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22nd February 2012, 04:12 AM  #2611  
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Join Date: Jun 2007

Quote:
Thanks for the feedback. The following formula is used in the Hornresp phase unwrapping algorithm. It was kindly provided by JeanMichel Le Cléac'h. PUNW(N) = PHASE(N) + Pi * Round((2.0144 * PUNW(N  1)  1.0144 * PUNW(N  2)  PHASE(N)) / Pi) On some occasions, particularly when resonances are unmasked, the formula can produce undesired results. This is the case for the examples you have quoted, resulting in a significant difference in the offset delay correction applied to the displayed phase (0.0 msec versus 1.4 msec). I am not aware of the theoretical basis of the above phase unwrapping formula  specifically, I don’t know how the multiplying factors 2.0144 and 1.0144 are arrived at. If 2.0144 is changed to 2.016 however, then the anomaly you have identified can be eliminated, but this may possibly just shift the problem somewhere else. Using your two examples to illustrate how the phase unwrapping formula can break down under certain conditions: EXAMPLE 1  Vrc = 261 litres N = 249 Freq = 347.0345 hertz PHASE(N) = 0.4271 radians PUNW(N  1) = 5.0184 radians PUNW(N  2) = 4.8949 radians PUNW(N) = 6.7103 radians EXAMPLE 2  Vrc = 262 litres N = 249 Freq = 347.0345 hertz PHASE(N) = 0.4387 radians PUNW(N  1) = 5.0181 radians PUNW(N  2) = 4.8946 radians PUNW(N) = 3.5803 radians Notice how that at 347 hertz the PHASE(N), PUNW(N 1) and PUNW(N  2) values are quite similar for the two examples quoted (as expected), but that the resulting PUNW(N) unwrapped phase values are quite different. This ultimately means that the corrected phase chart results will also be quite different, as you have observed. If JeanMichel happens to read this post, he may care to comment further on the theory behind the phase unwrapping technique, and if there is anything that can possibly be done to improve it. Kind regards, David
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22nd February 2012, 08:15 AM  #2612 
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Join Date: Oct 2005

Hello MaVo and David,
As David said what differs between the 2 cases is that a resonance due to the undamping of the rear load has slightly moves and changes in its intensity. In such undamped closed box the resonances as calculated by Hornresp can be more than +6dB above the max of the "normal response of the driver". How this plays on the phase as displayed by default in Hornresp is because we choose to define a useful response interval limited (at 6dB) by 2 frequencies F1 and F2. In MaVo'case F2 which is the upper frequency limit of the "useful part of the response" is sensitive to the presence of a strong resonance in the rear close volume this explains that the mean group delay calculation (which should be 0 in the case of closed box without front horn) is fooled and given around 1.5ms. Notice that when the graph phase is diplayed and if in both cases (261l and 262l) you set the delay at 0 the 2 graphs phase become the same. This means that the problem concerns only the default presentation of the phase graph but all the phases values in every cases are OK. In the same time you can also verify that the Impulses Responses in both cases are the same. I'l try to find a solution to the problem but it is possible that the only will be to remove the resonances in the closed box. About the value of 2.0144 and 1,0144 in the unwrapping formula, those coefficients are due to the fact that the interpolation has to be performed on a logarithmic frequency scale (the relation between a given frequency F(n) and the previous frequency F(n1) at the time I wrote this unwrapping module was : F(n) = 1.0144 * F(n1) Note to David : if you happened to modify that frequency multiplier then the values 1.0144 and 2.0144 in the unwrapping formula has to be modified too. Best regards from Paris, France JeanMichel Le Cléac'h Last edited by Jmmlc; 22nd February 2012 at 08:36 AM. 
22nd February 2012, 08:32 AM  #2613  
diyAudio Member
Join Date: Mar 2010

Quote:
Is there a trick to sim a Le Cleach Horn this large? Also do those stand type bass stands just use a half of the cross section of the horn ? If that makes sense 

23rd February 2012, 07:35 AM  #2614  
diyAudio Member
Join Date: Jun 2007

Hi JeanMichel,
Many thanks for your comments, they have helped me to better understand how your phase unwrapping method works. Quote:
Imag1 = 100 and Imag2 = 297 in each example (F1 = 41.20 Hz and F2 = 689.46 Hz). Quote:
Quote:
I tried unwrapping the phase using the following simple technique. It seems to work reasonably well, but in some instances gives different results to those generated by your more sophisticated method. Offset = 0 For N = 1 To Nfreq If PHASE(N)  PHASE(N  1) > Pi Then Offset = Offset + 2 * Pi PUNW(N) = PHASE(N)  Offset Next N Quote:
Just for your information, the exact coefficient for 1.0144 is actually 10 ^ (1 / 161) in Hornresp  not that it makes any real difference to the overall results :). Kind regards, David
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www.hornresp.net Last edited by David McBean; 23rd February 2012 at 08:04 AM. 

23rd February 2012, 07:50 AM  #2615  
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Join Date: Jun 2007

Hi 3GGG,
Quote:
Quote:
Kind regards, David
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23rd February 2012, 10:17 AM  #2616 
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Join Date: Mar 2010

great, that update is helpful. Just need it to run on a Mac now

23rd February 2012, 10:29 AM  #2617 
diyAudio Member
Join Date: Mar 2010

Jean Michel, how do i sim this type of horn? How does the height of the horn influence the horn. Looking to use a PD 2450
quoted from 2454 but the picture didn't come for the ride 
23rd February 2012, 12:41 PM  #2618  
diyAudio Member
Join Date: Jun 2007

Quote:
Now that I better understand the methodology of your phase unwrapping technique, I think that I may have found an error in your formula :). Working from first principles, I derived the following expression: PUNW(N)=PHASE(N)+2*Pi*Round(((A1*PUNW(N1)(A2A1)*PUNW(N2))/(2*A1A2)PHASE(N))/(2*Pi)) Where: A1=10^(1/161)=1.0144 A2=A1*A1 The difference comes about because you use coefficients 1.0144 and 1+1.0144, whereas I am using 1.0144 and 1.0144*1.0144, which I think should be more correct. Using the new formula I now get effectively the same result for the two examples provided by Mathias, as shown in the attachment. I would appreciate your comments. Kind regards, David
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23rd February 2012, 02:09 PM  #2619 
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Join Date: Oct 2005

Hello David,
Yes, this seems most correct! Best regards from Paris, France JeanMichel Le Cléac'h 
24th February 2012, 07:00 AM  #2620 
diyAudio Member
Join Date: Jun 2007

Hi JeanMichel,
Unfortunately when I doublechecked my derivation today, I found that I had also made an error :). I am using known values of PUNW(N2) and PUNW(N1) to find an approximate value for PUNW(N) using linear extrapolation. If A1=1.0144, A2=1.0144^2 and frequency at N2 = F, then: (PUNW(N)PUNW(N1))/(A2*FA1*F)=(PUNW(N1)PUNW(N2))/(A1*FF) Cancelling out the common F and rearranging / simplifying gives the following approximate value for PUNW(N): PUNW(N)=(A21)*PUNW(N1)(A2A1)*PUNW(N2))/(A11) Resulting in the following final formula: PUNW(N)=PHASE(N)+2*Pi*Round((((A21)*PUNW(N1)(A2A1)*PUNW(N2))/(A11)PHASE(N))/(2*Pi)) Unfortunately this new expression does not give the result we are looking for. Can you see where I am going wrong? Thanks. Kind regards, David
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