> Thanks again Jerry. Please bear with me a bit longer!
>
> That gives pole locations at sqrt(j), sqrt(-j), -1/sqrt(j), -1/sqrt(-j).
>
> But, aren't these the pole locations for a second order butterworth?
> There are four poles but we only use the left half plane, so this only
> gives two poles so this is a second order filter, right?
>
> Also, it's not obvious to me how the Q values come from the pole
> locations. Can you please enlighten me?
I omitted a sqrt. Go with Clay's formulation.
> Clay, thanks for the link. I wish I'd had that reference when I started!
> Just a quick question: to transform the lpf into hpf, all I do is change
> the '1' in each numerator to 's^2', right?
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by ●April 9, 20092009-04-09
On Apr 8, 1:45�pm, Jerry Avins <j...@ieee.org> wrote:
> esfield wrote:
> > Yeah, I just realized that.... so what are the other constraints on these
> > values that force them to be the ones Vlad gave?
>
> Math. All low-pass Butterworth filters have all derivatives zero at f=0.
> That gives a 4th-order filter the response 1 + x^4 where x = w/w0. Solve
> for the roots.
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> �����������������������������������������������������������������������
Actually a small correction is needed.
A Nth order Butterworth filter has a response of 1/sqrt(1+omega^2N)
So we actually start with 2*N roots of -1 and just use the roots on
the left hand plane to form our polynomial.
For example, a 2nd order butterworth
We start with the 4 roots of -1 on the unit circle. These are at
angles
90,135,225, and 315 degrees.
The ones on the LHP are simply at 135 and 225 degrees or in radians
these are 3pi/4 and 5pi/4
So we combine the poles in the left hand plane as conjugate pairs so
as to result in a polynomial with real valued coefs. For our 2nd order
case we only have one pair to combine. But our poly is simply the
product of (s-root) terms. For our specific case:
(s-exp(j3pi/4))(s-exp(j5pi/4)
which simplfies to
s^2+sqrt(2)s+1
For odd order cases you see that one of the roots is -1 so we have a
(s-(-1)) term. Of course this simplifies to s+1 which is a factor of
all odd order normalized Butterworth filters.
IHTH,
Clay
Reply by ●April 9, 20092009-04-09
On Apr 8, 3:33�pm, "esfield" <ethanstubblefi...@hotmail.com> wrote:
> Thanks again Jerry. �Please bear with me a bit longer!
>
> That gives pole locations at sqrt(j), sqrt(-j), -1/sqrt(j), -1/sqrt(-j).
>
> But, aren't these the pole locations for a second order butterworth?
> There are four poles but we only use the left half plane, so this only
> gives two poles so this is a second order filter, right?
>
> Also, it's not obvious to me how the Q values come from the pole
> locations. �Can you please enlighten me?
>
> Clay, thanks for the link. �I wish I'd had that reference when I started!
> Just a quick question: to transform the lpf into hpf, all I do is change
> the '1' in each numerator to 's^2', right?
I look at it as replacing each "s" by a 1/s - that inverts the
frewquency response.
For example
1/(1+s) -> transforms to 1/(1+1/s) now mult top and bot by s to
clean up the expresion and you get s/(s+1) which is now your 1st order
Butterworth highpass filter.
IHTH,
Clay
Reply by esfield●April 8, 20092009-04-08
Thanks again Jerry. Please bear with me a bit longer!
That gives pole locations at sqrt(j), sqrt(-j), -1/sqrt(j), -1/sqrt(-j).
But, aren't these the pole locations for a second order butterworth?
There are four poles but we only use the left half plane, so this only
gives two poles so this is a second order filter, right?
Also, it's not obvious to me how the Q values come from the pole
locations. Can you please enlighten me?
Clay, thanks for the link. I wish I'd had that reference when I started!
Just a quick question: to transform the lpf into hpf, all I do is change
the '1' in each numerator to 's^2', right?
Reply by ●April 8, 20092009-04-08
On Apr 8, 12:34�pm, "esfield" <ethanstubblefi...@hotmail.com> wrote:
> *update*
> Okay, I found my mistake in the math, so I can see now that a0 and a2 are
> indeed the only coefficients that change. �Which seems to confirm what
> others have said, namely that the fc will be the same for each filter and
> only the Q will differ.
>
> Is there an easier way to do this? �My design procedure was:
>
> 1)figure out what the pre-warped pole locations in the s-domain are
> 2)take each complex conjugate pole pair to create two second order
> sections
> 3)apply BLT to each biquad, with results that the filters are the same
> except for the Q factor.
>
> Since the Q factor is the only thing that changes, is there a way to
> figure out what the Q for each of these biquads will need to be to get a
> fourth-order butterworth, without going through all the tedium described
> above????
>
> Thanks again,
> Ethan
> Yeah, I just realized that.... so what are the other constraints on these
> values that force them to be the ones Vlad gave?
Math. All low-pass Butterworth filters have all derivatives zero at f=0.
That gives a 4th-order filter the response 1 + x^4 where x = w/w0. Solve
for the roots.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by esfield●April 8, 20092009-04-08
Yeah, I just realized that.... so what are the other constraints on these
values that force them to be the ones Vlad gave?
Thanks, Ethan
Reply by Jerry Avins●April 8, 20092009-04-08
esfield wrote:
> Thanks for the help, Vladimir!
>
> Wow.
>
> So all I had to do was find values of Q such that when multiplied
> together, they yeild 1/sqrt(2)??????
There an infinite number of such Q pairs. Vlad cave you the only ones
that work right.
> It seems so obvious now.
>
>
> So painfully, painfully obvious.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by esfield●April 8, 20092009-04-08
Thanks for the help, Vladimir!
Wow.
So all I had to do was find values of Q such that when multiplied
together, they yeild 1/sqrt(2)??????
It seems so obvious now.
So painfully, painfully obvious.
Ethan
Reply by Vladimir Vassilevsky●April 8, 20092009-04-08
Oh, what a lazy idiot.
Enjoy:
Q0 = 0.541196
Q1 = 1.306563
VLV
esfield wrote:
> *update*
> Okay, I found my mistake in the math, so I can see now that a0 and a2 are
> indeed the only coefficients that change. Which seems to confirm what
> others have said, namely that the fc will be the same for each filter and
> only the Q will differ.
>
> Is there an easier way to do this? My design procedure was:
>
> 1)figure out what the pre-warped pole locations in the s-domain are
> 2)take each complex conjugate pole pair to create two second order
> sections
> 3)apply BLT to each biquad, with results that the filters are the same
> except for the Q factor.
>
> Since the Q factor is the only thing that changes, is there a way to
> figure out what the Q for each of these biquads will need to be to get a
> fourth-order butterworth, without going through all the tedium described
> above????
>
> Thanks again,
> Ethan