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post #181 of 514 Old 01-15-2007, 04:12 AM
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Originally Posted by greeniguana00 View Post

I don't know much about spectrum analyzers, but I would assume you would get 1 for the amplitude of the 100hz wave and 1/3 for the amplitude of the 300hz wave. To my knowledge, the spectrum analyzer assumes everything is a sine wave, so it has circuitry that will find what the original sine waves must have been.

The spectrum analyzer is going to display an input signal as amplitude vs. frequency. It doesn't assume anything is a sine wave. It's just measuring a signal and telling us at which frequencies energy lies. Think of the little display on a graphic EQ. Lower frequencies are indicated by more lights being lit on the left end of the display, and higher frequencies light up on the right side of the display.

So, yeah, assuming that the spectrum analyzer had the resolution, you would see a spike at 100 Hz and one at 300 Hz (and the one at 10kHz).

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If the crossover was set at 200hz, they would be mostly separated. This is because the resistance of the speaker drops when the voltage fluctuates at frequencies above 200hz, allowing more current to flow in parallel to the woofer, even though the RMS voltage is lower.

As someone else pointed out, they would be perfectly separated if the crossover were ideal. Trying to rationalize that there are HF fluctuations in the voltage of the signal itself is going to get you into trouble here, though. Think frequency content.

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From here onwards, the crossover would need to much more complicated.

It can be as complicated as you like. Analog (passive) or digital. Doesn't matter.

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By the way, thanks for taking the time to ask questions.

Sure.

Now, continuing with our function generator and mixer example, let's add a couple more function generators. One at 500 Hz and one at 700 Hz, with 1/5 and 1/7 amplitudes, respectively. And let's forget about the 10kHz signal. And let's leave our "ideal" crossover at 200 Hz. Can we still separate the 100 Hz signal from the 300, 500 and 700 Hz signals?

--Otto
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post #182 of 514 Old 01-15-2007, 04:35 AM
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Originally Posted by greeniguana00 View Post

The main idea is still true. More power is consumed by the speaker.

No, you can't make a statement that is this general. You still haven't taken into account the total impedance function of the system (amp, crossover, drivers). You're trying to look at one single component and generalize the performance of the system - this is very poor engineering practice.
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post #183 of 514 Old 01-15-2007, 04:52 AM
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Originally Posted by greeniguana00 View Post

People are incorrectly using this to say that a crossover will separate these harmonics.

It's not incorrect. A crossover WILL absolutely separate harmonics.

Look at it this way: if you have a 100 Hz sine wave and a 3100 Hz sine wave that are mixed, sent through an amp, and then into a passive crossover set to split at 800 Hz, it is clear that the wave at 100 Hz will go to the woofer, while the wave at 3100 Hz will go to the tweeter. This should be apparent - if it isn't, then there is a major understanding gap here.

It is also clear that 3100 Hz is one of the harmonics of 100 Hz, and would be present in a 100 Hz square wave.

Now, there is absolutely NO difference between the 3100 Hz signal by itself or acting as a component of the 100 Hz square wave. The crossover will push the energy in the 3100 Hz signal to the tweeter either way, with no difference in the net effect.

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A sine wave also IS the sum of an infinite series of square waves (that I haven't quite figured out yet).

It's not an infinite series, per se, but you can can generate a sine wave with square waves followed by a reconstruction filter acting at the Nyquist frequency. This is a totally different principle, however, from the understanding of the harmonic content of a waveshape, and how a signal processor will act on it. [A passive crossover is a rudimentary signal processor, and the math is the same.]

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I don't know what you mean by "intensity" but I hope you are using it in the right way.

I mean the energy present in the input signal in that frequency range. This is a basic integral function.

Pink noise is a great example. Pink noise is generated specifically to have equal energy in each octave - this makes system tuning much easier. Now, if you start at 20 Hz, it's fairly obvious that you have ten octaves' worth of information before you get to 20 kHz. Thus, if you have a 100W total power pink-noise signal, there will be 10W of power in each octave. Here's the kicker: Since the energy is equal-per-octave, the bands from 20-2,560 Hz will contain 70 W of that energy, while the remaining 30 W will be spread from 2560-20kHz.
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post #184 of 514 Old 01-15-2007, 05:11 AM
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Originally Posted by MauneyM View Post

It is also clear that 3100 Hz is one of the harmonics of 100 Hz, and would be present in a 100 Hz square wave.

Now, there is absolutely NO difference between the 3100 Hz signal by itself or acting as a component of the 100 Hz square wave. The crossover will push the energy in the 3100 Hz signal to the tweeter either way, with no difference in the net effect.

You're absolutely right, of course. But greeniguana00's next post is going to be a plot of V vs time for these two frequencies summed together. He will then point out that you can "see" the high frequency transitions in the time-based plot. It's the high frequency transitions that "get through" to the capacitor, you see. We've been going around and around with this for a while now, and it's quite amusing.

However, I think I may be making some progress. I think I got him to agree that we could separate 100 Hz and 300 Hz signals. My example has now added 500 and 700 Hz signals. Let's see if he can agree that we can separate the 100 Hz signal from those higher. We'll go from there.

--Otto
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post #185 of 514 Old 01-15-2007, 05:24 AM
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Originally Posted by MauneyM View Post

It's not incorrect. A crossover WILL absolutely separate harmonics.

Look at it this way: if you have a 100 Hz sine wave and a 3100 Hz sine wave that are mixed, sent through an amp, and then into a passive crossover set to split at 800 Hz, it is clear that the wave at 100 Hz will go to the woofer, while the wave at 3100 Hz will go to the tweeter. This should be apparent - if it isn't, then there is a major understanding gap here.

This is obvious. I am not arguing against this.

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It is also clear that 3100 Hz is one of the harmonics of 100 Hz, and would be present in a 100 Hz square wave.

Yes.

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Originally Posted by MauneyM View Post

Now, there is absolutely NO difference between the 3100 Hz signal by itself or acting as a component of the 100 Hz square wave. The crossover will push the energy in the 3100 Hz signal to the tweeter either way, with no difference in the net effect.

I won't disagree with this. The problem comes when you add everything else in between. Look at this graph:

You can see which graph is which on the right. The green wave just has the sum of a sine wave and another sine wave at 1/5 the amplitude and 5 times the frequency. The blue wave contains the missing wave at 1/3 the amplitude and 3 times the frequency of the original sine wave. You can see how the information isn't all there anymore.

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Originally Posted by MauneyM View Post

I mean the energy present in the input signal in that frequency range. This is a basic integral function.

Pink noise is a great example. Pink noise is generated specifically to have equal energy in each octave - this makes system tuning much easier. Now, if you start at 20 Hz, it's fairly obvious that you have ten octaves' worth of information before you get to 20 kHz. Thus, if you have a 100W total power pink-noise signal, there will be 10W of power in each octave. Here's the kicker: Since the energy is equal-per-octave, the bands from 20-2,560 Hz will contain 70 W of that energy, while the remaining 30 W will be spread from 2560-20kHz.

Thanks. There are just so many uses of the word "intensity" that I didn't know what you meant. I also didn't know you were talking about octave bands.

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post #186 of 514 Old 01-15-2007, 05:29 AM
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Originally Posted by OttoSpiral View Post

You're absolutely right, of course. But greeniguana00's next post is going to be a plot of V vs time for these two frequencies summed together. He will then point out that you can "see" the high frequency transitions in the time-based plot. It's the high frequency transitions that "get through" to the capacitor, you see. We've been going around and around with this for a while now, and it's quite amusing.

However, I think I may be making some progress. I think I got him to agree that we could separate 100 Hz and 300 Hz signals. My example has now added 500 and 700 Hz signals. Let's see if he can agree that we can separate the 100 Hz signal from those higher. We'll go from there.

You will not get an accurate representation of the 100hz wave.

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post #187 of 514 Old 01-15-2007, 05:31 AM
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I also didn't know you were talking about octave bands.

I don't mean to speak for MauneyM, but I don't think he was talking about octaves in general. That just came out of the pink noise discussion. Don't get stuck thinking in octaves now.

--Otto
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post #188 of 514 Old 01-15-2007, 05:33 AM
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Originally Posted by greeniguana00 View Post

You will not get an accurate representation of the 100hz wave.

Would you agree that the spectrum analyzer will give an accurate representation of each of these four signals (100, 300, 500, 700)?

--Otto
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post #189 of 514 Old 01-15-2007, 05:36 AM
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What frequency ranges are you talking about, MauneyM?

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post #190 of 514 Old 01-15-2007, 05:37 AM
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Originally Posted by OttoSpiral View Post

Would you agree that the spectrum analyzer will give an accurate representation of each of these four signals (100, 300, 500, 700)?

That depends on how it works, which I don't know much about. I would assume so. If it is a spectrum analyzer where if you put in a square wave, you will get a bunch of sine waves, then yes, it probably would give an accurate representation.

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post #191 of 514 Old 01-15-2007, 05:45 AM
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Originally Posted by greeniguana00 View Post

That depends on how it works, which I don't know much about. I would assume so.

In general, this type of device uses some type of implementation of band pass filters or FFT to quantify frequency content. If it were implemented with passive band pass filters (old school, no doubt, but still effective), it would be very akin to a speaker type crossover, no?

This whole conversation revolves around the UNDERSTANDING of the FACT that time domain signals have a direct, calculable and measurable representation in the FREQUENCY DOMAIN. By "representation" I don't mean that there's been anything done to modify the signal, just a different way of looking at it. They are one and the same.

One question -- do you understand the relationship between the time domain and the frequency domain?

--Otto
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post #192 of 514 Old 01-15-2007, 05:46 AM
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Originally Posted by greeniguana00 View Post

I won't disagree with this. The problem comes when you add everything else in between. Look at this graph:

No, it doesn't change anything.

I think you're having some trouble understanding the difference between the time domain and the frequency domain. All of the charts you are putting up reference the time domain, but what you are talking (frequency response and filters) about should be looked at in the frequency domain.

Bluntly, you're using the wrong model, so the visualizations you are coming up with aren't relevant to theproblem you're trying to solve.

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Thanks. There are just so many uses of the word "intensity" that I didn't know what you meant. I also didn't know you were talking about octave bands.

I was only referencing octave-based measurements in the context of the pink-noise/RTA example I was giving, but it is useful to make a point. What I was trying to get across is the difference between looking at a discrete waveshape in the TIME domain (scope trace - your charts) and looking at the frequency components in the FREQUENCY domain (RTA trace, sampling bins).

You cannot look directly at the waveshape and assume that it gives you the information necessary to understand the individual frequency components. For that you need a Fourier transform function, and what it returns is frequency domain information. The two are not interchangeable, although they are necessarily related in a steady-state condition.

[EDIT: looks like OttoSpiral and I hit the time/frequency thing at the same time....]
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post #193 of 514 Old 01-15-2007, 05:49 AM
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Originally Posted by greeniguana00 View Post

What frequency ranges are you talking about, MauneyM?

I know you asked MM, but I'll give my answer: It doesn't matter! All of this theoretical discussion of square waves, Fourier transforms, passive filters and so on applies at ANY frequency. Forget about "tweeters" being damaged and just consider it as some component. If you have a component that can deal with X watts of power at frequency F, then don't exceed X. Doesn't matter if F is 10 kHz or 10 GHz as long as F is within the specs of the component (edit: I mean the component's applicable bandwidth).

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post #194 of 514 Old 01-15-2007, 05:51 AM
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I do understand the difference, but I think it makes the most sense to look at this in the time domain. I can look at a square wave and say it's just one frequency; does this mean the crossover won't separate it into sine waves?

When you talk about Fourier transforms, you are talking about breaking it up into it's sine wave components, right?

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post #195 of 514 Old 01-15-2007, 05:55 AM
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Originally Posted by OttoSpiral View Post

I know you asked MM, but I'll give my answer: It doesn't matter! All of this theoretical discussion of square waves, Fourier transforms, passive filters and so on applies at ANY frequency. Forget about "tweeters" being damaged and just consider it as some component. If you have a component that can deal with X watts of power at frequency F, then don't exceed X. Doesn't matter if F is 10 kHz or 10 GHz as long as F is within the specs of the component (edit: I mean the component's applicable bandwidth).

Yes, but I wan't to know if these frequency bands he is talking about are equal.

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post #196 of 514 Old 01-15-2007, 05:56 AM
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Originally Posted by greeniguana00 View Post

I do understand the difference, but I think it makes the most sense to look at this in the time domain.

No, it doesn't make sense at all - that's the problem. That's rather like looking at your shock absorbers when the problem is that your engine is misfiring.

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When you talk about Fourier transforms, you are talking about breaking it up into it's sine wave components, right?

No. Fourier transforms divide the total energy in an input wave into frequency 'bins', not discrete sine waves. For example, you might have an FFT set up to give you bins of:

20-40 Hz
40-80 Hz
80-160 Hz
160-320 Hz

Each output bin will give you the amount of energy in the designated frequency range. I have used octaves because the math is easier, and this is the wa most RTAs atually work (though they would probably use 1/12 or 1/24 octave for any real application).
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post #197 of 514 Old 01-15-2007, 05:58 AM
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Originally Posted by greeniguana00 View Post

I do understand the difference, but I think it makes the most sense to look at this in the time domain.

When you are considering frequency based filters in makes THE MOST SENSE to consider looking at the signal's frequency content, doesn't it? It does NOT make sense to look at the signal in the time domain.

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When you talk about Fourier transforms, you are talking about breaking it up into it's sine wave components, right?

Sure.

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post #198 of 514 Old 01-15-2007, 05:59 AM
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Okay, so a square wave would be one frequency then, right? And this would mean it wouldn't be separated by a pure frequency based filter.

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post #199 of 514 Old 01-15-2007, 05:59 AM
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Originally Posted by greeniguana00 View Post

Yes, but I wan't to know if these frequency bands he is talking about are equal.

Equal what? A bin (frequency range) is a transform output value, nothing more.
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post #200 of 514 Old 01-15-2007, 06:00 AM
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I meant is it 20-30hz, 30-40hz, 40-50hz, etc. or 20-40hz, 40-80hz, 80-160hz, etc.

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post #201 of 514 Old 01-15-2007, 06:04 AM
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Originally Posted by greeniguana00 View Post

Okay, so a square wave would be one frequency then, right? And this would mean it wouldn't be separated by a pure frequency based filter.

A square wave has duty cycle that is often referred to as a frequency, especially in clocked systems. However, it's composed of the series of sine waves, as we all know. Only the sine wave is a "pure" frequency.

You could create a sine wave by passing a square wave through a low pass filter. Does that make sense?

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post #202 of 514 Old 01-15-2007, 06:07 AM
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Originally Posted by OttoSpiral View Post

A square wave has duty cycle that is often referred to as a frequency, especially in clocked systems. However, it's composed of the series of sine waves, as we all know. Only the sine wave is a "pure" frequency.

So, it IS based on sine waves then, not just frequency alone? A square wave has just one frequency of voltage fluctuation in it.

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You could create a sine wave by passing a square wave through a low pass filter. Does that make sense?

Yes. The squared edges would just bypass the woofer, leaving something roughly resembling a sine wave.

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post #203 of 514 Old 01-15-2007, 06:12 AM
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Those squared edges would bypass the woofer and damage the tweeter. If the signal had continued into a full sine wave, the tweeter would stil be damaged, as would the woofer. The crossover only knows what is happening at one moment, so to it, the steep slope could be from a clipped wave or the full sine wave during the time when the tweeter would be damaged.

If the tweeter is moving from that steep slope in a clipped wave, it will also be moving just as much if the wave isn't clipped.

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post #204 of 514 Old 01-15-2007, 06:16 AM
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Originally Posted by greeniguana00 View Post

So, it IS based on sine waves then, not just frequency alone? A square wave has just one frequency of voltage fluctuation in it.

I thought we agreed that the square wave is based on sine waves a long time ago???

I don't know what you mean by "frequency alone."

Yes, in the time domain, a square wave has just one frequency of voltage fluctuation in it. But that doesn't tell the whole story -- it does not relate anything about its frequency content.

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Yes. The squared edges would just bypass the woofer, leaving something roughly resembling a sine wave.

I don't know what "squared edges would just bypass the woofer" means. If everything was ideal (esp the low pass filter), it wouldn't be "something resembling a sine wave." I would BE a sine wave at the "frequency" of the square wave (assuming a 50% duty cycle).

Forget speakers for a minute. If I had a PC clock square wave at 3 GHz, I could pass through the appropriate low pass filter and get a 3 GHz sine wave.

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post #205 of 514 Old 01-15-2007, 06:19 AM
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Originally Posted by greeniguana00 View Post

Those squared edges would bypass the woofer and damage the tweeter. If the signal had continued into a full sine wave, the tweeter would stil be damaged, as would the woofer. The crossover only knows what is happening at one moment, so to it, the steep slope could be from a clipped wave or the full sine wave during the time when the tweeter would be damaged.

If the tweeter is moving from that steep slope in a clipped wave, it will also be moving just as much if the wave isn't clipped.

I love how all this is stated as fact.

--Otto
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post #206 of 514 Old 01-15-2007, 06:23 AM
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Originally Posted by OttoSpiral View Post

I thought we agreed that the square wave is based on sine waves a long time ago???

I don't know what you mean by "frequency alone."

Yes, in the time domain, a square wave has just one frequency of voltage fluctuation in it. But that doesn't tell the whole story -- it does not relate anything about its frequency content.

I could make the same arguement about a sine wave composed of square waves. Votlage fluctuations are the only way to tell if a frequency exists if you are not assuming everything is composed of sine waves.

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Originally Posted by OttoSpiral View Post

I don't know what "squared edges would just bypass the woofer" means. If everything was ideal (esp the low pass filter), it wouldn't be "something resembling a sine wave." I would BE a sine wave at the "frequency" of the square wave (assuming a 50% duty cycle).


Assume this is a sine wave and the square wave is perfect. The part of the square wave outside the sine wave that has those sharp corners would not get through the low pass filter, leaving only the sine wave.

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post #207 of 514 Old 01-15-2007, 06:27 AM
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Originally Posted by OttoSpiral View Post

I love how all this is stated as fact.

Are you saying a crossover can somehow remember what happened in the past, or predict what will happen in the future? All the crossover can know is everything about a small section of the wave. Most of the important information that the crossover has to work with is the derivative of the signal, which usually helps in separating high and low frequency content.

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post #208 of 514 Old 01-15-2007, 07:03 AM
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Originally Posted by greeniguana00 View Post

Are you saying a crossover can somehow remember what happened in the past, or predict what will happen in the future? All the crossover can know is everything about a small section of the wave. Most of the important information that the crossover has to work with is the derivative of the signal, which usually helps in separating high and low frequency content.

You're trying to rationalize all this, and it ain't working. Of course, I never said anything about the future. But there is something to do with the past. You can't only consider the signal as an instantaneous point. There has to be history to it, and, yes, that does matter to the crossover.

What I am saying is this -- in general, filters act on signals based on their frequency content, regardless of what the voltage vs. time signal looks like. Do you agree with that?

--Otto
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post #209 of 514 Old 01-15-2007, 08:10 AM
 
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I would assume so. If it is a spectrum analyzer where if you put in a square wave, you will get a bunch of sine waves,

Why don't you wiki "spectrum analyser", since you don't have a clue what it is, and hold wikipedia as the ultimate fact repository?
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post #210 of 514 Old 01-15-2007, 08:12 AM
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Originally Posted by greeniguana00 View Post

Are you saying a crossover can somehow remember what happened in the past

Yes. Capacitors and inductors work in the time realm, so the response is dependent on what has happened in the past. Charge on a cap (or in an inductor's core) is determined not be the present input state, but by it's historic state.

This is the difference between working in the time and frequency domains.


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All the crossover can know is everything about a small section of the wave.

Not true - see above. Also keep in mind that if a crossover is working in a frequency are as low as, say, 100 Hz, it must by definition deal with many cycles' worth of information at higher frequencies. For example, in the single cycle at 100 Hz, there are 100 cycles of a 10 kHz wave.
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