I added an example video clip in my above post #37...
Thank you for sharing this diagram.We also know that most of the energy in music is found at lower frequencies. Here's the spectrum of the AES75 Music-Noise signal, that is supposed to represent an average of the energy levels in music.
View attachment 402712
Peaks show up in spectra as multiple frequencies, look at an impulse response in the freq domain, so your question dosnt relate to the OPs question.This is averaged. You miss the peak values, very valuable for knowing the power needed.
It needs to handle both, the accumulated plot shows how hot the amp/voice coils will get.Ok lets give it an other try: The OP asks about the amplifier power needed for different parts of the spectrum.
Assuming the goal is to have the amplifier handling this without clipping, it needs to be able to handle the peaks in the selected parts of the spectrum.
This is very different from the accumulated spectrum which gives an average per frequency component.
For instance the whole spectrum of Steely Dan (Normal CD):
View attachment 402523
and one part at 48.3 sec :
View attachment 402526
As you can see there is a peak at 200Hz at -19dB
in the accumulated spectrum this down by more than 10 dB.
His amp surely needs to be able to handle that -19dB peak and not the level in the accumulated plot
Your woofers are 25db more sensitive than the tweeter! And your filtering the woofers below 60hz, so no surprise and not realistic.What you could to is to play a few favorite tracks at your preferred listening level, and record the actual signal level (voltage) that goes to each loudspeaker driver using a sound card and some DAW software like Audacity. You may need to pad down the signal level, and also find and set a reference level. As an example you could set 4V to equal -20 dBFS. Then 0 dBFS will equal 40V, or 200W @ 8 ohms.
I've done this, and my findings were quite interesting... Just look at the very large peaks on my tweeter! Top of the scale is 100V. The tweeter's highest peak is at ~9 dB, which is ~36V, or ~450W @ 3 ohms !! Poor tweeter...
View attachment 402529
Some context is needed:
Woofer : XO at 60 / 450 Hz. Sensitivity ~105dB (four 15")
Lower mid: XO at 450 / 3kHz. Sensitivity ~90 dB (Magnepan 3.7)
Upper mid: XO at 3k / 6 kHz. Sensitivity ~80 dB (Magnepan 3.7)
Tweeter: XO at 6 kHz. Sensitivity ~80 dB (Magnepan 3.7)
Subwoofers not included here.
Music was Yello - Junior B played back with the volume control cranked up to eleven.
Thats low freq heavey. Very little music (of course theres EDM and the odd other thing) has anything below 30hz.We also know that most of the energy in music is found at lower frequencies. Here's the spectrum of the AES75 Music-Noise signal, that is supposed to represent an average of the energy levels in music.
View attachment 402712
Thank you Chris. But you dont need to add "room sound" to calculation. Absolutely you cant do it, since everybody has diffirent rooms.There isn't a simple formula that relates efficiency of loudspeakers (in SPL) below 100 Hz because the difference in efficiencies between types of bass bins vary by more than a factor of a hundred. Horn bass bins, like corner horns, show over 10% efficiency (i.e., conversion of electrical power into acoustic power, in watts) down to their cutoff frequencies, while closed box subwoofers typically have less than 0.1% efficiency.
There are formulas for equal loudness curves available from ISO 226:2023 which relate loudness level in phons (perceived human loudness) to actual decibels (dB):
MODELING THE ISO 226:2023 EQUAL-LOUDNESS-LEVEL CONTOURS BY STANDARDIZED LOUDNESS METHODS
This allows you to get a handle on the human hearing part.
The combination of these two efficiencies would yield a formula for electrical power input to achieve acoustic watt output. But then room acoustics begin to take over the picture, with in-room reflections and cancellations introducing extremely large variations in SPL vs. location that the listener experiences simply by moving his/her head a few inches to a new location.
So what you are asking for has basically no utility for calculation purposes since the "noise" variables completely swamp any attempt at calculation of SPL output at the listener's ears as a function of electrical power input--for low frequencies. Even higher frequencies experience rather large deviations in SPL vs. position and even time (the so-called "dense mode region" of the room--generally above 200 Hz for typical home hi-fi sized listening rooms.
In the sparse mode region below 200 Hz (in my room, the calculated Schroeder frequency based on RT30 reverberation time is about 100 Hz), deviations in SPL due to interior acoustic reflections can be greater than 100:1 in terms of SPL (linear scale).
Note that I used to model problems in like manner for a living (operations analysis), and build operational simulations that made use of these type of constructed relationships mathematically, so I do feel comfortable laying out the basic relationships in mathematical formulae. But in this case, what you want and what you are going to be able to achieve are really too far apart.
It's better to calculate the rule-of-thumb efficiencies of the human hearing system separately from the aggregated electrical-acoustic efficiencies of the various bass loudspeaker types: closed box, bass reflex, and horn-loaded. Then calculate a room gain vs. frequency based on the positioning of the bass bins relative to the room wall and corner positions (i.e., half-space, quarter space, eighth space room acoustic loadings vs. frequency), since these details in the problem space yield a difference in SPL in-room of over 18 dB.
Chris
It's really interesting that nobody couldn't invent a general formula which can be apllied to all!It needs to handle both, the accumulated plot shows how hot the amp/voice coils will get.
You want the answer to the OPs question look at the 3 drivers in a 3 way speaker (that goes down to 20hz). Which driver is built for the most power? Another hint, why are built in subwoofer amps so large?
Its Flethcher Munson. Low freqs dont sound as loud as mids so they are mixed louder.
We also know that most of the energy in music is found at lower frequencies.
My thoughts:
Although the "power" may be less for high for high frequencies, in a single amplifier configuration (no crossover before the amp) the high frequencies may represent the extreme swings of the signal, as the highs ride the lows, the amplifier must be able to swing full voltage and current and correctly modulate that high voltage and high current as the "low power" high frequencies ride the "high power" lows.
Example:
The high frequency swing pushes the limits of the voltage range, riding on the low frequency swing.
Full range and 10k high pass of the signal:
View attachment 403170
Do you have a source for this statement? Seems excessive.Yeah, SPL is ultimately about volume displacement, simplified it means cone area * excursion. To maintain same SPL octave lower in frequency, the volume displacement needs to quadruble.
Below is taken from this Linkwitz Lab page.Do you have a source for this statement? Seems excessive.
Can you give a concrete example with a real speaker/subwoofer?Below is taken from this Linkwitz Lab page.
View attachment 403269Loudspeaker system design
www.linkwitzlab.com
In the highlighted formula to estimate SPL from a speaker driver:
x is peak-to-peak displacement amplitude in m, SPL is proportional to 20 log10(x)f is frequency in Hz, SPL is proportional to 40 log10(f)d is piston diameter in m, SPL is proportional to 40 log10(d)r is listening distance in m, SPL is proportional to -20 log10(r)
SPL change is proportional to 40 log10(f). If we have frequencies of f and f/2, we get 12 dB reduction:
At f, SPL = ... + 40 log10(f) + ...At f/2, SPL = ... + 40 log10(f/2) + ... = ... + 40 log10(f) - 40 log10(2) + ... = ... + 40 log10(f) - 12 + ...
Since SPL is only proportional to 20 log10(x), to compensate for the 12 dB reduction, x has to increase to 4x.
So using an active speaker with a dedicated separate amps for tweeters and woofers works much better then a passive speaker + power amp(feeding tweeters and woofers simultaneously) combo. ? (Amps have same power)My thoughts:
Although the "power" may be less for high for high frequencies, in a single amplifier configuration (no crossover before the amp) the high frequencies may represent the extreme swings of the signal, as the highs ride the lows, the amplifier must be able to swing full voltage and current and correctly modulate that high voltage and high current as the "low power" high frequencies ride the "high power" lows.
Example:
The high frequency swing pushes the limits of the voltage range, riding on the low frequency swing.
Full range and 10k high pass of the signal:
View attachment 403170
Linkwitz showed an example of how he used the nomographs (you can do the same with the formulas he provided too) to size the drivers of a speaker system. I'd suggest that you read through it first.Can you give a concrete example with a real speaker/subwoofer?
For instance, At 20hz for 100db/1mt