A different question on wattage.

[JerryLove]

Something I realize I don't know.

Let's say I have a 2-way speaker. 90db sensitive, crossover at 1khz.

If I want to play a 5khz tone so that it's 90db in volume from 1m, I'm going to have to provide 1w of power (yes, there's some nuance about volts and current).

What if I want to simultaneously play a 500Hz tone at the same volume? Is the SPL (out of the speaker) still 90db? Am I now consuming 1w (since 90db) or 2w (I'm actually driving two separate drivers that are each 90db)?

Are the rules the same if a third tone is added? Say 5Khz?

 

[lovinthehd]

Offhand if you were actually getting 1 w each to each driver thru the passive crossover with 2w power you'd have an extraordinary crossover Or if each driver were actually outputting 90dB would combined be 93-96dB? Is the sensitivity rating even good for 5khz or 500hz? Good question tho!

 

[JerryLove]

Offhand if you were actually getting 1 w each to each driver thru the passive crossover with 2w power you'd have an extraordinary crossover Or if each driver were actually outputting 90dB would combined be 93-96dB? Is the sensitivity rating even good for 5khz or 500hz? Good question tho!

Technically: nothing says that I'm getting 1w to either driver. The sensitivity of a speaker takes the crossover into account.

So that example would me "1w into the crossover = 90db from the driver". The driver must be at least that efficient; but may well be more.

I think I can picture how summation occurs electrically. If you imagine a 2Hz and 3Hz sine wave on the same signal... then one wave will ride on the other and ever other peak on the 3Hz signal will be "doubled" where both peaks meet.

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.

It's not like we even mention the complexity of the source signal when discussing power needs.

 

[lovinthehd]

Technically: nothing says that I'm getting 1w to either driver. The sensitivity of a speaker takes the crossover into account.

So that example would me "1w into the crossover = 90db from the driver". The driver must be at least that efficient; but may well be more.

I think I can picture how summation occurs electrically. If you imagine a 2Hz and 3Hz sine wave on the same signal... then one wave will ride on the other and ever other peak on the 3Hz signal will be "doubled" where both peaks meet.

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.

It's not like we even mention the complexity of the source signal when discussing power needs.

LOL I said it was off the top of my head....yeah, the acoustic coupling rules I'm not sure of particularly but think it's like when you add speakers....overall I tend to think x amount of electricity will get you x amount of dB, maybe not 1:1 depending on content, why your question appealed to me originally as well. I've always wondered but don't have the chops to confirm it one way or the other.

 

[TLS Guy]

There is a fundamental error of thinking and understanding here.

Speaker sensitivity is measured at 1 meter with a drive of 2.83 volts at I KHz. This frequency is not written in stone, and sometimes an average of dreadings at three or occasionally more frequencies is used. However I KHz is usually the measured frequency. Now if the impedance is 8 ohms at I Khz then the speaker will draw 1 watt at I Khz.

So how much power is drawn at a given frequency is dependent on the impedance at that frequency. As you know a speaker's impedance fluctuates tremendously with impedance. So you have to calculate the power draw from at any given frequency from the impedance. So if the impedance of your speaker was 16 ohm at 5 KHz, not improbably, it would draw half a watt. If the impedance was 4 ohm at 500 Hz it would draw 2 watts.

 

[JerryLove]

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.

 

[Irvrobinson]

Surely I don't need 1000w to reproduce 1000 frequencies at 90db.

Jerry, I am getting the impression you ask what sounds like brain teaser questions that you already know the answer to, but I'll play along. Think about how a white noise test works at 90db. Assuming the speaker under test has approximately flat frequency response (a big assumption, but bear with me),white noise played at 90db one meter from the speaker will take the same voltage (notice I said voltage) as the speaker would take to produce 90db at one meter playing only 1KHz.

 

[Steve81]

What does that math actually look like (even heavily simplified)?

This is the (not so simple) math involved dealing with complex waveforms vs discrete frequencies:
https://en.wikipedia.org/wiki/Fourier_analysis
@PENG is probably a good guy to do a deep dive on that one

 

[highfigh]

Something I realize I don't know.

Let's say I have a 2-way speaker. 90db sensitive, crossover at 1khz.

If I want to play a 5khz tone so that it's 90db in volume from 1m, I'm going to have to provide 1w of power (yes, there's some nuance about volts and current).

What if I want to simultaneously play a 500Hz tone at the same volume? Is the SPL (out of the speaker) still 90db? Am I now consuming 1w (since 90db) or 2w (I'm actually driving two separate drivers that are each 90db)?

Are the rules the same if a third tone is added? Say 5Khz?

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.

You could measure the impedance at each frequency and average the results but as Mark wrote, 2.83VAC into an 8 Ohm load calculates to about 1W. It becomes a complex waveform and the impedance is measurable through voltage drop, but it's not very specific or accurate unless a specific time sample is used and averaged.

 

[lovinthehd]

This is the (not so simple) math involved dealing with complex waveforms vs discrete frequencies:
https://en.wikipedia.org/wiki/Fourier_analysis
@PENG is probably a good guy to do a deep dive on that one

Didn't Peng already mention something along those lines (of doing some Fourier calcs) in the other discussion about sub/bass amp loads?

 

[mtrycrafts]

Something I realize I don't know.

Let's say I have a 2-way speaker. 90db sensitive, crossover at 1khz.

If I want to play a 5khz tone so that it's 90db in volume from 1m, I'm going to have to provide 1w of power (yes, there's some nuance about volts and current).

What if I want to simultaneously play a 500Hz tone at the same volume? Is the SPL (out of the speaker) still 90db? Am I now consuming 1w (since 90db) or 2w (I'm actually driving two separate drivers that are each 90db)?

Are the rules the same if a third tone is added? Say 5Khz?

I would think the same when an amp is measure for power output at full band and at only 1 kHz? The 1 kHz is about 10% to 15% more?

 

[JerryLove]

Jerry, I am getting the impression you ask what sounds like brain teaser questions that you already know the answer to, but I'll play along. Think about how a white noise test works at 90db. Assuming the speaker under test has approximately flat frequency response (a big assumption, but bear with me),white noise played at 90db one meter from the speaker will take the same voltage (notice I said voltage) as the speaker would take to produce 90db at one meter playing only 1KHz.

I think I probably knew this at one point, but I don't remember.

Maybe if I turn the question around a little.

I have two speakers. They consume 1w when generating a sinewave @90db @1m for [really, every frequency] but the case of this argument 100Hz and 2000Hz. They have an 8ohm resistance [again, across every freqency unless this is a logical impossibility for some reason].

If I run a 2khz sine wave on speaker A, that takes 1 watt (by definition above).
If I run a 100Hz sine wave on speaker B, that takes 1 watt (again by definition). Meaning generating both of these sine waves takes me 2w.

If I were to measure SPL (assume the listening position is equidistant from both speakers); would I measure 90db? 93db? 96db? Something else?

Or do I need to understand Furior Transforms to get even a "back of napkin" understanding?

 

[Irvrobinson]

I think I probably knew this at one point, but I don't remember.

Maybe if I turn the question around a little.

I have two speakers. They consume 1w when generating a sinewave @90db @1m for [really, every frequency] but the case of this argument 100Hz and 2000Hz. They have an 8ohm resistance [again, across every freqency unless this is a logical impossibility for some reason].

If I run a 2khz sine wave on speaker A, that takes 1 watt (by definition above).
If I run a 100Hz sine wave on speaker B, that takes 1 watt (again by definition). Meaning generating both of these sine waves takes me 2w.

If I were to measure SPL (assume the listening position is equidistant from both speakers); would I measure 90db? 93db? 96db? Something else?

Or do I need to understand Furior Transforms to get even a "back of napkin" understanding?

Ignore watts, focus on volts. Like Mark said, real speakers have electrical characteristics that vary by frequency.

If you have two speakers in an anechoic chamber (the only way to run the test; rooms have acoustics), and you had two identical speakers fed with, say, 2.83 volts, playing two different frequencies at 90db at one meter from the microphone, you'd have twice the power being converted to sound with equal efficiency, so assuming linear output behavior in the speakers you'd get 3db more loudness, or 93db total loudness.

 

[Irvrobinson]

Ignore watts, focus on volts. Like Mark said, real speakers have electrical characteristics that vary by frequency.

If you have two speakers in an anechoic chamber (the only way to run the test; rooms have acoustics), and you had two identical speakers fed with, say, 2.83 volts, playing two different frequencies at 90db at one meter from the microphone, you'd have twice the power being converted to sound with equal efficiency, so assuming linear output behavior in the speakers you'd get 3db more loudness, or 93db total loudness.

Upon reflection, the second paragraph is incorrectly worded. My assertion is: if you have two identical speakers, and each produces 90db at one meter at one of two different frequencies with 2.83 volts of input in an anechoic chamber, the total loudness at measured by a microphone at one meter from both speakers will be 93db, because you'd have twice the acoustic power.

 

[PENG]

I think I probably knew this at one point, but I don't remember.

Maybe if I turn the question around a little.

I have two speakers. They consume 1w when generating a sinewave @90db @1m for [really, every frequency] but the case of this argument 100Hz and 2000Hz. They have an 8ohm resistance [again, across every freqency unless this is a logical impossibility for some reason].

If I run a 2khz sine wave on speaker A, that takes 1 watt (by definition above).
If I run a 100Hz sine wave on speaker B, that takes 1 watt (again by definition). Meaning generating both of these sine waves takes me 2w.

If I were to measure SPL (assume the listening position is equidistant from both speakers); would I measure 90db? 93db? 96db? Something else?

Or do I need to understand Furior Transforms to get even a "back of napkin" understanding?

No, you do not need to understand Fourier, you just need to read Irv's carefully worded post #14.

 

[highfigh]

Upon reflection, the second paragraph is incorrectly worded. My assertion is: if you have two identical speakers, and each produces 90db at one meter at one of two different frequencies with 2.83 volts of input in an anechoic chamber, the total loudness at measured by a microphone at one meter from both speakers will be 93db, because you'd have twice the acoustic power.

Two speakers producing the same output by receiving any equal voltage will produce 6dB more- one speaker receiving doubled power produces 3dB more and when you add the output from the second speaker, another 3dB is the result. This can be re-tested with one channel or two; connect one 8 Ohm speaker to the amplifier and send 2.83VAC to find the output and connect another speaker that's theoretically identical, in series- the output will be the same as one 8 Ohm speaker because the cone area has doubled even though the impedance has also doubled. Connect them in parallel and the output will be 6dB higher because the power and cone area have doubled, which is the same .result from using two channels- two speakers fed 2.83V, two 8 Ohm motor assemblies, same acoustic energy from each speaker.

 

[highfigh]

I think I probably knew this at one point, but I don't remember.

Maybe if I turn the question around a little.

I have two speakers. They consume 1w when generating a sinewave @90db @1m for [really, every frequency] but the case of this argument 100Hz and 2000Hz. They have an 8ohm resistance [again, across every freqency unless this is a logical impossibility for some reason].

If I run a 2khz sine wave on speaker A, that takes 1 watt (by definition above).
If I run a 100Hz sine wave on speaker B, that takes 1 watt (again by definition). Meaning generating both of these sine waves takes me 2w.

If I were to measure SPL (assume the listening position is equidistant from both speakers); would I measure 90db? 93db? 96db? Something else?

Or do I need to understand Furior Transforms to get even a "back of napkin" understanding?

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.

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.

 

[Irvrobinson]

Two speakers producing the same output by receiving any equal voltage will produce 6dB more- one speaker receiving doubled power produces 3dB more and when you add the output from the second speaker, another 3dB is the result. This can be re-tested with one channel or two; connect one 8 Ohm speaker to the amplifier and send 2.83VAC to find the output and connect another speaker that's theoretically identical, in series- the output will be the same as one 8 Ohm speaker because the cone area has doubled even though the impedance has also doubled. Connect them in parallel and the output will be 6dB higher because the power and cone area have doubled, which is the same .result from using two channels- two speakers fed 2.83V, two 8 Ohm motor assemblies, same acoustic energy from each speaker.

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.

 

[PENG]

Two speakers producing the same output by receiving any equal voltage will produce 6dB more- one speaker receiving doubled power produces 3dB more and when you add the output from the second speaker, another 3dB is the result. This can be re-tested with one channel or two; connect one 8 Ohm speaker to the amplifier and send 2.83VAC to find the output and connect another speaker that's theoretically identical, in series- the output will be the same as one 8 Ohm speaker because the cone area has doubled even though the impedance has also doubled. Connect them in parallel and the output will be 6dB higher because the power and cone area have doubled, which is the same .result from using two channels- two speakers fed 2.83V, two 8 Ohm motor assemblies, same acoustic energy from each speaker.

Irv is correct about the 3 dB gain in this case because the two speakers are playing non coherently, i.e. different frequency.

 

[Irvrobinson]

Irv is correct about the 3 dB gain in this case because the two speakers are playing non coherently, i.e. different frequency.

Excellent point, and one I forgot to mention. Two identical speakers side-by-side in an anechoic chamber playing the same frequencies do indeed produce more than 3db of additional output due to constructive interference:

http://www.phys.uconn.edu/~gibson/Notes/Section5_2/Sec5_2.htm