Post by: Zoesch on June 16, 2004, 07:36:51 PM
| Nika Aldrich wrote on Thu, 17 June 2004 03:25 |
Transducers are indeed infinite impulse response filters in that the impulse response of them has infinite characteristics, and when convolved with a stimulus the response does ring infinitely. Indeed the devices listed all have infinite impulse responses and they all are convolved with stimuli. |
To which I say, no they are not... infinite impulse response means that the response to an impulse exhibits infinite oscillations, this is what you would expect from a full-feedback system with no damping.
A transducer won't exhibit that behaviour, if excited with an impulse it will show a finite number of oscillations until it reaches equilibrium (Goes back to zero). If excited with a step function it will experience a finite number of oscillations until it reaches equilibrium.
This is consistent with control theory, if electromechanical transducers had infinite impulse responses simple things like a gearbox wouldn't work
| Quote: |
Since this seems wholy unrelated to Chuck's issue, perhaps this part of the discussion should be taken offline? I'd rather keep the topic focussed and helpful. |
It is related, IIR's weren't chosen on the reconstruction filter because of their impulse response characteristics, they were chosen because they are computationally efficient and easy to implement in a low power low cost device.
Post by: Nika Aldrich on June 16, 2004, 08:02:13 PM
| Zoesch wrote on Thu, 17 June 2004 00:36 |
A transducer won't exhibit that behaviour, if excited with an impulse it will show a finite number of oscillations until it reaches equilibrium (Goes back to zero). |
It never reaches equilibrium at a quantum level. It may APPEAR to reach equilibrium, but transducers do indeed have infinite impulse responses. Think of the forces at work against the transducer, like friction. Friction is a constant, so if you calculate the rate of decreasing displacement of a transducer it will continue to get smaller in perpetuitum but never actually reach the asymptote of equilibrium. That's not complex math. Transducers most definitely have infinite impulse responses.
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If excited with a step function it will experience a finite number of oscillations until it reaches equilibrium. |
How do you propose to calculate the exact number of oscillations? And what happens at the last oscillation - does it just SNAP into equilibrium?
| Quote: |
It is related, IIR's weren't chosen on the reconstruction filter because of their impulse response characteristics, they were chosen because they are computationally efficient and easy to implement in a low power low cost device. |
If you don't have an IIR filter then your waveform will never conform to Nyquist. If the waveform is time-limited it inherently has infinite bandwidth. Since infinite bandwidth is illegal we have to have a time-unlimited waveform - ergo it must be an IIR.
It has nothing to do with "computations." The IIR filter as after the conversion occurs - in the analog world.
Nika.
Post by: Zoesch on June 16, 2004, 11:59:17 PM
| Nika Aldrich wrote on Thu, 17 June 2004 10:02 |
It never reaches equilibrium at a quantum level. |
Nothing is at equilibrium at quantum level, but that's not what we're discussing here.
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It may APPEAR to reach equilibrium, but transducers do indeed have infinite impulse responses. |
I can prove to you that no, that's never the case, a system will reach equilibrium once the energy contributions prior and after the impulse, over time, are equal.
| Quote: |
Think of the forces at work against the transducer, like friction. Friction is a constant, so if you calculate the rate of decreasing displacement of a transducer it will continue to get smaller in perpetuitum but never actually reach the asymptote of equilibrium. |
It will once it reaches thermal equilibrium, WTF are you onto here? At one point in time the oscilation amplitude will be equal than the amplitude of the thermal movements of the material's molecules, how fast does that happen depends on the material, the temperature, the amplitude of the impulse and so on, but it WILL reach equilibrium, there's no running from the Laws of Thermodynamics.
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That's not complex math. Transducers most definitely have infinite impulse responses. |
This is basic physics, math can substantiate anything, even erroneous assumptions.
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How do you propose to calculate the exact number of oscillations? And what happens at the last oscillation - does it just SNAP into equilibrium? |
I don't propose, that's why control theory and impulse response characterization were invented/discovered.
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If you don't have an IIR filter then your waveform will never conform to Nyquist. If the waveform is time-limited it inherently has infinite bandwidth. Since infinite bandwidth is illegal we have to have a time-unlimited waveform - ergo it must be an IIR |
Nope, you can have a FIR filter for reconstruction, look up Quadrature Modulation Filters and see for yourself.
Again, it's all about efficiency... computational efficiency.
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It has nothing to do with "computations." The IIR filter as after the conversion occurs - in the analog world. |
Can you show me how you implement such a large summing network in analog? DAC's have a digital brickwall filter (IIR) that is followed by an analog LPF, so the IIR filter is in the boundary between digital and analog signals.
Post by: Nika Aldrich on June 17, 2004, 12:11:13 AM
| Zoesch wrote on Thu, 17 June 2004 04:59 |
I can prove to you that no, that's never the case, a system will reach equilibrium once the energy contributions prior and after the impulse, over time, are equal. |
If they continue to dissipate at a fixed rate then how do they ever reach equilibrium? Equilibrium is the asymptote.
| Quote: |
Nika: It has nothing to do with "computations." The IIR filter as after the conversion occurs - in the analog world. |
| Quote: |
Zoesch: Can you show me how you implement such a large summing network in analog? DAC's have a digital brickwall filter (IIR) that is followed by an analog LPF, so the IIR filter is in the boundary between digital and analog signals. |
The analog LPF is an IIR filter. A simple feedback loop is an IIR filter - a filter with an infinite impulse response.
Also, typically the digital LPF is a linear phase FIR, but that's beside the point.
Nika.
Post by: Erik on June 17, 2004, 12:27:41 AM
| Zoesch wrote on Wed, 16 June 2004 23:59 |
DAC's have a digital brickwall filter (IIR) that is followed by an analog LPF, so the IIR filter is in the boundary between digital and analog signals. |
1) DACs commonly use an FIR
2) An IIR can be represented as two FIRs
So really, what's the point of this thread other than to confuse newbies and make George nauseated by the drivel?
--Erik
Post by: Zoesch on June 17, 2004, 12:55:23 AM
| Erik wrote on Thu, 17 June 2004 14:27 |
So really, what's the point of this thread other than to confuse newbies and make George nauseated by the drivel? |
To make you come in and add to the drivel of course...
Post by: Zoesch on June 17, 2004, 01:37:42 AM
| Nika Aldrich wrote on Thu, 17 June 2004 14:11 |
If they continue to dissipate at a fixed rate then how do they ever reach equilibrium? Equilibrium is the asymptote. |
No, Equilibrium is the original state before exitation, you are getting complete equilibrium (Which you can only achieve at absolute zero) with thermal equilibrium. Thermal equilibrium will eventually happen as there's no more energy to dissipate.
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The analog LPF is an IIR filter. A simple feedback loop is an IIR filter - a filter with an infinite impulse response. |
A simple negative feedback loop on an ideal opamp that has zero losses is an IIR system (it's a buffer)... however, show me a perfect substractive device with no losses?
And again, this is not a transducer, there's no negative feedback on a speaker cone, there is lossy negative feedback on a speaker cabinet.
Post by: steve parker on June 17, 2004, 06:19:26 AM
" Equilibrium is the original state before exitation, you are getting complete equilibrium (Which you can only achieve at absolute zero) with thermal equilibrium. Thermal equilibrium will eventually happen as there's no more energy to dissipate."
if this were true would it not cut out the possibility of any IIR in the real world?
is this not just a case of modelling an "ideal" in which (as with most modelling) real-world things like friction are ignored?
steve parker.
Post by: Zoesch on June 17, 2004, 07:09:45 AM
But negative feedback systems ARE NOT IIR systems, which is what's being tossed around as being the case
Post by: Nika Aldrich on June 17, 2004, 02:33:54 PM
Sorry, we are not in agreement on this, the other thread. Can you just give me a formula that models the behavior of a transducer and the forces that act on it? Make it as simple as you'd like. I'd like to run the model and figure out when the result really becomes "0."
Nika.
Post by: Nika Aldrich on June 17, 2004, 02:55:41 PM
>I would say that all natural resonant system are IIR in nature
>up to the point where the oscillations become unmeasurable due
>to noise over time. Also resonant systems
>in the natural world DO have feedback - always. I.e, a guitar
>string or a pipe in an organ has mechanical and acoustic
>feedback. There is no such thing in the natural
>world as resonance without feedback of some kind. The FIR is an
>unnatural filter in the real world that can only exist within
>the bounds of math and signal processing.
I would just add that resonant systems don't cease their impulse responses just because they become unmeasurablely low amongst the noise. The behavior exists whether we have the tools to measure it or not.
Nika.
Post by: Zoesch on June 17, 2004, 03:45:07 PM
And that's where the all transducers are IIR system assumption is wrong.
Want to model it for yourself?
Model the speaker excursion, find the point where it becomes zero... that would be the point where all forces are at equilibrium (That's Newton, not me)... find the value of the tensor, input it back into the formula for V to SPL and find the voltage that matches that tensor.
There's your system at equilibrium
Not when the response becomes zero.
I can have a system whose electrical impulse response is infinite yet its mechanical response is finite, not asymptotically approaching zero.
And this is the danger of math without physics.
Post by: Nika Aldrich on June 17, 2004, 04:00:43 PM
| Zoesch wrote on Thu, 17 June 2004 20:45 |
And that is correct, |
then:
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I can have a system whose electrical impulse response is infinite yet its mechanical response is finite |
How can Paul's quote be correct - that any resonant device in nature has an infinite impulse response and that a finite impulse response is a creation of mathematics that can't happen in the natural world...
... yet you have a device with a finite impulse response when stimulated?
A cymbal - finite or infinite impulse response? According to you it has a finite impulse response. According to Paul it can't. How can you both be correct? And how can you be correct that Paul is correct when he disagrees with what you say?
Nika.
Post by: Zoesch on June 17, 2004, 05:01:40 PM
Not really.
Again, let's go back to the basics of a cymbal...
It gets struck by a stick (It's impulse response) and it vibrates in response to that impulse... it's surface will exhibit modes as the vibrations travel the surface of the cymbal and yet the cymbal itself will move in an elliptic fashion following the force and direction of the impact.
So far so good right?
The cymbal has to displace air in order for it to be heard, air which has a weight, air which due to its properties will do its damn best to go back and fill the space that was left void.
Air who also acts as a dampener, because it has friction, because work has to be done for the cymbal to displace that air, work which dissipates as heat, heat which is not infinite because neither is the work nor the initial force.
So as you dissipate heat, you have less energy to work, and after each oscillation you'll have less and less energy, until you reach thermal equilibrium... you have no more energy to dissipate, your initial hit, which exerted a work potential on your system has dissipated.
And so far you are in agreement with Thermodynamics.
Your cymbal will return to its original state before the impulse, it might be a long time, it might be a short time, it all depends on the efficiency of heat transfer and the work efficiency.
So far you are in agreement with basic Newtonian physics.
There's no feedback on the system.
But what if you were actually playing against a wall? Wouldn't reflections act as feedback? No, simply because their force contribution would be significantly less than those of the system and they will also diminish with time.
If a butterfly batters its wings in china, do I feel it on my cymbals? Sure, does it change the state of the system? No...
Likewise with a speaker cone, a microphone diaphragm, and even a tuned pipe on an acoustic organ.
And here's where people get their cables crossed.
If I model a speaker, I can simply model its frequency response as the product of two bandpass filters (One high, the tweeter and one low, the woofer), if I take that model and look at its impulse response it will have an infinite impulse response... sure, and that's not contradictory, it's a model.
But if you want a real world extension of your model you need to add losses, non-linear behaviour of the speaker cones, breakup modes on the cone and so on...
It's no longer the product of two band pass filters. It becomes a complex differential non-linear system, whose impulse response is not infinite.
So Paul is right, and I am right, and so far neither of us are contradicting each other.
To the experiment yourself, grab a speaker cone, wire it to a toggle switch and press it, recording the output with a measurement microphone, leave it a long time...
You won't be able to measure below the noise floor of the microphone however, but if you have a spectral interferometer and a couple of lasers handy it makes for a very fun experiment.
Post by: Nika Aldrich on June 17, 2004, 05:47:39 PM
| Zoesch wrote on Thu, 17 June 2004 22:01 |
Are Paul and I disagreeing? Not really. Again, let's go back to the basics of a cymbal... It gets struck by a stick (It's impulse response) and it vibrates in response to that impulse... it's surface will exhibit modes as the vibrations travel the surface of the cymbal and yet the cymbal itself will move in an elliptic fashion following the force and direction of the impact. So far so good right? |
Yes. And just for the record, I really am genuinely interested in understanding your perspective on this. I do believe that you are incorrect, but I am interested in seeing if our perspectives meet at a mutual understanding. So yes, I'm following along, and we hit the cymbal with an impulse...
| Quote: |
The cymbal has to displace air in order for it to be heard, air which has a weight, air which due to its properties will do its damn best to go back and fill the space that was left void. ... So as you dissipate heat, you have less energy to work, and after each oscillation you'll have less and less energy, until you reach thermal equilibrium... you have no more energy to dissipate, your initial hit, which exerted a work potential on your system has dissipated. And so far you are in agreement with Thermodynamics. |
Check. We're on the same page so far.
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Your cymbal will return to its original state before the impulse, it might be a long time, it might be a short time, it all depends on the efficiency of heat transfer and the work efficiency. |
This is where we disagree. The rate in which the energy is converted from the cymbal's behavior into heat is a constant, and the resonation of the cymbal decreases in amplitude at a fixed percentage over like moments in time. Kinetic friction is a constant force, and as a constant force it will decrease the movement of the cymbal a like percentage in like amounts of time. As such it will never return to a fixed state but will rather always be in motion. Start with an impulse of amplitude 1 and use any force constant to ascertain the point at which it's movement ceases and you will find that it never does.
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There's no feedback on the system. |
The feedback has to do with the mechanical structure of the device acting on itself, but I don't want to go down that path. I'm not as comfortable thinking of it that way - I'll leave that to Paul. It was his quote that all resonant devices have mechanical feedback in them.
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But what if you were actually playing against a wall? Wouldn't reflections act as feedback? No, simply because their force contribution would be significantly less than those of the system and they will also diminish with time. |
Key word here is "diminish." Yes, the movement of all mechanical devices will diminish over time, but it never completely stops.
| Quote: |
But if you want a real world extension of your model you need to add losses, non-linear behaviour of the speaker cones, breakup modes on the cone and so on... It's no longer the product of two band pass filters. It becomes a complex differential non-linear system, whose impulse response is not infinite. |
I agree that it becomes a complex, differential, non-linear system, but all forces acting on the speaker are still constant and thus the impulse response remains infinite. Take all of the combined forces acting on the speaker in all directions and add them all together and then run the speaker. It still has constant forces acting on it in the end. And as long as forces are constant its movement will never cease.
| Quote: |
So Paul is right, and I am right, and so far neither of us are contradicting each other. |
I'm not here to battle who is wrong and who is right, but you are insisting that a speaker has a finite impulse response and Paul and I are both saying that no such thing exists in the natural world, and that all mechanical devices - namely all devices with elasticity exhibit infinite behaviour as the response to an impulse. I don't possibly see how we can all be right about this issue together.
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To the experiment yourself, grab a speaker cone, wire it to a toggle switch and press it, recording the output with a measurement microphone, leave it a long time... You won't be able to measure below the noise floor of the microphone... |
Right! But not being able to measure it does not mean that the behavior ceases to continue. The original transient still has an effect on the cymbal even though the amplitude of the movement becomes less than the amplitude of the random movement of the measuring device, the air around it, etc. The impulse's effect is still present if we could only measure it. We humans, for example, can measure the amplitude of waveforms that are as small as 1/16 the amplitude of random noise that it exists within, as that is the limitation of our temporal masking. More precise devices can measure the amplitude with filters far deeper into the noisefloor than that. The behavior exists whether we can measure it or not. It is still an infinite impulse response.
Nika.
Post by: Nika Aldrich on June 17, 2004, 06:50:57 PM
Two questions really quickly:
1. Put an impulse into a physical space like a room and let the reverberant field decay. Would you agree that the sound never completely decays and that the resulting sound has an effect on the air pressure behavior in the room ad infinitum?
2. Take a speaker that you believe has a finite impulse response. Put an impulse into it. What would potentially be the amplitude of the last peak excursion before the speaker reaches a 100% rested state at equilibrium. What would be the potential amplitude of, say, the last peak before that one? Maybe .00000038M? and .00000076M?
Nika.
Post by: Daniel_Dettwiler on June 17, 2004, 07:35:45 PM
iir systems (infinite impulse response) works, because in practise they don't resonate for ever. The easiest digital lowpass filer is a iir filter (I don't recall the formula, I am sure nika or someone can bring it...). Theoreticly if you give an impulse in this filter it would sound (resonate) for ever. Practically after a very very short time, the sound is gone into the last bit. So while the sound does not anymore exist in practise, in theorie this filter would sound for ever. Thats why the filter is an IIR Filter. It does not matter if in practise the sound goes a way, by any sort of limitations fo the system.
Most analog filters (if not all) are IIR (if im right), but same here, the signal is very fast down to the noise floor. But theoretically it would resonate for ever, thats why it is called IIR .
Divide 100 by two as many times as you want, it will never be zero. It will always be half as little as the number before.
A piano string (or any string for that reason) is an IIR System as well. After you hit the string, it would sound for ever. It does not matter if in practis at some point you think that the string does not anymore resonate because of some limitations of the system. In theorie a string would resonate for ever, and that is what matters.
So I think Loudspeaker are IIR Systems.
Daniel
Post by: archtop on June 17, 2004, 11:42:07 PM
You know I been tryin' to follow along ( I have no idea why)
Physics in my tiny high school involved dodgeball and running
Nika, you are sayin' the cymbal never stops,
at the molecular level maybe, but I don't hear that.
so for me, it does stop, the piano string too, thats why we hit 'em again.
is my zipper open, I feel funny, why am I reminded of the old saying
"Better to remain silent, and thought a fool than to open your mouth and remove all doubt"
alright, I'll get back outta the way now
Richard Williams
Post by: steve parker on June 18, 2004, 01:09:43 PM
if this were true would it not cut out the possibility of any IIR in the real world?
is this not just a case of modelling an "ideal" in which (as with most modelling) real-world things like friction are ignored?
sorry to quote myself (bad form and all that!)
thanks for your reply stefan but...are my statements above where your disagreements with nika are or am i missing the point somewhere?
where i'm struggling is that the real world always includes factors that negate infinities in mathematical models of anything. that doesn't stop the models being good ones and i can't think of any examples where the application is considered qualitatively different from the model, just because the application doesn't perform to the mathematical ideal.
or...do you disagree that even models of speakers and strings etc are IIRs?
all the best!
steve parker.
Post by: Nika Aldrich on June 18, 2004, 01:21:09 PM
| fuze wrote on Fri, 18 June 2004 18:09 |
where i'm struggling is that the real world always includes factors that negate infinities in mathematical models of anything. |
Mike,
No. See Paul Frindle's post on the other thread. The real world does not have factors that negate infinities. It just adds more complex forces that make the equations more complex, but forces are always constant and thus infinity is still in play. The real world also adds noise, but the addition of noise does not mean that the behavior doesn't still exist, at an amplitude that is lower than that of the noise.
Indeed IIR filters are NOT just an ideal model of the real world. Transducers in the real world really ARE IIR filters. Our models of them are often simplified and don't really fully demonstrate their movement. We often simplify and only show one force acting. But even if we show ALL of the forces acting, the transducer doesn't actually ever stop. It just decreases its movement in amplitude much faster than the model shows.
Make sense?
Nika.
Post by: Zoesch on June 18, 2004, 05:02:27 PM
Are you implying that no system returns to its original state?
You were asking me about what amplitude will the ripple have before it gets cancelled, which is easy, again, just calculate where the speaker motion is equal to the atmospheric pressure, your transducer can't go any lower than that.
Atomic resonance has nothing to do with mechanical feedback... just like potential wells have nothing to do with friction. Every atomic structure will resonate at a specific frequency that is given by its structural composition and energy levels, and this is the energy level in which energy transfer is maximized.
Atoms are oscillators in perpetual motion, but they also lose energy over time, shed electrons, become ionized and so on.
Once the impulse response amplitude is down at the original motion of the material the impulse response becomes finite, again, you can't generate energy out of nowhere, unless you are implying that every real world exitation results in atomic fission.
Is the measured time for the system to reach equilibrium inconmensurably large? Yes, well, not really but let's assume so.
Is it infinite? No.
Post by: Nika Aldrich on June 18, 2004, 05:30:05 PM
| Zoesch wrote on Fri, 18 June 2004 22:02 |
Nika, are you implying that the phenomena continues at a subatomic level? |
Yes. Just like a single bit can represent infinite dynamic range, though much of it is well below the noise floor.
| Quote: |
Are you implying that no system returns to its original state? |
Yes.
| Quote: |
You were asking me about what amplitude will the ripple have before it gets cancelled, which is easy, again, just calculate where the speaker motion is equal to the atmospheric pressure, your transducer can't go any lower than that. |
If the forces on it are a constant it will never equal the atmospheric pressure of its environment.
Did you read Paul's post last night on the tremolo thread? He seems to ardently disagree with you.
Nika.
Post by: Zoesch on June 18, 2004, 09:32:46 PM
Take potential wells/barriers, in a Josephson junction for example (Which has a boundary condition that violently disagrees with most of the basic principles of system equilibrium) there's a measurable current output as temperature lowers even when there's no potential difference across the electrodes, but it's because both superconductors having different ranges need to maintain phase coherency at the barrier.
In semiconductive materials the electrons need to have enough energy to traverse the potential barrier, so yes, once your energy is below that barrier you are done,a dn if you lower the temperature the material becomes less conductive.
In non-conductive material the potential barrier is much higher, your electrons must have a very high enery level to traverse the barrier or be exited with a tremendous amount of energy.
All materials have potential wells and barriers, some are lower than others.
Now in your impulse response dilemma, the energy of the impulse will eventually fall below the potential barrier of the material, the result? No more contributions to the material's movement and the response.
On resonances... all atoms resonate, not all molecules resonate (And some who resonate change states after resonation), why do you think that is happening?
I read Paul's posts, I don't see the ardent part... he's arguing this:
We have experimentally concluded that within observable intervals the energy decreases exponentially, ergo the energy in the system decreases exponentially and thus never approaches equilibrium.
That's the conclusion out of observable phenomena, what happens where we can't measure we can only infer with a mathematical model.
Again, what we know of the system is that it approaches a state where the phenomena is unobservable and it resembles the original state.
I'm not saying that the model is wrong, it's a model, but the model doesn't return to equilibrium, and that's a modellic assumption.
Say you want to make the model more complex, add all the forces, mechanical and electric, that affect your system, add losses, add leakage, add feedback and add random variations on those forces, does it mean that the observation of exponential loss is valid? Only on those intervals where it is observable and quantifiable... where you can't measure you can only infer.
Post by: Nika Aldrich on June 18, 2004, 09:41:41 PM
And explain to me how this differs from an IIR filter in the electrical world? Is an electrical IIR filter a "true" IIR filter based on your understanding of the atomic level physics involved, any more so than a transducer is?
Nika.
Post by: Erik on June 19, 2004, 12:46:36 AM
After giving up on the basic math proof, I argued everything from meteorology to the fact that black people don't rely on samplers to put the pump in the thump for bass lines -- they use synths.
He responded with more of the high school debate club antics you're seeing here. You can keep pointing out the obvious until you're blue in the face -- but he'll squirm and suddenly start talking about how lift isn't bernoulli, it's newton.
In these types of discussions (and especially that 'tremolo' thread), it's always fun to take some things people are saying when they obviously don't understand what they mean, put them between quote marks, and then paste it into Google.
Turns the sites they're plagiarizing right up. (Works great on the DUC, too.)
--Erik
Post by: Zoesch on June 19, 2004, 02:15:16 AM
BTW Erik, before you enter a debate make sure you at least understand what an inductor is... since your models are so exact that not even convolution can touch them I assume that you understand impedance and inductance.
Ta!
Post by: Nika Aldrich on June 19, 2004, 02:34:34 AM
Still waiting on an answer to the question...
| Zoesch wrote on Sat, 19 June 2004 07:15 |
BTW Erik, before you enter a debate make sure you at least understand what an inductor is... since your models are so exact that not even convolution can touch them I assume that you understand impedance and inductance. |
Hmm. Seems totally unrelated to me. I fail to see how choosing the appropriate mathematical modelling to simulate a non-linear process requires an understanding of inductors. Frankly, I don't understand inductance at all, but I know where convolution is appropriate and where it is not. I also don't know the proper methods for making an electro-optical circuit, but I know how to count in two's complement. So?
I also fail to see how any of this is related to the slowly-ending mystery of whether or not transducers are IIR filters.
Nika.
Post by: Zoesch on June 19, 2004, 04:20:35 AM
| Nika Aldrich wrote on Sat, 19 June 2004 16:34 |
Zoesch, Still waiting on an answer to the question... |
So am I, I asked you ages ago to prove that the system after a period of time isn't at the same state as the system before the exitation... so far, I'm still waiting. I also asked you where's the feedback loop on a speaker cone or a microphone capsule, still no answer on that one either.
| Quote: |
Hmm. Seems totally unrelated to me. I fail to see how choosing the appropriate mathematical modelling to simulate a non-linear process requires an understanding of inductors. Frankly, I don't understand inductance at all, but I know where convolution is appropriate and where it is not. I also don't know the proper methods for making an electro-optical circuit, but I know how to count in two's complement. So? |
So? I know both, I ... and one thing that you don't know is the context in which that conversation happened and of this remark, which as far as I can see wasn't directed to you.
And again, put emphasis on the word appropriate... is the model appropriate because it is 100% correct or because it is 100% solvable (And those two might be mutually exclusive)... nevermind, I think this is heading down the path of arguing that Carbon, Graphite and Diamonds are exactly the same.
BTW so it seems that the new trend is to only understand what you do, grab a mathematical model of an inductor and assume is correct without understanding how one in real life work. Choose your poison.
| Quote: |
I also fail to see how any of this is related to the slowly-ending mystery of whether or not transducers are IIR filters. Nika. |
Don't know, you tell me...
Post by: Nika Aldrich on June 19, 2004, 09:38:13 AM
| Zoesch wrote on Sat, 19 June 2004 09:20 | ||
So am I, I asked you ages ago to prove that the system after a period of time isn't at the same state as the system before the exitation... so far, I'm still waiting. |
I have tried to prove it to you in any way I know how to no avail.
| Quote: |
I also asked you where's the feedback loop on a speaker cone or a microphone capsule, still no answer on that one either. |
And I told you that Paul Frindle would be able to answer that better than I. I simply don't see the significance of the presence of feedback on an IIR filter like Paul does. If the impulse response of a device is infinite it is an IIR filter - regardless of what causes it to be that way. Feedback loops are how we mimic that behavior in the electrical world.
But again, if you argue that a speaker is not a true IIR filter because the physics breaks down at some sub-atomic level, then I ask how that is different from an electrical IIR filter? Are either of them "true" IIR filters? It appears to me that you are avoiding this question.
Nika.
Post by: Zoesch on June 19, 2004, 01:27:03 PM
Post by: Nika Aldrich on June 19, 2004, 02:11:58 PM
| Zoesch wrote on Sat, 19 June 2004 18:27 |
I told you before that the answer to both is no, they are not... \ |
Then WHY IN THE WORLD ARE YOU PICKING THIS BLOODY FIGHT??
If you want to rename IIR filters to recognize that at the subatomic level there is still some mystery then that is a completely separate discussion. But by the definitions in use for IIR filters both transducers and electrical feedback systems are IIR filters, granted that at some nanoscopic level both fade into question. Either transducers and feedback systems are both FIR filters or they are both IIR filters, but they are the same, and you are intentionally obscuring that fact because you realize you made a mistake several pages ago and now you're just being stubborn for the sake of procrastinating an acknowledgement that you were out of line. You've now had both myself and Paul tell you that yet you persist with the same "when they reach equilibrium" now, wasting my time for pages. Of course if they "reach" equilibrium at a steady state then they are the same. They will never reach equilibrium, per the fact that forces are constants.
Can you just apologize for the error and let us drop this ridiculous thread before it gets even more embarassing?
Transducers, as with ANY natural resonating object, are IIR filters. FIR filters, per Paul's statement that you said you agree with, are only figments of mathematical limitations that do not accurately model anything in the natural world. You said you agree with Paul, so great, we'll put that to rest. Transducers are NOT FIR filters.
Great.
I'm done.
Nika.
Post by: ajmogis on June 19, 2004, 02:46:52 PM
Think about a digital IIR filter. At some point it will stop resonating and all you're left with is digital zero and the dither bubbling away. It might in some way still be mathematically resonating to infinity (and minus infinity for that matter!), but that's neither terrribly useful or important. There's still a point at which there's no more signal. You run out of bits.
In the "real" world it is the same. At some point the cymbal stops resonating and you're left with the thermal vibrations of the molecules. Infinity is still there, it's just not important. The definition of something having an IIR or an FIR has more to do with the other characteristics of how they work and how they're constructed, not whether or not they REALLY go on forever. It's just an easy way to name the two varieties so you can talk about them.
As far as I know FIRs exist only in the digital world.
-AJ
Post by: Johnny B on June 20, 2004, 03:01:02 AM
If I bang the hell out of a guitar that's
plugged into a Marshall half stack and get
screaming feedback with it, but then turn
the amp off, unplug it from the wall, and
come back 3 months later, are any of the
Greenback speakers still vibrating?
My guess would be, "No."
But I could be wrong, I often am.
Post by: steve parker on June 20, 2004, 05:40:06 AM
plugged into a Marshall half stack and get
screaming feedback with it, but then turn
the amp off, unplug it from the wall, and
come back 3 months later, are any of the
Greenback speakers still vibrating?"
stefan would say no....
nika might say yes.........(actually i doubt it)..
the only answer i'm happy with is "no - but due only to outside influences - friction etc."
steve parker
Post by: Zoesch on June 20, 2004, 06:33:21 AM
| Nika Aldrich wrote on Sun, 20 June 2004 04:11 |
Then WHY IN THE WORLD ARE YOU PICKING THIS BLOODY FIGHT?? |
Sorry, I didn't get the memo that this was a fight... I intend to keep my composture, I'd suggest you do the same.
| Quote: |
If you want to rename IIR filters to recognize that at the subatomic level there is still some mystery then that is a completely separate discussion. |
That's not the issue, you said on the other thread "Oh, BTW, transducers are IIR systems" and I said no... if you had said "Oh, BTW, transducer mathematical models exhibit an infinite impulse response, in real life they will return to the initial pre-impulse state due to the rejection properties of materials" and this thread wouldn't exit.
Call me anal retentive, but last thing that I want is people to believe that their speakers and microphone capsules are still vibrating after they have been left alone.
| Quote: |
But by the definitions in use for IIR filters both transducers and electrical feedback systems are IIR filters, granted that at some nanoscopic level both fade into question. |
Micro and macroscopic, again, take a speaker cone, you have a free field magnetic coil that pushes an elastic cone, this elastic cone exhibits inertia and you need a tensor for it to move, any force below this tensor value will not make the speaker cone move, comprende? The coil might be oscillating ad infinitum (It won't) but those oscillatons won't affect the speaker cone at the macro and microscopic levels.
The transducer is a system to convert electrical energy into mechanical energy and has to be treated as such... if you want to use a simplified model, be my guest, make sure you label it as such.
That's like taking the small signal transistor model of one resistor and one current source to be the exact behavior of the transistor for all inputs.
| Quote: |
Either transducers and feedback systems are both FIR filters or they are both IIR filters, but they are the same, and you are intentionally obscuring that fact because you realize you made a mistake several pages ago and now you're just being stubborn for the sake of procrastinating an acknowledgement that you were out of line. |
Excuse me, but I'm not intentionally or stubbornly obscuring anything... I'm being crystal clear with you... does the model exhibit an infinite impulse response? Yes, does the physical system exhibit that behavior? No.
| Quote: |
You've now had both myself and Paul tell you that yet you persist with the same "when they reach equilibrium" now, wasting my time for pages. |
And mine as well...
| Quote: |
Of course if they "reach" equilibrium at a steady state then they are the same. |
Thank you, that would've saved you pages of debate... but moving on...
| Quote: |
They will never reach equilibrium, per the fact that forces are constants. |
Oh, but they are not, if you blast air into a room expect a change in room pressure for a while, if your sound wave hits a wall expect a reflection... hardly constant right?
If you take a simple second order low pass filter and measure its output with no input you get a measurement right? That's your equilibrium, and that's not zero, so any exponential decay function will converge to it.
You are stubbornly going on about equilibrium being "zero" and I'm stubbornly going on about equilibrium being that initial measurement.
| Quote: |
Can you just apologize for the error and let us drop this ridiculous thread before it gets even more embarassing? |
Didn't know I owed you an apology or anything.
| Quote: |
Transducers, as with ANY natural resonating object, are IIR filters. FIR filters, per Paul's statement that you said you agree with, are only figments of mathematical limitations that do not accurately model anything in the natural world. |
FIR filters are not implementable in the analog domain... big difference between that an your statement.
| Quote: |
You said you agree with Paul, so great, we'll put that to rest. Transducers are NOT FIR filters. |
What Paul and I are agreeing with is not what you and I are disagreeing with, you are standing on a mathematical premise that has no resemblance to physical behavior, but so be it.
| Quote: |
Great. I'm done. Nika. |
Great, so am I... George can decide whether to leave this thread up or delete it as he sees fit.
Post by: PP on June 20, 2004, 08:31:11 AM
There is an extremely useful lesson to be learnt here.
This thread eloquently demonstrates the advantages of allowing people to be able to post their views whilst others (with a differing view) can post theirs separately, giving their views and opinions.
This allows those interested enough in this forum to read it allowing them the respect of being able to utilise their own brains, experience and personal judgement in order to decide the relative merits between the respective views.
We have an excellent moderator, highly acclaimed throughout the entire world for his far reaching contributions to the professional recording industry that does intervene very decisively, when he feels it is necessary and properly merited.
None of us should presume ourselves worthy to usurp the moderator’s role and overstep our authority in this forum by doing so ourselves.
Enduing upon ourselves an authority we do not in fact possess, and then using it to effectively censure others whose views differ from our own.
This is of concern to me, as several internationally regarded engineers have commented that they would indeed have liked to participate in this forum at times.
But do not wish to have their views pulled apart piece meal by others with a lot more time to spare, that are far less experienced, and a whole way further down the food chain to boot!
They want to be able to express their opinions without the NECESSITY of engaging in a time consuming argument involving complex matters where semantics alone can give a completely false representation of their views.
They don’t want line by line dissections of their postings reposted and twisted against other unrelated comments they have made, that give a wholly unrepresentative view of their opinion.
This is not unreasonable.
They just want a chance to have their say without a wrangle. This drives contributors away from the forum as regards to being willing to post here themselves.
Surely,
This is DEMOCRACY!
This particular threads debate reminds me of a conversation I had with someone a while ago regarding aerodynamics.
Basically the young man (thankfully working now for Mercedes Benz) was seeking to explain to me that if a 747 flew through the sky creating spiralling patterns of air from its wing edges, those spiralling patterns would continue in infinity right round the earth’s atmosphere and go on ad infinitum.
Theoretically, this may be so,
Real life is not theoretical though is it?
Other air currents will eventually dominate and control the flow of air and at times that air might even be completely still.
But let’s not linger in the doldrums.
Perhaps this will help.
http://www.whale.to/m/cathie.html
Let’s lighten up a little and allow others to have their view expressed.
The world will NOT spontaneously combust if we do.
Best Wishes to all
Peter
Peter Poyser
Post by: Paul Frindle on June 20, 2004, 07:58:53 PM
| ajmogis wrote on Sat, 19 June 2004 19:46 |
Think about a digital IIR filter. At some point it will stop resonating and all you're left with is digital zero and the dither bubbling away. It might in some way still be mathematically resonating to infinity (and minus infinity for that matter!), but that's neither terrribly useful or important. There's still a point at which there's no more signal. You run out of bits. In the "real" world it is the same. At some point the cymbal stops resonating and you're left with the thermal vibrations of the molecules. Infinity is still there, it's just not important. The definition of something having an IIR or an FIR has more to do with the other characteristics of how they work and how they're constructed, not whether or not they REALLY go on forever. It's just an easy way to name the two varieties so you can talk about them. As far as I know FIRs exist only in the digital world. -AJ |
Yes this illustrates a good point. The digital filter IIR model you quote is actualy NOT an accurate IIR in a natural sense because you have allowed it to 'run out of bits' - it has therefore been allowed to stop - i.e. it's no longer infinite.
And this is the hub of the misunderstanding.
In reality what SHOULD happen with a digital IIR (and in fact we always ensure it does) is that the dither (very very importantly) will prevent the filter from stopping and therefore the transfer charateristic of the IIR remains in permanence - like natural physical systems in the real world. What this means is the the freq response that the filter applies to the noise - or dither - or ANY possible stimulus however small, will remain true to the IIR implementation. I.e. it does not need an external stimulus to 'become' a filter and it never stops being one
As would be the case with say a cymbal - i.e. its transfer characteristic is essentially constant (leaving out the possibility of material being modified by the force of a loud smash) whether responding to a smash, resonating with external ambient sound in the environment, or just being moved by the noise of the Brownian motion of air molecules due to temperature. I.e. it does not stop having resonance or the ability to modify signal at ANY time.
However long you left it in the quiet you would theoretically be able to extract it's resonances and freq responses - provided you could do a long enough FFT analysis - it's just that it would appear to tend towards becoming constant energy (noise filtered by the cymbals response) in the end cos the noise would eventually overwhelm the exponential decay of the original stimulus.
Even if the cymbal were placed in a complete vaccuum without any losses (and no external stimuli by definition) it would still produce it's charateristic response due to molecular movement because it was not completely cold - it would just be somewhat quieter (i.e. the molecular noise would have the response of the cymbal imposed on it). The only time it would ever come to a complete (absolute) rest and cease being a filter of some kind (or in fact even a cymbal at all) would be at absolute zero temperature - i.e. infinitely cold, i.e. electrons at rest with no atomic movement - no longer matter as we know it - and impossible by definition to achieve fully.
And yes you are right in saying that an Finite IR filter (with an impulse response that ends after the calculation has finished) is an oddity that can only exist within a mathematical environment and has no parallel in the physical world AFAIK - because nature does not 'calculate' within the human paradigm of a 'finite math precision' representation. It is only humans that evaluate by 'counting' and give up looking when things get beyond our reasonable perceptions and therefore seek to percieve reality as 'numerically quantised'.
Nature has no need to. And in fact neither do we ultimately
Post by: Paul Frindle on June 20, 2004, 08:18:20 PM
| Peter Oxford wrote on Sun, 20 June 2004 13:31 |
Gentlemen! <snip) Let’s lighten up a little and allow others to have their view expressed. The world will NOT spontaneously combust if we do. Best Wishes to all Peter Peter Poyser |
Here here
In fact this argument is a philosophical one - but none the less it is becoming one of great importance IMO, since it is exposing the thinking concepts and perceptual mutual exclusions that underpin much of the popular unease with digital processing.
Post by: Nika Aldrich on June 21, 2004, 10:43:20 AM
Nika.
Post by: Zoesch on June 21, 2004, 03:19:20 PM
| Nika Aldrich wrote on Tue, 22 June 2004 00:43 |
Thank you, Paul. Great post. I think that sums that up fairly well. FIR systems can only occur in the mathematical realm and cannot occur in the natural realm. All devices that resonate in the natural world are filters of an IIR variety, including, therefore, transducers. Nika. |
I hope you have the presence of mind to realize that what Paul is saying and what you are saying aren't the same. Paul is talking about the cymbal's transfer function, being infinite (Which it is, as it exists for all points in time), that its properties being continuous (Which they are, as they are continuous for all points in time) and that if it is a filter, it never stops being one (Which it won't, of course). He is not saying that the phenomena will continue below the noise floor, again, unless you have measured it you are only inferring that it happens.
This is a hell of a lot different than saying All transducers (Including those on free field configurations?) are IIR Filters.
Post by: Eliott James on June 21, 2004, 03:54:28 PM
| Nika Aldrich wrote on Mon, 21 June 2004 10:43 |
Thank you, Paul. Great post. I think that sums that up fairly well. FIR systems can only occur in the mathematical realm and cannot occur in the natural realm. All devices that resonate in the natural world are filters of an IIR variety, including, therefore, transducers. Nika. |
Do you realize that all we have in reality IS the natural realm? All your mind can conceive of math and any other thing is part of the natural realm.
Post by: Nika Aldrich on June 21, 2004, 04:20:39 PM
| Zoesch wrote on Mon, 21 June 2004 20:19 |
I hope you have the presence of mind to realize that what Paul is saying and what you are saying aren't the same. Paul is talking about the cymbal's transfer function, being infinite, that its properties being continuous and that if it is a filter, it never stops being one . He is not saying that the phenomena will continue below the noise floor, again, unless you have measured it you are only inferring that it happens. This is a hell of a lot different than saying All transducers (Including those on free field configurations?) are IIR Filters. |
Well let's see. The following are quotes from Paul's messages on this board (both under this thread and the tremolo thread.
After excitation, the decay of all these systems is exponential meaning they never actually reach zero action. The rate at which this exponential decay happens is determined by the loss - but it's still exponential - so still never 'gets there' theoretically. We give up and say it stopped when we can no longer percieve it - for electical signal this 'might' be when the action drops below noise due to temperature.
This means that after an impulse is put into the device the device never reaches equilibrium.
What is a pendulum, what is a tuning fork, what is a piano string, what is a simple L/C network, what is a glass tumbler etc etc I.e. anything with resonance?
All are systems with mechanical, acoustic or electrical feedback - with a gain of something less then unity - which prevents them being oscillators
This means that all resonant devices have feedback loops, like IIR filters.
The fact the we CAN model them with IIR representations (analogue or digtial) does not mean they are NOT IIR systems fundamentally. It's just that some exist in domains other than electricity or signal processing
He is saying that IIR filters in the electrical world are simulating IIR transfer functions in the natural world.
In reality what SHOULD happen with a digital IIR (and in fact we always ensure it does) is that the dither (very very importantly) will prevent the filter from stopping
and then
As would be the case with say a cymbal - i.e. its transfer characteristic is essentially constant (leaving out the possibility of material being modified by the force of a loud smash) whether responding to a smash, resonating with external ambient sound in the environment, or just being moved by the noise of the Brownian motion of air molecules due to temperature. I.e. it does not stop having resonance or the ability to modify signal at ANY time.
This means that a cymbal continues to resonate even when its motion wrt to the initial impulse is lower than the noisefloor of the environment it is in.
The only time it would ever come to a complete (absolute) rest and cease being a filter of some kind (or in fact even a cymbal at all) would be at absolute zero temperature
He says that the cymbal is a filter, and that it continues to filter any transient played into it until it is at a state of rest, and the state of rest only happens at 0degrees Kelvin.
And yes you are right in saying that an Finite IR filter (with an impulse response that ends after the calculation has finished) is an oddity that can only exist within a mathematical environment and has no parallel in the physical world AFAIK
I thought this was pretty black and white but you seem to have a hard time with this one. This means that a cymbal, being a filter, is not an FIR filter. Ergo, if it is not a FINITE impulse response filter it is inherently an INfinite impulse response filter. I think Paul and I are basically saying the same thing, there. But just to make sure that Paul really meant what he said, let's go here, from page 1 of this thread:
The FIR is an unnatural filter in the real world that can only exist within the bounds of math and signal processing.
Also, from the same post, regarding feedback again,
Also resonant systems in the natural world DO have feedback - always. I.e, a guitar string or a pipe in an organ has mechanical and acoustic feedback. There is no such thing in the naturalworld as resonance without feedback of some kind.
And finally, with respect to the essential question, Paul writes:
I would say that all natural resonant system are IIR in nature up to the point where the oscillations become unmeasurable due to noise over time.
Again, you wrote:
| Quote: |
This is a hell of a lot different than saying All transducers are IIR Filters. |
Hmmm. I'm led to think that there is NOT a hell of a lot of difference between what Paul is saying and what I'm saying.
Nobody necessarily said Paul possessed absolute truth on all of this, and you can certainly disagree with him, but as I see it now, Paul is pretty clear that any transducer (ANY resonant device, for that matter) is, as a filter, an IIR filter.
Nika.
Post by: Zoesch on June 21, 2004, 07:54:54 PM
Once you get this we can continue talking about the subject, but if you keep insisting that equilibrium means that the forces are zero we won't get anywhere.
Post by: Paul Frindle on June 21, 2004, 07:58:02 PM
| Zoesch wrote on Mon, 21 June 2004 20:19 |
I hope you have the presence of mind to realize that what Paul is saying and what you are saying aren't the same. Paul is talking about the cymbal's transfer function, being infinite, that its properties being continuous and that if it is a filter, it never stops being one . He is not saying that the phenomena will continue below the noise floor, again, unless you have measured it you are only inferring that it happens. This is a hell of a lot different than saying All transducers (Including those on free field configurations?) are IIR Filters. Nobody necessarily said Paul possessed absolute truth on all of this, and you can certainly disagree with him, but as I see it now, Paul is pretty clear that any transducer (ANY resonant device, for that matter) is, as a filter, an IIR filter. Nika. |
Oh guy's this is surely getting silly. I am nothing more than a guy who works with this stuff and thinks about it all the time - that's all. But to take this one step further:
The theoretical timing window for a filter is not defined by us deciding ourselves 'it's done sounding'
The fact that you stop playing and/or unplug an instrument and walk away at the moment YOU decide it's gone completely quiet, does not make it an FIR system. We cannot impose the condition of everything n the natural world being 'practically finite' cos we have given up looking and so decide - as this cannot wash - because there are other factors that give the game away. A fundamental difference between an IIR filter and an FIR is that the FIR only computes it's response over a limited range of time (its time window). Therefore the only reason the impulse response is 'finite' is that once the impulse has been calculated for all the terms contained in the FIR filter - no more calculation is done and no more output results. Can anyone think of any natural system where there is resonance and filtering where this is true - surely not?
The other imnportant side effect of finite time filters is that because they work over a limited period of the passing signal, they produce an error function in the filtering characteristic. I.e. so even if the excitation signal were perfect, starting at - infinity and going on for + infinity, the FIR acting on it will exhibit transitional freq response distortions because IT looks at the signal only during the finite length of the filter calculation. We can obviously make this better by windowing the coefficients (i.e. smoothing the beginning and end at the boundaries of the filters time window so as to reduce the effects of the transistion), but unless the FIR has infinite length these artefacts will always exist to some degree or other. If it HAD infinite length of course, it would no longer be an FIR but an IIR because it would possess an infinite number of calculating terms.
Can anyone think of ANY natural system that is 'aware' of the number of total 'operations' it has made within a given period and decides to end when a certain fixed number have passed - surely not?
And can anyone think of a natural system that produces boundary disturbances in it's freq response because it 'looks' at the signal only over a limited, fixed and finite time window? I certainly can't think of any that would apply to musical instrument mechanisms.
If your guitar string were an FIR mechanism it would cease producing it's natural tone after a fixed length of time - however hard or soft you plucked it! Can anyone think of any instrument that exhibits this anomaly - surely not?
Honsetly guy's I can't understand what the argument is all about, why people are getting so het-up and why it continues
Are you sure that its not simply that someone did a typo or got their FIRs and IIRs mixed up earlier in this thread?
Post by: Paul Frindle on June 21, 2004, 09:01:13 PM
| Zoesch wrote on Tue, 22 June 2004 00:54 |
I'm going to hold any long response until you understand the basics of equilibrium... the SUM of all forces on the system is zero, not the forces, the sum of all the forces acting on the system. Once you get this we can continue talking about the subject, but if you keep insisting that equilibrium means that the forces are zero we won't get anywhere. |
No this cannot be, because you are not specifying the time period within which this summation is being conducted. Over an infinite period, any combination of forces could be said to have been in equilibrium during that period - provided that they consisted of an equal spread of integrally equivalent +ve and -ve events.
Actually this reminds me of a story from an analogue design I did around 20 years ago. I got a job as a design engineer for a high end console company trying to make a digitally controlled analogue assignable desk (SSL). One of the new components around at the time that looked promising was the MDAC - a kind of bipolar digital divider that allowed you to change the gain of an analogue signal with a digital code. Problem was that the damned thing clicked because at every change of its digital code it let out a very fast pulse (around 1 to 2uS long) of varying size and polarity depending on the code. This was optimistically termed as 'charge injection' in the data sheet. Since the pulse was essentially unipolar it could not be filtered out, since this only stretched the energy dissipation over a longer period - making it even more audible in some cases.
When I turned up various elaborate schemes had been hatched and tried to avoid this, including cross fade schemes bewteen 2 devices and 'silent period' generators that cut the signal during the changes etc etc - all of which were variously flawed and the project was stuck because of it. One very respected engineer from Danish Radio (I think) had reasonably stated at the time that "it would always be impossible to change the gain of the device whilst listening to it, for the simple reason that the energy of the transition could never be lost"!
My trick was realising (lateral thinking) that although the energy could indeed never be lost - it COULD be turned into something the integral of which over time would tend towards zero polarity - and therefore could be filtered out in the audio freq band by conventional means. So I got the darned pulse and subjected it to a resonant L/C network. The result then was that the energy pulse caused HF 'ringing' which produced a sine wave that dropped exponentially over time as the energy was dissipated. Point being, that because the ringing energy was bipolar going equally +ve and -ve, (and it was a high freq >100KHz) it could be filtered out of the audio band, simply because the average energy over time (its integral) tended towards zero polarity. The click duly disappeared below -80dB (despite dire predictions by people much brighter than myself) and it worked just fine after that
The (on topic) moral of the (long) story is that the concept of 'equilibrium' is something that requires definition in the time domain - or misunderstanding can result and very important and useful things can get missed.
In this case, that the summation of forces over time apparently equaling zero is not a reliable indicator of 'nothing significant happening'
Post by: Johnny B on June 22, 2004, 03:11:54 AM
I loved the part where you challenged
all those bright people who made the
dire predictions of doom.
Going off topic a little,
I once had a long discussion
with a Buddhist monk, of all people.
He carefully explained that everything
in the universe is vibrating at some level,
even objects that appear solid, and our duty
was to bring ourselves into harmony with
these mysterious universal vibrations.
I guess this may partially explain
why so many musicians such as Tina Turner and
Herbie Hancock practice Buddhism.
And it makes me wonder about a dumb idea
that it might be cool to experiment with
'verters that were speed and filter adjustable
in an attempt to bring it more into the
harmony of the programme material.
Goofy idea probably, but fun to contemplate.
Post by: Curve Dominant on June 23, 2004, 12:24:56 AM
| Quote: |
posted by Paul Frindle: A fundamental difference between an IIR filter and an FIR is that the FIR only computes it's response over a limited range of time (its time window). Therefore the only reason the impulse response is 'finite' is that once the impulse has been calculated for all the terms contained in the FIR filter - no more calculation is done and no more output results. Can anyone think of any natural system where there is resonance and filtering where this is true - surely not? |
The reason I re-posted that, is:
1) in the sincere hope that someone actually READS it, and
2) in the sincere hope that someone actually seeks to UNDERSTAND it.
Three whole pages of bandwidth on GM's forum deserves to be justified by some level of UNDERSTANDING what one is reading.
Zoesch may have infinite time on his hands to compose circular debates which lead nowhere, but some of us do not. Simply debating for the sake of debating is not an excercise which serious engineers waste their time with.
The subject of filters, be they infinite or finite, DOES have some bearing on our craft.
Confusing the issue with a muddy mess of half-baked pseudo-technical babble, does not.
It might help us respect the existence of this thread if there was some real engineering issue which inspired it, some problem which needed to be resolved, or some opportunity which could be addressed. A reason, a why.
Something like, "Of the infinite properties of transducers and other physical phenomena...because I have this mix which isn't quite gelling for some reason..." or,
"Of the infinite properties of transducers and other physical phenomena...because I'm not getting a satisfactory cymble sound..." or,
"Of the infinite properties of transducers and other physical phenomena...because I built this DAC and it sounds kinda crappy - help!"
In other words...
What's the question??
Post by: Nika Aldrich on June 23, 2004, 12:56:09 AM
The question had to do with the tremolo thread on this site. Chuck had stated that we can't ever reconstruct a waveform because it needs infinite time in order to be band-limited (true). I said that that's what IIR filters do - they give infinite time so that the waveform can be band limited. I then explained that all transducers do this, including speakers, ear drums, microphones, etc. Zoesch challenged that I was wrong and that transducers are FIR devices. I said that, in the spirit of keeping Chuck's thread on topic, perhaps Zoesch and I had better move this subject off to another thread and hash it out without confusing the issues in Chuck's thread. Thus, a new thread was started to discuss whether or not transducers really are IIRs. As of yet I think just about everyone in the thread agrees that transducers are IIR filters and that FIR filters can only exist in the world of mathematics except for Zoesch.
I hope that explains the nature of the thread. The issue is actually somewhat important with regards to foundational knowledge of why Nyquist works, IMHO. Chuck's thread has been some degree of evidence of that.
Nika.
Post by: Curve Dominant on June 23, 2004, 02:02:46 AM
| Quote: |
posted by Nika: Eric, .... Zoesch challenged that I was wrong and that transducers are FIR devices. |
Ouch.
It took 3 pages to correct THAT??!??
Aye aye aye!
Sorry guys, but I got work to do...
Post by: Zoesch on June 23, 2004, 03:11:29 AM
| Nika Aldrich wrote on Wed, 23 June 2004 14:56 |
As of yet I think just about everyone in the thread agrees that transducers are IIR filters and that FIR filters can only exist in the world of mathematics except for Zoesch. |
Do me one favor and don't misquote me:
1)I have never said that FIR filters exist in the real world, they are only valid mathematically
2)I am challenging not the mathematical approximation of transducers IN A FEEDBACK CONFIGURATION to be IIR filters, I am challenging the blind belief of the "physics impaired" that a)equilibrium does not exist, b)that equilibrium means absolute zero, c)that inferred behavior is the same as observed behavior and e)That the model is equivalent to the simplified model.
3)BTW, so far, I haven't seen you come and describe how a force of 1 mN can move a 20 gram paper cone @ 1 p.a.
Of course, this would've been simplified if we had stuck to the part of the impulse response where you incorrectly assume that the system is still oscillating beyond measurable parameters (Seriously, I want to see that theory supported on say the potential barrier between Boron and Silicon Oxide) and I am stating that the response is constant and unmesurable.
Maybe you have never spent any time behind measurement equipment, I don't know your background and I won't assume either way.
To me, unsurprisingly this all comes from the standpoint of mathematic "correctness" which is a dangerous position to take, specially considering the tag team you seem to be forming here.
As I don't know what you two are about maybe you want to step down from the spin machine soapbox and display a behavior more fit to the forum's intent and forms and the discussion at hand.
Post by: Nika Aldrich on June 23, 2004, 08:44:39 AM
| Zoesch wrote on Wed, 23 June 2004 08:11 |
Do me one favor and don't misquote me: 1)I have never said that FIR filters exist in the real world, they are only valid mathematically |
OK.
| Zoesch wrote on Thu, 17 June 2004 00:36 | ||
To which I say, no they are not... infinite impulse response means that the response to an impulse exhibits infinite oscillations, this is what you would expect from a full-feedback system with no damping. A transducer won't exhibit that behaviour, if excited with an impulse it will show a finite number of oscillations until it reaches equilibrium (Goes back to zero). If excited with a step function it will experience a finite number of oscillations until it reaches equilibrium. |
Nika
Post by: davidstewart on June 23, 2004, 08:59:18 AM
Zoesch wrote:
"Actually, there's no feedback loop in acoustic instruments (Or in normal transducers) so their behaviour is completely different from that of an IIR filter..."
Anyone in their right mind would conclude that "completely different from that of an IIR filter..." means that it MUST be an FIR filter. There is no in between choice. (And of course that would be the wrong conclusion.)
It seems (to me anyway) you have confused this issue and, for some reason, are reluctant to simply back away from it. IMHO this "dancing" you are doing is serving to muddy an otherwise pretty clear and easily understood reality. I don't see it as helpful. Perhaps this discussion should die.
Sorry to pile on, but it's gotten beyond ridiculous at this point.
Post by: Paul Frindle on June 23, 2004, 11:25:25 AM
| Zoesch wrote on Wed, 23 June 2004 08:11 |
As I don't know what you two are about maybe you want to step down from the spin machine soapbox and display a behavior more fit to the forum's intent and forms and the discussion at hand. |
Spin machine soapbox!!
Ok - if I am one of the people respectfully refered to as 'you two' I am more than happy to 'step down', got better things to do anyway. I was honestly only trying to help. Sorry if this constitutes a behaviour which is 'unfit to the forum's intent'.
Please understand respectfully that I absolutely NEVER waste my valuable time peddling dishonest 'spin', nor do I go out of my way to gratuitously upset people
Consider me dismissed - err - gone
Post by: Zoesch on June 23, 2004, 02:34:48 PM
| davidstewart wrote on Wed, 23 June 2004 22:59 |
Anyone in their right mind would conclude that "completely different from that of an IIR filter..." means that it MUST be an FIR filter. There is no in between choice. (And of course that would be the wrong conclusion.) |
No, but thanks for the reminder. I'm not going to repeat myself so cf. Above.
Post by: Nika Aldrich on June 23, 2004, 03:03:03 PM
And according to that, a transducer is indeed "very different" from an IIR filter while at the same time not an FIR filter in that it is actually an IIR filter in its new name.
From all of the dialogue and all of the discourse this is, I'm afraid, the only conclusion I can draw. Good try, David, but I believe your post still left just enough wiggle room through which Zoesch can slip out.
It still seems to me that Zoesch realizes he made a mistake several pages ago and rather than fess up to it and move on and say "thanks for teaching me something - I hadn't looked at it that way. Wow, there's always time to learn something new!" and gain the respect of the onlookers in the thread for appearing human, he instead continues to dig himself deeper and redirect and wiggle to try to "have been right all along" about this topic. At any moment now I'm expecting a "yup, I agree, that's exactly what I've been saying" comment. I'm expecting the same in about a month on the delta-sigma modulator thread elsewhere.
Nika.
Post by: Zoesch on June 23, 2004, 03:58:23 PM
All of a sudden you seem to have rewritten the laws of physics and you seem to imply that the mechanical aspects of the transducer are to be obvied, so please prove it.
BTW y(n)=Ax(n)+Bx(n-1)? I suppose you actually meant y(n)=Ax(n)-Bx(n-1).
I'll deal with the other things in the appropriate places and at the appropriate time.
Post by: Nika Aldrich on June 23, 2004, 04:20:58 PM
| Quote: |
y(n)=Ax(n)+Bx(n-1)? |
... is a first order low pass FIR filter.
| Quote: |
I suppose you actually meant y(n)=Ax(n)-Bx(n-1) |
No. That's a first order high pass FIR filter.
What I meant is what I wrote - y(n)=Ax(n)+By(n-1), which is a first order low pass IIR filter (unless you're renaming it because it isn't actually "infinite" but rather "unknown").
Nika.
Post by: davidc on June 23, 2004, 04:45:17 PM
| Paul Frindle wrote on Wed, 23 June 2004 16:25 | ||
Spin machine soapbox!! Ok - if I am one of the people respectfully refered to as 'you two' I am more than happy to 'step down', got better things to do anyway. I was honestly only trying to help. Sorry if this constitutes a behaviour which is 'unfit to the forum's intent'. Please understand respectfully that I absolutely NEVER waste my valuable time peddling dishonest 'spin', nor do I go out of my way to gratuitously upset people Consider me dismissed - err - gone |
Paul,
Please don't go. Some of us really do appreciate your input. It would be a big loss if you were not to contribute to the forum.
Best Regards
David C
Post by: Zoesch on June 23, 2004, 04:59:14 PM
| Nika Aldrich wrote on Thu, 24 June 2004 06:20 | ||||
... is a first order low pass FIR filter.
No. That's a first order high pass FIR filter. What I meant is what I wrote - y(n)=Ax(n)+By(n-1), which is a first order low pass IIR filter (unless you're renaming it because it isn't actually "infinite" but rather "unknown"). Nika. |
Fair enough and yes that is correct then.
Post by: George Massenburg on June 24, 2004, 09:04:35 AM
George