Need help with filter implementation

2ndhannes · Mar 3, 2005, 03:15 PM

Hi,

I need help with the implementation of a digital inverse RIAA (phono)
equalizer filter. In a forum-search I found two old threads on this topic,
but no solution

I found filter coefficients for the RIAA-curve in a post of one Frederick
Umminger at

http://www.musicdsp.org/showone.php?id=122

the coefficients are

[color=red] b=[0.02675918611906 -0.04592084787595 0.01921229297239]
a=[1.00000000000000 -0.73845850035973 -0.17951755477430][/color]

but unfortunately, these values form the original curve and not the
inverse curve needed to deemphasize the signal for playback.

Now I am trying to invert the filter, but when I normalize the values so
a0=1, then abs(a1) will become > 1:

[color=Red] b=[0.5839116305884448 -0.4311945070669476 -0.1048223881275120]
a=[1.0000000000000000 -1.7160778983199925 0.7179700042784745]
[/color]
How can I implement this filter in Dane when the absolute value of a coefficient
is > 1? Is there any way to implement an unnormalized filter with a0 <> 1?

Any help will be appreciated!

Thanks,
Hannes

Russ · Mar 4, 2005, 12:51 AM

One possibilty is to scale the coefficient value down (i.e. multiply it by .1), so that it is between -1 and 1, and then (at a point where the expected result of using the unscaled values would result in a number bewteen -1 and 1) scale the result back up (i.e. multiply by 10). Of course you would probably lose some precision doing it this way.

Another option is to use two registers. One register for the integer part of the number, and another register for the fractional part of the number. If the absolute value of the expected result of the math operations is > 1, then you will need two registers for the result as well (do the integer math, and fractional math seperatley), and only combine the values into one register at the point where the expected result is between -1 and 1, etc. Of course this can get tricky, depending on what type of math operations you need to do using these values, etc.

I am not an expert or anything, but hopefully this will give you some ideas to get you started.

Tril · Mar 4, 2005, 10:49 PM

I'll try to help but I can't promise you I'll be successful. From the information you've posted it seems the the data you have is to be used in a IIR filter. I can't be 100% because I don't know much about it. I'll try to make a IIR filter with the values you gave. It should not be much harder than making a FIR filter. I think that the hardest part with IIR filters is determining the coefficients.

I tried to implement a FIR filter to do the equalization but it needed too many coefficients to have a good curve. IIR is supposed to take less coefficients than FIR. If it works, it might give a acceptable curve.

On a sidenote, I'm having problems with copy&pasting code in Dane and it's slowing my coding efforts. If you can help me, answer to the thread
http://www.hardwareheaven.com/effects-dsp/69774-error-invalid-plugin-library-when-copy-pasting-dane.html

Soeren_B · Mar 5, 2005, 08:34 AM

Hi,

You can use the EQs in the kX package (lowpass, highpass, peaking... all beginning with EQ). They all use the same DSP code which is a biquad implementation, and you should be able to replace the default coefficients in the DSP code with scaled versions of the ones mentioned above except a0 (if I've read it correct)

In the existing EQs the coefficients are downscaled with a factor of 2 (divided by 2 in the GUI code, so you should do the same) and is upscaled before output. If you need to scale it even further to get the coefficients between +/- 1 increase "sca" accordingly.

HTH
Soeren

2ndhannes · Mar 5, 2005, 10:30 AM

Hi everyone, thanks for the replies so far!

@ Soeren:
As I understood it, simply scaling is not allowed: in all formulas I found so far,
the coefficient a0 is assumed to be '1'. When I scale the coefficients, all values
are affected - a0, too. I am not sure what the filter implementation is supposed
to do in this case, cause a0 does not appear in the normal biquad code.

@ Tril:
Yes, this is supposed to be a IIR filter with b[] defining the zeroes and a[]
defining the poles of the magnitude response H(z). You are right that the
hardest part is determining the coefficients - without mathlab or some other
expensive software, this is really difficult

If you want to design a filter and test the response, I can post some c code
here that calculates the magnitude response of the digital filter and reports
the error vs. the analogue reference riaa filter.

@ Russ: The idea with the two registers sounds promising, I did not think of
this so far. I will try it and report here...

Bye,
Hannes

Soeren_B · Mar 5, 2005, 07:56 PM

I am not sure that Frederick Umminger makes sense in saying that:

All you should need to do to get the complementary curve is swap the a and b
vectors, and then multiply both vectors by 1/a(0) to normalize. That will
give the coefficients for the inverse filter.

(http://aulos.calarts.edu/pipermail/m...er/017849.html - I assume that is where you get your inversion method from)

It seems that a0 of the coefficients he calculates has already been normalized to 1, so I don't it makes sense to just inverse the coefficients and normalize them again (correct me if I am wrong).

Using Umminger's coefficients for the cutter gives some kind of aggressive highpass filter, but I have no idea if it is correct.
He doesn't ever seem to actually test these coefficients, though.

/Soeren

Tril · Mar 5, 2005, 09:53 PM

This Frederick Umminger said t[color=white]hat he used http://www.hagtech.com/pdf/riaa.pdf as a reference.[/color]
[color=#ffffff][/color]
[color=#ffffff]I've looked at that document and it looks well made and professionnal. There should not be any errors in the document. [/color]
[color=#ffffff][/color]
At http://www.hagtech.com/ on the main page you can read :

Quote:

Who We Are
Hagerman Technology LLC is a small company focused on the design and manufacture of audiophile-grade electronics. We offer unique products that provide true value through innovation and elegance - from the fertile and creative imagination of designer Jim Hagerman. All products are handcrafted in the USA and come with a lifetime warranty. Free shipping on all half-kits!

[color=#ffffff]If you want to make an RIAA filter, I think you should use this pdf as a starting point. [/color][color=#f0f0f0]Frederick Umminger might have made errors.[/color]

2ndhannes · Mar 7, 2005, 02:18 PM

Hi,

you are right in saying that it's probably best to go back to the reference filter...
The coefficients of Mr. Umminger might have be off... one small problem is quite
obvious: the posted filter coefficients for 44.1 and 48 kHz are the same - so I
guess one of the sets is not quite right anyway.

However, I had big difficulties finding any easy methods to design an IIR filter
from an analog response curve. This seems to be a problem that can only be
solved numerically, and I have no software that will do this for me...

That's the reason why I am still trying to make use of the Frederick's coefficients - after all, his non-inverted filter curve looked good in a plot, and it would save me the work of having to design my own filter.

Maybe tomorrow I will have some time and try Russ' idea with the coefficient-
splitting...

Bye,
Hannes

eyagos · Mar 7, 2005, 05:35 PM

If you know the curve properties and the type of IIR filter, there are many pages on Internet with java apps that calculates the coeffeccients. The program Matlab also has a very good module for IIR filter design.

2ndhannes · Mar 7, 2005, 06:54 PM

Hi everyone,

I got stuck with my daywork, so I fiddled with the riaa eq again. Using the
hints from Russ, I decided to divide the large coefficient a1 by two and add
the term y[n-1]*a1 two times in the summation.

Result: working inverse RIAA filter, and sounds good, too!

I know that it's bad practise to do heavy signal amplification (+42dB) in the
dsp, but I am right now listening to my turntable, connected to the line-in of
my audigy 2 zs.

If anyone wants to test the filter, you can get it here:

http://www.buchrecycling.de/kx/inverse_riaa_v1.zip

Oh, I tested the response vs the reference curve: max deviation is -0.6dB
at 5500Hz. I think that's ok... I didn't test phase response, though...

Edit: here is a picture of the response curve (measured with 'HobbyBox')

Now, back to my other project (diploma thesis

)

Bye,
Hannes

jiiteepee · Mar 4, 2007, 01:28 PM

Old thread but, I implemented this RIAA filter as a VST plugin recently. The RIAA EQ curve looks good ...

http://img73.imageshack.us/my.php?im...gnitudegb6.png

but, ... what 'bout phase results

http://img297.imageshack.us/img297/9...aaphaseiv7.png

jiitee

jiiteepee · Mar 30, 2007, 07:17 PM

These coefficients are maybe a bit more accurate than those F. Ummingers which ones Hannes used in this implementation (and these are surely made for 48kHz

):

a0= 1.00000000000000000
a1=-1.73276550000000020
a2= 0.73455344362250008
b0= 1.00000000000000000
b1=-0.75555209999999995
b2=-0.16462571129709999

Maximum error from 0.00 Hz TO 20000.00 Hz is 0.1395898dB
Maximum Phase error from 0.00 Hz TO 20000.00 Hz is ~± 24 degrees

Hmm.. do you really need to 'split' the a1 and then calculate twice in formula ? ... this is not necessarily needed in VST implementation.

EDIT: These coefficients gives +12.5dB higher level than the standard is so you need to divide b0, b1 and b2 with ~4.3 to get the 0dB line crossed @ 1kHz. It's not necessary to do this 'transfer' if you don't start getting too high signal levels.

jiitee

Maddogg6 · Mar 30, 2007, 10:36 PM

Quote:

Hmm.. do you really need to 'split' the a1 and then calculate twice in formula ? ... this is not necessarily needed in VST implementation.

'split' - I thought you need to 'scale' all those coefficients down to be < 1.. ??
Or is that what you mean? - but I think you do that before writing DSP code.. ??

Or are you hinting at some way of making use of those coefficients that are > 1 with out needing to 'scale' them??
As I imagine scaling the coefficients would cause additional errors in tests you posted??

jiiteepee · Mar 31, 2007, 12:54 AM

Quote:

Originally Posted by Maddogg6

'split' - I thought you need to 'scale' all those coefficients down to be < 1.. ??
Or is that what you mean? - but I think you do that before writing DSP code.. ??

Or are you hinting at some way of making use of those coefficients that are > 1 with out needing to 'scale' them??
As I imagine scaling the coefficients would cause additional errors in tests you posted??

Split means just what Hannes have done ... he did split (divide) "a1" into two parts to get the "abs(a1/2)<1" and then changed the formula a bit to get both of these stored (check the code, it's there).

I've read that "normalizing is a convenience, not an inherent need" written by some DSP engineer who designs/builds filters as his profession. I've also read that "scaling is used to reduce the possibility of overflows when your filter operates in fixed-point arithmetic mode". It looks that there are couple scaling methods you could use but, those maybe effects negatively on SNR.

I don't know how the SB DSP works internally or/and acts with these 'oversized' coefficients but, the filter as software version does not pay attention for this. It is all the same if those a's and b's are in range -1 and 1 or in range -100 and 100?
As we're calculting only value of y[n] ... if the formula b0*x[n] + b1*x[n-1] + b2*x[n-2] - a1*y[n-1] - a2*y[n-2] results abs(y[n])<1 then the filter should be stable. It's all the same what size those a/b coefficients were ... the only important factor is the result for y[n] (and of course that those coefficients produces a stable filter either as normalized or not).

So, in software filter, these coefficients

b = [ 20.5245880391493 -17.4865098741392 -2.24152539814392]
a = [ 1.00000000000000 -1.86300261413672 0.86250080415614]

used in formula y[n] = b0*x[n] + b1*x[n-1] + b2*x[n-2] - a1*y[n-1] - a2*y[n-2]

results equally with these

b = [ 1.00000000000000 -0.851978604432149 -0.10921171201431]
a = [ 1.00000000000000 -0.93150130706836 0.86250080415614] // a1/2 -> twice in formula

used in formula y[n] = b0*x[n] + b1*x[n-1] + b2*x[n-2] - a1*y[n-1] - a1*y[n-1] - a2*y[n-2] (this is the method Hannes used (though, he also swapped the sign for a1 and a2 in his code so he could use adding only)).

Those 1st coefficients just gains the output ~+26.7dB from ~+14.6dB which the latter coefficients gain is (i.e. tot. gain becomes ~+41.3dB).

I have not tried how scaling would effect in these.

NOTE: Actually, above F.Ummingers coefficients for 88.2kHz won't produce a stable filter so it won't work in this case (but it's not because of the values are out of range ">-1" - "<1" but because of [- "a1" - "a2" > 1] --> after a while y[n] > 1 and starts to grow very fast) (--> as by the coefficient signs in this case, b0+b1+b2--a1-a2 = b0+b1+b2+a1-a2).

Since I don't know the SB DSP, you could easily test if those 'oversized' coefficients for 48kHz filter, which I gave in my previous post, works in Hannes code (remember change those signs for a1 and a2).

jiitee

Maddogg6 · Mar 31, 2007, 01:16 AM

Quote:

Since I don't know the SB DSP, you could easily test if those 'oversized' coefficients for 48kHz filter, which I gave in my previous post, works in Hannes code (remember change those signs for a1 and a2).

well without trying - Im 99.99% certain the kX DSP being 'fixed' point will choke on any #'s > 1.0 in magnitude (-1.0 included) - unless they are integers.

Now my point with scaling affecting test results - I imagine (read: Im an idiot and don't know anything for certain) - that the simple act of multiplying/scaling those coefficients - will cause some rounding in arithmetic and thus a loss of accuracy in the end - which I also imagine would have to equate to *some* sort of difference - most likely in the form of error.

But you answered my original Q about the split - I don't understand completely - but I get the idea anway.

jiiteepee · Mar 31, 2007, 09:53 AM

Quote:

Originally Posted by Maddogg6

well without trying - Im 99.99% certain the kX DSP being 'fixed' point will choke on any #'s > 1.0 in magnitude (-1.0 included) - unless they are integers.

You know this better than I do. By the documentation, the Emu10k1 is a custom 32 bit fixed point digital signal processor and accepts input in range -1 - 1. The E-MU E-DSP™ is a 32-bit DSP with 67-bit accumulator (double precision w/ 3 headroom bits). By the kX Project compatibility list this E-MU DSP is a "Pro" version of 10k2 (used in Audigy 2/4 serie cards)? Does the "kX DSP" work in fixed point form w/ 10k2 too or does the 10k2 still use fixed point arithmetic internally?

Quote:

Now my point with scaling affecting test results - I imagine (read: Im an idiot and don't know anything for certain) - that the simple act of multiplying/scaling those coefficients - will cause some rounding in arithmetic and thus a loss of accuracy in the end - which I also imagine would have to equate to *some* sort of difference - most likely in the form of error.

I think the normalization won't be harmful if you just have enough numbers after decimal point. In calculation of this RIAA filter, IIRC, those 'base' values used as starting values were 0.00318, 0.000075 and 0.000318 (max 6 decimals). I have compared the coefficient data accuracy effect using original 17 decimal value against 7,6,5,4,3, and 2 decimal values and the magnitude did not differ much in graphic presentation until less than five decimal accuracy were used. Using 2 decimal value made filter unstable (drifting magnitude response):

http://img134.imageshack.us/img134/4356/dectestxv4.png

Quote:

But you answered my original Q about the split - I don't understand completely - but I get the idea anway.

Maybe this was just because of the DSP being fixed point and didn't accept the -1.7xxxx as one whole value. Actually, you can divide those coefficient into as many parts as you like but then you need to be sure that
- the sum of those keeps equal w/ the original value,
- you multiply the right sample with all these and
- the ratio keeps equal (as for an example, a0:a1:a2).

jiitee

Maddogg6 · Mar 31, 2007, 10:30 AM

Quote:

Does the "kX DSP" work in fixed point form w/ 10k2 too or does the 10k2 still use fixed point arithmetic internally?

Well - my reference to kX DSP - was toward the EMU10kX DSPs
AFAIK - they (10K1 and 10K2) operate the same way.. ?? but - as I mention before - I'm not certain about anything.

Now, I always thought the differences between 10K1 (live/APS) and 10K2 (Audigy) where resources (instructions/cycle, GPR's, TRAM and I/O - for 7.1) Microcode looks identical - BUT its also 'compiled' or loaded (or what ever??) - I'm confident we're not exposed to 'raw' DSP assembly code either... if there are differences - Im not aware of anything more than I mentioned.

The 'PRO' - I assume your referring to the EMU APS - AFAIK uses the same 10K1 DSP as SBLives - in fact those cards are sooo much like SBLives that a hacked EMU APS driver was made that allowed SBLives to use the APS drivers. (re: APS-Live drivers - I used very happily with my SBLive Value OEM CT4832 before I found KX - I found kX as a referral from the old APSLive web site here: http://freenet-homepage.de/apslive/)
The differences between SBLives and APS were in codecs and connectivity, but electronically very similar.

jiiteepee · Mar 31, 2007, 10:41 AM

Quote:

Originally Posted by Maddogg6

The 'PRO' - I assume your referring to the EMU APS - AFAIK uses the same 10K1 DSP as SBLives
...

I meant those pro cards from Creative E-MU: 0404, 1212M, 1616/1616M, 1820/1820M
- http://kb.kxproject.lugosoft.com/list.html

jiitee

Maddogg6 · Mar 31, 2007, 10:53 AM

Quote:

Originally Posted by jiiteepee

I meant those pro cards from Creative E-MU: 0404, 1212M, 1616/1616M, 1820/1820M
- http://kb.kxproject.lugosoft.com/list.html

jiitee

Ahh - that list has been updated... I know very little about those cards
Assuming the '10K2 Pro grade' - means they are diffrent enough that kX wont work with them, but imagine they otherwise operate the same internally as fixed point. My guess the only difference is in information released and subtle changes to prevent another 'APSLive hack' from happening on EMUs 10K2 based cards drivers for the CL 10K2 based cards... ??
I would think that if say; the 'Pro' 10K2 handled 24/96 directly in the DSP (and eliminated the P16V 'hack' to support 24/96 - or something significant like this) - they would have named the chip differently than 10K2... ??

Mar 3, 2005, 03:15 PM	#1
2ndhannes DriverHeaven Junior Member Join Date: Jan 2005 Posts: 37 Rep Power: 0	Need help with filter implementation Hi, I need help with the implementation of a digital inverse RIAA (phono) equalizer filter. In a forum-search I found two old threads on this topic, but no solution I found filter coefficients for the RIAA-curve in a post of one Frederick Umminger at http://www.musicdsp.org/showone.php?id=122 the coefficients are [color=red] b=[0.02675918611906 -0.04592084787595 0.01921229297239] a=[1.00000000000000 -0.73845850035973 -0.17951755477430][/color] but unfortunately, these values form the original curve and not the inverse curve needed to deemphasize the signal for playback. Now I am trying to invert the filter, but when I normalize the values so a0=1, then abs(a1) will become > 1: [color=Red] b=[0.5839116305884448 -0.4311945070669476 -0.1048223881275120] a=[1.0000000000000000 -1.7160778983199925 0.7179700042784745] [/color] How can I implement this filter in Dane when the absolute value of a coefficient is > 1? Is there any way to implement an unnormalized filter with a0 <> 1? Any help will be appreciated! Thanks, Hannes

Mar 7, 2005, 06:54 PM	Thread Starter #10
2ndhannes DriverHeaven Junior Member Join Date: Jan 2005 Posts: 37 Rep Power: 0	Hi everyone, I got stuck with my daywork, so I fiddled with the riaa eq again. Using the hints from Russ, I decided to divide the large coefficient a1 by two and add the term y[n-1]a1 two times in the summation. Result: working inverse RIAA filter, and sounds good, too! I know that it's bad practise to do heavy signal amplification (+42dB) in the dsp, but I am right now listening to my turntable, connected to the line-in of my audigy 2 zs. If anyone wants to test the filter, you can get it here: http://www.buchrecycling.de/kx/inverse_riaa_v1.zip Oh, I tested the response vs the reference curve: max deviation is -0.6dB at 5500Hz. I think that's ok... I didn't test phase response, though... Edit: here is a picture of the response curve (measured with 'HobbyBox') Now, back to my other project (diploma thesis ) Bye, Hannes Last edited by 2ndhannes; Mar 9, 2005 at 12:16 PM.*

Mar 4, 2007, 01:28 PM	#11
jiiteepee HardwareHeaven Senior Member Join Date: Apr 2005 Location: FI Posts: 400 Rep Power: 0	Old thread but, I implemented this RIAA filter as a VST plugin recently. The RIAA EQ curve looks good ... http://img73.imageshack.us/my.php?im...gnitudegb6.png but, ... what 'bout phase results http://img297.imageshack.us/img297/9...aaphaseiv7.png jiitee __________________ Last edited by jiiteepee; Mar 30, 2007 at 07:07 PM.

Mar 30, 2007, 07:17 PM	#12
jiiteepee HardwareHeaven Senior Member Join Date: Apr 2005 Location: FI Posts: 400 Rep Power: 0	These coefficients are maybe a bit more accurate than those F. Ummingers which ones Hannes used in this implementation (and these are surely made for 48kHz ): a0= 1.00000000000000000 a1=-1.73276550000000020 a2= 0.73455344362250008 b0= 1.00000000000000000 b1=-0.75555209999999995 b2=-0.16462571129709999 Maximum error from 0.00 Hz TO 20000.00 Hz is 0.1395898dB Maximum Phase error from 0.00 Hz TO 20000.00 Hz is ~± 24 degrees Hmm.. do you really need to 'split' the a1 and then calculate twice in formula ? ... this is not necessarily needed in VST implementation. EDIT: These coefficients gives +12.5dB higher level than the standard is so you need to divide b0, b1 and b2 with ~4.3 to get the 0dB line crossed @ 1kHz. It's not necessary to do this 'transfer' if you don't start getting too high signal levels. jiitee __________________ Last edited by jiiteepee; Mar 31, 2007 at 08:20 PM.

Mar 4, 2005, 12:51 AM	#2
Russ HardwareHeaven Extreme Member Join Date: Jan 2005 Posts: 5,561 Rep Power: 62	One possibilty is to scale the coefficient value down (i.e. multiply it by .1), so that it is between -1 and 1, and then (at a point where the expected result of using the unscaled values would result in a number bewteen -1 and 1) scale the result back up (i.e. multiply by 10). Of course you would probably lose some precision doing it this way. Another option is to use two registers. One register for the integer part of the number, and another register for the fractional part of the number. If the absolute value of the expected result of the math operations is > 1, then you will need two registers for the result as well (do the integer math, and fractional math seperatley), and only combine the values into one register at the point where the expected result is between -1 and 1, etc. Of course this can get tricky, depending on what type of math operations you need to do using these values, etc. I am not an expert or anything, but hopefully this will give you some ideas to get you started.

Mar 4, 2005, 10:49 PM	#3
Tril S-3D enthusiast Join Date: Sep 2004 Location: Canada Posts: 1,676 Rep Power: 64 System Specs	I'll try to help but I can't promise you I'll be successful. From the information you've posted it seems the the data you have is to be used in a IIR filter. I can't be 100% because I don't know much about it. I'll try to make a IIR filter with the values you gave. It should not be much harder than making a FIR filter. I think that the hardest part with IIR filters is determining the coefficients. I tried to implement a FIR filter to do the equalization but it needed too many coefficients to have a good curve. IIR is supposed to take less coefficients than FIR. If it works, it might give a acceptable curve. On a sidenote, I'm having problems with copy&pasting code in Dane and it's slowing my coding efforts. If you can help me, answer to the thread http://www.hardwareheaven.com/effects-dsp/69774-error-invalid-plugin-library-when-copy-pasting-dane.html

Mar 5, 2005, 08:34 AM	#4
Soeren_B kX Project DSP Engineer Join Date: Dec 2002 Location: Denmark Posts: 94 Rep Power: 0	Hi, You can use the EQs in the kX package (lowpass, highpass, peaking... all beginning with EQ). They all use the same DSP code which is a biquad implementation, and you should be able to replace the default coefficients in the DSP code with scaled versions of the ones mentioned above except a0 (if I've read it correct) In the existing EQs the coefficients are downscaled with a factor of 2 (divided by 2 in the GUI code, so you should do the same) and is upscaled before output. If you need to scale it even further to get the coefficients between +/- 1 increase "sca" accordingly. HTH Soeren

Mar 5, 2005, 10:30 AM	Thread Starter #5
2ndhannes DriverHeaven Junior Member Join Date: Jan 2005 Posts: 37 Rep Power: 0	Hi everyone, thanks for the replies so far! @ Soeren: As I understood it, simply scaling is not allowed: in all formulas I found so far, the coefficient a0 is assumed to be '1'. When I scale the coefficients, all values are affected - a0, too. I am not sure what the filter implementation is supposed to do in this case, cause a0 does not appear in the normal biquad code. @ Tril: Yes, this is supposed to be a IIR filter with b[] defining the zeroes and a[] defining the poles of the magnitude response H(z). You are right that the hardest part is determining the coefficients - without mathlab or some other expensive software, this is really difficult If you want to design a filter and test the response, I can post some c code here that calculates the magnitude response of the digital filter and reports the error vs. the analogue reference riaa filter. @ Russ: The idea with the two registers sounds promising, I did not think of this so far. I will try it and report here... Bye, Hannes

Mar 5, 2005, 07:56 PM	#6
Soeren_B kX Project DSP Engineer Join Date: Dec 2002 Location: Denmark Posts: 94 Rep Power: 0	I am not sure that Frederick Umminger makes sense in saying that: All you should need to do to get the complementary curve is swap the a and b vectors, and then multiply both vectors by 1/a(0) to normalize. That will give the coefficients for the inverse filter. (http://aulos.calarts.edu/pipermail/m...er/017849.html - I assume that is where you get your inversion method from) It seems that a0 of the coefficients he calculates has already been normalized to 1, so I don't it makes sense to just inverse the coefficients and normalize them again (correct me if I am wrong). Using Umminger's coefficients for the cutter gives some kind of aggressive highpass filter, but I have no idea if it is correct. He doesn't ever seem to actually test these coefficients, though. /Soeren

Mar 7, 2005, 02:18 PM	Thread Starter #8
2ndhannes DriverHeaven Junior Member Join Date: Jan 2005 Posts: 37 Rep Power: 0	Hi, you are right in saying that it's probably best to go back to the reference filter... The coefficients of Mr. Umminger might have be off... one small problem is quite obvious: the posted filter coefficients for 44.1 and 48 kHz are the same - so I guess one of the sets is not quite right anyway. However, I had big difficulties finding any easy methods to design an IIR filter from an analog response curve. This seems to be a problem that can only be solved numerically, and I have no software that will do this for me... That's the reason why I am still trying to make use of the Frederick's coefficients - after all, his non-inverted filter curve looked good in a plot, and it would save me the work of having to design my own filter. Maybe tomorrow I will have some time and try Russ' idea with the coefficient- splitting... Bye, Hannes

Mar 7, 2005, 05:35 PM	#9
eyagos DriverHeaven Addict Join Date: Dec 2002 Posts: 259 Rep Power: 0	If you know the curve properties and the type of IIR filter, there are many pages on Internet with java apps that calculates the coeffeccients. The program Matlab also has a very good module for IIR filter design.