Reply by Fred Marshall June 10, 20052005-06-10
"Jerry Avins" <jya@ieee.org> wrote in message 
news:OoidnR53xuVmXTTfRVn-qQ@rcn.net...
> Fred Marshall wrote: > > ... > >> Then replace the zero coefficient at the center with 0.5. >> Result: a perfect minimax halfband filter. > > Neat! Given the usual tools, which way is easier? >
Jerry, Well, the code provided by Juha seems pretty darned easy to use if you have Matlab or Scilab. The code I keep posted will write the coefficients to a file - not a big deal but perhaps convenient. I just tried using the standard P-M followed by using the trick Juha's code uses. That seems a little more trouble because you have to grab and manipulate the interim result from PM. One might check the results to see how they compare. Probably all pretty close. I believe in any case, when the stopband attenuation peaks get to be around 200dB, then numerical problems make convergence a problem. Does anybody care? :-) Fred
Reply by Jerry Avins June 10, 20052005-06-10
Fred Marshall wrote:

   ...

> Then replace the zero coefficient at the center with 0.5. > Result: a perfect minimax halfband filter.
Neat! Given the usual tools, which way is easier? jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by Fred Marshall June 10, 20052005-06-10
"robert bristow-johnson" <rbj@audioimagination.com> wrote in message 
news:BECEA26D.81CF%rbj@audioimagination.com...
> in article 5JqdnR5udeMrETXfRVn-vw@rcn.net, Jerry Avins at jya@ieee.org > wrote > on 06/09/2005 14:38: > >> Middle coefficient non zero, but all other odd-numbered coefficients. >> Round-off errors usually cause computer programs to return very small >> numbers instead of zero. Ignore those. > > also, make awfully damn sure that the gain about Nyquist/2 is perfectly > odd > symmetry and that the weighting about Nyquist/2 is perfectly even > symmetry. > then you will get virtually half of the FIR coefs to be very small numbers > that, as Jerry sez, should be set to zero. >
No need really guys. That bit of code provided by Juha takes care of everything and yields zero coefficients where they belong without cheating. You can do the same with the P-M program: Define an even-length lowpass filter response at band edges [0 .4 .5 .5] with band values [0.5 0] Then upsample the fiter by a factor of 2 (i.e. intersperse zero coefficients) Then replace the zero coefficient at the center with 0.5. Result: a perfect minimax halfband filter. Fred
Reply by Fred Marshall June 10, 20052005-06-10
"I. R. Khan" <ir_khan@hotmail.com> wrote in message 
news:3gsbmbFe3s7qU1@individual.net...
> Fred, thanks for the code. > > Just a comment. While matlab can easily import coefficients generated by > the code, Mathematica has some problems due to use of E. You might > consider using plane %g format for lower order designs. > > Ishtiaq. > > > "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message > news:h9ydnfZxKcaFyTXfRVn-iQ@centurytel.net... > >> >> There's a PC executable that I keep posted here: >> >> ftp://ftp.mission-systems-inc.com/outgoing/Halfband/ >> >> Fred >> >> >> >
Ishtiaq, It's been a while.... I think the E format only shows up in the examples and the actual input required is a bit more forgiving. Fred
Reply by Fred Marshall June 10, 20052005-06-10
"I. R. Khan" <ir_khan@hotmail.com> wrote in message 
news:3gsb31Fe2bocU1@individual.net...
> Thanks Juha for the code. > > I don't understand some statements. Is [0 2*passbandEdge 1 1] in remez > statement correct?
I found this statement interesting. It sugggests something I'd not tried using the Parks-McClellan program - which is to define a zero-length "band" with "x x", in this case "1 1". I tried these same parameters in the P-M program and it worked. The remez program in Matlab uses the normalization of fs/2=1 instead of the more common fs=1. Since the passbandEdge is stated as 0.4 (which is typical for halfband filters), meaning 0.4*fs, that's why it shows up as 2*passbandEdge. I also found the algorithmic approach interesting. First, the minimax filter response above is found for a filter of odd order / even length. Now, a filter of even length is either registered in time so that it's not even around zero or, to make it symmetrical around zero time one has to double the sample rate of the filter by inserting zeros. I like to do this so that the FFT is purely real making it easier to examine. So, this short bit of code first finds an antisymmetric lowpass filter of even length. Because the filter is even in time and of even length, the frequency response is only made of of odd harmonics of cosines. This means that the frequency response at fs/2 is zero (all odd cosines are zero at fs/2). Then, it intersperses N-1 zeros in time to double the sample rate as above. This results in frequency response zeros at the new fs/4 and 3fs/4 and a frequency response with zero integral. Then, it adds 0.5 which makes the antisymmetric response equal to 0.5 at fs/2 and 3fs/4, the passband value wiggling around 1.0 and the stopband value wiggling identically around zero(because the stopband is a mirror image of the passband). Probably this is the same trick for using P-M or "remez" rather directly for halfband filters that fred harris mentions in his book. The filter is symmetrical in time so length 24 gives 12 degrees of freedom and the passband ripple peaks count to N+1 or 13 as expected. Because the stated desired response at fs/2 is zero, the same as what the resulting response will be for a symmetrical, even-length filter, there is no error peak at fs/2. Fred
Reply by Juha June 10, 20052005-06-10
I. R. Khan wrote:
> Thanks Juha for the code. > > I don't understand some statements. Is [0 2*passbandEdge 1 1] in remez > statement correct? > Could you please send the coefficients for one or two different small > orders, say for filterOrder = 5 and/or 10, so that I can match it with > the results I am getting. > > For filterOrder = 6, I get > h = [0.0755, 0, -0.0433, 0, 0.0495, 0.5000, 0.4483, 0, 0.0495, 0, > -0.0433, 0, 0.0755} > and reponse does not seem to be a halfband lowpass. > > Ishtiaq. >
Jes, it is valid statement. The peculiar syntax is due to reason that there was a bug in matlab 6.5.1. Btw the "filterOrder" shoud be odd number. -- BR, Juha
>> ver
------------------------------------------------------------------------------------- MATLAB Version 6.5.1.199709 (R13) Service Pack 1 Operating System: OSF1 V5.1 732 alpha Java VM Version: Java 1.1.8-5 from Compaq Computer Corp. -------------------------------------------------------------------------------------
>> %%%%% Correct design in matlab 6.5.1 >> clear all >> filterOrder=5; passbandEdge=0.4; >> [initH, ripple] = remez(filterOrder, [0 2*passbandEdge 1 1], ...
[0.5 0.5 0 0], {256}) initH = 0.0537 -0.0915 0.3132 0.3132 -0.0915 0.0537 ripple = 0.0509
>> h((filterOrder+1):filterOrder:(2*filterOrder+1)) = [0.5 0]
h = Columns 1 through 7 0 0 0 0 0 0.5000 0 Columns 8 through 11 0 0 0 0
>> h(1:2:end) = initH(1:end)
h = Columns 1 through 7 0.0537 0 -0.0915 0 0.3132 0.5000 0.3132 Columns 8 through 11 0 -0.0915 0 0.0537
>> %%%%%% This doesn't work on 6.5.1 (an excellent error message!!) >> filterOrder=5; passbandEdge=0.4; >> [initH, ripple] = remez(filterOrder, [0 2*passbandEdge], [0.5 0.5],
{256}) at the Nyquist frequency. The order is being increased by one.ero
> In /commercial/matlabr13.1/toolbox/signal/signal/remez.m at line 204
*** FAILURE TO CONVERGE *** Probable cause is machine rounding error. Number of iterations = 3 Warning: If the number of iterations exceeds 3, the design may be correct, but should be verified with freqz. initH = 0.0000 0 0.0000 0.5000 0.0000 0 0.0000 ripple = 1.7270e-17
>> %%%%% But it works on matlab 7.0 >> ver
------------------------------------------------------------------------------------- MATLAB Version 7.0.4.352 (R14) Service Pack 2 Operating System: Linux 2.4.29-pre2-cs1 #2 SMP Mon Dec 20 11:59:34 EET 2004 i686 Java VM Version: Java is not enabled -------------------------------------------------------------------------------------
>> >> filterOrder=5; passbandEdge=0.4; >> [initH, ripple] = remez(filterOrder, [0 2*passbandEdge], [0.5 0.5],
{256}) initH = 0.0537 -0.0915 0.3132 0.3132 -0.0915 0.0537 ripple = 0.0509
>> h((filterOrder+1):filterOrder:(2*filterOrder+1)) = [0.5 0]
h = 0 0 0 0 0 0.5000 0 0 0 0 0
>> h(1:2:end) = initH(1:end)
h = 0.0537 0 -0.0915 0 0.3132 0.5000 0.3132 0 -0.0915 0 0.0537
Reply by robert bristow-johnson June 10, 20052005-06-10
in article 5JqdnR5udeMrETXfRVn-vw@rcn.net, Jerry Avins at jya@ieee.org wrote
on 06/09/2005 14:38:

> Middle coefficient non zero, but all other odd-numbered coefficients. > Round-off errors usually cause computer programs to return very small > numbers instead of zero. Ignore those.
also, make awfully damn sure that the gain about Nyquist/2 is perfectly odd symmetry and that the weighting about Nyquist/2 is perfectly even symmetry. then you will get virtually half of the FIR coefs to be very small numbers that, as Jerry sez, should be set to zero. -- r b-j rbj@audioimagination.com "Imagination is more important than knowledge."
Reply by I. R. Khan June 9, 20052005-06-09
Fred, thanks for the code.

Just a comment. While matlab can easily import coefficients generated by the 
code, Mathematica has some problems due to use of E. You might consider 
using plane %g format for lower order designs.

Ishtiaq.


"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message 
news:h9ydnfZxKcaFyTXfRVn-iQ@centurytel.net...

> > There's a PC executable that I keep posted here: > > ftp://ftp.mission-systems-inc.com/outgoing/Halfband/ > > Fred > > >
Reply by I. R. Khan June 9, 20052005-06-09
Thanks Juha for the code.

I don't understand some statements. Is [0 2*passbandEdge 1 1] in remez 
statement correct?
Could you please send the coefficients for one or two different small 
orders, say for filterOrder = 5 and/or 10, so that I can match it with the 
results I am getting.

For filterOrder = 6, I get
h = [0.0755, 0, -0.0433, 0, 0.0495, 0.5000,  0.4483, 0, 0.0495, 0, -0.0433, 
0,  0.0755}
and reponse does not seem to be a halfband lowpass.

Ishtiaq.

> filterOrder=23; passbandEdge=0.4; > [initH, ripple] = remez(filterOrder, [0 2*passbandEdge 1 1], ... > [0.5 0.5 0 0], {256}) > h((filterOrder+1):filterOrder:(2*filterOrder+1)) = [0.5 0]; > h(1:2:end) = initH(1:end); > freqz(h)
Reply by Andor June 9, 20052005-06-09
Fred Marshall wrote:
...
> There's a PC executable that I keep posted here: > > ftp://ftp.mission-systems-inc.com/outgoing/Halfband/
Fred, thanks! Andor