```Yes, I got it.

There was problem with Center Frequency.

I have to prewarp both (low and high) band frequencies and then make their
geometric mean wcGM=sqrt(wcL*wcH). But it is not all. There is necessary do
wcGM1=2*arctan(wcGM/2).

With this wcGM1 it gives precious result on all decimal places.

Thank you very much.
```
```On 02/21/2011 05:09 AM, Blaazen wrote:
>> I was confused a bit by your terminology -- you're using nominator and
>> denumerator, while the usual terminology is numerator (the 'top part')
>> and denominator (the 'bottom part').
>>
>> So I didn't realize that your concern was purely over normalization.
>>
>> I can think of two things to try:
>>
>> One: do the bilinear transformation on the numerator as well as the
>> denominator.  This should get you close.
>>
>> Two: Test the filter gain at the center frequency, by plugging in z =
>> e^(j * th), with th = the bandpass frequency in radians/sample.  Then
>> adjust the gain as necessary to get what you want.
>>
>> --
>
> Sorry for confusing. My english is not perfect :-).
>
> Probably you mean: th = sqrt(wch*wcl)  (center freq. is geometric mean of
> pre-warped low and high band cutoff frequencies)
>
> I use this simple algorithm:
> It simply transform one pole of prototype into two poles of band-pass.
>
> Pole:=0.5*(Spoles[i]*BW);
> help:=csqrt(1.0 - sqr(W0 / Pole));
> Spoles[i]:=Pole*(1.0 + help);
> Spoles[pairs+i]:=Pole*(1.0 - help);
>
> It works well, then I do BLT-Z transform and my denominators are fine.
>
> But how can I apply this on zeros. For example Butterworth filter prototype
> has all zeros in infinity.
>
> Well, 4-order Bandpass filter has 4 (complex) zeros on z-plane:
> 1+i*0
> 1+i*0
> -1+i*0
> -1+i*0
> so numerator of H(z) become: [1, 0, -2, 0, 1]
>
> When numerator=denominator at center freqency then it is normalized.
>
> e^(j * th)= cos(th) + i*sin(th)
>
> Is it the right way?

Yes. It may be better to ask that |numerator| = |denominator| at the
center frequency, although I think the phase shift will, indeed, be zero.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html
```
```>I was confused a bit by your terminology -- you're using nominator and
>denumerator, while the usual terminology is numerator (the 'top part')
>and denominator (the 'bottom part').
>
>So I didn't realize that your concern was purely over normalization.
>
>I can think of two things to try:
>
>One: do the bilinear transformation on the numerator as well as the
>denominator.  This should get you close.
>
>Two: Test the filter gain at the center frequency, by plugging in z =
>e^(j * th), with th = the bandpass frequency in radians/sample.  Then
>adjust the gain as necessary to get what you want.
>
>--

Sorry for confusing. My english is not perfect :-).

Probably you mean: th = sqrt(wch*wcl)  (center freq. is geometric mean of
pre-warped low and high band cutoff frequencies)

I use this simple algorithm:
It simply transform one pole of prototype into two poles of band-pass.

Pole:=0.5*(Spoles[i]*BW);
help:=csqrt(1.0 - sqr(W0 / Pole));
Spoles[i]:=Pole*(1.0 + help);
Spoles[pairs+i]:=Pole*(1.0 - help);

It works well, then I do BLT-Z transform and my denominators are fine.

But how can I apply this on zeros. For example Butterworth filter prototype
has all zeros in infinity.

Well, 4-order Bandpass filter has 4 (complex) zeros on z-plane:
1+i*0
1+i*0
-1+i*0
-1+i*0
so numerator of H(z) become: [1, 0, -2, 0, 1]

When numerator=denominator at center freqency then it is normalized.

e^(j * th)= cos(th) + i*sin(th)

Is it the right way?
```
```On 02/20/2011 02:09 PM, Blaazen wrote:
>> Can you cite a reference for how you're doing this with complex numbers?
>>
>> Since I'm not sure exactly what you're doing I can't say for sure -- but
>> why don't you try transforming each bandpass stage to a 2nd-order
>> filter, then doing your complex pole magic on that?
>>
>> --
>>
>> Tim Wescott
>> Wescott Design Services
>> http://www.wescottdesign.com
>>
>> Do you need to implement control loops in software?
>> "Applied Control Theory for Embedded Systems" was written for you.
>> See details at http://www.wescottdesign.com/actfes/actfes.html
>>
>
> I made mistake in my previous message. Unnormalized nominator of each 4-th
> order BP filter is [1, 0, -2, 0, 1] of course.
> I exactly do (example for 2-nd order low-pass prototype):
> 1) I calculate 2 complex poles of Butterworth low-pass prototype (it is 1
> conjugate pair)
> 2) I transform them into 4 complex poles (2 conj. pairs) together with
> frequency transform - now it is Bandpass filter
> 3) I apply Bilinear-Z transform for these 2 conj. pairs and I have 4
> complex poles on Z-plane now.
> 4) Now I calculate denumerator of H(z): (1-2*Re+Re*Re+Im*Im) etc.
> This works well for all even prototypes.
>
> But I don't know how to get nominator. I feel I am very near of my goal but
> still missing last piece of puzzle :-)
>

I was confused a bit by your terminology -- you're using nominator and
denumerator, while the usual terminology is numerator (the 'top part')
and denominator (the 'bottom part').

So I didn't realize that your concern was purely over normalization.

I can think of two things to try:

One: do the bilinear transformation on the numerator as well as the
denominator.  This should get you close.

Two: Test the filter gain at the center frequency, by plugging in z =
e^(j * th), with th = the bandpass frequency in radians/sample.  Then
adjust the gain as necessary to get what you want.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html
```
```Mistake again.

>4) Now I calculate denumerator of H(z): (1-2*Re+Re*Re+Im*Im) etc.
>This works well for all even prototypes.

I meant:
2 poles on Z-Plane [Re1, Im1] and [Re2, Im2] ->  -> ->
[1, -2*Re1, Re1^2+Im1^2]*[1, -2*Re2, Re2^2+Im2^2]
(of course only 2 of 4 poles are necessary because they are conjugate)
```
```>Can you cite a reference for how you're doing this with complex numbers?
>
>Since I'm not sure exactly what you're doing I can't say for sure -- but
>why don't you try transforming each bandpass stage to a 2nd-order
>filter, then doing your complex pole magic on that?
>
>--
>
>Tim Wescott
>Wescott Design Services
>http://www.wescottdesign.com
>
>Do you need to implement control loops in software?
>"Applied Control Theory for Embedded Systems" was written for you.
>See details at http://www.wescottdesign.com/actfes/actfes.html
>

I made mistake in my previous message. Unnormalized nominator of each 4-th
order BP filter is [1, 0, -2, 0, 1] of course.
I exactly do (example for 2-nd order low-pass prototype):
1) I calculate 2 complex poles of Butterworth low-pass prototype (it is 1
conjugate pair)
2) I transform them into 4 complex poles (2 conj. pairs) together with
frequency transform - now it is Bandpass filter
3) I apply Bilinear-Z transform for these 2 conj. pairs and I have 4
complex poles on Z-plane now.
4) Now I calculate denumerator of H(z): (1-2*Re+Re*Re+Im*Im) etc.
This works well for all even prototypes.

But I don't know how to get nominator. I feel I am very near of my goal but
still missing last piece of puzzle :-)

```
```On 02/20/2011 12:15 PM, Blaazen wrote:
> Hello everybody,
>
> I have problem with nominator of H(z) of bandpass filter.
> I develop musical open-source software (non-commercional) and I wrote my
> filter-design algorithms based on polynomials (without complex aritmetic).
> I have all working well (Bessel, Butterworth, Chebyshev, inverse Chebyshev
> and Cauer's elliptic filters Low, High Band pass and band Reject).
> Now I decided do rewrite my algorithms to zero-pole design with complex
> numbers. I started with Butterworth filters. I have LP&  HP and works well
> (design is almost 8 times faster, filters are more stable and ussualy I can
> get higher order of it).
> But here comes the band-pass problem. In my previous solution I made
> frequency transform at s-plane polynomial (s ->  ( s^2+w0^2)/(s*dw) ). But
> how can I do this when I have no polynomial but only transformed poles?
> For example, unnormalized z-nominator of 4-th order BP seems like: [1, -2,
> 1]
> But how can I make it normalized?
> (Note that denumerators (poles) are OK (compared to another software and
> also to my previous solution).)

Can you cite a reference for how you're doing this with complex numbers?

Since I'm not sure exactly what you're doing I can't say for sure -- but
why don't you try transforming each bandpass stage to a 2nd-order
filter, then doing your complex pole magic on that?

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html
```
```Hello everybody,

I have problem with nominator of H(z) of bandpass filter.
I develop musical open-source software (non-commercional) and I wrote my
filter-design algorithms based on polynomials (without complex aritmetic).
I have all working well (Bessel, Butterworth, Chebyshev, inverse Chebyshev
and Cauer's elliptic filters Low, High Band pass and band Reject).
Now I decided do rewrite my algorithms to zero-pole design with complex
numbers. I started with Butterworth filters. I have LP & HP and works well
(design is almost 8 times faster, filters are more stable and ussualy I can
get higher order of it).
But here comes the band-pass problem. In my previous solution I made
frequency transform at s-plane polynomial (s -> ( s^2+w0^2)/(s*dw) ). But
how can I do this when I have no polynomial but only transformed poles?
For example, unnormalized z-nominator of 4-th order BP seems like: [1, -2,
1]
But how can I make it normalized?
(Note that denumerators (poles) are OK (compared to another software and
also to my previous solution).)

Thanks for any help.

```