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. |
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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. |
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). |
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If excited with a step function it will experience a finite number of oscillations until it reaches equilibrium. |
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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. |
Nika Aldrich wrote on Thu, 17 June 2004 10:02 |
It never reaches equilibrium at a quantum level. |
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It may APPEAR to reach equilibrium, but transducers do indeed have infinite impulse responses. |
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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. |
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That's not complex math. Transducers most definitely have infinite impulse responses. |
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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? |
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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 |
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It has nothing to do with "computations." The IIR filter as after the conversion occurs - in the analog world. |
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. |
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Nika: It has nothing to do with "computations." The IIR filter as after the conversion occurs - in the analog world. |
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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. |
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. |
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? |
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. |
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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. |
Zoesch wrote on Thu, 17 June 2004 20:45 |
And that is correct, |
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I can have a system whose electrical impulse response is infinite yet its mechanical response is finite |
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? |
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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. |
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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. |
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There's no feedback on the system. |
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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. |
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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. |
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So Paul is right, and I am right, and so far neither of us are contradicting each other. |
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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... |
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. |
Zoesch wrote on Fri, 18 June 2004 22:02 |
Nika, are you implying that the phenomena continues at a subatomic level? |
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Are you implying that no system returns to its original state? |
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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. |
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. |
Nika Aldrich wrote on Sat, 19 June 2004 16:34 |
Zoesch, Still waiting on an answer to the question... |
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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? |
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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. |
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. |
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I also asked you where's the feedback loop on a speaker cone or a microphone capsule, still no answer on that one either. |
Zoesch wrote on Sat, 19 June 2004 18:27 |
I told you before that the answer to both is no, they are not... \ |
Nika Aldrich wrote on Sun, 20 June 2004 04:11 |
Then WHY IN THE WORLD ARE YOU PICKING THIS BLOODY FIGHT?? |
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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. |
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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. |
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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. |
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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. |
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Of course if they "reach" equilibrium at a steady state then they are the same. |
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They will never reach equilibrium, per the fact that forces are constants. |
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Can you just apologize for the error and let us drop this ridiculous thread before it gets even more embarassing? |
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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. |
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You said you agree with Paul, so great, we'll put that to rest. Transducers are NOT FIR filters. |
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Great. I'm done. Nika. |
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 |
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 |
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. |
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. |
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. |
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This is a hell of a lot different than saying All transducers are IIR Filters. |
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. |
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. |
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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? |
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posted by Nika: Eric, .... Zoesch challenged that I was wrong and that transducers are FIR devices. |
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. |
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 |
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. |
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. |
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.) |
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y(n)=Ax(n)+Bx(n-1)? |
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I suppose you actually meant y(n)=Ax(n)-Bx(n-1) |
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 |
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. |