A different question on wattage.

TLS Guy

TLS Guy

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#21
yes, there's nuance about volts and current (resistance); but this is a question about principle not about some specific real-world speaker.

The basic question is, all else being equal, how does power draw scale as more frequences are added at a given volume. Does this depend on how far apart the frequencies are? I mean: would 1000hz and 1001hz sum similarly to 1000hz and 5000hz (assuming a flat-resistance speaker)? What does that math actually look like (even heavily simplified)?

Surely I don't need 1000w to reproduce 1000 frequencies at 90db.
If the impedance was constant, then yes you would. However the sound would be impossibly loud, and burn out the driver fast.
 
TLS Guy

TLS Guy

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#22
No, you do not need to understand Fourier, you just need to read Irv's carefully worded post #14.:D
Yes, that is correct. However as I pointed out this is all confounded by impedance of the speaker.

Yes, that really goes to the heart of the issue. As I keep stating we continue to neglect the area where power really is required.

Here is the impedance curve for my speakers in the lower great room.



So the impedance is varying from 32 to 4 ohms. Even so it is a pretty easy load. The only adverse phase angles are at points where the impedance is very high. Note that the impedance is high in the range of a sub. This is common for ported speakers, and again reinforces my assertion that subs do not significantly off load receivers. I know there are some speakers that do have very adverse loads, but you can avoid that. So you have the tuning peaks and the peaks of impedance associated with the 400 Hz and 4 Khz crossover points. So most power is required for a target spl at 700 Hz and 2.5LHz. In fact X8 the power at those frequencies compared to 40 Hz.

This all goes back to your previous post and especially the really interesting ncbi article cited by Peng. For a lot of music those curves support my contention that the real power of music is in that 100 Hz to 500 Hz range and actually extending to 2.5 KHz. That is certainly true for most of the music I listen to. Those curves are exactly what I have observed on the spectrum meter within wave lab.

I think this is important as in the classical, choral opera environment, it is my contention that a lot of speakers do not have the power handling in that range.

In the concert hall one is impressed with the heft of a large cello and double bass section pulling hard. If you add the bass brass and woodwinds the power required is enormous. In my experience most speakers can not reproduce the shear heft of that sound, which is really vital for imparting the proper impact of the music.

So that is why I devote a lot of resources to that band. So in the mid lines there are two 7" high power drivers with TL loading and a high powered 10" driver handling the BSC which is roughly half the power below 500 Hz. So in each speaker two 7" drivers TL loaded with dedicated amp and the 10" driver below that with its own amp. Both 10 " drivers handle duty below 60 Hz plus LFE effects. The 10" drivers are loaded by the large TL.

That really does properly reproduce the full heft of large musical forces.

So my point is there is far too much attention paid to powerful subs and not nearly enough attention paid to the region where a lot of power is truly required.
 
JerryLove

JerryLove

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#23
No speaker has a ruler-flat response, impedance or phase curve, so you need to look at the anechoic output & impedance at each frequency and because of that, the output can be related to its impedance.
I am aware; but that only adds ambiguity. Since it's not a logical requirement for a speaker (merely a physical reality of design),it's simplest to assume "flat".

Look a the math used to determine the fall of a dropped object when teaching gravity. We first expressly state "in a vacuum"; but then we gloss over everything else, not because it's not always present, but because it adds confusion to that level of abstration. The Earth isn't a point-mass, the dropped object will pull the Earth towards it. The presence of inertia from spin comes into play. The rest of the universe will also exert forces slightly differently on both objects.

I say this because people keep repeating that resistance isn't constant. I understand that. But it's not *really* related to my question. I'm asking for some idealized speaker. We can add in the complications *after*.

If you want to test this, The Loudspeaker Design Cookbook had a description of the equipment and method needed- my copy is close to 40 years old, so it's one of the early editions and I'm sure it has been updated.
No. I just want to understand it.

I take 100 audio channels in my Non-linear editor (sorry, not sure what the audio-only equivalent is called). Each channel has a sinewave at a frequency different from all others but at the same amplitude.

I feed one channel into a speaker such that I consume 1w and get 90db of output.
Now I add in the second channel. What happens to consumption and SPL.
Now I add in 50 more channels. What happens.

Assuming a speaker that was utterly flat in FR and ohm load across all frequencies.

When I think of it electrically: it seems like the frequencies would sum; such that, rather than having 1w constant output, I'd have dips to 0w/0db (where the two signals are 180-degrees out of phase) and 2w/93db (where they are entirely in phase).

But then that thinking falls apart when I consider two things.
1) the thought that an audio source might have 100 different frequencies overlapping.
2) If the crossover just happens to pass these to different drivers: wouldn't each driver play each of the sinewaves individually? Thus no summation (at least prior to being turned into sound).
 
JerryLove

JerryLove

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#24
If the impedance was constant, then yes you would. However the sound would be impossibly loud, and burn out the driver fast.
But I see some rather flat areas in the impedance curve your provided. What if my frequencies are all contained in that flat area?

There must be some negative-re-enforcement effect I'm missing that stops orchestras from breaking speakers.
 
P

PENG

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#25
But I see some rather flat areas in the impedance curve your provided. What if my frequencies are all contained in that flat area?

There must be some negative-re-enforcement effect I'm missing that stops orchestras from breaking speakers.
Can you clarify your question, and is there a specific concern?
 
TLS Guy

TLS Guy

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#26
But I see some rather flat areas in the impedance curve your provided. What if my frequencies are all contained in that flat area?

There must be some negative-re-enforcement effect I'm missing that stops orchestras from breaking speakers.
Well, first of all thank you for noting that the impedance curve of those speakers is compared to a lot relatively flat. This are my Raymond E Cooke memorial speakers. Both Raymond Cooke of KEF and Peter Walker of Quad had good evidence to believe that adverse impedance curves were a bad thing. So I used Ray's techniques to extensively reduce the adverse effects of difficult amp loading. We do test amps under resistive loads and should not. Peter had persuasive evidence that amps are by no means equal driving a variety of speakers. His amp designs did do better under difficult loads. They had to as his speakers were highly reactive loads! I use Quad amps exclusively and with very good reason.

Now as to your negative reinforcement. The limits are set by the available amp power with onset of clipping and the power handling of the speaker before thermal compression sets in and or burnout.

So to put it simply as you try to add the same power at multiple frequencies the limit of amps and speakers will reduce power at said frequencies as you add them.
The power of any system is finite and can not be exceeded. I think I have alluded to the fact that this does limit the realism of reproduction in many instances.

The trick in design, as I keep pointing out, is to really asses where the large power requirements are, and design for it. My contention is that this is not properly done in the design of most systems. I don't just do things differently just to be different but for well thought out reasons.
 
Irvrobinson

Irvrobinson

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#27
There must be some negative-re-enforcement effect I'm missing that stops orchestras from breaking speakers.
There isn't. Achieving a given level of loudness with multiple frequencies does not require a linearly additive amount of voltage. The input *power* (not voltage, assuming a relatively flat frequency response) varies by the electrical characteristics Mark discusses, but the power required is not additive. That's why I brought up the white noise scenario. 90db of white noise does not mean every frequency included in the noise spectrum is played at 90db and requires whatever volts of input. You can run this test yourself; anyone can. Play a white noise test file from the internet, like:

https://www.audiocheck.net/testtones_whitenoise.php

Play it back at a reasonable volume (which in my opinion is about 75db),measuring with a sound level meter in unweighted mode, and see that your amplifier and speakers don't blow up, as the additive power consumption hypothesis would suggest.
 
TLS Guy

TLS Guy

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#28
There isn't. Achieving a given level of loudness with multiple frequencies does not require a linearly additive amount of voltage. The input *power* (not voltage, assuming a relatively flat frequency response) varies by the electrical characteristics Mark discusses, but the power required is not additive. That's why I brought up the white noise scenario. 90db of white noise does not mean every frequency included in the noise spectrum is played at 90db and requires whatever volts of input. You can run this test yourself; anyone can. Play a white noise test file from the internet, like:

https://www.audiocheck.net/testtones_whitenoise.php

Play it back at a reasonable volume (which in my opinion is about 75db),measuring with a sound level meter in unweighted mode, and see that your amplifier and speakers don't blow up, as the additive power consumption hypothesis would suggest.
Yes, that is exactly the point. The power is shared among the frequencies, but that sharing is directed by program and the characteristics of the load.

The only speakers are know that present anything like a truly resistive load are the Magnepans.
 
highfigh

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#29
I am aware; but that only adds ambiguity. Since it's not a logical requirement for a speaker (merely a physical reality of design),it's simplest to assume "flat".

Look a the math used to determine the fall of a dropped object when teaching gravity. We first expressly state "in a vacuum"; but then we gloss over everything else, not because it's not always present, but because it adds confusion to that level of abstration. The Earth isn't a point-mass, the dropped object will pull the Earth towards it. The presence of inertia from spin comes into play. The rest of the universe will also exert forces slightly differently on both objects.

I say this because people keep repeating that resistance isn't constant. I understand that. But it's not *really* related to my question. I'm asking for some idealized speaker. We can add in the complications *after*.


No. I just want to understand it.

I take 100 audio channels in my Non-linear editor (sorry, not sure what the audio-only equivalent is called). Each channel has a sinewave at a frequency different from all others but at the same amplitude.

I feed one channel into a speaker such that I consume 1w and get 90db of output.
Now I add in the second channel. What happens to consumption and SPL.
Now I add in 50 more channels. What happens.

Assuming a speaker that was utterly flat in FR and ohm load across all frequencies.

When I think of it electrically: it seems like the frequencies would sum; such that, rather than having 1w constant output, I'd have dips to 0w/0db (where the two signals are 180-degrees out of phase) and 2w/93db (where they are entirely in phase).

But then that thinking falls apart when I consider two things.
1) the thought that an audio source might have 100 different frequencies overlapping.
2) If the crossover just happens to pass these to different drivers: wouldn't each driver play each of the sinewaves individually? Thus no summation (at least prior to being turned into sound).
For a theoretical model, OK, let's assume the impedance is absolutely constant- mixing frequencies will have to result in phase cancellations- I'm not sure we can call this a 'complication' but you did address it with your example of two signals with reversed polarity. With multiple frequencies, some amount of phase shift will occur, which we would see in an RTA display as comb filtering.

Would your 'non-linear editor' be similar to an audio mixer?

The crossover may cause the different frequencies to go to different drivers, but if it's a passive design, there's always going to be some amount of phase shift and that complicates things- if they're absolutely phase-aligned, the frequencies may contribute their own phase shift.

Your comment about gravity seems to assume the act of dropping an object makes it fall to the surface when, in reality, it's falling toward the center of the body. The dropped object may draw the Earth toward it through mutual attraction, but only to the extent of its mass. (what happens if we make a feather out of lead?)

The effects of the rest of the universe would mostly be negligible because of the proximity of the object & Earth, distance, etc. The Sun and Moon would be the obvious exceptions.

Maybe looking at the question and list more assumptions, as a way to clarify what we're dealing with, to eliminate the existing points of confusion.

Good question to ponder!
 
Irvrobinson

Irvrobinson

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#30
Yes, that is exactly the point. The power is shared among the frequencies, but that sharing is directed by program and the characteristics of the load.

The only speakers are know that present anything like a truly resistive load are the Magnepans.
Frankly, you're confusing me about what your position is sometimes, as in your response in post #21. The impedance and phase characteristics of a given speaker are irrelevant to answer Jerry's original question - which is whether or not reproducing multiple frequencies simultaneously for a given total loudness level had additive power requirements, and it doesn't. I understand your strong feelings about optimum speaker design, and why the loads presented by various design choices are important to achievable performance in real systems, but that's really another set of considerations altogether.
 
highfigh

highfigh

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#31
Your conjecture seems to violate the law of conservation of energy. Assuming the frequency response is identical at a given voltage and the speakers are in an anechoic chamber, I don't see how it's possible to get 4x the acoustic power from 2x the devices at the same output level.
I took your comment to mean two speakers, each driven by 1W- the second driver adds its output and the added power does the same. Did you mean 1W total, to both speakers?
 
Irvrobinson

Irvrobinson

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#32
I took your comment to mean two speakers, each driven by 1W- the second driver adds its output and the added power does the same. Did you mean 1W total, to both speakers?
No, since we're talking about different frequencies I prefer to use voltage, so my assertion would be one volt to each of two identical speakers, each playing different frequencies, would result in 3db more loudness.
 
Phase 2

Phase 2

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#33
You'll never get an (exact) math answer. (Measurements) of audio isn't an exact science. Add in tone control -/+ than you just changed up the whole math equation
 
highfigh

highfigh

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#34
You'll never get an (exact) math answer. (Measurements) of audio isn't an exact science. Add in tone control -/+ than you just changed up the whole math equation
That's the reason I made the comment about listing the assumptions- even changing the temperature or humidity will change the acoustic results.
 
P

PENG

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#35
The thing is: I'm having trouble seeing how this works when there are potentially hundreds of frequencies and sub harmonics simultaneously without eating tons of power.
Think about pink and white noise. For example, the test tone of say a D&M avr is a pink noise, it has many frequencies, but each time you increase the volume by 3 dB your spl increase by 3 dB, the voltage will increase by 1.414 times, power draw will increase by 2X. That's for calibrated pink noise, if you pick a bunch of random signal of different frequencies, that the result in measured dB spl will not be the same as the sensitivity will change, but the measured spl will still double with 1.414X V (or 2X P),so it won't be eating tons of power just because there are tons of frequency to play. It only takes higher voltage/power if you need more spl, not just more frequencies.
 
JerryLove

JerryLove

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#38
You'll never get an (exact) math answer. (Measurements) of audio isn't an exact science. Add in tone control -/+ than you just changed up the whole math equation
I'm not after an exact measure.

I have a reasonable understanding how a sine wave becomes sound.
I understand how the power/volume relationship works with a sine wave.
I have some idea how multiple sines add electrically; but I am unsure of my knowledge there.

It seems impossible that, if two sine waves played separately [separate speakers on separate amps] use 2X power, when played together [one speaker, one amp] they can use X power.

From that I conclude that power consumption must rise as the number of frequencies rise... at least at peak (where the sine peaks meet).

My problem with that is that there are essentially an infinite number of frequencies in real sound. It's more of a continuum than discrete, perfect tones. [correct me if that's wrong]

Any not-basically-zero number X any almost-infinity number = insane power.

But that's not the reality. I can play an orchestra without disabling the state's power grid.

So I'm missing something... but I'm not sure what.

I start trying thought experiments (what if I dedicated one speaker to each frequency, or each instrument... how does the power scale and how can I apply that to a single speaker); but they lead me down rabbit holes.
 
P

PENG

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#39
But that's not the reality. I can play an orchestra without disabling the state's power grid.

So I'm missing something... but I'm not sure what.
So you obviously know what the reality is, just missing the technical reason?

I think if you consider the fact (that seems like the only thing you could have missed) that a moving coil loudspeaker's average sound pressure level output just follow the input voltage and does not care much about the signal waveform (against, always some minor variations at different frequencies..),basically a voltage sensitive device. By the way that's why 90 dB/2.83V/1m makes more sense than 90 dB/1W/1m, because given the driving voltage, you can predict the spl without having to know the impedance at each frequency.

You can have 1,2,3,....n sine wave signals of different frequencies, as long as you fix the voltage, the speaker will give you X dB regardless, at 2.83 V driving one load, whatever the impedance is, you obvious won't affect any power grid right?

If you use the same 2.83 V to drive two similar speakers, then now you have two speakers making sound, so it bound to give more spl and draw more total power, not because of the 2 vs 1 frequency, but only because two speakers vs one speaker. I know this sounds redundant, but it really is that simple.
 
JerryLove

JerryLove

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#40
So you obviously know what the reality is, just missing the technical reason?
Within some limits; but not entirely.

I know that power requirements don't spiral towards infinity; but I don't know if the SPL is constant or changed, nor how power requirements change.

You can have 1,2,3,....n sine wave signals of different frequencies, as long as you fix the voltage, the speaker will give you X dB regardless, at 2.83 V driving one load, whatever the impedance is, you obvious won't affect any power grid right?

If you use the same 2.83 V to drive two similar speakers, then now you have two speakers making sound, so it bound to give more spl and draw more total power, not because of the 2 vs 1 frequency, but only because two speakers vs one speaker. I know this sounds redundant, but it really is that simple.
How does this play out when a passive breaks apart a signal?

Assuming all drivers are involved, can I multiply the number of channels on the crossover by the voltage? Or is the voltage reduced by the crossover when splitting (let's assume, for argument, all drivers in the speaker are equally efficient and no deliberate inefficiency is added by the crossover).

It can't just halve the voltage (as that would reduce the volume of an individual wave). But if it maintains the voltage to (say) 2 drivers (separated by a crossover),isn't that functionally the same as two separate speakers?
 

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