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#1 |
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diyAudio Member
Join Date: May 2008
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I can't seem to find any useful info on the power consumption of individual speakers based on frequency in a 2 or 3 way. For example, How much power does a tweeter crossed above 3000 Hz use in comparison to a woofer crossed below 500 Hz?
Any formulas or links to this info. I appreciate anyone's help. |
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#2 |
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diyAudio Member
Join Date: Nov 2003
Location: Brighton UK
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Hi,
The 50% power point is ~ 350Hz for rock, a little higher for classical. Treble wise peak and average powers are rarther different, average levels above 3kHz are low, but peak levels not so much. Mid/treble split around 3Khz of the remaining 50% is ~ 30%/20%. For bi-amping with the same amplifiers active bass/mid and passive mid/treble (possibly with active BSC) is an attractive option, see : http://www.musicanddesign.com/HybridDesign.html
__________________
There is nothing so practical as a really good theory - Ludwig Boltzmann When your only tool is a hammer, every problem looks like a nail - Abraham Maslow |
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#3 | |
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diyAudio Member
Join Date: Dec 2004
Location: Novi, Michigan
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Quote:
When an amp clips it shifts the spectrum upwards and this changes the "typical" spectrum dramatically. This is why a short blast of a clipped signal can take out a tweeter almost instantly. In real terms, tweeters take a lot more power than woofers. Consider a woofer from 0 - 1000 Hz. and a tweeter from 1000 Hz up to 20 kHz. The tweeter is getting 19 times more power than the woofer. |
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#4 |
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diyAudio Member
Join Date: May 2008
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Thanks to both of you for your replies. They were both very helpful. While I don't quite understand why a tweeter would possibly use 19 x's the power of a woofer above 1000 Hz, it's more logical for me to see the the power to be 30/20 %(Mid/Tweet) above 350 Hz. However I'm not an expert and I respect all responses.
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#5 | |
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diyAudio Member
Join Date: Nov 2003
Location: Brighton UK
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Quote:
__________________
There is nothing so practical as a really good theory - Ludwig Boltzmann When your only tool is a hammer, every problem looks like a nail - Abraham Maslow |
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#6 | |
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diyAudio Member
Join Date: Dec 2004
Location: Novi, Michigan
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Quote:
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#7 | |
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diyAudio Member
Join Date: Nov 2003
Location: Brighton UK
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Quote:
You said it without qualification, I didn't. It only applies to white noise which is irrelevant. The statement is completely wrong.
__________________
There is nothing so practical as a really good theory - Ludwig Boltzmann When your only tool is a hammer, every problem looks like a nail - Abraham Maslow |
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#8 |
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diyAudio Member
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While I have not done a comprehensive study of this, when trying to answer the same question I have written a program in Matlab that takes a chunk of a song in, breaks it up into octave-sized frequency bands, and then computes the rms power in each of those bands. After sending in about a dozen "typical" songs (reasonably uncompressed rock songs), I found that the energy per octave is slightly downward-tilted in frequency and falls off pretty quick below 30 Hz. If I break the signal up into say, 20-80 Hz, 80-250 Hz, 250-1.5k Hz, and 1.5-20 kHz the power going into each of these bands is pretty close.
The EIA 426B standard noise is designed to approximate the power distribution in music. The frequency content of this is as such: 4th order BW highpass at 40 Hz, half order lowpass (-3 dB/oct) at 1 kHz. What I have observed is "on average" consistent with this. |
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#9 |
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diyAudio Member
Join Date: Dec 2004
Location: Novi, Michigan
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This is quite reasonable. Had the other "gentleman" not been so rude, I would have pointed out that my claim was for noise as implied by the context of the note. With a -3 dB slope to the noise (i.e. pink) the power frraction above about 1 kHz will be about 1/2 the total. At -0dB it with be 19/20 the the total just as I claimed. |
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#10 |
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diyAudio Member
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If you mean the 426B noise, the filter is applied to pink noise, not to white, so the power per octave is itself decreasing above 1 kHz.
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