New version of Porous Absorber spreadsheet

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New version of Porous Absorber spreadsheet

Postby ChrisW » Thu May 05, 2005 11:15 am

Hi folks

I have finally finished an update to my Porous Absorber spreadsheet.  It now contains an extra sheet called Microperforated panel in which the absorption curve of a micro-perforated panel is calculated.  Please discard any and all previous versions.

The beauty of this design is that no layer of porous material is required in order for it to function as an absorber!  In other words, you can make a see through absorber using clear perspex or glass!

http://www.whealy.com/acoustics/Porous.html

My initial problem with implementing this calculation, was that I needed to find the zero'th and first order Bessel function values of complex numbers, and the Analysis ToolPak functions supplied with Excel only permit real numbers as parameters.  (See Cox & D'Antonio, Eq 6.36, pg 181)

Therefore, I would like to take this opportunity to thank Gordon Everstine who donated his Fortran implementation of the Bessel function taking a complex parameter.  After some jiggery-pokery with Excel, and some performance problems caused by the ATP's horribly inefficient data type conversion, I have reimplemented this code in VBA.

Brief explanation of microperforated panel absorption
When air flows across any substance of a different density, then a boundary layer is created.  It is within this layer that resistance (friction) and turbulence occur.

This effect can be seen on a massive scale when two air masses of different temperatures collide.  The cold air, being denser, is forced underneath the lighter warm air.  Add lots of moisture to the warm air - and you've got yourself the recipe for a storm system!  The turbulence and energy released within such a boundary layer is the driving force behind the frequent storm activity seen in the central and southern US States.

But I digress...

Imagine you have a section of half inch pipe and you wanted to pour treacle or molasses down it.  You would have a real problem because the treacle is so thick and sticky, that it would stick to the sides of the pipe and eventually block the flow altogether.  What's the problem here?  Viscosity.  Treacle is such a viscous liquid that it won't flow through such a small pipe without being pumped.

Now lets repeat the same experiment with water.  The water flows though the pipe without any problem.  Why? Because its viscosity is much lower.  (The viscosity of water is actually a problem for fire fighters.  If you raise the pressure (and therefore the velocity) under which water is pumped through a fire hose to too high a level, then the friction created between the water and the walls of the hose creates turbulence and the water won't flow properly.)

The point here is that when ever one substance flows past another substance of a different density, a boundary layer is formed.  The thickness of the boundary layer is proportional to the viscosity of the fluid(s), and the resistance created by the boundary layer is proportional to fluid velocity - up to a point.  There is a cut off point after which the boundary layer becomes turbulent, and the resistance ceases to be linear.

Treacle has a very high viscosity, so its boundary layer is actually thick enough to block its movement through the pipe.  Water still forms a boundary layer, but its viscosity is low enough that it flows without a problem.

This is the very reason why porous absorber materials such as Rockwool or glassfibre absorb sound energy.  The gaps between the fibres are small enough that the air literally sticks to the fibres because of its own viscosity.

OK, back to the plot...

In all these calculations, a gas like air is treated as a very low density fluid.  The boundary layer for air flowing over a hard flat surface at low velocity (a few mm per second) is only about 0.2mm thick.  The object of the exercise here is to harness the resistance found within the boundary layer, and use it to act as an acoustic absorber.  However, we don't want to use a layer of porous material.  Therefore, the solution is to perforate a panel with holes sufficiently small that boundary layer resistance becomes significant.

This is why the holes should be less than about 1mm in diameter.

Hopes this is simple enough without being too simplistic.

Regards

Chris W
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Postby Eric.Desart » Thu May 05, 2005 12:00 pm

Chris

Many thanks for such damned valuable and unique document.

I adjusted the topic in the "Calculation Tools" section too.
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Postby ChrisW » Thu May 05, 2005 12:39 pm

:)

The next task is to get my brain around the planar hybrid surfaces described in Cox & D'Antonio chapter 11.  I think it will take me a little longer to build a useful tool here.  Not only is the maths more involved, but implementing a solution in Excel will also be more of a challenge.

Regards

Chris W
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Postby avare » Thu May 05, 2005 1:43 pm

Chris:

I don't know how this this clearly.  This is second time this wek that you have impresssed me with your interest in calculation tools and sharing them with others.  The first was the absorption vs.incident angle for spaced homgenous absorbers.

What I can say clearly is thank you.

Andre
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Postby ChrisW » Thu May 05, 2005 3:00 pm

Hi Andre

You're welcome (all donations gratefully received :D)

Whenever I study a technical subject, I want to make sure that I understand it correctly.  And the best way to prove (to myself) that I understand a subject is to implement a solution.  I believe this is the best way of learning because I have had to put what I know into practice - and make it work.  So I built the spreadsheets.  At first, these were for my own use, but I then I thought, "Why shouldn't others benefit from the tools as well".  So I did a bit polishing to make them look pretty, and released them.

If anyone can find an error in one of my calculations (and they have been known :wink:), then please let me know.  Explain what the error is, and I'll correct it.  That way the calculation tool improves and I increase my understanding of the subject.  For me that's a win-win situation.

Regards

Chris W
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Postby Bob » Sun Sep 04, 2005 6:00 pm

Hi Chris:

FYI
LarryChannin wrote:from http://www.avsforum.com/avs-vb/showthre ... ost6138925
I tried applying the Perforated Panel calculator to takeaim's question regarding his soffit construction in the Broadband bass soffit construction thread and seemed to get useful results.

However, I haven't had any luck in getting useful results when applying the Slotted Panel calculator to predict what would happen if we cut slots into his existing seating riser.
Regards
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Postby ChrisW » Mon Sep 05, 2005 8:12 am

Hi Bob

Thanks for the info.  I've had a quick look at the thread, and it does appear that the riser construction being modelled is not what my spreadsheet was designed to work on.

My porous absorber spreadsheet contains the calculations to predict the absoption of a two or three layer construction against a backing of infinite impedence.  If other words, the quations in the spreadsheet allow for the following combinations:

1) Backing layer of infinite impedence + air gap + porous absorber + slotted/perforated panel
2) Backing layer of infinite impedence + porous absorber + air gap + slotted/perforated panel
3) Backing layer of infinite impedence + porous absorber + slotted/perforated panel

Larry Chanin's thread in the AVS forum describes a riser in which two 8" porous layers are separated by a layer of plywood.  Unfortunately, my spreadsheet cannot account for the presence of this middle layer of plywood.  And to be honest, it would be very difficult to produce a predictive tool that would produce anything like useful results for this type of construction.  The plywood panel would have to be treated as an edge mounted diaphragm, and the properties of the plywood would then have to be entered by the user, and...  you get where I'm heading?

I don't think such a tool would ever produce results that could be described as "useful".

However, having looked at the "No air gap" plot my spreadsheet is producing on the Slotted Panel spreadsheet, I suspect that there could be a bug in the calculations here.

I'll have to investigate and get back to you.

Regards

Chris W
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Postby jtvrdy » Mon Sep 05, 2005 1:38 pm

Chris,
my Excel is in Spanish I have the tool pack installed but the spreadsheet don't work...

I need to install something more or it only works on English versions ?

thanks.
Josep Tvrdy
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Postby ChrisW » Mon Sep 05, 2005 1:57 pm

Hi Josep

I have had several people ask me about using the spreadsheet in non-English versions of Excel.  I had hoped to be able to produce a single, multi-lingual version of the spreadsheet. But time is against me here.

Eric Desart wrote an excellent solution to this problem, but in order for me to test my implemention, I would need a copy of Excel installed in all the relevant languages (and also the necessary language skills to translate the text).  Eric's tool contains a wizard that converts the function names found in the cell formulae from the foreign language to the local language.

So what I am asking is the following.  If I send you an unlocked version of the spreadsheet, would you be willing to do the translation work of both the function names (using Eric's tool) and the English text?  The translated spreadsheet would then be returned back to me and distributed via my website with full accreditation to yourself for the translation work.

If you (or any other person who has experienced this problem) is interested in doing this, then please contact me (chris@whealy.com) directly and I will send you the spreadsheet.

Regards

Chris W
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Postby jtvrdy » Mon Sep 05, 2005 2:15 pm

I will try to transalte it..,where i can find this tool for the funtion names?

you can send it to tvrdy99@hotmail.com
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Postby Larry Chanin » Mon Sep 05, 2005 6:29 pm

ChrisW wrote:Hi Bob

Thanks for the info.  I've had a quick look at the thread, and it does appear that the riser construction being modelled is not what my spreadsheet was designed to work on.

My porous absorber spreadsheet contains the calculations to predict the absoption of a two or three layer construction against a backing of infinite impedence.  If other words, the quations in the spreadsheet allow for the following combinations:

1) Backing layer of infinite impedence + air gap + porous absorber + slotted/perforated panel
2) Backing layer of infinite impedence + porous absorber + air gap + slotted/perforated panel
3) Backing layer of infinite impedence + porous absorber + slotted/perforated panel

Larry Chanin's thread in the AVS forum describes a riser in which two 8" porous layers are separated by a layer of plywood.  Unfortunately, my spreadsheet cannot account for the presence of this middle layer of plywood.  And to be honest, it would be very difficult to produce a predictive tool that would produce anything like useful results for this type of construction.  The plywood panel would have to be treated as an edge mounted diaphragm, and the properties of the plywood would then have to be entered by the user, and...  you get where I'm heading?

I don't think such a tool would ever produce results that could be described as "useful".

However, having looked at the "No air gap" plot my spreadsheet is producing on the Slotted Panel spreadsheet, I suspect that there could be a bug in the calculations here.

I'll have to investigate and get back to you.

Regards

Chris W


Hi Chris,

Bob was kind enough to refer me to your calculators.  They are extremely impressive.

It's quite possible that I have misapplied the Slotted Panel calculator, or made some other dumb operator error.  I hope you haven't misinterpreted my remarks about not obtaining "useful" results as being critical of your fine effort.  You'll notice that I close the posting with the statement:

I'm pretty sure that Dennis uses slotted risers all the time, so I guess I'm probably doing something wrong in applying the spreadsheet.
 

And in this thread Broadband bass soffit construction I praise your work.

The problem as I mentioned in this thread, and over in the Bass Trap made out of a Seating Platform thread, is that Mr. Everest's formulas don't show us the effect of using insulation of any kind, nor as you pointed out, do they calculate absorption. Chris Whealy's spreadsheets do provide this and he gives credit to four professional acousticians who have reviewed his work. That's good enough for me.


To clarify, I am just looking for results for the top section of the seating riser, discounting any effect of the lower riser cavity.  That would mean condition #3 above, with an 8" cavity and a panel thickness of 1-1/2".  Then, varying the slot width and spacing, I am shooting for a resonator that would have a fairly wide bandwidth with a peak of roughly 200 Hz.  I was able to achieve this using your Perforated Panel calculator using the same cavity depth and panel thickness.

If you have the time to review this I would appreciate it.

Thanks.

Larry
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Postby ChrisW » Mon Sep 05, 2005 11:32 pm

Hi Larry

Ok, I didn't know you were only trying to model the top section of the riser, so in that case you are using my spreadsheet correctly.  And thanks for the compliments.  These spreadsheets represent a large investment of time and effort.  All donations gratefully received!  :D

Now down to business, why doesn't my spreadsheet give you useful/helpful results?

Well I've checked the calculations in the spreadsheet against the methods and equations listed in section 6.3 of Cox and D'Antonio, and I can't find any errors in the calculation.  So I entered the values quoted in section 6.4.1 (page 178) and got pretty much exactly the same curve as shown in the book in figure 6.18.

Then I realised what was going on.  What value were you using for the flow resistivity?  If you're using a typical value for Rockwool or medium density fibreglass (e.g. 16,500), then you will not get the useful results you require.

Change the flow resistivity down to about 450, and then the "no air gap" curve jumps up to useful values.  If you don't know what the flow resistivity value represents, then I would recommend you read the sheet of the same name in the spreadsheet.  This is the most important value for predicting the absorption of a porous material.

The higher the number, the denser the porous absorber.  The denser the porous material, the fewer air gaps it will contain and therefore it will offer greater resistance to the flow of air through its fibres.  On the other hand, the lower the value, the more open the fibre structure is (or to put it in non-technical terms, its all light and fluffy), therefore, air can flow more easily through the porous material.

If you take a look at the attached screen shot, you will see that I have been able to jig the numbers so that you get your broad absorption peak centred at around 200Hz.  Using the following values, you should get the performance you're after:

Panel thickness = 38.1mm (1.5")
Distance between slots = 50.8mm (2")
Slot width = 19.1mm (3/4")

Cavity depth = 203.2mm (8")
Absorber thickness = N/A
Absorber flow resistivity = 450 rayls/m

Your riser needs to be filled with a very low density porous absober such a loft insulation batts that you have ripped apart to increase the fluffiness.  When you do this, you please wear protective gloves, goggles, a breathing mask and keep your arms and legs covered; glass fibre is evil stuff!

According to IR761, 65mm glassfibre batts at 11.7Kg/m3 have an average flow resistivity of about 3,600 rayls/m.  Therefore, you're really going to need to break this stuff up to lower this value.  Try changing the flow resistivity from 450 to 3,600 and see how (without an air gap) it just dumps the performance. 8O

Alternatively, continue using your higher density absorbent material, and just leave an air gap of a known depth above the absorbent layer... (this does sound simpler than ripping up glass fibre!) :)

Hope that helps.

Regards

Chris W
Attachments
Riser Absorption.jpg
(130.94 KiB) Downloaded 208 times
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Postby Larry Chanin » Tue Sep 06, 2005 4:38 pm

ChrisW wrote:Hi Larry

Ok, I didn't know you were only trying to model the top section of the riser, so in that case you are using my spreadsheet correctly.  And thanks for the compliments.  These spreadsheets represent a large investment of time and effort.  All donations gratefully received!  :D

Now down to business, why doesn't my spreadsheet give you useful/helpful results?

Well I've checked the calculations in the spreadsheet against the methods and equations listed in section 6.3 of Cox and D'Antonio, and I can't find any errors in the calculation.  So I entered the values quoted in section 6.4.1 (page 178) and got pretty much exactly the same curve as shown in the book in figure 6.18.

Then I realised what was going on.  What value were you using for the flow resistivity?  If you're using a typical value for Rockwool or medium density fibreglass (e.g. 16,500), then you will not get the useful results you require.

Change the flow resistivity down to about 450, and then the "no air gap" curve jumps up to useful values.  If you don't know what the flow resistivity value represents, then I would recommend you read the sheet of the same name in the spreadsheet.  This is the most important value for predicting the absorption of a porous material.

The higher the number, the denser the porous absorber.  The denser the porous material, the fewer air gaps it will contain and therefore it will offer greater resistance to the flow of air through its fibres.  On the other hand, the lower the value, the more open the fibre structure is (or to put it in non-technical terms, its all light and fluffy), therefore, air can flow more easily through the porous material.

If you take a look at the attached screen shot, you will see that I have been able to jig the numbers so that you get your broad absorption peak centred at around 200Hz.  Using the following values, you should get the performance you're after:

Panel thickness = 38.1mm (1.5")
Distance between slots = 50.8mm (2")
Slot width = 19.1mm (3/4")

Cavity depth = 203.2mm (8")
Absorber thickness = N/A
Absorber flow resistivity = 450 rayls/m

Your riser needs to be filled with a very low density porous absober such a loft insulation batts that you have ripped apart to increase the fluffiness.  When you do this, you please wear protective gloves, goggles, a breathing mask and keep your arms and legs covered; glass fibre is evil stuff!

According to IR761, 65mm glassfibre batts at 11.7Kg/m3 have an average flow resistivity of about 3,600 rayls/m.  Therefore, you're really going to need to break this stuff up to lower this value.  Try changing the flow resistivity from 450 to 3,600 and see how (without an air gap) it just dumps the performance. 8O

Alternatively, continue using your higher density absorbent material, and just leave an air gap of a known depth above the absorbent layer... (this does sound simpler than ripping up glass fibre!) :)

Hope that helps.

Regards

Chris W


Hi Chris,

Thanks very much for the response.

Bob and I are trying to respond to questions from a friend who is in the process of building a small home theater.  Unfortunately, he has already completed the construction of his seating riser.  So his flow resistivity is limited to what he has had installed.  I'm guessing that 8" of batt insulation has a flow resistivity of 5,733 rayls/m.  So using this value confirms your comments that we shouldn't expect useful results for a Slotted Panel.  I'm going to attempt to attach a screen shot of a resulting curve.

Larry
Attachments
Slotted Riser.JPG
8" deep riser, 1.5" top, filled with batt insulation
(102.22 KiB) Downloaded 191 times
<a href="http://mysite.verizon.net/res8ycu4/index.html">
Larry's Home Theater</a>
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Postby Larry Chanin » Tue Sep 06, 2005 4:52 pm

Hi Chris,

Following up on my previous posting, I attempted to run an absorption curve for the same riser, but instead of using slots in the top, I used your Perforated Panel calculator to design a perforated top.  

In this case, using the same value for flow resistivity, I received useful results for a riser filled with batt insulation.

I'm frankly surprised that the same structure, filled with the same insulation, works for a perforated top and doesn't work for a slotted top.  Do you have any insights regarding this phenomena?

I've attached a curve for the perforated top.

Thanks.

Larry
Attachments
Perforated 8 inch Riser.JPG
8" riser, 1.5" perforated top, filled with insulation
(91.96 KiB) Downloaded 182 times
<a href="http://mysite.verizon.net/res8ycu4/index.html">
Larry's Home Theater</a>
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Postby ChrisW » Wed Sep 07, 2005 6:29 am

Hi Larry

Thats' a good question about why the change from slots to perforations should alter the absorption so much.  This is where I'll have to check up on the derivation of the theories behind the calculations.

I'm using a type of calculation called a transfer matrix.  This involves calculating the impedence of the layer closest to the backing plane, and then factoring that value into the calculation for the next layer, and so on.  In this manner you can calculate the impedence of a structure built from multiple layers.  The impedence of the top layer is then converted to an absorption value and this is what is plotted on the graph.

In the case of your riser, the value calculated for the impedence of the porous layer is identical in both the perforated and the slotted cases (as you'd expect), but the difference then comes in the impedence calculation of the perforated or slotted layer.

I can't account for the difference right now, and I'm still not ruling out the possibility of a bug.

I'll have to look into it and get back to you.

Regards

Chris W
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Postby andrebrito » Thu Dec 22, 2005 7:04 pm

I got this curve while I was playing with your calculator.,.. funny to see a second peak ressonance

I was trying to compare the matlab scripts with your excel file but the matlab are not running well, I have to check what is wrong

http://www.acoustics.salford.ac.uk/rese ... cripts.htm
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Postby andrebrito » Thu Dec 22, 2005 7:18 pm

Chris I'm not sure if this is my problem or not but I get different results from your calculator and from Trevor's matlab sheets for a microperforated panel

check the figure attached, peak absorptions are much higher

Perhaps could be something that is setup differently, viscosity ?  T.Cox usex viscosity = 1.85e-5

My other idea is that his script is not using thirdband octaves but really calculation for each Hz, that might be a reason

Andre
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Postby ChrisW » Fri Dec 23, 2005 2:22 pm

Hi Andre

Yes I have noticed that there are discrepancies between the results of Trevor's MATLAB scripts and my spreadsheet.

Here is what I have found.

1) There is a bug in my spreadsheet because I have used the hole radius as if it were a diameter! (Doh!!)  So until I issue a correction, double the value of the "Hole radius" field (cell C6).

2) The MATLAB script uses a value of 340m/s for the speed of sound, where as my spreadsheet calculates the speed of sound from first principles and for a pressure of 1atm and 20 dec C (which in turn uses the specific heat ratio and the gas constant), comes out at a speed of 343.38 m/s.  This may seem like only a small difference (~1%), but it has a significant impact on height of the secondary absorption peaks.  Fiddle around which the value in MATLAB and you'll see what I mean.

3) There are computational differences between MATLAB and Excel.  For instance, the array holding the impedance values for the top of the air cavity in MATLAB contains for items 1 to 10:

0-7222.3i 0-6446.7i 0-5710.1i 0-5123.9i 0-4552.0i 0-4057.0i 0-3599.3i 0-3210.1i 0-2858.4i 0-2530.7i

Where as in Excel, the same items in the same array hold:

0-7223.0j 0-6433.2j 0-5729.3j 0-5101.9j 0-4542.7j 0-4044.2j 0-3599.7j 0-3203.3j 0-2849.7j 0-2534.2j

These differences appear small, but I think they contribute to some of the discrepancies that are observed.

4) My biggest problem though is that the implemenation in the MATLAB scripts do not follow the equations quoted in the book.  My spreadsheet has implemented the equations quoted in the book, not simply re-implemented the MATLAB scripts in Excel.

I have noticed discrepancies in several places where the MATLAB scripts have terms in them that cannot be accounted for given the text and equations they are supposed to refer to.  For instance, script 6_2.m for a slotted absorber calculates the panel's open area using the term

eta = (0.0465*4*d)/(pi*0.05^2)      Where d is the slot width

To me, this seems to have several fudge factors built into it.  What is the 0.0465*4 there for? And the (pi*0.05^2) term in the denominator is dividing by the area of a circle or radius 0.05!?  Circle?  This is a slotted absorber we're talking about here, not a perforated panel.  A  value of 0.05 is not used as a dimension anywhere in the preceding coding.  To me, it looks like insufficient time was taken to correlate the quoted equations with the MATLAB coding.  Yet the graphs shown in the text are the actual output of the MATLAB scripts.  Something's not right here...


The microperforated panel script 6_3.m also has one of these implementation "anomalies" if you will.

In Script 6_3, equation 6.36 has not been implemented exactly as printed on page 181.  In the text, it shows a term in which a Bessel function of the first order is divided by the product of (k prime * sqrt(-j)) and a Bessel function of the zero'th order.

In the MATLAB script however, the k prime term is not used in the denominator.  The original MATLAB script implements Eq 6.36 as follows:

Code: Select all
z1 = -j*rho*c*cot(k*l);   %Impedance, top of cavity;

%Impedance of covering sheet
kd = a*sqrt(rho*w/viscosity);
s = kd*sqrt(-j);
z2 = j*w*rho*t./(1 - 2*besselj(1,s)./(s.*besselj(0,s)));



But if the MATLAB script follows Eq 6.36, then the whole script should read

Code: Select all
f = [62.5,70,79,88,99,111,125,140,157,177,198,223,250,281,315,354,397,445,500,561,630,707,794,891,1000,1122,1260,1414,1587,1782,2000,2245,2520,2828,3175,3564,4000,4490,5040,5657,6350,7127,8000,8980,10079,11314,12699,14254,16000]
nf = length(f);
w = 2*pi*f;
c = 343.38;
k = w/c;
rho = 1.204;
viscosity = 1.85e-5;
eta = pi*a^2/(D^2);   %Open area
impedence = rho * c;

zair = -j*rho*c*cot(k*l);   %Impedance, top of cavity;

%Impedance of covering sheet
kprime = a*sqrt(rho*w/viscosity);

z1 = j*w*rho*t./(1 - 2*besselj(1,kprime*sqrt(-j))./(kprime*sqrt(-j).*besselj(0,k*sqrt(-j))));
z2 = z1/eta   + zair + j*w*1.7*rho*a/eta + sqrt(2*w*rho*viscosity)/(2*eta);

R=(z2-rho*c)./(z2+rho*c);   %reflection factor
anormal=1-abs(R).^2;   %absorption coefficient

hold on
plot(f,anormal,'g')
xlabel('f (Hz)')
ylabel('abs. coeff')


You should notice several things about my modified script.

First, I have not used the linspace() function to generate the frequency array f.  Instead, I have hardcoded the same freqency values I used in my spreadsheet.

Second, I have changed the values of the speed of sound and air density to be the same as my spreadsheet.

Third, I have implemented Eq 6.36 exactly as printed in the book, and I get a similar shaped curved (with two peaks), but the absorption is lower (peaking at 0.35 instead of the original 0.9).

Conclusion
I have implemented the equations listed in the book as carefully as I can, and yet there are still differences between my spreadsheet and the MATLAB scripts.  All I can put this down to is that Trevor Cox has (for whatever reason) not exactly implemented his own equations.

I will correct the known bugs in my spreadsheet and reissue it; however, there will still be differences I cannot resolve.

Chris W
Attachments
Modified Script 6_3.gif
Plot from modified MATLAB Script 6_3. This script implements Eq 6.36 exactly as quoted in Trevor Cox's book. Notice that the shape is the same as the original plot, but the absorption peaks are lower.
Modified Script 6_3.gif (7.03 KiB) Viewed 15727 times
--
The voice of ignorance speaks loud and long,
but the words of the wise are quiet and few.
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ChrisW
 
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Postby bert stoltenborg » Fri Dec 23, 2005 2:59 pm

Hehehe!

:mrgreen:
If you view life with the knowledge that there are no problems, only opportunities, you are a marketing manager.......this is my personal philosophy
bert stoltenborg
 
Posts: 4569
Joined: Sun Apr 18, 2004 11:03 am
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Postby ChrisW » Fri Dec 23, 2005 3:03 pm

Hi Bert

Why the laughter?

Chris W
--
The voice of ignorance speaks loud and long,
but the words of the wise are quiet and few.
                                        --
ChrisW
 
Posts: 153
Joined: Tue Mar 30, 2004 12:42 pm
Location: Brentwood, Essex, UK

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