I am comparing all-pass to high-pass to get at the OP's question. High-pass has lower peaks in this case. Not that it is significant.
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Does a high-pass filter lower the needed power for a speaker?
- Thread starter jfburk
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I selected the peakier part of the tune, since only 30 seconds is permitted.
Original (all pass) has a peak of -6.2dBfs, High Pass a peak of -0.8dBfs.
If I can remember, the peak of the whole original was -4.1dB. The high-pass was something like -4.6dB. Just eyeballing your highlighted section it looks like the high-pass has slightly reduced peaks there as well. So I am scratching my head at your high-pass number of -0.8dB.
Very strange. Could you sum low and high pass again and compare it with the original?
There's some phase shift in the filtered tracks so it won't come out "the same"
Example:
60+120Hz sine, low pass at 80HZ, and high pass at 80Hz, and the sum:
80Hz square, same filters:
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Here is what I get, using your highlighted part of the 2L track (high-passed). So -4.87dB, not -0.8dB. The original has a peak of -4.85dB in that section so hardly any difference.
Things can get counter-intuitive when we look at the crest factor. Here is my simulation of the pathological example of a square wave. The square wave frequency is 50 Hz. It separately goes through a pair of LR4 high pass and low pass filters (100 Hz cut-off). Both the low passed signal and high passed signals have voltage swings higher than the original!
[Edit:] Ray is ahead of me!
[Edit:] Ray is ahead of me!
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What filter parameters?
Nevermind.
Ok, maybe a mistake on my end, will try again.
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My premise that the low frequencies are at a lower level than the high frequencies remains...
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Was away from the internet for awhile. I've tried a handful more selections, and to my surprise the high passed version is the one with a higher RMS value. So the high passed version will use more power than the low passed versions. However, in either case, you can get by with less power high passed than full range. So while I thought the low passed version would use more power in answer to the OP's question, it is still true a less powerful amp will work for high passing vs full ranging.
It looks more like an 80 hz high pass lets you get by with a 100 watt instead of 125 watt amp. Which is so small a difference as not to matter. Power in class D is cheap enough it isn't a big deal.
It looks more like an 80 hz high pass lets you get by with a 100 watt instead of 125 watt amp. Which is so small a difference as not to matter. Power in class D is cheap enough it isn't a big deal.
Take a square wave. If you remove the higher-order harmonics i.e. low-pass-filter it, you get overshoots.
And if you high-pass-filter it at the same frequency, you get the opposite stuff. A high rise that forms the start of the square wave followed by a shoot into the negative which keeps the wave square. Ringing at around the 0 mark, if you would.
Because the high order and low order harmonics work together to give you that final instantaneous voltage. But individually, they can be more. When the two ringings add together, you get a flat square wave.
Fourier transform is powerful isn't it.
Yes. Power is lower. But crest factor may increase , and thus can require a bigger voltage swing to give the lower RMS power.
But I see higher peaks... because, uh, because, uh, uh...
If clipping is the concern I think Ray is right that you can't get by with lower power amp, assuming lower power means lower voltage.
Can't explain what is right or wrong about the examples above, but just thinking about this logically... If you filter out part of the spectrum (any part), the acoustic power produced by the speaker will be lower. Since the efficiency of the speaker has not changed, the electrical power required from the amplifier must be lower - conservation of energy. However, as Ray points out, the peak voltage requirement will not necessarily change, and since voltage swing is is the primary determinant of the amplifier power rating (for a given impedance) the necessary power rating of the amplifier may not change. However, you may be able to get by with a less "robust" amplifier. Granted, some hand-waving in that last statement.
Because the individual lower frequency components actually have higher peak amplitude than that of the square wave.
I think because we measure amplifier power with a single frequency sine wave, we often lose sight of the crest factor of the signal. The crest factor of a sine wave is only 3 dB. Music is much higher (see clipping from CTA-2034A, page 52). That's why we often under-estimate the power (or more precisely, peak voltage and current) required of an amplifier. A SPL of 90 dB of music can require 9 dB or more voltage and current swing than a 90 dB SPL single tone sine wave.
I can make things even more complicated for ya... but such complication is usually necessary
Ok, continuing from the previous post: Each individual component frequency of the square wave can have more higher peak amplitude than that of the square wave. Yet, when I sum the individual components together, I get a lower peak voltage, and consequently, a lower power too? And that confuses people.
The magic here is: The frequencies are antiphase with each other. Hence they cancel out.
Does that mean since they cancel out = their power or voltage or whatever requirement also cancel out? Not exactly.
Because we have to also consider the amp as something that can provide different amounts of power at different frequencies too.
An amp's max output power / an amp's output impedance is frequency dependent. Heck, a power supply's output power / output impedance is frequency dependent. More power supply capacitors = less LF impedance for example, more decoupling capacitors = less HF impedance etc.
And supplying power at frequency A may not affect the ability to supply power at frequency B. Although, of course, in this context, there is a high % of effect. At 100kHz vs 10Hz, then maybe not.
But point is, even the different frequency components may cancel out, it doesn't change the fact that the amp is trying to push power at different frequencies, each demanding performance from the amp in a different way. By HPF-ing, you are removing some of it.
Say we have to produce a wave. F1 (lower frequency) + F2 (higher frequency) gives a lower peak amplitude than F1 alone because F2 is antiphase with F1.
F1 is drawing power from the electrolytic caps, F2 is drawing power from the ceramic caps (or whatever decoupling blah). Even though the frequencies may be antiphase, the energy that would have flowed back to the electrolytic caps can't, because it is too high frequency, so it flows to the ceramic caps.
So even though the final voltage may be lower, I still need my amp to be able to produce each individual voltage component properly, and each individual voltage component can still be higher than the final voltage.
Or in other words,
The same amp that is able to reproduce one sine wave at amplitude X, may only be able to reproduce one square wave at amplitude Y, where Y is lower than X. So we're not breaking any physics rule nor layman thinking here. We don't need more power despite filtering, we just happen to need less in certain frequencies and the leftovers appear like that, but that doesn't mean we don't have to supply for these leftovers in the first place.
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