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Old 15th February 2007, 12:22 PM   #1
teemuk is offline teemuk  Finland
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Default T/S conversion from acoustic to electrical lumped model?

I've been studying the Thiele-Small papers. In “Vented-Box Loudspeaker systems Part I” Small “presents” a conversion routine from acoustic model to electric model. Small describes this as “taking the dual of” acoustic model “and converting all impedance elements to their electrical equivalents by the relationship” Ze=Bl^2/(ZaSd^2).

It looks like all series connections are converted to parallel connections and all parallel connections to series connections. Also, C converts to L, and vice versa while R stays the same.

English is not my native tongue so I’m not quite sure what Small meant by “taking the dual”. The papers do not explain this process either as it’s explained (in detail) in some of the reference articles – which I do not have. Could someone answer why the conversion process goes this way with more “common folk” terms or direct me to a link of an article explaining this.

I am not a total newbie on T/S parameter stuff so I know which parameters represent enclosure and which parameters represent the driver. Small also ignores the effect of voice coil inductance so it’s not shown in the figures. What I’m generally interested is not the details but the theory of why the conversion process goes this way.
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Old 15th February 2007, 12:51 PM   #2
MJK is offline MJK  United States
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The best book I have ever seen that explains these conversions between different types of circuit models (mechanical, acoustic, and electrical) is Beranek's Acoustics text. He does a great job of explaining the advantages of these circuit manipulations, if you read his text the Thiele-Small papers become much easier to understand.

If I remember correctly, taking the dual of a circuit is done by placing a node in each loop and then connecting them to form the outline of the dual circuit. This is a concept that can be found in any undergraduate Electrical Engineering AC circuits text. Every circuit element is inverted so that inductors become caps and the caps become inductors while resistors are 1/resistance. I think voltage sources become current sources also. The advantage of doing this is that the equivalent mechanical circuit uses velocity as a current and force as a voltage while the equivalent acustic circuit uses volume velocity as a current and pressure as a voltage. I have probably butchered the explanation, I am writing from memory based on reading the text about 20 years ago. If you are going to buy one book to learn and understand acoustics and speaker design buy Beranek's Acoustics text. To understand the detailed derivations in Thiele and Small's papers in my opinion it is required reading.

Hope that helps,
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Old 15th February 2007, 01:08 PM   #3
SY is offline SY  United States
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If you find Beranek confusing (as I did my first time through), it may be worthwhile to start with Harry Olson's "Dynamical Analogies." Once I understood that, Beranek became much clearer.
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Old 15th February 2007, 01:29 PM   #4
teemuk is offline teemuk  Finland
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Quote:
Originally posted by MJK
If I remember correctly, taking the dual of a circuit is done by placing a node in each loop and then connecting them to form the outline of the dual circuit. This is a concept that can be found in any undergraduate Electrical Engineering AC circuits text.
Wow. That was fast. And a pretty comprehensive reply too I guess. During my studies I have taken few courses in electronics but it was practically just the basic stuff, AC circuits etc. being the most "technical" part of the thing. We were never taught about the concept of "taking a dual" of a circuit. Practically, I'm pretty self-taught in electronics so this is definitely a thing I have to study more now. Thank you for the explanation, I shall look into it.

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Originally posted by SY
If you find Beranek confusing (as I did my first time through), it may be worthwhile to start with Harry Olson's "Dynamical Analogies." Once I understood that, Beranek became much clearer.
I have to see if I can fid Beranek's text somewhere. Olson's book I have as .pdf. I shall look into it. Thanks.
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Old 15th February 2007, 01:35 PM   #5
MJK is offline MJK  United States
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Sy, that is a great book too. You can download a copy of Olsen's text at :

http://www.pmillett.com/tecnical_books_online.htm

When I read Beranek I came with a background in computer modeling and simulations so it clicked for me right away. It is what I do for a living.
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Old 15th February 2007, 08:51 PM   #6
Svante is offline Svante  Sweden
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The dual transformation originates in the analogy that is made between the mechanical and electical worlds. A mechanical impedance is defined as Zm=F/v, ie force divided by velocity. The similarity with an electrical impedance is obvious where ohm's law states that Ze=U/I. From this, it is natural to make an analogy between the mechanical and electrical entities. Force corresponds to voltage and velocity to current, and (surprise) impedance to impedance. Masses become inductors, springs become capacitors and mechanical resistance (viscous losses) become resistors.

However, looking at the conversion from electrical to mechanical by the loudspeaker, the force is instead proportional to current via the force factor. It is "the other way around", and this results in that the transducer can be seen as a "dual transformer" or a gyrator. It turns out that an impedance on one side of the gyrator is seen as T²/Z, ie impedance seen on the electrical side is Ze=T²/Zm.

Testing this equation on a mass, which has Zm=jwM results in the electrical impedance Ze=T²/(jwM), which is identical to the impedance of a capacitor with C=M/T². Likewise, the electrical impedance seen from a spring, with the mechanical impedance 1/(jw*Cm) results in an electrical impedance of Ze=T²*jw*Cm, ie it looks as an inductance L=Cm * T².

If the impedance is a series connection Zm=Zm1+Zm2, then the resulting electrical impedance is Ze=T²/(Zm1+Zm2)=1/(1/(T²*Gm1)+1/(T²*Gm2)), ie a parallel connection of two admittances T²*Gm1 and T²*Gm2. In a similar way, a parallel connection gets transformed to a series connection.

...and it all originates from the relation Ze=T²/Zm.

It is possible to make admittance analogies as well. In these, the voltage corresponds to the velocity and the current corresponds to force. In this case springs correspond to inductors and masses to capacitors and no dual transformation is nessecary. The methods are equivalent, but for some reason almost everywone uses the first method.

And yes, there is a transformation from mechanical to acoustical too, that is not a dual transformation but an ordinary one, which corresponds to a multiplication/division by Sd².

HTH
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Old 15th February 2007, 11:04 PM   #7
Ron E is offline Ron E  United States
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Quote:
Originally posted by teemuk
I have to see if I can fid Beranek's text somewhere. Olson's book I have as .pdf. I shall look into it. Thanks.
If you want to buy it, the best place is the acoustical society of america's website. You can buy it there for $33. Even if you have to pay an import duty it is probably still cheaper than buying it overseas.
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Old 15th February 2007, 11:26 PM   #8
teemuk is offline teemuk  Finland
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“Masses become inductors, springs become capacitors and mechanical resistance (viscous losses) become resistors.”

This much I figured. And to continue, in acoustic model air volume becomes capacitor, duct with gas inside becomes an inductor and air resistance becomes resistor.

According to my view on Olson’s book it seems the circuit on the right (the attachment of my first post) is a combination of electrical, rectilinear (mechanic) and acoustic models. Beranek’s book shows a similar model but with transformers in between the different analogous elements (fig 3.43 p. 81). The book also describes a concept of impedance inverting transformer but the circuit in concerned figure doesn’t seem to invert impedance as the transformers can be omitted with certain component value conversions (fig 3.44 p. 84). Am I right?

Mechanic and acoustic analogies seem to be “interchangeable” with each other but what Beranek’s book also introduces are “mobility” and “impedance” type analogies where the impedances seem to be inverted. Am I right that the described impedance inversion is needed when converting between these two analogies? Is this what is done for the lumped circuits shown in the attachment of the first post and if yes, why? This is the part, which I find extremely hard to follow in this whole concept.

If my questions seem stupid try to bear me, this stuff is pretty new to me.
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Old 15th February 2007, 11:59 PM   #9
MJK is offline MJK  United States
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Quote:
Beranek’s book shows a similar model but with transformers in between the different analogous elements (fig 3.43 p. 81). The book also describes a concept of impedance inverting transformer but the circuit in concerned figure doesn’t seem to invert impedance as the transformers can be omitted with certain component value conversions (fig 3.44 p. 84). Am I right?
To get to either the acoustic circuit on the left or the electrical circuit on the right the complete circuit in Fig 3.43 is simplified by replacing the transformers, and the components on the side being eliminate, by equivalent components modified by the turns ratio of the transformer. So starting with the complete circuit in Fiigure 3.43 you could roll the acoustic elements through the transformer into the mechanical circuit and then roll this result through the transformer into the electrical circuit. Then using some simplifying assumptions, remove circuit elements that don't contribute very much to form the simplified circuit on the right side of your picture in the first post.

Quote:
Mechanic and acoustic analogies seem to be “interchangeable” with each other but what Beranek’s book also introduces are “mobility” and “impedance” type analogies where the impedances seem to be inverted. Am I right that the described impedance inversion is needed when converting between these two analogies? Is this what is done for the lumped circuits shown in the attachment of the first post and if yes, why? This is the part, which I find extremely hard to follow in this whole concept.
I think you have it right. The reason engineers might chose the mobility circuit over the impedance circuit has to do with what quantity can be measured on the equipment being modeled. Remember the goal of the model is to study the behavior of the device without having to run many tests and trying to interpret the results, the model helps in the understanding and prediction of the device's response to excitation. So if you attach an accelerometer to a structure you can measure how much it is moving and detemine the velocity, it is hard to measure the force applied internally between structural parts. For a sound generating device, an acoustic test, it is easy to measure the SPL or pressure of the sound waves and very difficult to measure the air velocity. The type of model used in the analysis can make it easier to correlate with measured test data. So for mechanical systems the mobility model is often selected. See Section 3.3 for a discussion of this concept and Section 3.8 for conversions using the dual of a circuit.

Hope that helps,
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Old 16th February 2007, 04:31 PM   #10
Svante is offline Svante  Sweden
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I wrote a bit about this in the techdoc of Basta!. Maybe this image clarifies a bit, it shows the three domains in one diagram. Electrical to the left, a gyrator (or "dual transformer", = the speaker motor) mechanical in the middle, a transformer (=the membrane) and the acoustic domain, here without any loading impedance.

Click the image to open in full size.

Scroll down to "Loudspeaker" on this page:

http://www.tolvan.com/basta/Basta!TechDoc.htm
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