Still new digital filter structures to be found? [DSP]

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From: robert bristow-johnson on 9 Mar 2010 02:07

On Mar 9, 12:37 am, spop...(a)speedymail.org (Steve Pope) wrote:
> Tim Wescott <t...(a)seemywebsite.now> wrote:
>
> >Rune Allnor wrote:
> >> On 5 Mar, 19:54, "gretzteam" <gretzt...(a)yahoo.com> wrote:
> >>> Now does anybody know of a group doing active research on finding new
> >>> structures?
> >> Why would anyone want to? The lattice / ladder structures
> >> date back at least to the '60s / '70s; possibly a lot further.
> >> If there is anything at all going on, it would be in the realm
> >> of Kalman'ish filters, like uncented KFs, H_inf or particle
> >> filters.
> >> This stuff on filter structures is *ancient*.
> >You still see papers in the IEEE Circuits & Systems transactions, mainly
> >having to do with clever ways to implement them in full-custom silicon.
>
> This has nothing to do with filter structures, but recently I've been
> looking into ways of synthesizing all-pole bandpass filters that meet
> given design constraints. I'm very sure I'm not breaking new
> territory here, OTOH I haven't seen the method I'm using written up
> anyway either. Basically, apply a window to a sinusoid that is the
> weighted sum of a Hamming window, and a rectangular window;

the Hamming is a weight sum of rectangular and Hann.

> then do linear prediction on the result;

the resulting *what*?

> then after some diddling, you have
> your bandpass filter coefficients.

i'm intrigued, but still have nearly no idea of what is happening,
mathematically.

> The relative weight of the Hamming and rectangular components
> controls the Q of the resulting filter in a straightforward way,
> and being all-pole, it is cheaper than a filter with both
> poles and zeros.

i sorta think, that with only two degrees of freedom (the angle and
magnitude of the conjugate poles) that you can come up with a closed
for expression for the bandpass function (in log) you get for a biquad
with some metric for Q.

> But I'm sure there is some textbook way of achieiving the same result,
> that I simply haven't bothered to look up...

this is interesting, but i have little idea about what you're saying.

curiously,

r b-j

From: Vladimir Vassilevsky on 9 Mar 2010 10:41

Steve Pope wrote:

Glad to see you again. It has been a while since your last posting.
Would you be able to attend the COMP.DSP conference in April?

> This has nothing to do with filter structures, but recently I've been
> looking into ways of synthesizing all-pole bandpass filters that meet
> given design constraints. I'm very sure I'm not breaking new
> territory here, OTOH I haven't seen the method I'm using written up
> anyway either. Basically, apply a window to a sinusoid that is the
> weighted sum of a Hamming window, and a rectangular window;

This controls sidelobes vs main lobe. BTW, Hamming is already a sum of
Hanning and rectangular windows corresponding to minimal sidelobes.

> then do
> linear prediction on the result; then after some diddling, you have
> your bandpass filter coefficients.
>
> The relative weight of the Hamming and rectangular components
> controls the Q of the resulting filter in a straightforward way,
> and being all-pole, it is cheaper than a filter with both
> poles and zeros.

So, the problem with classic bandpass Butterworth/Chebyshev/etc. is
because BLT will introduce zeroes at 0 and Nyquist? May be, just drop
those zeroes out of the filter?

> But I'm sure there is some textbook way of achieiving the same result,
> that I simply haven't bothered to look up...
>
>
> Steve

Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com

From: Steve Pope on 9 Mar 2010 14:39

From: Steve Pope on 15 Mar 2010 17:05

robert bristow-johnson <rbj(a)audioimagination.com> wrote:

>On Mar 9, 12:37�am, spop...(a)speedymail.org (Steve Pope) wrote:

>> This has nothing to do with filter structures, but recently I've been
>> looking into ways of synthesizing all-pole bandpass filters that meet
>> given design constraints. �I'm very sure I'm not breaking new
>> territory here, OTOH I haven't seen the method I'm using written up
>> anyway either. �Basically, apply a window to a sinusoid that is the
>> weighted sum of a Hamming window, and a rectangular window;
>
>the Hamming is a weight sum of rectangular and Hann.
>
>> then do linear prediction on the result;
>
>the resulting *what*?
>
>> then after some diddling, you have
>> your bandpass filter coefficients.
>
>i'm intrigued, but still have nearly no idea of what is happening,
>mathematically.

So more exactly I did the following:

Parameters are:

center frequency
N = window size (I used 2048 samples, for Fc = about Fs/60)
0 < alpha < 1 which controls the Q

Procedure is:

(1) Create a sinusoid X(t) that is N samples long at the center
frequency

(2) Create H(t), a Hamming-windowed version of X

(3) Create Y(t) = alpha * H(t) + (1-alpha) * X(t)

(4) Extract LPC coefficents of the desired order from Y
(no, I did not use the Burg method...).

(5) Take the real part of the LPC coefficeients and use
them as the filter coefficients of an all-pole filter.

I'm not sure if this is scientific; there is a little warping
of the center frequency, and the relationship between alpha
and Q is something I observed only heuristically. But it
did yield useful filters.

Steve

From: Steve Pope on 15 Mar 2010 17:09

Vladimir Vassilevsky <nospam(a)nowhere.com> wrote:

>Steve Pope wrote:

>Glad to see you again. It has been a while since your last posting.
>Would you be able to attend the COMP.DSP conference in April?

I am actively considering it, and will decide soon.

>> This has nothing to do with filter structures, but recently I've been
>> looking into ways of synthesizing all-pole bandpass filters that meet
>> given design constraints. I'm very sure I'm not breaking new
>> territory here, OTOH I haven't seen the method I'm using written up
>> anyway either. Basically, apply a window to a sinusoid that is the
>> weighted sum of a Hamming window, and a rectangular window;
>
>This controls sidelobes vs main lobe. BTW, Hamming is already a sum of
>Hanning and rectangular windows corresponding to minimal sidelobes.

Right, I now see this.

>> then do
>> linear prediction on the result; then after some diddling, you have
>> your bandpass filter coefficients.
>>
>> The relative weight of the Hamming and rectangular components
>> controls the Q of the resulting filter in a straightforward way,
>> and being all-pole, it is cheaper than a filter with both
>> poles and zeros.

>So, the problem with classic bandpass Butterworth/Chebyshev/etc. is
>because BLT will introduce zeroes at 0 and Nyquist? May be, just drop
>those zeroes out of the filter?

Good idea, I haven't looked at this.

Steve

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