[music-dsp] Filter design question

Luigi Castelli superbigio at yahoo.com
Mon Oct 15 17:14:54 EDT 2007


Thank you Robert and Charles.

I managed to find my formula and to clear up my understanding of
digital filters.

Gotta love this forum.

Best.

- Luigi





--- Charles Henry <czhenry at gmail.com> wrote:

> On 10/13/07, Luigi Castelli <superbigio at yahoo.com> wrote:
> > ok,
> >
> > when I plug: z^-1 = e^(-jw) I get:
> >
> > H(w) = { [ (1 - alpha) / 2 ] * [ (1 + cos(w) - jsin(w) ) / (1 -
> alpha *
> > cos(w) + alpha * jsin(w) ) ] }^K
> >
> > where do I go from here ?
> 
> At the cutoff frequency, you're only looking for magnitude, right?
> So, you take the magnitude of the top and bottom separately.  The
> real/imaginary parts give the phase response:
> 
> so, it becomes
> |H(w)|^2={ [ (1 - alpha) / 2 ] * [ (1 + cos(w))^2 + sin(w)^2 ) / ((1
> -
> alpha *cos(w))^2 + (alpha * sin(w) )^2 ] }^K
> 
> -3dB mean half power ( |H(w)|^2=0.5)
> 
> so you get
> 0.5^(-K)=[(1-alpha)/2] * [ 2 + 2cos(w)]/[1+alpha^2-2*alpha*cos(w)]
> 
> now, keep in mind that in this version of things, w is valid between
> (-pi,pi)
> Good luck solving!
> 
> Chuck
> 
> 
> 
> 
> 
> 
> 
> >
> > - Luigi
> >
> >
> >
> >
> > --- robert bristow-johnson <rbj at audioimagination.com> wrote:
> >
> > >
> > > > ----- Original Message -----
> > > > From: "Luigi Castelli" <superbigio at yahoo.com>
> > > > To: music-dsp at music.columbia.edu
> > > > Subject: [music-dsp] Filter design question
> > > > Date: Fri, 12 Oct 2007 09:45:29 -0700 (PDT)
> > > >
> > > >
> > > > Hi,
> > > >
> > > > the transform function of a high order IIR lowpass filter is
> given
> > > by:
> > > >
> > > > H(z) = { [ (1 - alpha) / 2 ] * [ (1 + z^-1) / (1 - alpha*z^-1)
> ]
> > > }^K
> > > >
> > > > where K is the filter order, alpha is the filter parameter
> which
> > > can be
> > > > determined by the 3dB cutoff frequency wc, i.e. |H(wc)|^2 = 0.5
> > > >
> > > > How do I derive the close-form formula for alpha ?
> > >
> > > plug & chug!
> > >
> > > plug:  z^-1 = e^(-jw) = cos(w) - j*sin(w)
> > >
> > > collect real terms together and imag terms together.
> > >
> > > then get an expression for
> > >
> > >    |H(e^(jw))|^2
> > >
> > > and then set it to 1/2.  if you apply some common trig
> identities,
> > > you'll see that the sin(w) terms get squared and and added to
> > > cos^2(w) terms (go to 1) and you'll be left with only cos(w)
> terms
> > > that you can solve for.  fortunately the ^K power is applied to
> > > *both* numerator and denominator, otherwise you would have a
> bitch of
> > > a time solving a high order polynomial.  the magnitude of the
> stuff
> > > inside the {..}^K gets set to (1/2)^(1/K), so a closed form
> solution
> > > is still doable.
> > >
> > > > Thank you.
> > >
> > > i hope pointers are acceptable in lieu of just doing this for
> you.
> > >
> > > --
> > >
> > > r b-j                  rbj at audioimagination.com
> > >
> > > "Imagination is more important than knowledge."
> > >
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> > >
> >
> >
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