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)

and

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.

Resolution, Part 1: White Noise

This is the start of what will be a series of posts that are an attempt to answer a question about the pro’s and con’s of implementing a volume control in the digital domain. When I first thought about how to answer this question, I thought I could do it in a couple of sentences – but the more I thought about it, the more I realised that the answer is complicated…

There’s no doubt in my mind that I’m making this answer more complicated than necessary, but, as Carl Sagan once said, “If you wish to make apple pie from scratch, you must first create the universe.”

So, to begin, we have to define what “noise” is from the point of view of audio engineering.

On the one hand, we can define it simply. “Noise” is a random signal. We can be more accurate and say that this means that the amplitude of a noise signal cannot be predicted using a knowledge of what has come before in time.

If I flip a coin, it will be either heads or tails. I can’t predict this. It will be random. If I flip it 100 times, and, by some strange coincidence, I get 100 “tails”, there is still a 50% chance of getting a tails on the 101st flip. What has happened before can, in no way, be used to predict what is about to happen.

Of course, what is about to happen on the 101st flip has a limited number of possible outcomes. I cannot flip the coin and get “dog” as a result… (this sounds silly, but it will come in handy later…) Just like I cannot roll two dice and get a 13…

In LPCM digital audio, a noise signal is one where each individual sample in the signal has a random value that is in no way related to any of the previous samples. Its range (the set of possible values from which we can pick our random number) may be limited (depending on the specific characteristics of the noise signal and what may have come before), but it will be random.

Typically, when you are talking to someone in audio about noise, they describe it using a colour as the first descriptor. So, you’ll hear of “white noise” and “pink noise”, as the two most popular examples. For the purposes of this series of postings, we’ll only be talking about white noise. So, what is this?

One definition that you’ll see thrown around a lot says something like “white noise is a random signal that has equal energy per linear bandwidth” or “… equal energy per hertz” or “…equal intensity at different frequencies” or something like this. These descriptions are sort of true if you don’t want to get into temporal details, which, unfortunately, is exactly where we’re headed…

The good thing about those definitions is that they describe a general characteristic of white noise. If you take a white noise signal, and you measure the intensity of (or the energy in) the signal for a given bandwidth (say, a bandwidth of 100 Hz ranging from 200 Hz to 300 Hz) then it will be the same in another frequency range with the same bandwidth (say, a bandwidth of 100 Hz ranging from 1,000 Hz to 1,100 Hz). Note that these two bandwidths are the same in hertz – not in a multiplier like octaves or semitones or decades. So, if you have white noise that has a total bandwidth of 0 Hz to 20,000 Hz, then you will have the same amount of energy in the 0 – to – 10,000 Hz band as you will in the 10,000 – to – 20,000 Hz band. In other words (to us humans), there is as much energy in the top octave of the signal as in the rest of the bandwidth combined.

This is why white noise sounds like “bright” and “hissy” (similar to the “ss” sound in “hissy”) and not “darker” like the “sh” sound in “ash” (as they incorrectly claim here…). Since white is a “bright” colour, then we use the word “white” to describe the frequency-dependent energy distribution of “white” noise.

However, this is not really true. The truth is that a white noise signal has an equal probability per bandwidth of having the same energy level. This little detail is usually left out, partly because it’s complicated, and partly because it doesn’t matter in most cases in the real world. However, in our case, it does.

Let’s look at an example. I made a white noise signal in Matlab using the statement
rand(SignalLength, 1) – rand(SignalLength, 1)
where SignalLength is the length of the noise signal in samples, and the 1 means that I’m doing this for 1 audio channel…. mono is so retro…

You may be wondering why I did a
rand() – rand()
instead of just a
rand().
the simple answer for now was that I wanted to make the signal “balanced” on either side of the zero line and the rand() function in Matlab has a range of 0 to 1.

I know… I could have done this by saying
2 * (rand(SignalLength, 1) – 0.5)
but there is another reason that we’ll get into later…)

I then used a DFT to find the magnitude response of this signal. The result – both the signal and its magnitude response – are shown below in Figure 1.

Figure 1: A random signal shown in the time domain (top plot) and its magnitude response (bottom plot).

Some additional information that is really not important: The sampling rate of this signal is 2^16 (65,536 Hz), and I did a 2^16 point DFT, so I have one frequency bin per hertz. (If this last bit of information is confusing, but interesting, please start reading this…)

You may notice that the magnitude is “flat” – meaning that it generally doesn’t slope upwards or downwards. However, you will also notice that it is certainly not “flat” – meaning that it is not a perfectly straight line. In fact, if we zoom in on both plots, we can see Figure 2.

Figure 2: A portion of Figure 1, zoomed in to show some details.

Notice that we do NOT have an equal amount of energy per hertz… if we did, then the bottom plot would be a flat line.

If I do all of that again – make a new noise sample the same way (with a new set of random numbers) and plot the result, and a zoomed in version, I get Figures 3 and 4.

Figure 3: The result of running the same code that generated Figure 1. However, this is a new set of random numbers. Notice that, on first glance, it is the same as Figure 1, but if you look carefully, it is completely different.
Figure 4: A detail from Figure 3. Notice that, on first glance, it is the same as Figure 2, but if you look carefully, it is completely different.

Compare Figures 1 and 3 or Figure 2 and 4. You’ll notice that they have similar characteristics overall – but not only are they NOT identical, they are completely different (on a sample-by-sample or a DFT bin-by-bin comparison).

Let’s say that I run this code and generate a white noise signal 1 second long, and I then calculate the magnitude response of that noise signal and store it. Then, I’ll repeat this, and average the new magnitude response to the first one. Then, I’ll do it again, each time, including the magnitude response to the average of all of the magnitude responses that I’ve done….

For each 1-second slice of time, the noise signal does not have equal energy per bandwidth – however, it is certainly white noise.

This is because, each time I do this, the average magnitude response will get flatter and flatter… and eventually, after doing this an infinite number of times, it will be a flat line.

This means that, white noise will have an equal amount of energy per bandwidth only if I wait long enough. The question is “how long is long enough?” The answer to that question depends on what you’re doing with it.

Another way to look at this…

In the each of the examples above, I made 1 second-long white noise signals and used the entire signal – all 65,536 samples – to calculate the magnitude response.

What happens if I have a one-second long signal, but only a portion of it is a burst of white noise, and the rest is silence? For example, look at Figure 5.

Figure 5: A 1-second long signal that contains a burst of white noise for 0.5 seconds.

Figure 5’s magnitude response looks similar to the ones we’ve seen before (apart from being a little lower overall than the plots in Figures 1 and 3 – because there’s less energy overall in 0.5 sec of noise than there is in 1 second of noise). I’ll keep going to show what happens if we take this to an extreme.

Figure 6. 1/8-second of noise in a 1-second signal
Figure 7. A detail from Figure 6. Notice how smooth the magnitude response has become…

The magnitude response shown in Figure 7 looks very different from the ones we’ve seen before. It’s much smoother… We’ll keep going…

Figure 8. 16 samples of white noise in the middle of a 65,536 sample long signal of silence. Notice that, even when not “zoomed in”, the magnitude response is smooth

Figure 8 is very different again… The total magnitude response, even when not “zoomed in” is smooth. It’s important here to note that the actual response that we see there will be different every time I run the random generator again. For example, look at Figure 9, which is also a 16-sample long white noise signal.

Figure 9. 16 samples of white noise in the middle of a 65,536 sample long signal of silence. Notice that, even when not “zoomed in”, the magnitude response is smooth – but it is not identical to the one in Figure 8 because the random numbers are not the same.

If we keep getting shorter and shorter, eventually we’ll get down to a single sample with a random value. However, since it’s a single sample (that is very probably non-zero) in a long string of zeros, then its magnitude response will be completely flat. It will not be noise – it will be an impulse with a random level. And it won’t sound like noise – it will sound like a click.

Summary

There are two basic important things to know at this point.

  • White noise has the frequency content you expect only if you average over time.
  • The shorter the time the noise is present, the less energy you will have, overall.

The discussion continues in Part 2.

P.S.

Thanks to David for emailing and pointing out that it’s “Hz” and “hertz” but not “Hertz”. I’ve corrected the text above… Being reminded of this reminds me of a Steven Wright joke – “I’m having amnesia and déjà vu at the same time. I think I’ve forgotten this before…”