BeoLab + 3rd party source with a high-level output

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

This week, I was asked a very specific question about connecting an older pair of BeoLab loudspeakers to a stereo preamp from another company. Specifically, the owner was wondering why the pairing wasn’t working out too well – and he had already had a theory that the problem had something to do with the sensitivity of his BeoLab 9’s.

To be honest, I don’t really know what the problem is with this specific customer’s system – but I made a guess and I figured that the answer might be useful to someone else…

For starters, let’s do some sensitivity training. More accurately, let’s talk about loudspeaker sensitivity. This is a measure of how loud the acoustical output of a loudspeaker is for a given electrical input. Since BeoLab loudspeakers are active (meaning, in part, that the amplifiers are built-in) this means that we are talking about an output level in dB SPL for a given input in volts.

For most BeoLab loudspeakers, you will get an output of 88 dB SPL for an input of 125 mV RMS if you measure the loudspeaker on-axis in a free field. There are some exceptions to this, most notably BeoLab 1, 9, and 5, which will produce 91 dB SPL instead.

So, this tells us how loud the loudspeaker will be for a given input. But my guess is that this had nothing to do with the customer’s problem.

Most customers connect their BeoLab loudspeakers to a Bang & Olufsen source using something called a “Power Link” connection. This is a little bundle of wires that contains two audio channels (probably left and right) as well as a data channel (telling the loudspeaker things like the volume setting, for example) and a 5 V DC on/off signal.

Power Link is specified to have a maximum level of 6.5 V RMS, assuming that the signal is a sine wave. This means that a device with a Power Link output can produce no more than 9.2 V Peak. It also means that a device with a Power Link input (like a BeoLab loudspeaker) will clip (and therefore distort) at its input if you feed it with more than 9.2 V Peak.

(If you do some math, you can calculate that 20*log10(6.5 V RMS / 125 mV RMS) = 34.3 dB. Therefore, if a BeoLab 9 loudspeaker will produce 91 dB SPL with a 125 mV RMS input, then it should produce 91+34.3 dB = 125.3 dB SPL for a maximum accepted input of 6.5 V RMS. Of course, this is not possible – but it’s because the loudspeaker is limited by its drivers, amplifiers, and power supply – not the input maximum input level.)

Back to the question: The customer in question mentioned his stereo preamp’s brand and model number. A little Duck-Duck-Go-ing helped me to find the manual for that particular device, and in the back of that document, I found out that the maximum output level of the preamp was 29 V RMS – which is a lot…

So, the problem is very likely that his preamp is overloading the input stages of the BeoLab 9. So, if he turns the volume knob on the preamp up to maximum, and he’s playing a tune that is mastered to be loud on the playback media, then the BeoLab 9’s input will be clipped. Changing the sensitivity of the loudspeaker could make it quieter – but it will still be clipped… So the distortion won’t get better – everything will just get quieter.

There are some different solutions to this problem. The easiest one is to not turn up the volume on the preamp – but this is not the best solution, because it means that he’s not using the full dynamic range of the preamp (probably), and therefore that the noise from the preamp is higher in level than it needs to be at the input of the BeoLab 9.

There is, however, a very cheap and simple solution, and that is to attenuate the output of the preamp so that when it is set to its maximum output level, it is just hitting the maximum input level of the loudspeaker’s input.

How do we do this? the first question is to find out what the attenuation should be.

Maximum output level = 29.0 V RMS

Maximum input level = 6.5 V RMS

20*log10 (6.5 / 29.0) = -12.99 dB

This is the same as a linear gain of 0.2241.

Now we’re going to build a voltage divider. This is device made of two resistors, placed in series (end-to-end) and connected to the output of the source. The point where the two resistors connect together is used as the output to the loudspeaker, resulting in a schematic as shown below.

As you can probably see in the schematic, the grounds of the two devices (which are connected to the exterior casings of the RCA Phono plugs) are connected together. As the voltage on the pin of the source goes up and down, the voltage on the pin of the loudspeaker also goes up and down – but by less. How much less is determined by the values of the resistors.

For example, if the resistors are equal (R1 = R2) then the output will be half of the input. If R2 is one tenth of the total of R1+R2, then the output will be one tenth of the input. You can calculate this gain yourself with a simple equation:

Linear Gain = R2 / (R1 + R2)


Gain in dB = 20 * log10 (Linear Gain)

So, for example, if R1 = 8,000 Ω and R2 = 2,000 Ω, then the gain will be

2000 / (8000 + 2000) = 0.2

which is equal to 20*log10 (0.2) -13.98 dB.

Unfortunately, if you want to do this with only two resistors, you can’t be too choosy about their resistances. There are standard resistor values, and you’ll have to pick from that list.

Also, it’s a good “rule of thumb” to try and keep the resistance “seen” by the source around 10,000 Ω (or 10 kΩ) – just to keep it happy. If you make the value too low, then you will be asking for it to deliver too much current (and its maximum output level will drop). If you make it too high, you might create and antenna and result in some extra noise…

So, I want to make R1 +R2 about 10,000 Ω, and I want R2 / (R1+R2) to be about 0.2241 (because I’m trying to convert 29 V RMS to 6.5 V RMS). So, I go to a list of standard resistor values like this one and I start trying to simultaneously fulfill those two requirements.

After some trial and error, I find out that if I make R1 = 8.2 kΩ and R2 = 2.4 kΩ, I can come pretty close.

2400 / (8200 + 2400) = 0.2264 = -12.902

close enough. Now I just need to get a soldering iron and a bit of wire, and put it all together…

The details…

However, if you clicked on that list of standard resistor values, you might notice that it says ±5% at the top of the table. This is normal. If you go to your local resistor store and you buy a 1 kΩ resistor – it probably won’t be exactly 1,000.000000000 Ω. But it will be close. If you buy from the ±5% stack, then any resistor in that bunch will be within 5% of the stated value. So, for a 1 kΩ resistor ±5%, it will be somewhere between 950 Ω and 1050 Ω.

So, then the question is, for the resistors that I just picked, how bad can it get, and is that good enough?

Well… this can be calculated. I just put in the worst-case values for my two resistors into the math, and do it over and over until I get all the possible answers. This would look like this:

If we look at this in terms of how far away we are from the target – the gain error, then it looks like this:

So, if we randomly choose resistors out of the bag, the worst that can happen is that we will be 0.6 dB below the target or 0.8 dB above the target.

This means that, if we’re not careful, and we’re unlucky, then we can get a mismatch between the two channels of 1.4 dB (assuming that one channel was a worst-case low and the other is a worst-case high). This is enough to be audible as about a 15% shift towards the louder loudspeaker – which is probably not acceptable.

So, the moral of the story is that you should measure your resistors before soldering them into the circuit.

Note, however, that it’s not necessary to make the gains perfect to improve the imaging. You just need to make them equal in the two channels for that…

Speaking Passively

The circuit I show above is called a “passive” circuit. This means that it doesn’t require any external power source (like a battery or a power supply) to work. However, it also means that it can’t make things louder – no matter what resistor values you choose, the output will always be less than the input.

There are lots of reasons why this is a useful little circuit. It’s cheap, it’s easy to make, it’s small (you could hide it inside one of the RCA connectors), and it will prevent you from overloading the input of the downstream device (in this case, a loudspeaker). Not only that, but it will also attenuate the noise generated by the source – so not only will the customer’s system no longer clip, it will also (probably) have a lower noise floor.

DFT’s Part 2: It’s a little complex…

Links to:
DFT’s Part 1: Some introductory basics

Whole Numbers and Integers

Once upon a time you learned how to count. You were probably taught to count your fingers… 1, 2, 3, 4 and so on. Although no one told you so at the time, you were being taught a set of numbers called whole numbers.

Sometime after that, you were probably taught that there’s one number that gets tacked on before the ones you already knew – the number 0.

A little later, sometime after you learned about money and the fact that we don’t have enough, you were taught negative numbers… -1, -2, -3 and so on. These are the numbers that are less than 0.

That collection of numbers is called integers – all “countable” numbers that are negative, zero and positive. So the collection is typically written

… -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 …

Rational Numbers

Eventually, after you learned about counting and numbers, you were taught how to divide (the mathematical word for “sharing equally”). When someone said “20 divided by 5 equals 4” then they meant “if you have 20 sticks, then you could put those sticks in 4 piles with 5 sticks in each pile.” Eventually, you learned that the division of one number by another can be written as a fraction like 3/1 or 20/5 or 5/4 or 1/3.

If you do that division the old-fashioned way, you get numbers like this:

3 1 = 3.000000000 etc…

20 5 = 4.00000000 etc…

5 4 = 1.200000000 etc…

1 3 = 0.333333333 etc…

The thing that I’m trying to point out here is that eventually, these numbers start repeating sometime after the decimal point. These numbers are called rational numbers.

Irrational Numbers

What happens if you have a number that doesn’t start repeating, no matter how many numbers you have? Take a number like the square root of 2 for example. This is a number that, when you multiply it by itself, results in the number 2. This number is approximately 1.4142. But, if we multiply 1.4142 by 1.4142, we get 1.99996164 – so 1.4142 isn’t exactly the square root of 2. In fact, if we started calculating the exact square root of 2, we’d result in a number that keeps going forever after the decimal place and never repeats. Numbers like this (π is another one…) that never repeat after the decimal are called irrational numbers

Real Numbers

All of these number types – rational numbers (which includes integers) and irrational numbers fall under the general heading of real numbers. The fact that these are called “real” implies immediately that there is a classification of numbers that are “unreal” – but we’ll get to that later…

Imaginary Numbers

Let’s think about the idea of a square root. The square root of a number is another number which, when multiplied by itself is the first number. For example, 3 is the square root of 9 because 3*3 = 9. Let’s consider this a little further: a positive number muliplied by itself is a positive number (for example, 4*4 = 16… 4 is positive and 16 is also positive). A negative number multiplied by itself is also positive (i.e. 4*-4 = 16).

Now, in the first case, the square root of 16 is 4 because 4*4 = 16. (Some people would be really picky and they’ll tell you that 16 has two roots: 4 and -4. Those people are slightly geeky, but technically correct.) There’s just one small snag – what if you were asked for the square root of a negative number? There is no such thing as a number which, when multiplied by itself results in a negative number. So asking for the square root of -16 doesn’t make sense. In fact, if you try to do this on your calculator, it’ll probably tell you that it gets an error instead of producing an answer.

For a long time, mathematicians just called the square root of a negative number “imaginary” since it didn’t exist – like an imaginary friend that you had when you were 2… However, mathematicians as a general rule don’t like loose ends – they aren’t the type of people who leave things lying around… and having something as simple as the square root of a negative number lying around unanswered got on their nerves.

Then, in 1797, a Norwegian surveyor named Casper Wessel presented a paper to the Royal Academy of Denmark that described a new idea of his. He started by taking a number line that contains all the real numbers like this:

Figure 1: The number line containing all real numbers.

He then pointed out that multiplying a number by -1 was the same as rotating by an angle of 180º, like this:

Figure 2: Multiplying a number by -1 is the same as rotating by 180º

He then reasoned that, if this were true, then the square root of -1 must be the same as rotating by 90º.

Figure 3: This means that the square root of -1 must be the same a rotating by 90º.

This meant that the number line we started with containing the real numbers is the X-axis on a 2-dimensional plane where the Y-axis contains the imaginary numbers. That plane is called the Z plane, where any point (which we’ll call ‘Z’) is the combination of a real number (X) and an imaginary number (Y).

Figure 4: The Z-plane, where X-values are real and the Y-values are imaginary.

If you look carefully at Figure 4, you’ll see that I used a “j” to indicate the imaginary portion of the number. Generally speaking, mathematicians use and physicists and engineers use so we’ll stick with j. (The reason physics and engineering people use j is that they use i to mean “electrical current”.)

“What is j?” I hear you cry. Well, is just the square root of -1. Of course, there is no number that is the square root of -1

and therefore

Now, remember that = -1. This is useful for any square root of any negative number, you just calculate the square root of the number pretending that it was positive, and then stick a after it. So, since the square root of 16, abbreviated sqrt(16) = 4 and sqrt(-1) = j, then sqrt(-16) = 4j.

Complex numbers

Now that we have real and imaginary numbers, we can combine them to create a complex number. Remember that you can’t just mix real numbers with imaginary ones – you keep them separate most of the time, so you see numbers like


This is an example of a complex number that contains a real component (the 3) and an imaginary component (the 2j). In some cases, these numbers are further abbreviated with a single Greek character, like α or β, so you’ll see things like

α = 3+2j

In other cases, you’ll see a bold letter like the following:

Z = 3+2j

A lot of people do this because they like reserving Greek letters like α and ϕ for variables associated with angles.

Personally, I like seeing the whole thing – the real and the imaginary components – no reducing them to single Greek letters (they’re for angles!) or bold letters.

Absolute Value (aka the Modulus)

The absolute value of a complex number is a little weirder than what we usually think of as an absolute value. In order to understand this, we have to look at complex numbers a little differently:

Remember that j*j = -1.

Also, remember that, if we have a cosine wave and we delay it by 90º and then delay it by another 90º, it’s the same as inverting the polarity of the cosine, in other words, multiplying the cosine by -1. So, we can think of the imaginary component of a complex number as being a real number that’s been rotated by 90º, we can picture it as is shown in the figure below.

Figure 5. The relationship bewteen the real and imaginary components for the number (2 + 3 j). Notice that the X and Y axes have been labeled the “real” and “imaginary” axes.

Notice that Figure 5 actually winds up showing three things. It shows the real component along the x-axis, the imaginary component along the y-axis, and the absolute value or modulus of the complex number as the hypotenuse of the triangle. This is shown in mathematical notation in exactly the same way as in normal math – with vertical lines. For example, the modulus of 2+3j is written |2+3j|

This should make the calculation for determining the modulus of the complex number almost obvious. Since it’s the length of the hypotenuse of the right triangle formed by the real and imaginary components, and since we already know the Pythagorean theorem then the modulus of the complex number (a + b j) is

Given the values of the real and imaginary components, we can also calculate the angle of the hypotenuse from horizontal using the equation

This will come in handy later.

Complex notation or… Who cares?

This is probably the most important question for us. Imaginary numbers are great for mathematicians who like wrapping up loose ends that are incurred when a student asks “what’s the square root of -1?” but what use are complex numbers for people in audio? Well, it turns out that they’re used all the time, by the people doing analog electronics as well as the people working on digital signal processing. We’ll get into how they apply to each specific field in a little more detail once we know what we’re talking about, but let’s do a little right now to get a taste.

In the previous posting, that introduces the trigonometric functions sine and cosine, we looked at how both functions are just one-dimensional representations of a two-dimensional rotation of a wheel. Essentially, the cosine is the horizontal displacement of a point on the wheel as it rotates. The sine is the vertical displacement of the same point at the same time. Also, if we know either one of these two components, we know:

  1. the diameter of the wheel and
  2. how fast it’s rotating

but we need to know both components to know the direction of rotation.

At any given moment in time, if we froze the wheel, we’d have some contribution of these two components – a cosine component and a sine component for a given angle of rotation. Since these two components are effectively identical functions that are 90º apart (for example, a cossine wave is the same as a sine that’s been delayed by 90º) and since we’re thinking of the real and imaginary components in a complex number as being 90º apart, then we can use complex math to describe the contributions of the sine and cosine components to a signal.


Let’s look at an example. If the signal we wanted to look at a signal that consisted only of a cosine wave, then we’d know that the signal had 100cosine and 0sine. So, if we express the cosine component as the real component and the sine as the imaginary, then what we have is:

1 + 0 j

If the signal were an upside-down cosine, then the complex notation for it would be (1 + 0 jbecause it would essentially be a cosine * -1 and no sine component. Similarly, if the signal was a sine wave, it would be notated as (0 1 j).

This last statement should raise at least one eyebrow… Why is the complex notation for a positive sine wave (0 1 j)? In other words, why is there a negative sign there to represent a positive sine component? (Hint – we want the wheel to turn clockwise… and clocks turn clockwise to maintain backwards compatibility with an earlier technology – the sundial. So, we use a negative number because of the direction of rotation of the earth…)

This is fine, but what if the signal looks like a sinusoidal wave that’s been delayed a little? As we saw in the previous posting, we can create a sinusoid of any delay by adding the cosine and sine components with appropriate gains applied to each.

So, is we made a signal that were 70.7sine and 70.7cosine. (If you don’t know how I arrived that those numbers, check out the previous posting.) How would you express this using complex notation? Well, you just look at the relative contributions of the two components as before:

0.707 – 0.707 j

It’s interesting to notice that, although this is actually a combination of a cosine and a sine with a specific ratio of amplitudes (in this case, both at 0.707 of “normal”), the result will look like a sine wave that’s been shifted in phase by -45º (or a cosine that’s been phase-shifted by 45º). In fact, this is the case – any phase-shifted sine wave can be expressed as the combination of its sine and cosine components with a specific amplitude relationship.

Therefore (again), any sinusoidal waveform with any phase can be simplified and expressed as its two elemental components, the gains applied to the cosine (or real) and the sine (or imaginary). Once the signal is broken into these two constituent components, it cannot be further simplified.

Link to DFT’s Part 3: The Math

Fc ≠ Fc

I was working on the sound design of a loudspeaker last week with some new people and software – so we had to get some definitions straight before we messed things up by thinking that we were using the same words to mean the same thing. I’ve made a similar mistake to this before, as I’ve written about here – and I don’t being reminded of my own stupidity repeatedly… (Or, as Stephen Wright once said “I’m having amnesia and deja vu at the same time – I think I’ve forgotten this before…”)

So, in this case on that day, we were talking about the lowly 2nd-order Low Pass Filter, based on a single biquad.

If you read about how to find the cutoff frequency of a low-pass filter, you’ll probably find out that you find the frequency where the gain is one half of the power of that in the bandpass portion of the filter’s response. Since 10*log10(0.5) = -3.01 dB, then this is also called the “3 dB down point” of the filter.

In my case, when I’m implementing a filter, I use the math provided by Robert Bristow-Johnson to calculate my biquad coefficients. You input a cutoff frequency (Fc), and a Q value, and (for a given sampling rate) you get your biquad coefficients.

The question then, is: is the desired cutoff frequency the actual measurable cutoff frequency of the system? (Let’s assume for the purposes of this discussion that there are no other components in the system that affect the magnitude response – just to keep it simple.)

The simple answer is: No.

For example, if I make a 2nd-order low pass filter with a desired cutoff frequency of 1 kHz (using a high enough sampling rate to not introduce any errors due to the bilinear transform) and I vary the Q from something very small (in this example, 0.1) to something pretty big (in this example, 20) I get magnitude response curves that look like the figure below.

Magnitude responses of 2nd order low pass filters with Q’s ranging from 0.1 to 20.

It is probably already evident that these 25 filter responses plotted above that they do not all cross each other at the 1 kHz line. In addition, you may notice that there is only one of those curves that is -3.01 dB at 1 kHz – when the Q = 1/sqrt(2) or 0.707.

This begs the question: what is the gain of each of those filters at the desired value of Fc (in this case, 1 kHz)? This is plotted as the red line in the figure below.

The actual gain value of the filters at the desired Fc, and the maximum gain at any frequency.

This plot also shows the maximum gain of the filters for different values of Q. Notice that, in the low end, the maximum value is 0 dB, since the low pass filters only roll off. However, for Q values higher than 1/sqrt(2), there is an overshoot in the response, resulting in a boost at some frequency. As the Q increases, the frequency at which the gain of the filter is highest approaches the desired cutoff frequency. (As can be seen in the plot above, by the time you get to a Q of 20, the gain at Fc and the maximum gain of the filter are the same.)

It may be intuitively interesting (or interestingly intuitive) to note that, when Q goes to infinity, the gain at Fc also goes to infinity, and (relatively speaking) all other frequencies are infinitely attenuated – so you have a sine wave generator.

So, we know that the gain value at the stated Fc is not -3 dB for all but one value of Q. So, what is the -3 dB point, if we state a desired Fc of 1 kHz and we vary the Q? This is shown in the figure below.

The -3 dB point of a 2nd order 1 kHz low pass filter as a function of Q.

So, varying the Q from 0.1 to 20 varies the actual Fc (or, at least, the -3 dB point) from about 104 Hz to about 1554 Hz.

Or, if we plot the same information as a function (or just a multiple) of the desired Fc, you get the plot below.

So, if you’re sitting in a meeting, and the person in front of you is looking at a measurement of a loudspeaker magnitude response, and they say “could you please put in a low pass filter with a cutoff frequency of 1 kHz and a Q of 0.5” you should start asking questions by what, exactly, they mean by “cutoff frequency”… If not, you might just wind up with nice-looking numbers but strangely-sounding loudspeakers.