Find impedance - circuit substitution?

Discussion in 'Homework Help' started by cfuttrup, Feb 26, 2011.

  1. cfuttrup

    Thread Starter New Member

    Feb 26, 2011
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    Hi all.
    This is not really homework - more like a hobby, about loudspeakers, about a model for damping material in cabinets.

    My problem is that although the circuit consist of resistors, capacitors and inductors, calculation of the impedance is not straightforward. I need an analytical solution to put into a microsoft excel spreadsheet - not a numerical solution.

    I've tried to consult my old school book "Circuits, Devices and Systems" by Smith and Dorf... but I guess I just can't wrap my head around it anymore... my mechanical engineering degree is 14 years back (this class even further), but I like to stay sharp on my math.

    Well, you can see the circuit in the attached PDF file - page 1 - where I also started solving 12 equations with 12 unknown... one way of doing it.

    Here comes the question - I suddenly thought I could substitute one circuit for another and make it easier to solve. The substituted circuit is shown on page 3 of the same PDF file.

    Is this legal, or not?

    Here is my line of thought:

    In the substituted circuit I have taken the node between Lua and Laf, removed the string to the right side of the circuit, then terminated the right side with two strings to the top. These two strings represents the two ways the current can flow to the top (or bottom, if you like). One of these strings is another Lua, the other is through Laf and Rb.

    ... Friends of mine (some electrical engineers), could say for sure and recommended to simplify the circuit instead - but I can do that, remove Rb and my trouble is solved (or just hook Rb in parallel with Lua, for example).

    Your help will be highly appreciated - even just a direction for an answer.

    Best regards,
    Claus
     
  2. Jony130

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  3. cfuttrup

    Thread Starter New Member

    Feb 26, 2011
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    Hi Jony130 - thanks for the comments, and the suggestion to go for Laplace transformation. I've printed your lectures (at a first quick glance, looks quite good).

    P.S. I'd like to do a numerical Laplace transformation in Microsoft Excel, maybe another good thing will come out of the lecture notes. Thanks!
     
  4. Jony130

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    But you don't need to do any laplace transformation.
    All you have to do is to find Z(s) with help of a simply circuit theory and a lot of algebra.
    Next you substituted jω for s=jω , and now you end up with complex number solution.

    Zc(s) = 1/(s*C)
    Zr(s) = R
    Z_L(s) = s*L
     
    Last edited: Feb 26, 2011
  5. cfuttrup

    Thread Starter New Member

    Feb 26, 2011
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    OK, will try that.

    (P.S. I know that I just wrote L and C in the PDF file ... but of course I have to calculate their impedances as jwL and such). Still need the trick how to calculate the impedance of the circuit.
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    When faced with a circuit as complex as the one you have been given, I replace all of the elements be they resistor, capacitor, or inductor with the label Z1, Z2, Z3, ..., etc. I then solve the circuit using parallel and series combination of impedances. Once I have obtained the final expression I then go back and replace the capacitors with 1/sC and the inductors with sL and then simplify the expression.

    Just like jony130 suggested.

    hgmjr
     
  7. cfuttrup

    Thread Starter New Member

    Feb 26, 2011
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    Hi hgmjr.

    Yes, this concept works for the circuit in page 3 of the PDF file... but the circuit in page 1 is not so easy to get around.

    Can you explain how you would do it for the circuit in page 1 of the PDF file?

    Best regards,
    Claus
     
  8. hgmjr

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    Jan 28, 2005
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    The technique should work on all circuits. The two circuits look almost identical to me.

    hgmjr
     
  9. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    Finding analytical (symbolic) expressions for circuits as complicated as this usually gives a result that is very complex. Inspecting such a complex expression rarely gives insight.

    You are attempting to use the branch method, and you have 12 unknowns. Such a system will be impossible to solve symbolically by hand. Even with the assistance of a modern mathematical software, the result will be very complex.

    Using the nodal method, you will have only 5 unknowns (nodes). I've redrawn the circuit, labeling the 5 nodes. Strictly speaking there should be a node between CMF and RAF, but since you probably don't want the impedance there that node need not be made explicit.

    You haven't said at which node you want the impedance, but I'm guessing that it's the very top node in the schematic.

    The way to solve this is to set up the nodal matrix (admittance or Y matrix) and then invert the Y matrix, which gives a Z (impedance) matrix. The elements on the diagonal of the Z matrix are the (driving point) impedances at each of the nodes. Since I've numbered the top node as node number 1, the impedance you want will be the (1,1) element of the Z matrix.

    The result is shown in the second image, with the s (Laplace) variable shown. To get a numeric result, substitute 2 Pi j f for s. Then a value for f (frequency) can substituted and the impedance expression evaluated at that frequency. The result will be a complex number, Z1. To get the magnitude of the impedance, calculate the square root of Z1 times the conjugate of Z1.

    The third image shows the impedance expression with the substitution s = 2 Pi j f made. A special symbol is used for j, sqrt(-1).

    In the second and third images, the numerator of the impedance expression is shown in (mostly) red and the denominator in blue.

    You can see how complex the full symbolic impedance expression is.

    One way to get a simpler expression is to allow some of the variables to be constant numerical values and allow only 1 or 2 variables to be symbolic. Then you can see how the impedance varies with respect to the 1 or 2 symbolic variables.
     
  10. The Electrician

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    FYI, the admittance matrix for your circuit is shown in the attachment.
     
  11. cfuttrup

    Thread Starter New Member

    Feb 26, 2011
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    Hi "The Electrician" - thank you VERY much for your inputs. I will go study immediately. You mention admittance circuit (and conversion) ... a game I never understood before - but your input seems so convincing at a first glance - convincing that there's a solution here, that i look forward to learning something in the process. Again, thank you!
     
  12. cfuttrup

    Thread Starter New Member

    Feb 26, 2011
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    Hi "The Electrician" - your input has at least given me insight into how complex my problem actually is. Cheating a little bit might be the solution - i.e. letting Rb be in parallel with Lua only - means that impedances can be added up. Since Lua and Laf are connected (depending on amount of damping material in the speaker box) and since Rb has largest influence when Lua is large and Laf is small, making the change is not so detrimental.

    I am still thinking about getting the solution out of the 5-node problem you setup for me ... but the other solution is also spinning in my head.

    Again, thanks for your help. It is very informative.

    Best regards,
    Claus
     
  13. The Electrician

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    Oct 9, 2007
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    Why don't you post a schematic of your proposed change (letting Rb be in parallel with Lua only) so it's perfectly clear what the full schematic is, and I'll make the change to the admittance matrix (which I've saved) and get the solution. Even though it might be easier to get a solution without using the nodal method after the change you suggest, it's also possible (as it always is) to solve with the nodal method. Since I've saved the earlier admittance matrix, to make a small change is very easy.

    Also, give me some numeric values for the various parameters and I'll substitute them in the matrix and solve for a numeric value of the impedance you asked for.
     
  14. cfuttrup

    Thread Starter New Member

    Feb 26, 2011
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    Hi there - I should have given you an update earlier :) ...

    The scientific paper I submitted to the "Audio Engineering Society" sketches the "true" solution as a diagram - but no mathematics. It simply becomes too complicated. I mention in the paper text that Rb could be re-located for simplicity. This is what I will implement.

    I have already successfully measured a loudspeaker in free air + in a box and determined that the losses increase significantly, similar to Rb = 172 ohm. Then I have measured with damping material in the box and found that the losses in the box increase significantly by this. I plan to present the results at the AES Convention in London in May.

    A friend of mine made a SPICE model for me with the simplified circuit (as sketched in my first email) as well as the full circuit ... but my inability to run LT_SPICE means I couldn't use it for much. At last - I dropped the idea originally presented in my first email (making a circuit with different configuration, but same impedance - noone seems to be able to tell me if this would actually work or not).

    The model (with Rb parallel to Lua) will be implemented in a Microsoft Excel spreadsheet, which you can find here (currently in preliminary state):
    http://www.scan-speak.dk/toolbox.htm

    ... I have already made the changes in the spreadsheet, but would like the original creator of that spreadsheet (a good friend, former colleague from Dynaudio) to at least check that I did the changes correctly, before posting a new version in the toolbox section.

    Best regards,
    Claus
     
  15. The Electrician

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    Claus,

    I sent you an email through the forum. Did you get it?
     
  16. cfuttrup

    Thread Starter New Member

    Feb 26, 2011
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    Yes I did. See private mail.

    BR,
    Claus
     
  17. cfuttrup

    Thread Starter New Member

    Feb 26, 2011
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    Dear The Electrician.

    Thank you very much for your help. The AES paper is now public. The presentation went well - the paper shows the "real" circuit, but the presentation shows the simplified circuit. The Excel spreadsheet toolbox uses the simplified circuit. All because of your help. Thank you. The paper has been presented one week later at the Munich High-End show. Further more mentioned to press and important people around the globe, including Japan. I hope people will notice the quality of the work.

    Best regards,
    Claus
     
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