Music energy curve
Sat, 05 Aug 2006 19:48:13 GMT
How does a pop-rock music energy curve distribute over different
If you talk about energy per octave, it would probably be pretty equal over
the entire range. Maybe a slight falloff over the entire range towards the
That would be, essentially, the same spectral distribution as
pink noise. And, in fact, most music has a spectral distribution
that is VERY different than pink noise. Viewing it as energy
Spectral analysis on pop/rock music really does result in a pink noise alike
curve. If you select a broad time interval (couplet-refrain-bridge) and,
let's say, 1/3 octave bands. There is no specific area that peaks out or
dips in extremely.
per constant percentage bandwidth (i.e., per octave or per
fractional octave), it's quite the opposite.
EV's work showed IIRC a fairly flat distribution up to ~ 4kHz after which it
dropped at ~ 6dB/octave I think. This was based on 'electric rock' of the time (
highs. The refrains will usually have a strong falloff after 15-16 kHz
because of the closed hihats, less use of cymbals and clean guitars (which
produce a lot of harmonics above 16 kHz).
Where is the highest peak most likely to be?
Depends on taste, but I prefer a nice, equally spread response. Most
In between wich frequencies would, let's say 70% of the energy stay within?
Most acoustic instruments (in other words: everything but synthesizers)
won't go much lower than 50 Hz and higher than 12 kHz. Beyond the 12 kHz
border, you will have natural harmonics that will go as far as 20 kHz, but
it's energy is much lower (which doesn't mean that is is not important, but
you won't have to do anything about it cause it's just there and it's nice).
Nice question. What's the purpose of it, if I may ask?
Some (as myself) may want speakers that has reasonable HiFi and still can
produce a strong volume. All along with a tiny cost.
In such a case the XO may be set according to energy handeling between
drivers, not the optimum HiFi (at low dB's).
I was also thinking that this most likely set a theoretical XO depending
only of the size of a driver (everything else excluded). An ideal XO
according to driver size.
A guess, as example according to power handeling (distorsion and
8" woofer - 5½" mid - XO ~1KHz
10" " - 5½ mid - XO ~1,5Khz
...which interesting enough goes in the opposit direction (?) compared to
the usual frequency preferred XO.
but, then again when expanding the mid driver the traditional change takes
10" " - 6" mid - XO ~1Khz.
The other day I saw a speaker with 2x10" + 1x 4½"mid (XO ~800Hz /5KHz) which
was said to "play loud". I would guess that the mid would set a rather low
dB-limit for that design. The ad was pretty close to a real lie.
Why would the size of the mid-range cone determine how loud it goes?
Loudness is determined by electro-acoustic conversion efficiency and
power handling, neither of which are determined by the diameter of the
cone or dome.
When the diameter exceeds the wavelength, there is a gain factor,
as well as dispersion. The greater the cone are, the more efficient is the
Unless you want to tie yourself up into some very
complicated knots, I'd advise you to limit your
assumptions to the piston region of operation.
Otherwise, as you hint, things get VERY complicated.
process of conversion to sound. Small drivers also have less power dissapation limiting overall
But this depends on frequency, as large cones beam high frequencies, as
you noted in your first paragraph. Consequently, 'speakers are normally
operated such that their size is smaller than the wavelengths at which
Small drivers also have less power dissapation limiting overall
If all other things are kept equal, no.
The piston band electroacoustic efficiency of a speaker
is determined as:
n0 = k (Bl)^2 / (Re Sd^2 Mas)
k = p0 /(2 pi c), p0 = 1.18 kg/m^3 and c=342m/s
B = gap flux desnity
l = length of voice coil winding in gap
Re = DC resistance of voice coil
Sd = effective area of piston
Mas = effective acoustical moving mass of piston
So, from this relation, it would APPEAR that there is an area
term in the equation. But wait! there are two very interesting
things. First, you will notice that the efficiecny n0 goes as
the RECIPROCAL of Sd, siggesting that the larger the
radiating area, the LOWER the efficiency. That seems
completely out of whack.
However, when we examine another parameter, Mas, the
effective acoustic mass of the diaphragm, we find that is
Mas = Mms/Sd^2
where Mms is the actual mechanical moving mass of the
diaphragm. So, taking our original expression and substituing
Mas with Mms/Ds^2, we end up with:
n0 = k (Bl)^2 / (Re Mms)
In other words, a releation for deternmining piston band
electroacoustic efficiency based solely on mechanical
terms, and which shows that radiating area is not a
That may seem still to be couterintuitive, but here's
the explanation. Increasing the area, if there was no other
effect, would increase the amount of air moved by the cone,
and thus would suggest that doing so increases the output.
But by increasing the area and causing more air to be
moved, you are ALSO increasing the acoustic mass being
driven, and, quite logically, increasing the mass decreases
efficiency. SInce both go as Sd (actually, Sd^2), the result
is to remove area as a factor in electroacoustic efficiency.
output capabilities. You can also use 8 4-1/2 inch drivers in
This is a function of the voice coil and how it is cooled. Frequency
again comes into the equation as large heavy voice coils have too much
inertia to be used at high frequencies, and they will also resonate with
their suspension compliance so have to be used at lower frequencies only.
You can also use 8 4-1/2 inch drivers in
line, and get over 100 dB sensitivity, and get large power handling.
Yes, you can use multiple drivers to increase power handling and thus
total loudness, but I don't see how multiple drive units are more
efficient. Also, multiple drivers will increase the beaming effect as
they work as a line array.
A fairly extensive study along these lines can be found in
Greiner and Eggera, "The Spectral Amplitude Distribution
of Selected Compact Discs," J. Audio Eng. Soc., vol 37,
no 4, 1989 April.
Overall, it indicates that rock music, as one example, has
an overal distribution which is reasonably flat
over the range of about 125 to about 2 kHz or so, with
a 6 dB/oct rollowff at low frequencies and between a
6 and 12 dB/octave rolloff abof that range.
It will vary according to style etc.
Well, apparently not HUGELY so. All styles, even including
classical orchestral and jazz, show overal similar shapes,
though jazz and, especially solo classical instrumental
music showing that the jhigh frequency break occurs at
One overall trend that is quite apparent is that for the most
the energy drops fairly precitously below 40 Hz almost every
time. Notable exceptions include large-scale romantic French
organ works and the Telarc 1812 overture release.
A second overall trend is a similar dropoff above 8 kHz, with
almost NO exceptions (some very small fragments of the
Telarc 1812 overture showed essentiall flat energy
distribution t 16 kHz and beyond: esception those specific,
short and rare passages, the rest of the recording followed
the model seen everywhere else).
Event considering what one might thing has a lot of high-
frequency energy, like harpsichord, above 8 kHz, energy
dropped fairly rapidly and consistently.
If dropped a few samples onto my odds and ends web site - have a look
Electro-Voice did some work on this a long time back in fact. If I can find the
old data I'll see if I can post it for you. Do you have access to binary groups
Electrically, most of it goes towards bass. The exact range depends on
the music and system EQ. A car system puts a lot of power into the 60Hz
range while a home system will put the power into a wider range of bass.
A BOOM-car system, maybe.
Not all car systems are geared toward BOOM.