-emphasis”. This is a process where a signal is distorted (here, I use the word “distorted” to mean “changed”, not “clipped”) at the recording or encoding process to counter-act the effects of something that will happen at playback. One example of this is a RIAA equalisation that applies an overall bass-heavy tilt to the frequency response at playback, and therefore the signal is given the opposite tilt when it’s cut onto the vinyl master. Dolby noise reduction for analogue magnetic tape follows a similar philosophy.
Another type of intentional distortion applied to an audio signal is based on assumptions of what happens at playback. Mixing engineers for television often emphasise lower frequency bands, assuming that everyone’s television loudspeakers needed some help. Pop and rock recording engineers check the mix on a low-quality mono loudspeaker and may make adjustments to the mix – to make sure it survived a clock radio or a portable Bluetooth loudspeaker (depending on which decade we’re talking about). Stereo vinyl records can’t have big low-frequency differences in the two audio channels otherwise the needle will hop out of the groove, so they’re mixed and mastered accordingly.
I’ve been reading “The RCA Victor Dynagroove System”, by Harry F. Olsen, published in the April 1964 issue of the Journal of the Audio Engineering Society. In it, he describes the entire recoding chain, including something that piqued my interest called a “Dynamic Styli Correlator” which is a distortion that is applied to the audio at almost the last stage of the signal path before it reaches the cutter head of the lathe that creates the lacquer master. You can see it here in Figure 1 from the article (I drew the red box around it).
Cool name; almost worthy of Dr. Heinz Doofenshmirtz (although it’s missing the “-inator”). But what is it?
One of the problems with playing back a vinyl record is that the shape of the needle on your turntable is not the same shape as the cutting stylus on the lathe. Consequently, the path that the needle tracks is not exactly the same as the path of the stylus. The result of this mis-match is that the electrical input signal that is used to make the master (the original recording) is not the same as the electrical output signal that comes out of your turntable (what you hear).
The idea behind the Dynamic Styli Correlator was that the actual path of the playback needle could be predicted, and the groove cut by the stylus could be modified to ensure that the output was correct. In other words, the distortion caused by the playback needle was estimated, and a distorted groove was cut to make the needle behave. This is shown graphically in Figure 29 of the article:
This is a great idea if the system works and if the prediction of the playback needle’s path is correctly predicted. However, neither of these two assumptions is guaranteed; so a number of things can go wrong here, and if anything can go wrong, it probably will.
However, it does mean at least as a start, that if you play an old RCA Victor Dynagroove record with a stylus shape that wasn’t invented yet in 1964 (say, a contact line stylus made for CD-4 Quadraphonic records, for example). Then you might wind up doing a much better job of reproducing the distortion that RCA created in the first place, instead of what they thought you were supposed to hear.
Back when I was at McGill, one of my fellow Ph.D. students was Mark Ballora, who did his doctorate in converting heart rate data to an audible signal that helped doctors to easily diagnose patients suffering from sleep apnea.
