Closed-Box Loudspeaker with a Series Capacitor |
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snowflake
Old Croc Joined: 29 December 2004 Location: Bristol Status: Offline Points: 3118 |
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Posted: 22 January 2022 at 1:02am |
thiele's paper
I have successfully followed the procedure to calculate the box parameters from the driver parameters for the four example transfer functions in table 1 - except that the second line for 0.5dB Chebyshev doesn't match what I get. Either the calculated values in the table are wrong, or the values of x3,x2,x1 in equation 25 are wrong. I can't quite follow how x3,x2,x1 are derived from the specified value of K so can't check. Anyone understand the bit about normalisation on p.579?
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snowflake
Old Croc Joined: 29 December 2004 Location: Bristol Status: Offline Points: 3118 |
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tried taking coefficients from these tables and rescaling so that x0 term is 1. but that's not the answer
bbm:978-0-306-47953-3/1.pdf (springer.com) |
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toastyghost
The 10,000 Points Club Joined: 09 January 2007 Location: Manchester Status: Offline Points: 10920 |
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It seems like the normalisation is just the process of identifying the maximum and minimum value for the excursion from the calculations, and using these to determine the normalised value for any value within the data series?
That's usually done by: x_(norm) = (x – x_(min)) / (x_(max) – x_(min)) Where x in this case would be the calculated excursion at a given frequency. That normalises values as 0 to 1, if you want it to go from -1 to 1 then just multiply the 0 to 1 values by 2 and subtract 1: x_(norm) = (2*(x – x_(min)) / (x_(max) – x_(min)))-1 If you’re feeling lazy, Matlab (and therefore I presume Octave) has a dedicated function, unsurprisingly named normalize. https://uk.mathworks.com/help/matlab/ref/double.normalize.html As for the Chebychev seeming to be wrong, have you checked for any "comments on" follow ups to the paper? One thing the AES doesn't seem well set up for is updating papers with corrections. Edited by toastyghost - 22 January 2022 at 6:08pm |
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snowflake
Old Croc Joined: 29 December 2004 Location: Bristol Status: Offline Points: 3118 |
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a value of K is chosen to give the desired ripple. maximum value is 1 and minimum is
1/[1+K^3/(4+3K)^2] ripple in dB = 10.log(min) I guess you put that value of K into eq22 but how do you get from there to the coefficients in eq25 and eq26?
can't find any comments on the paper - and remarkably little discussion of the technique anywhere. |
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stevie
Registered User Joined: 16 March 2008 Location: Dorchester Status: Offline Points: 425 |
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You might be overthinking this. The technique works on speaker systems that are underdamped, i.e, peaking around 100Hz or so. The series capacitor flattens the peak and extends the LF response somewhat. It works on reflex boxes too. Capacitor values tend be be somewhere between 400 and 800uf.
With reflex boxes, it increases power handling at low frequencies, although I've never been able to figure out how much power the capacitors can handle. With values that high, electrolytic are the only choice. I've never bothered to calculate the capacitor values. It's easier to try a few values and measure what they do. |
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snowflake
Old Croc Joined: 29 December 2004 Location: Bristol Status: Offline Points: 3118 |
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yes, if you are decided on a driver then there are only two variables - Vb and capacitor size - so it doesn't take long to find a combination that gives the desired response. I've been modelling in Hornresp but ended up with a much smaller box than suggested using the maths in this paper.
Using trial and error for a vented system with capacitor means at least three variables - the chance of finding the best combination without working to a mathematical alignment are slim and time-consuming. |
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turbo7
Registered User Joined: 15 March 2009 Status: Offline Points: 233 |
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This looks interesting but i cant have a look a the paper, paywall..
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snowflake
Old Croc Joined: 29 December 2004 Location: Bristol Status: Offline Points: 3118 |
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if anyone wants a copy of the paper then send me a PM with your email address
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