Double Bass Four-letter word beginning with "B"

[seamonkey]

I wouldn't mind seeing all speaker makers publish sensitivity figures, like EA does, but I'm still gonna play a cabinet or an amp before I buy it, so does it really mattter?

I see this point a lot. Yes - you should always try a cabinet. You need to know what the specs actually mean to your ears.

But - You can't possibly try a cabinet in a store, or in 30 days, and duplicate all the venues you may typically play in. You can take the specs and compare it to your actual playing experience. You'll know what you typically like the next time you buy a cabinet.

Another example, Also say you need a louder cabinet, you have a 200w amp. You know your cabinet at a SPL of 110 at 150W, so you know to look for cabinets that has a higher SPL at 150W.

BTW - I have a Behringer ADI 21 preamp and the plastic case is very rugged - but I haven't stress tested it. Time will tell. Molded plastic is everywhere now days so I expect they've worked out some of the kinks.

 

[drurb]

All good points. I will give a technical reason why I care about THD. Maybe this only applies to solid state amps. An amp reaches "hard" clipping at a certain power level. Going beyond hard clipping, the distortion goes up, and also the RMS output power. Driving the amp from just below clipping to 1% THD does not increase the output power very much, so a 1% THD spec represents the actual hard clipping limit of the amp.

I think you are confusing the transfer of energy to higher frequencies (as a result of clipping) with an increase in RMS power.

 

[WalterBush]

I think you are confusing the transfer of energy to higher frequencies (as a result of clipping) with an increase in RMS power.

I'm not positive, but I believe that it is an increase in RMS power--that's why meters reading actual RMS voltage are more accurate than meters that only read peak-to-peak. Peak-to-peak measurements will be the same whether the wave measured is a perfect sine wave or a clipped square wave, but a power amp is working much harder to reproduce the clipped square wave, and will show a higher RMS voltage. This is one of the many reasons why square waves will fry your bass rig.

 

[drurb]

I'm not positive, but I believe that it is an increase in RMS power--that's why meters reading actual RMS voltage are more accurate than meters that only read peak-to-peak. Peak-to-peak measurements will be the same whether the wave measured is a perfect sine wave or a clipped square wave, but a power amp is working much harder to reproduce the clipped square wave, and will show a higher RMS voltage. This is one of the many reasons why square waves will fry your bass rig.

No, this has nothing to do with true-RMS meters. I believe what fdeck has done is to calculate the strength of the additional harmonics as if they are ADDED to the fundamental. Clipping produces a redistribution of energy. You do not get more than you would have gotten if you could have amplified the fundamental cleanly.

 

[WalterBush]

No, this has nothing to do with true-RMS meters. I believe what fdeck has done is to calculate the strength of the additional harmonics as if they are ADDED to the fundamental. Clipping produces a redistribution of energy. You do not get more than you would have gotten if you could have amplified the fundamental cleanly.

OK, OK. Please note, I'm not trying to argue, I genuinely want to learn more here. I thought that the redistribution of energy was precisely what RMS calculations of voltage were about, i.e. RMS voltage was a measurement of what voltage the amp would be putting out if the clipped signal wasn't clipped, so the RMS voltage of a clipped signal is higher than one of a non-clipped signal, although the peak-to-peak voltages are the same.
I realise that this theory probably displays a basic lack of understanding regarding RMS measurements, and I would like a brief explanation, if that's possible.
Ordinarily I wouldn't want to hijack a thread like that, but since we're discussing specs and how easily they're misinterpreted, I though it might be germaine to the discussion...

 

[fdeck]

I think you are confusing the transfer of energy to higher frequencies (as a result of clipping) with an increase in RMS power.

I am assuming that RMS output power is the power dissipation of a resistance equal to the rated load. In other words, measure the output voltage with a true-RMS voltmeter while driving a resistive load. Report the outptu power as the square of the RMS output voltage divided by the load resistance. To the best of my knowledge, that's how amps are rated.

Or at least, how they should be

Another way of stating it, is as the time-averaged product of output voltage and current.

 

[drurb]

I am assuming that RMS output power is the power dissipation of a resistance equal to the rated load. In other words, measure the output voltage with a true-RMS voltmeter while driving a resistive load. Report the outptu power as the square of the RMS output voltage divided by the load resistance. To the best of my knowledge, that's how amps are rated.

Or at least, how they should be

Another way of stating it, is as the time-averaged product of output voltage and current.

True, but you don't get something for nothing! When you clip, you do not manufacture any power, of course, and the RMS voltage of the resultant waveform is no greater than what the fundamental (or the waveform of the original signal) would have been.

 

[drurb]

OK, OK. Please note, I'm not trying to argue, I genuinely want to learn more here. I thought that the redistribution of energy was precisely what RMS calculations of voltage were about, i.e. RMS voltage was a measurement of what voltage the amp would be putting out if the clipped signal wasn't clipped, so the RMS voltage of a clipped signal is higher than one of a non-clipped signal, although the peak-to-peak voltages are the same.
I realise that this theory probably displays a basic lack of understanding regarding RMS measurements, and I would like a brief explanation, if that's possible.
Ordinarily I wouldn't want to hijack a thread like that, but since we're discussing specs and how easily they're misinterpreted, I though it might be germaine to the discussion...

See my response above. I'll try to help further. We agree that limiting a waveform cannot produce additional power or energy. The math all works. Imagine starting with a sine wave. If you clip it, thus squaring off the tops, you create higher frequencies. Keep going and you'll essentially get a square wave. The power in the new clipped waveform is not greater than that in the original sine wave. That is, you cannot limit voltage and wind up with more power. In fact, you have less! If you compute the RMS voltage (and you are correct not to want peak voltage or anything like that) you will find that it is LESS than that of the original sine wave. Now, if we consider fdeck's case, where we are not starting with a sine wave and clipping it but, rather amplifying up to some voltage limit, you'll find that the best we can do is maintain the power that would have been in the original sine wave if we could amplify it cleanly. I really do think I've got this right.

 

[seamonkey]

I find this article explains it pretty well for me at least.

http://www.bcae1.com/2ltlpwr.htm

This is actually a car audio site - but it's pretty good.

The only thing I'd add is I don't think anyone would design a power supply that could supply full voltage for constant clipping. The voltage is going to drop, and you'll end up with lower power out.

 

[fdeck]

See my response above. I'll try to help further. We agree that limiting a waveform cannot produce additional power or energy. The math all works. Imagine starting with a sine wave. If you clip it, thus squaring off the tops, you create higher frequencies. Keep going and you'll essentially get a square wave. The power in the new clipped waveform is not greater than that in the original sine wave. That is, you cannot limit voltage and wind up with more power. In fact, you have less! If you compute the RMS voltage (and you are correct not to want peak voltage or anything like that) you will find that it is LESS than that of the original sine wave. Now, if we consider fdeck's case, where we are not starting with a sine wave and clipping it but, rather amplifying up to some voltage limit, you'll find that the best we can do is maintain the power that would have been in the original sine wave if we could amplify it cleanly. I really do think I've got this right.

(fdeck scrambling to get the cosmos back in order)... Ah, now I think we could actually both be right, but you are being precise and I was not. Let's start with an idealized amplifier fed by a sinewave at just a hair below the clipping limit. We could achieve clipping in two ways:

1. Reduce the clipping limit. I think this is what you are describing -- energy is transferred from the fundamental to the overtones, and the RMS power does go down.

2. Crank up the input amplitude beyond the clipping limit. Now the energy in the fundamental remains unchanged, but the grass starts growing in the overtones, and the RMS power goes up.

I am to blame for not specifying which of these I was talking about. I think you are using the first case, and are indeed correct about what is happening.

Maybe a way to rephrase my original "reason" for knowing the THD rating is simply that I want to know the output waveform that delivers the rated power of the amp. If the THD is low enough, i.e., roughly a percent, then I know what the waveform is -- a sinewave. If the THD gets into the double digits, then I have no idea what the waveform is, unless the manufacturer tells me.

 

[mje]

...
Maybe a way to rephrase my original "reason" for knowing the THD rating is simply that I want to know the output waveform that delivers the rated power of the amp. ....

If I understand you: You might drive an amp into clipping while the power supply may still be able to deliver more current. As the wave approaches a square wave, the area under the curve goes up- you're pushing a bit more energy through the load than if you were reproducing a sine wave. That is, a square wave at a given amplitude and frequency requires more power than does a sine wave. So you and DRURB are talking about different things.

 

[fdeck]

I think that we had complementary explanations. Perhaps a unifying concept is a RMS-to-peak ratio.

 

[drurb]

(fdeck scrambling to get the cosmos back in order)... Ah, now I think we could actually both be right, but you are being precise and I was not. Let's start with an idealized amplifier fed by a sinewave at just a hair below the clipping limit. We could achieve clipping in two ways:

1. Reduce the clipping limit. I think this is what you are describing -- energy is transferred from the fundamental to the overtones, and the RMS power does go down.

2. Crank up the input amplitude beyond the clipping limit. Now the energy in the fundamental remains unchanged, but the grass starts growing in the overtones, and the RMS power goes up.

I am to blame for not specifying which of these I was talking about. I think you are using the first case, and are indeed correct about what is happening.

Maybe a way to rephrase my original "reason" for knowing the THD rating is simply that I want to know the output waveform that delivers the rated power of the amp. If the THD is low enough, i.e., roughly a percent, then I know what the waveform is -- a sinewave. If the THD gets into the double digits, then I have no idea what the waveform is, unless the manufacturer tells me.

Yes, but I was considering the second case as well. In that instance, all I was saying is that the RMS is no greater in the clipping case than what it would have been for another amplifier that could have amplified that signal cleanly. We agree that the additional harmonics do not somehow violate conservation of energy. The cosmos is in order!

 

[drurb]

I find this article explains it pretty well for me at least.

http://www.bcae1.com/2ltlpwr.htm

This is actually a car audio site - but it's pretty good.

The only thing I'd add is I don't think anyone would design a power supply that could supply full voltage for constant clipping. The voltage is going to drop, and you'll end up with lower power out.

Two things: The site is correct for the most part but whomever wrote it should visit (or revisit) a EE text. The bit about the speaker-cone motion is, I believe, quite flawed as well. That's all I'll say about that. Second, a power supply will work pretty much as indicated. What's pictured is called "hitting the supply rails."

 

[drurb]

If I understand you: You might drive an amp into clipping while the power supply may still be able to deliver more current. As the wave approaches a square wave, the area under the curve goes up- you're pushing a bit more energy through the load than if you were reproducing a sine wave. That is, a square wave at a given amplitude and frequency requires more power than does a sine wave. So you and DRURB are talking about different things.

Yes and no, see above. I think we're all set on this! Now about Behringer and business ethics...

 

[seamonkey]

I did order the compressor and bass-eq pedals. When I get them I'll take them apart and post pictures. I can do a subjective listening report, but won't be able to do any scientific measurements for a couple of months. My music PC is being used for recording right now.