#1: You have to drive to a meet someone at a specific time. Let’s say that you only have to drive on one road to get there, and the speed limit is the same the whole way. You calculate the time it will take to get there on time, and you start driving – but there’s traffic. So, you wind up driving half the distance at half the speed, then the traffic disappears.
How fast do you have to go the rest of the way to arrive at the meeting on time?
#2: You’re driving on a two-lane highway where the speed limit is 70 km/h. You are driving 100 km/h, and you pull into the left lane to pass someone who is driving the speed limit. Everything about the car you’re passing is identical to yours – even the driver weighs the same as you do. At the instant that you are side-by side, a train appears across the road in front of you and stops. You both hit the brakes at exactly the same time to try and stop from hitting the train.
Luckily, the person in the other car stops just as his bumper touches the train, let’s say 1 mm before touching it… But, because you were driving faster, you cannot stop in time.
How fast are you going when you hit the train?
#1. Most people instinctively say “double the speed limit” to make up the lost time. However, this is not the right answer.
Let’s say that the meeting is 100 km away, and the speed limit is 100 km/h. Therefore, it should take you 1 hour to get to the meeting.
If you drive half the distance (50 km) at half the speed (50 km/h), then at the moment the traffic clears up, you should have been at your destination. So, you would have to drive infinity km/h to get there. However, since teleportation doesn’t exist yet, you might as well just call and tell them you’ll be late.
#2: This one is a little tougher, but it should be pretty intuitive for someone working in audio. A car’s brakes work by taking the energy in the car’s momentum, and converting that to heat in the brake discs. The key word there is energy.
So, the question is: if you consider the amount of energy removed from the car going 70 km/h, and take that out of the energy in the car going 100 km/h, how much energy is left?
The answer is 70 km/h. For someone in audio, this might look like a familiar answer, since 0.7 V has half of the power of 1.0 V (assuming identical loads). In the case of the cars, it’s because the amount of power (the amount of energy that’s transmitted over time – in this case, to heat the brakes) to bring the car from 70 km/h to 0 km/h is identical to the amount of power it takes to bring the same car from 100 km/h to 70 km/h. (An audio geek might joke that 70 km/h is 3 dB slower than 100 km/h.)
Slow down. You’re not going to make it to the meeting anyway, and driving a little bit faster means you’re going to hit the train much harder than you think.
These days, I’m spending a lot of time wrapping my head around the relationship between the frequency and the time responses of filters. In doing so, I’m digging into the concept of “Q”, of course. As a result, I’m reading my old books and some Internet sites, and I’m frequently presented with something like the following:
That, of course, is from the Wikipedia entry on “Q”.
However, in the Bell Telephone System Technical Publication – Monograph 2491, called “The Story of Q” by Estill I. Green ( published in the American Scientist, Vol 43, pp 584-594, in October 1955), it states:
“For a time, Johnson* designated the ratio of reactance to effective resistance of a coil by the symbol K. It was in 1920, while working the practical application of the wave filter which G. A. Campbell had invented some years before, that he for the first time employed the symbol Q for his parameter. His reason for choosing Q was quite simple. He says that it did not stand for ‘quality factor’ or anything else, but since the other letters of the alphabet had already been pre-empted for other purposes, Q was all he had left.”
So, if we’re going to be pedantic (which I love to be) there are two errors on that Wikipedia page. Firstly, Q does not stand for Quality. Secondly, it’s not the “Q factor”, it’s just the “Q”.
As an aside, that monograph is not only informative, it’s fun to read (depending, of course, on your definition of “fun”). For example, near the end of the paper, Green applies Q to rotating bodies (which is not a surprise, since an audio-wave oscillation is just a rotation represented in two dimensions). In that section, he points out that the rotation of the earth is slowing down due, in part, to tidal friction. Consequently, the length of a day is increasing at a rate of 0.00164 second per century, which would make the Q of the rotation of the earth equal to about 10,000,000,000,000 (10^13).
* K.S. Johnson worked in the Western Electric Company’s Engineering Department, which became Bell Telephone Laboratories in 1925.
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…
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…
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.