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# Speaker Wattage < Receiver Wattage - Page 3

Quote:
Originally Posted by Jack Hidley

Peter has it exactly correct. The two links he posted are exactly what the issue is here.

If you take a sine wave and clip it, you now have a waveform that contains harmonics that never existed in the original sine wave. I'm using the example of a sine wave to make things simple. The same is true of any waveform. If you make the top and/or bottom of any waveform flat (clip it), you are introducing lots of high frequency energy into the signal. It doesn't matter what circuit or amplifier is used to do this, it is a function of basic physics that this is true.

On Monday, I'll take the FFT of a sine wave and post the resulting spectrum. I'll then clip the sine wave and post the new spectrum.

If you do take the FFT of a pure sine wave you will get just one frequency (duh). If you clip it, you will get different frequencies. This does not mean that these frequencies will be separated by the crossover! Fourier transformation is just one way to break up a single function. DC can also be represented as the sum of two out of phase sine waves, but does this matter? No! It would take some serious effort to figure out how to make a crossover that would separate these different frequencies in a clipped wave.

A pure sine wave could also be represented as a clipped wave plus this:
If a crossover would separate the different frequencies in a clipped wave (which I don't think it can), then wouldn't it also be able to separate a pure sine wave into a clipped wave (harmonics and all) and the wave in the picture? Crossovers can deal with things other than pure sine waves without breaking them up into pure sine waves, just so you know.

Some instruments have some very strange waveforms, so a crossover that can't just pass a clipped wave would also probably mess up the sound of certain instruments.
Quote:

If a crossover would separate the different frequencies in a clipped wave (which I don't think it can)

Think only in the frequency domain. If you FFT a square wave you can see the high frequency components. It's there. Period. We're not talking about the tooth fairy here. The crossover is concerned with the frequency domain.

If we can agree that there is indeed high frequency content in the signal, and there is in a square wave, then that high frequency information will be passed by the high pass filter and cut off at whatever point below the crossover frequency. There. The crossover just separated the high frequency content from lower frequency content. That's all the crossover is doing. It's not decomposing any signals, it's just filtering frequency content.

I don't know what your picture above has to do with anything. Yeah, graphically you could "kind of" add that to a clipped sine wave, but I don't think it's that easy mathematically. A crossover is not breaking up time domain signals.

You're right. Most instruments do not make a perfect sine wave. Still, the shape of the wave, in the time domain, does not matter to the crossover. That signal's representation in the frequency domain is what matters, and the crossover will filter that signal if it falls into its applicable filter bandwidth.
Quote:
Originally Posted by greeniguana00

If you do take the FFT of a pure sine wave you will get just one frequency (duh). If you clip it, you will get different frequencies. This does not mean that these frequencies will be separated by the crossover!

Yes, it will.
Listen to OttoSpiral.
Quote:
Originally Posted by OttoSpiral

Think only in the frequency domain. If you FFT a square wave you can see the high frequency components. It's there. Period. We're not talking about the tooth fairy here. The crossover is concerned with the frequency domain.

If we can agree that there is indeed high frequency content in the signal, and there is in a square wave, then that high frequency information will be passed by the high pass filter and cut off at whatever point below the crossover frequency. There. The crossover just separated the high frequency content from lower frequency content. That's all the crossover is doing. It's not decomposing any signals, it's just filtering frequency content.

I don't know what your picture above has to do with anything. Yeah, graphically you could "kind of" add that to a clipped sine wave, but I don't think it's that easy mathematically. A crossover is not breaking up time domain signals.

You're right. Most instruments do not make a perfect sine wave. Still, the shape of the wave, in the time domain, does not matter to the crossover. That signal's representation in the frequency domain is what matters, and the crossover will filter that signal if it falls into its applicable filter bandwidth.

A high pass filter can be composed of just a single capacitor in series with the tweeter. Looking at a section of a clipped wave, starting at no voltage, the voltage increases then stops increasing and remains constant for a while before it finally starts decreasing again. What about this causes the voltage to the tweeter to fluxuate at high frequencies? The harmonic frequencies just aren't there to be filtered out! The reason the capacitor normally filters out these frequencies is because there are high frequency voltage fluxuations in the main signal. A clipped wave just doesn't have these fluxuations, since the voltage is constant, so they can't be filtered out. When the voltage is constant, the capacitor just charges and stops the flow of electricity.
OK, lets consider a simple cross over that consists of only a capacitor in series with a tweeter.
The capacitor blocks current associated with slow changes in voltage (i.e. low frequencies) from reaching the tweeter. Fast changes in voltage (i.e. high frequencies) can force current to appear to flow through the capacitor. A severly clipped wave contains very fast changes in voltage, which allows current to flow through the tweeter. Does this make sense?

If not, I'm afraid we're going to have to wait till you take a basic class in electronics. For now you'll just have to accept that a square wave can be represented as a series of sin waves, and that some of those waves have very high frequency.
Come on, now. Are you for real? I thought we almost ready to agree upon the FACT that a square wave can be decomposed into an infiinte series of sine waves of increasing frequency at odd multiples of the fundamental. Please refer back to the Fourier transform links. Can we agree on that? If so, then can you agree that there is high frequency information in the signal? Well, then, there's high frequency content there, and that's just the way it is.

You're doing too much thinking and hand-waving. Just because the signal is constant for a while does not mean that there is no high frequency component. You're trying to rationalize that square waves don't have high frequency because you can't "see" it with a drawing of a square wave. Do you understand the FT/FFT concept? Time domain vs. frequency domain? When the FFT is done, and there's high frequency content shown in the frequency spectrum, what do you think that is? And why would a crossover (a simple cap or otherwise) NOT be able to separate the low frequency content from the high frequency content?

Hey, here's an experiment for you -- download a signal generator that will run on your PC. Run a low frequency square wave through your system. Start at, say 100 Hz. Keep the volume low and carefully increase it. Now, do you hear anything above 100 Hz? You can do a reference with a 100 Hz sine wave to see what 100 Hz sounds like. Then, do it with the square wave again. Now does there appear, to your ear, that there is any frequency content above 100 Hz? Like maybe at 300 Hz? Or 500 Hz?
Quote:
Originally Posted by OttoSpiral

Come on, now. Are you for real? I thought we almost ready to agree upon the FACT that a square wave can be decomposed into an infiinte series of sine waves of increasing frequency at odd multiples of the fundamental. Please refer back to the Fourier transform links. Can we agree on that? If so, then can you agree that there is high frequency information in the signal? Well, then, there's high frequency content there, and that's just the way it is.

You're doing too much thinking and hand-waving. Just because the signal is constant for a while does not mean that there is no high frequency component. You're trying to rationalize that square waves don't have high frequency because you can't "see" it with a drawing of a square wave. Do you understand the FT/FFT concept? Time domain vs. frequency domain? When the FFT is done, and there's high frequency content shown in the frequency spectrum, what do you think that is? And why would a crossover (a simple cap or otherwise) NOT be able to separate the low frequency content from the high frequency content?

Hey, here's an experiment for you -- download a signal generator that will run on your PC. Run a low frequency square wave through your system. Start at, say 100 Hz. Keep the volume low and carefully increase it. Now, do you hear anything above 100 Hz? You can do a reference with a 100 Hz sine wave to see what 100 Hz sounds like. Then, do it with the square wave again. Now does there appear, to your ear, that there is any frequency content above 100 Hz? Like maybe at 300 Hz? Or 500 Hz?

I am not denying that square waves can be shown as the weighted sum of every odd harmonic. I am denying that there are any high frequency voltage fluxuations in a square wave or a clipped wave. The crossover picks out voltage fluxuations, not the Fourier series.

I am also pointing out that, even if these did exist, they would so minutely small that damaging a tweeter with them would require that the volume of the original unclipped signal be so high that it would have damaged the speakers the same amount or more had it not been clipped. (edit: with some exceptions)

Quote:

And why would a crossover (a simple cap or otherwise) NOT be able to separate the low frequency content from the high frequency content?

Because the high frequency content is canceled out by everything else until you are left with nothing but a square wave.

Have you ever heard a poorly mixed CD where you can't tell the separation between sounds? This is often due to something like this. If the crossover could separate the different frequencies when the frequencies are largely canceling each other out, then bad mixes would sound better. The same thing happens when two flutes are out of tune, the different sounds blend into something that sounds awful.
Quote:
Originally Posted by dknightd

OK, lets consider a simple cross over that consists of only a capacitor in series with a tweeter.
The capacitor blocks current associated with slow changes in voltage (i.e. low frequencies) from reaching the tweeter. Fast changes in voltage (i.e. high frequencies) can force current to appear to flow through the capacitor. A severly clipped wave contains very fast changes in voltage, which allows current to flow through the tweeter. Does this make sense?

If not, I'm afraid we're going to have to wait till you take a basic class in electronics. For now you'll just have to accept that a square wave can be represented as a series of sin waves, and that some of those waves have very high frequency.

If you are suggesting that because the wave is clipped at the top, there is a fast change in the voltage between when it is not clipped and clipped, then you need to take a basic class in calculus. The change in voltage is minutely small, the derivative is what is changing rapidly (from something very large to something very small).

If you are suggesting that the clipped wave has a very quick voltage change from when it has no voltage to when it reaches that plateau, then you should remember that if the wave wasn't clipped, that change would be just as fast!
greeniguana00, I think I understand your argument now. Correct me if I am wrong.

If I am correct, you are arguing that a tweeter gets damaged because as an amplifier is turned up to the point of clipping, the power delivered to the high frequencies will continue to increase in the part of the signal that is not being clipped. Since usually the low frequencies will clip first (which makes sense since that is where most of music power is - it is in the notes fundimental frequency, not the harmonics) this is a reasonable assumption. As you continue to increase amplification (even after clipping begins) the high frequency portion of the music will continue to contain more energy.

You argue that since the power of the harmonics generated by clipping decreases in amplitude as 1/f**2 that it is not important in the destruction of a tweeter.

Hopefully you will agree that a cross over does not care if the high frequencies are caused by the harmonics generated by a musical instrument, or if the harmonics are generated by driving an amp to clipping.

So the question is, are the harmonics generated by a clipped signal larger or smaller than the harmonics generated by a musical instrument? This I do not know. I suspect both effects act together - which is larger I do not know. Maybe clipping is a red herring and all that really matters is the larger high frequency power in the music that is not being clipped (as I think you are arguing).

I just found this web page

http://www.st-andrews.ac.uk/~jcgl/Sc...ing/page1.html

(click on the arrows at the bottom to see subsequent pages)

that I think nicely describes both your mechanism for tweeter damage (starting
on page 3 of the above link) and the clipping argument for tweater damage (which they get to first). They start discussing the role of the crossover on page 4. They don't really say which mechanism is dominant, but it seems like a good description and analysis.
That's basically it. My main argument is that if a clipping amp causes a speaker to be damaged, that an amp completely identical to it that doesn't clip (in most cases this means a more capable power supply) with will cause the same or more damage at the same volume setting in most cases. In other words, if an amp doesn't damage a speaker when it doesn't clip, it you replace the power supply with a less capable power supply (causing it to clip) it will continue to not damage the speaker. It is not the clipping that causes the damage, IMO, it is the things that clipping tends to indicate (ex: volume too loud) or cover up (ex: again, volume too loud). It's sort of like people saying that strange noises coming from an engine cause damage. In most cases the noises are just an indication of something else going wrong, not the cause of damage.

I agree that a cross over does not care if the high frequencies are caused by the harmonics generated by a musical instrument or a clipping amp. I just don't think that high frequency oscillations are generated during clipping.

Quote:

I agree that a cross over does not care if the high frequencies are caused by the harmonics generated by a musical instrument or a clipping amp. I just don't think that high frequency oscillations are generated during clipping.

I'm assuming that by "oscillations" you mean "components" Why throw this in there? They ARE generated by clipping. We've already been over this. Your comments have been all over the map. Why the hell should we believe ANYTHING you say about speaker damage, clipping, or anything else for that matter?
Quote:
Originally Posted by greeniguana00

I just don't think that high frequency oscillations are generated during clipping.

I'm fairly sure that they are. I don't see how else you can go from a nice single frequency sine wave, to a severly clipped sine wave (that resembles a square wave), without introducing higher frequencies than you started with. What I'm not sure about is if the generated harmonics are important to killing a tweeter.

I'll tell you what, I think I've got an old tweater in the basement that I have no use for.
If you are interested, I could send it to you, if you have a way of generating say a 400 hz (or whatever frequency you want to try) signal and feeding it to a moderately powerful amp. Then we could find out.

You could send a very low volume 400hz signal to the just the tweeter. The one I'm thinking about I think has a built in crossover, so you should hear very little coming out of it at 400hz. Then you could slowly turn up the volume. What little, if any, sound that this tweeter can make at 400hz should slowly increase, until your amp started clipping. Then you should start hearing something more suddenly start coming out of the tweeter. These would be the high frequencies generated when an amp starts clipping. Turn it up some more and I'll bet the tweeter dies even though it is only being fed a 400hz signal.

The one I'm thinking about actually has both a high midrange and tweeter in a single enclosure, it is a leftover from a previous car stereo. I used to drive this unit as part of a 100 watt/channel system - I'm pretty sure you could destroy it by sending it a clipped 400hz signal from a 100 watt amp (heck even 20 watt amp should do it). If you are interested in destroying a tweater for sake of science, PM me your address. I'll donate it and pay shipping. I'd try it myself, but, I'm a little afraid when the tweeter failed it may take my amp out with it (not likely, but possible) and doing it myself would not provide you the first hand experience of blowing up a tweater with what started as a low frequency signal.

Could be fun experiment I'll donate the tweeter and shipping if you are interested in trying it and telling us how it turned out.
I don't think I've taken any of these out of context, really. Chronologically I've moved them around, of course. If out of context, my apologies...

Quote:
Originally Posted by greeniguana00

I think this idea that high frequency noise is magically introduced when an amp is clipping is wrong.

Quote:
Originally Posted by greeniguana00

A clipped wave just doesn't have these fluxuations, since the voltage is constant, so they can't be filtered out.

Quote:
Originally Posted by greeniguana00

If you do take the FFT of a pure sine wave you will get just one frequency (duh). If you clip it, you will get different frequencies. This does not mean that these frequencies will be separated by the crossover!

Quote:
Originally Posted by greeniguana00

The harmonic frequencies just aren't there to be filtered out!

Quote:
Originally Posted by greeniguana00

I just don't think that high frequency oscillations are generated during clipping.

So are they there or not?

Quote:
Originally Posted by greeniguana00

Even if it could, the amount of power in the higher harmonics is so low compared to the clipped wave that damaging the tweeter with them would be difficult.

Quote:
Originally Posted by greeniguana00

There are cases where these harmonics (if they even exist) could cause damage.

Quote:
Originally Posted by greeniguana00

I am also pointing out that, even if these did exist, they would so minutely small that damaging a tweeter with them would require that the volume of the original unclipped signal be so high that it would have damaged the speakers the same amount or more had it not been clipped. (edit: with some exceptions)

So, if they exist, they might be able to cause damage sometimes because they might be big enough, except for in the cases where they are too small? You've done a lot of speculation to get to this point, so I'm just not sure. I'm not going to do the math to figure out what the real energy levels are, and I'm not really qualified to assert whether there is or isn't enough energy to do damage. I don't think you are either... Maybe the Director of Engineering will pop back up. You can argue with him...

Quote:
Originally Posted by greeniguana00

DC can also be represented as the sum of two out of phase sine waves

Let's see an equation for that one! Please!!!! And remember, you can't introduce a DC bias from the get-go.

Quote:
Originally Posted by greeniguana00

The crossover picks out voltage fluxuations, not the Fourier series.

The crossover is sensitive frequency content of a signal, regardless of what it looks like on a scope or spectrum analyzer.

Quote:
Originally Posted by greeniguana00

I just don't belive that a crossover is capable of separating these frequencies from just a clipped wave.

Quote:
Originally Posted by greeniguana00

It would take some serious effort to figure out how to make a crossover that would separate these different frequencies in a clipped wave.

Quote:
Originally Posted by greeniguana00

I agree that a cross over does not care if the high frequencies are caused by the harmonics generated by a musical instrument or a clipping amp.

So could a crossover separate the frequency spectrum of a square wave (or clipped signal) or not?

I think we agreed that a square wave can be described by the FT. There's high frequency content. Consider a pure tone from a function generator input to your system. A sine wave at 100 Hz, suppose. What does that sound like?

OK. Now, introduce a simple mixer and another function generator generating a sine wave 1/3 the size of the one at 100 Hz, and at 300 Hz. What does that sound like?

Add another function generator, creating a sine wave at 1/5 the amplitude of the one at 100 Hz, and set that frequency to 500 Hz.

Keep doing that.

See where I'm going with this? You are now inputting discrete sine waves into your system. Would those be separable by the crossover? You bet... If you look at each one individually, you see a pure tone at the different frequencies. Of course, why would the crossover NOT be able to separate those frequencies? Now, instead of looking at the outputs of the individual function generators, look at the output of the mixer -- the sum total of all the function generators. It will be something approaching a square wave. Yes! They are one and the same, and they are both separable by the crossover.

OK. Now I did tonight was I suggested you do. That is, I input a square wave to my system. However, I did slightly more than that.

First, I set my front speakers to small and my crossover point to 200 Hz. I'm using an IB sub and main speakers rated down to 30 Hz. I use a program called "Room EQ Wizard" that can generate sines, squares, pink, etc. So what did I do?

First I generate a 20 Hz tone (sine wave). BUUUMMMMMMMM. 20 Hz coming out of my sub. No problem. Sub and mains are crossed at 200 Hz using my Outlaw processor. I didn't get right up next to them, but as far as I could tell, there was nothing coming out of the mains. As we would expect with a crossover point of 200 Hz.

Next, I changed it to output a 20 Hz square wave. What do I get? I get stuff coming out of the sub, sure, but I now also have lots and lot of stuff coming out of the mains. This tells me that 1) the 20 Hz square wave has frequency components > 200 Hz and 2) that my crossover was able to separate those frequencies, sending some to the sub and some to the mains.

Yes, this was all performed using a digital crossover, but a passive (analog) crossover should behave the same in practice.

I'm afraid that I'm being a jerk now, so I'll just stop. My apologies for getting a little heated, but this is all a bunch of BS. You can say whatever you want about power supplies, clipping, frequency content and so on, and I guess we'll just have to agree to disagree. I also see that you've opted to agree with dknightd's overly generous consideration of your argument. That's the easy way out, and I still belive 100% that you are hypothesizing all of your response to him. I would submit that the compilation of your above quotes shows a series of direct contradictions and a haphazard argument. I'm sorry, but it just doesn't add up. Good luck.
These may look like they conflict, but they don't:
Quote:
Originally Posted by greeniguana00

If you do take the FFT of a pure sine wave you will get just one frequency (duh). If you clip it, you will get different frequencies. This does not mean that these frequencies will be separated by the crossover!

Quote:
Originally Posted by greeniguana00

I just don't think that high frequency oscillations are generated during clipping

In the first one, I am taking about taking an FFT (Fast Fourier Transform). This usually comes down to taking the signal and rewriting it as the sum of sine waves. You would get high frequencies in the FFT of almost any non-sine wave. I am saying that just because these frequencies are in the FFT doesn't mean they will be separated by the crossover.
In the second one I am stating my opinion (it's actually a fact IMO) that clipping doesn't introduce tangible voltage fluctuations.

One way of describing what happens when a wave is clipped is that harmonics are added to the original wave (this is what people are talking about when they refer to Fourier). These harmonics that are added are high frequency, but actual high frequency voltage fluctuations (I finally spelled it correctly) are not introduced. These high frequency harmonics that are added serve the sole purpose of canceling out the part of the wave that needs to be clipped.

Another way of describing a what happens when a wave is clipped is that an amp doesn't have enough power, so the part of the wave that requires the most voltage (the peak) is just not pushed as high as it should be pushed. It's not like the top of the wave is snipped off like is implied from the word describing it. It's also not like the wave just hits a ceiling on the way up. It can be described well as a car that's at its top speed; no matter how much faster the driver wants the car to go, it won't go any faster.

I think the mistake people make is combining these two descriptions in to a hybrid model that doesn't work.

Quote:
Originally Posted by OttoSpiral

You are now inputting discrete sine waves into your system. Would those be separable by the crossover? You bet... If you look at each one individually, you see a pure tone at the different frequencies. Of course, why would the crossover NOT be able to separate those frequencies?

Wrong. You are not inputting discrete sine waves into your system!!!!!!
Quote:
Originally Posted by Wikipedia

The word discrete comes from the 15th Century Latin word discretus which means separate.

http://en.wikipedia.org/wiki/Discrete
This is one of the points I made earlier. The sine waves cancel parts of each other out forming a square wave. They are not by any means discrete.

If you look at the sum of the powers contained in the harmonics in a square wave compared to the power contained in the square wave itself (assuming constant impedance) you will see that the square wave actually has less power. This is because the harmonics cancel parts of each other out, decreasing the overall power.

Talking about normal, non-clipped music for a second, the reason that the power dissipated by a speaker increases when high frequencies are added is not because the RMS voltage of the signal is increasing, but because the impedance of the speaker is decreasing (because the crossover is opening)

Quote:
Originally Posted by OttoSpiral

I'm not going to do the math to figure out what the real energy levels are

You don't need to, because I already did (if you look back). I would appreciate it, though, if you would double check to make sure all my calculations are correct. According to my calculations, if the high frequency harmonics did make their way to the tweeter for some reason, they would do no damage (with some exceptions). I already did the math to find an exception, and a few could exist (which is why I said "with some exceptions").
Quote:

I am saying that just because these frequencies are in the FFT doesn't mean they will be separated by the crossover

Do you have any comment on my experiment with the sine wave vs. square wave using a 200 Hz crossover point? That experiment empirically proved, to me, that 1) there is high frequency content in a square wave, and 2) that such high frequency content can be separated by a crossover. Please address this directly.

Quote:

In the second one I am stating my opinion (it's actually a fact IMO) that clipping doesn't introduce tangible voltage fluctuations.

OK, fine, whatever. The time-based signal doesn't appear to have any new high frequency voltage fluctuations to the human eye. But the act of clipping did introduce high frequency content regardless of what the scope trace looks like. It shows up in the spectrum analyzer.

Quote:

Wrong. You are not inputting discrete sine waves into your system!!!!!!

Wow, you really got me there. What I should have said, and is clear from the description that I previously wrote, is that you are inputting the sum of discrete sine waves. I was trying to do this so that you could agree that there is high frequency content that will be separated by the crossover. Your link to Wikipedia is silly; of course I know what "discrete" means.

Quote:

You don't need to, because I already did (if you look back).

Why would I trust your math after all this other fallacious reasoning?

Finally, I'm very interested to see the equation for DC voltages represented by two out of phase sine waves. You must have missed that request in my previous post. Please address this directly.
Quote:
Originally Posted by OttoSpiral

Do you have any comment on my experiment with the sine wave vs. square wave using a 200 Hz crossover point? That experiment empirically proved, to me, that 1) there is high frequency content in a square wave, and 2) that such high frequency content can be separated by a crossover. Please address this directly.

Well it proves that with your system, a square wave will cause the tweeters to move. For all I know, it could just be the fact that you have an active crossover that's causing this.

Quote:
Originally Posted by OttoSpiral

OK, fine, whatever. The time-based signal doesn't appear to have any new high frequency voltage fluctuations to the human eye. But the act of clipping did introduce high frequency content regardless of what the scope trace looks like. It shows up in the spectrum analyzer.

Since high frequency voltage fluctuations are the normal means of conveying high frequency information, I would assume a lack of them means a lack of high frequency information (whether intended or not). I am not arguing that you could view a clipped wave as having high frequency content, but I don't think the fact that a clipped wave contains high frequency content according to one decomposition of it means that it will pass the information to the tweeter. As an analogy, a digital signal also can have content at a different frequency then it actually is, but that doesn't matter much when you are running it though an analog amp and speakers.

Quote:
Originally Posted by OttoSpiral

Why would I trust your math after all this other fallacious reasoning?

I welcomed you to check it for errors. If it is correct, it is correct. If it is fallacious, then you should tell me how.

Quote:
Originally Posted by OttoSpiral

Finally, I'm very interested to see the equation for DC voltages represented by two out of phase sine waves. You must have missed that request in my previous post. Please address this directly.

In degrees, take the sum of these two: v(t) = sin(t) + 10 and v(t) = sin(t+180) +10 I know that this is not one of the things allowed by Fourier (neither are out of phase waves to my knowledge), but that wasn't the point. The point was that speakers, amps, and crossovers were not designed to give the Fourier series of a particular input. Fourier has just one way of breaking up different types of waves, but there are many others. Someone could do something similar to Fourier with square waves instead of sine waves. Fourier's decision to use sine waves as the fundamental building blocks doesn't mean that your crossover does.
Quote:
Originally Posted by greeniguana00

Well it proves that with your system, a square wave will cause the tweeters to move. For all I know, it could just be the fact that you have an active crossover that's causing this.

Haha. You're just real stubborn, aren't you. A passive crossover will work the same in practice. I already indicated that in my post of two prior. You simply do not know what you are talking about.

Quote:

Since high frequency voltage fluctuations are the normal means of conveying high frequency information, I would assume a lack of them means a lack of high frequency information (whether intended or not).

In the case of the square wave, it's the sharp edges that introduce the high frequency content. There are no time-domain-obvious high frequency fluctuations, but there is high frequency content.

Quote:

I don't think the fact that a clipped wave contains high frequency content according to one decomposition of it means that it will pass the information to the tweeter.

Yet crossover theory, the high frquency content in the spectrum and my experiment prove you wrong. Think what you want. You are wrong.

Quote:

As an analogy, a digital signal also can have content at a different frequency then it actually is, but that doesn't matter much when you are running it though an analog amp and speakers.

Well, any digital signal that will alias to a higher frequency will have that cut out, by a FILTER, prior to going through a DAC and hitting your speakers. Bad example and irrelevant.

Quote:

I welcomed you to check it for errors. If it is correct, it is correct. If it is fallacious, then you should tell me how.

The fallacy in your arguments is now laid out before you. You are simply wrong on several counts, and you refuse to admit it. To those who know what's going on, it's obvious. I'm sure you can impress your friends that aren't up to speed on this stuff, though.

Quote:

In degrees, take the sum of these two: v(t) = sin(t) + 10 and v(t) = sin(t+180) +10

You've introduced a DC bias with the "+10" part. I already said that that's an unacceptable answer in my previous post, but you chose to ignore that. Do you really think anyone will buy this? Do you really have a valid answer?

After perusing some of your other posts, you seem to be more of a computer guy dabbling in EE. There's another guy around like that, and he gets all mixed up, too. Don't feel bad, though. Keep reading and working at it. I guess I just don't get why you want to come out here with this stuff though.
I haven't seen it much in home theater but in car audio

ok then how do after market car audio speakers get blown by stock head units left & right?

15 -30 watts <<<< 55-125 rms

btw... i personally think its probably the abrupt stops in a square wave that damage the speaker.

this is just a shot in the dark basically if the speaker sounds like ass it could blow

and the 1 watt clipping isn't significant enough to cause damage to anything.
but when you're sending a clipped signal to a speaker thats moving about 3/4 or more of its throw... i would assume that damage could be done. while a clipped signal that doesn't even cause the speaker to move probably doesn't hurt it.
Quote:
Originally Posted by OttoSpiral

In the case of the square wave, it's the sharp edges that introduce the high frequency content.

This is a very uninformed comment. At least I do my reading about a topic I don't understand much about. I can tell you thought about it, but your thinking is wrong. It's best to think of it as the electricity actually being pushed. Water in pipes is often used as an analogy for electricity. The simplest way to look at an amplifier is as a really high pressure stream of water controlled by a valve which is controlled by a signal (let's just keep it to one polarity for the sake of simplicity). When this valve is moved, the pressure that the speaker receives varies. When the valve is closed, the pressure is zero. When the valve is all the way open, the pressure is as high as the valve allows though when fully opened.

A clipped signal would cause something like this: The valve would start opening and the pressure would increase from nothing to as high as is allowed. The pressure would then remain constant for a while, after which it would come back down. The "sharp edges" you see in the graph represent a very quick change in the rate of movement of the valve, not in the movement of the water
Quote:
Originally Posted by sumfoo1

I haven't seen it much in home theater but in car audio

ok then how do after market car audio speakers get blown by stock head units left & right?

15 -30 watts <<<< 55-125 rms

btw... i personally think its probably the abrupt stops in a square wave that damage the speaker.

this is just a shot in the dark basically if the speaker sounds like ass it could blow

and the 1 watt clipping isn't significant enough to cause damage to anything.
but when you're sending a clipped signal to a speaker thats moving about 3/4 or more of its throw... i would assume that damage could be done. while a clipped signal that doesn't even cause the speaker to move probably doesn't hurt it.

This is what I thought at first. Those abrupt stops you see don't actually happen to the speaker cone itself. The graph is not a graph of the position of the speaker cone, but of the amount that the amp is pushing the speaker cone.

The graph is like the graph of the spedometer of a car. An abrupt stop in the spedometer of a car doesn't mean that the car stopped, it means that the car stopped accelerating. The car only changes speed if the spedometer changes, not if it stops changing. The voltage is still positive (or negative depending on the phase) meaning it is still pushing the speaker, but it just can't push any harder because the amp it too weak.
Quote:

This is a very uninformed comment.

No it's not. You can think what you like, you can rationalize what you want, you can deny the experiment I did last night and you can use all the pipes and water examples you wish.

The introduction of the sharp edge of the square wave creates high frequency harmonics, as described by the Fourier transform. If you think otherwise, you are simply wrong. If you think those high frequency components cannot be separated by a crossover, you are wrong.

Reread the posts from ChrisWiggles and Jack Hidley (Director of Engineering at NHT). They are right, you are wrong.
There is no way he will understand, or admit to understand this. Give it up guys. I have!
You don't need to be an expert to be right. Take this: sin(x) + (1/3)*sin(3x) + (1/5)*sin(5x) + (1/7)*sin(7x) + (1/9)*sin(9x)+(1/11)sin(11x) and paste it into this online graphing utility: http://www.coolmath.com/graphit/index.html and hit "Eval". You can confirm this approximates a square wave. Look at the coefficients of each of the waves that is being added. It goes from 1 to 1/3 to 1/5 etc. Each of these coefficients represents the amplitude of each of these waves. Correct? To get the RMS voltage of each of these waves, multiplying by 0.707 should work. The RMS voltages are (rounded to the thousandth place) 0.707, 0.236, 0.141, 0.101, 0.079, and 0.064. Power, in watts, it volts squared divided by resistance. Assuming an impedance of 1 ohm, the harmonics would have wattages of 0.500, 0.056, 0.020, 0.010, 0.006, and 0.004. If we multiply all of these by 200 (to get the percentage of watts that each harmonic has compared to the original unclipped wave), we get 100.0%(original wave), 11.2%(3x frequency), 4.0%(5x frequency), 2.0%(7x frequency), 1.2%(9x frequency), and 0.8%(11x frequency).

Assuming I did everything correctly, you can now find the amount of power contained in each of the harmonics in a square wave by simply multiplying by wattage of the original unclipped wave by the percentage corresponding to the harmonic.

I am not going to argue with you anymore over whether these harmonics are separable by a crossover. I don't need to to show my original point. If you do a few example problems using the percentages I came up with, you will find that in most cases when the original unclipped wave wouldn't be damaging to the speaker, the harmonics won't be powerful enough to damage the tweeter, or they will be to low in frequency to pass through the crossover.

This may conflict with what some experts say, but I am fairly sure it is right. If you find any mistakes, I will correct them.

edit: I am using well known information and explaining every step of how I got these numbers. If I did everything correctly, then harmonics from clipping aren't the main cause of speaker failure.

You give the frequency of the wave that you are changing to a square wave. Then you give either the amplitude, RMS voltage, or wattage dissipated with a 1 ohm load. Then you give the number of harmonics you want to know the wattage of, including the original wave. It then gives you the wattages from high to low of the harmonics that would be in a square wave of the frequency you specified.

This way you can quickly and easily tell if it is at all possible for the tweeter to be damaged by the harmonics in the Fourier series of a particular square wave.

I based it off my post above, so if my post above has errors, please let me know so I can fix the utility.
It seems unlikely to blow a tweeter by clipping an 80 Hz beat. But clip an electric guitar and get into serious power to the tweeter. Somewhere in between, like a screaming voice, would pobably do it too. Most tweeters can't take much power.
Assuming a woofer is blown at anything above 100 watts and a tweeter is blown at anything above 10 watts, there are a few situations where the harmonics in the Fourier series of a square wave can destroy a tweeter. It you had a crossover set at 4000hz, if you tried to play a 100 watt 1000hz tone and it was clipped, you would get 20 watts at 5000hz and 14 watts at 7000hz, clearly something that would damage the tweeter.

I used this example specifically so that I could point out that if this speaker also had a mid-range that could handle 50 watts, the 1000hz tone would have blown the mid-range, clipped or not. Usually the extremely amplified high frequencies would cause damage before the harmonics do. Volume settings high enough to cause damage by clipping are dangerous to speakers, clipping or not. Don't pin it on clipping, pin it on people turning their amps up to higher settings that they should.

When talking about frequencies this close to potential crossover values and this loud, you should always be concerned, even with an amp that doesn't clip.
Quote:
Originally Posted by psgcdn

There is no way he will understand, or admit to understand this. Give it up guys. I have!

Words of wisdom. I give up, too. Good luck, greeniguana00.
Quote:
It you had a crossover set at 4000hz, if you tried to play a 100 watt 1000hz tone and it was clipped, you would get 20 watts at 5000hz and 14 watts at 7000hz, clearly something that would damage the tweeter.

OK OK OK. I admit, I just can't resist. Does this mean that you now acknowledge that a crossover can separate the harmonics generated by clipping? And that they exist at all?
I do admit that these harmonics exist... in the Fourier series of a square wave. I have admitted this all along. I still don't believe this means that they are nescesarily in the signal.

Before I believe that a crossover can separate the "harmonics generated by clipping" I will need a detailed, non-BS, well-supported explaination of the phenomenon that causes the tweeter to move when there are no high frequency voltage fluctuations. I am looking for something above just saying that sharp edges in the wave cause clipping. I want to know what exactly is causing a voltage fluctuation after the crossover does not exist before the crossover.

Maybe I am looking at this the wrong way? Maybe the source of the high frequencies is the amplifier, not the crossover. Amplifiers have THD ratings, and I assume they would increase when they are amplifing to a level that causes clipping.

Edit: Strange, I started reading about THD, and it seems to be in reference only to pure sine waves.
The square wave can be decomposed into a harmonic series.

A filter is required to do this.

A crossover is such a filter.
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