Reply by dbd April 13, 20082008-04-13
On Apr 13, 1:54 am, "marathonBoon" <sealb...@yahoo.com.sg> wrote:
> >Hi, > > after reading through the replies, i am still unable to solve the > mentioned question.How to proved the polyphase interpolation? > > Thanks, > Boon > > Hi, > > > > > This actually means that we are doing a poly-phase > > decomposition of the same filter. > > > Let us take an example of I = 2 and following filter > > > h(n) = [h0 h1 h2 h3 h2 h1 h0]; > > > this when delayed by one clock will result in > > > h(n-1) = [0 h0 h1 h2 h3 h2 h1 h0]; > > > now we need to perform decimation by 2 to get our phases > > > p0(n) = h(n/2) = [h0 h2 h2 h0]; > > p1(n) = h((n-1)/2) = [0 h1 h3 h1]; > > > you may want to perform all calculations using non-causal > > approach as equation talks about n = 0, +/- I ........ > > >Regards > >Bharat Pathak > > >Arithos Designs > >www.Arithos.com > > >DSP design consultancy and training company. > > >>Hi, > > >>The question is as follows: > > >>10.19 Prove the following expressions for an interpolator of order I. > >>(a) The impulse response h(n) can be expressed as > > >>h(n)=&#4294967295;&#4294967295;_(k=0)^(l-1)&#4294967295;&#4294967295;P_(k(n-k)) > > >>where > > >> P_k (n)=P_k (n/I),for n=0,&#4294967295;&#4294967295;I,&#4294967295;&#4294967295;2I&#4294967295;&#4294967295;.and 0 otherwise
Boon No one around seems to have the book and edition your problem comes from. If you want have people respond to the question you are going to at least be able to state it in a manner that can be understood without reference to definitions in your book. Towards that end, without the book, I don't know what "interpolator of order l" is defined as. Also, in the expression: h(n)=&#4294967295;&#4294967295;_(k=0)^(l-1)&#4294967295;&#4294967295;P_(k(n-k)) the character: &#4294967295;&#4294967295; as rendered by my newsreader's character set has no meaning I understand. Perhaps no one else can interpret it either. Of course, on comp.dsp you can't count on discussion disappearing because of a lack of definitions, but lack of definitions usually sets the discussion free to wander. And we have. Dale B. Dalrymple
Reply by marathonBoon April 13, 20082008-04-13
>Hi,
after reading through the replies, i am still unable to solve the mentioned question.How to proved the polyphase interpolation? Thanks, Boon Hi,
> > This actually means that we are doing a poly-phase > decomposition of the same filter. > > Let us take an example of I = 2 and following filter > > h(n) = [h0 h1 h2 h3 h2 h1 h0]; > > this when delayed by one clock will result in > > h(n-1) = [0 h0 h1 h2 h3 h2 h1 h0]; > > now we need to perform decimation by 2 to get our phases > > p0(n) = h(n/2) = [h0 h2 h2 h0]; > p1(n) = h((n-1)/2) = [0 h1 h3 h1]; > > you may want to perform all calculations using non-causal > approach as equation talks about n = 0, +/- I ........ > >Regards >Bharat Pathak > >Arithos Designs >www.Arithos.com > >DSP design consultancy and training company. > > > >>Hi, >> >>The question is as follows: >> >>10.19 Prove the following expressions for an interpolator of order I. >>(a) The impulse response h(n) can be expressed as >> >> >>h(n)=&sum;_(k=0)^(l-1)&#9618;P_(k(n-k)) >> >>where >> >> P_k (n)=P_k (n/I),for n=0,&plusmn;I,&plusmn;2I&hellip;.and 0 otherwise >> >
Reply by Jerry Avins April 11, 20082008-04-11
dbd wrote:

   ...

> Examples of sidelobeless windows: > > triangle to a power (2 or greater) > > Hanning-Poisson (for large alpha) See the fred harris 1978 Proceedings > paper
... Thanks much. I have more to play with. Insights are grand! Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Reply by dbd April 10, 20082008-04-10
On Apr 10, 4:35 pm, Eric Jacobsen <eric.jacob...@ieee.org> wrote:
> On Thu, 10 Apr 2008 15:53:55 -0400, Jerry Avins <j...@ieee.org> wrote: > >dbd wrote: > >> On Apr 9, 7:28 am, "bharat pathak" <bha...@arithos.com> wrote: > > >>> ... > > >>> In window based design you will always have same ripple in > >>> passband and stopband. hence the designed filter meets > >>> the half band criteria of perfect symmetry around fc = Fs/4. > > >>> ... > >>> regards > >>> bharat pathak > > >> In a window based design you will always have the same -error- in > >> passband and stopband. Ripple can be avoided by the use of > >> sidelobeless windows. >
> >What window is completely free of sidelobes? > > >Jerry > > Double hung? > > Casement? > > Sorry... ;) > > Eric Jacobsen > Minister of Algorithms > Abineau Communicationshttp://www.ericjacobsen.org
Doesn't that require that we determine an acceptable definition of 'sidelobeless' on the architectural domain? No more, or less sorry... :) Dale B. Dalrymple
Reply by Eric Jacobsen April 10, 20082008-04-10
On Thu, 10 Apr 2008 15:53:55 -0400, Jerry Avins <jya@ieee.org> wrote:

>dbd wrote: >> On Apr 9, 7:28 am, "bharat pathak" <bha...@arithos.com> wrote: >> >>> ... >>> >>> In window based design you will always have same ripple in >>> passband and stopband. hence the designed filter meets >>> the half band criteria of perfect symmetry around fc = Fs/4. >>> >>> ... >>> regards >>> bharat pathak >> >> In a window based design you will always have the same -error- in >> passband and stopband. Ripple can be avoided by the use of >> sidelobeless windows. > >What window is completely free of sidelobes? > >Jerry
Double hung? Casement? Sorry... ;) Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org
Reply by dbd April 10, 20082008-04-10
On Apr 10, 12:53 pm, Jerry Avins <j...@ieee.org> wrote:
> dbd wrote: > > On Apr 9, 7:28 am, "bharat pathak" <bha...@arithos.com> wrote: > > >> ... > > >> In window based design you will always have same ripple in > >> passband and stopband. hence the designed filter meets > >> the half band criteria of perfect symmetry around fc = Fs/4. > > >> ... > >> regards > >> bharat pathak > > > In a window based design you will always have the same -error- in > > passband and stopband. Ripple can be avoided by the use of > > sidelobeless windows. > > What window is completely free of sidelobes? > > Jerry > -- > Engineering is the art of making what you want from things you can get. > &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Examples of sidelobeless windows: triangle to a power (2 or greater) Hanning-Poisson (for large alpha) See the fred harris 1978 Proceedings paper For examples with applications: Kay, S. Smith, D. An optimal sidelobeless window Signal Processing, IEEE Transactions on Vol. 47 No. 9 Sept. p2542 1999 P. Depalle and T. H&#4294967295;elie. Extraction of spectral peak parameters using a short-time Fourier transform modeling and no sidelobe windows. In Proceedings of IEEE Workshop on Applications of Signal Processing to Audio and Acoustics, New Paltz, USA, 1997. Dale B. Dalrymple
Reply by Jerry Avins April 10, 20082008-04-10
dbd wrote:
> On Apr 9, 7:28 am, "bharat pathak" <bha...@arithos.com> wrote: > >> ... >> >> In window based design you will always have same ripple in >> passband and stopband. hence the designed filter meets >> the half band criteria of perfect symmetry around fc = Fs/4. >> >> ... >> regards >> bharat pathak > > In a window based design you will always have the same -error- in > passband and stopband. Ripple can be avoided by the use of > sidelobeless windows.
What window is completely free of sidelobes? Jerry -- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Reply by dbd April 10, 20082008-04-10
On Apr 10, 2:50 am, "bharat pathak" <bha...@arithos.com> wrote:

> if and only if the following two conditions are met > we get half band filter with even coefficients being > zero except at n = 0. > > 1. the passband freq fp and stop band freq fs are symmetric > w.r.t Fs/4, i think u agree on this.
It isn't just the values of the parameters fp and fs that is symmetric, it is the whole response curve.
> 2. passband deviation d1 and stopband deviation d2 are same.
It isn't just the deviation limits of the extremum of the response, it's a constraint on all points on the curve.
> here we are not characterizing half band filter. We are trying > to design half band filter with special requirement like even > coefficients being 0, except for n = 0.
Getting the symmetry correct at all points on the curve is what gets most of the even coefficients to be zero. The symmetry forces most of the even coefficients to zero, whether the response is rippled or monotonic. I've given examples of monotonic alternative approaches for windowed and optimized designs. Talking about the 'ripples' does not address the situation.
> a half band filter which is just symmetric w.r.t Fs/4 does > not offer any hardware advantage. specially true when designing > with remez design algo, kaiser or chebwin.
The problem with the straightforward application of remez to halfband filter specifications is that only the constraints on error peaks are symmetric, not all points on the response curve, which is why remez can give non-zero coefficients on the even terms. There are ways to use remez to generate true halfband filters. See: Satisfying the Haar condition in halfband FIR filter design Ansari, R. Acoustics, Speech, and Signal Processing, IEEE Transactions on On page(s): 123-124 Volume: 36, Jan 1988 This has been discussed on comp.dsp before.
> anyway the whole discusssion is pointless as the OP has given > the question which is different then what we are debating here.
If all material not immediately related to OP questions were to be removed from comp.dsp, there would be very little left.
> > rgds > bharat pathak >
Dale B. Dalrymple
Reply by bharat pathak April 10, 20082008-04-10
Hi,

  This actually means that we are doing a poly-phase
  decomposition of the same filter.

  Let us take an example of I = 2 and following filter

  h(n)   = [h0 h1 h2 h3 h2 h1 h0];

  this when delayed by one clock will result in

  h(n-1) = [0 h0 h1 h2 h3 h2 h1 h0];

  now we need to perform decimation by 2 to get our phases

  p0(n) = h(n/2)     = [h0 h2 h2 h0];
  p1(n) = h((n-1)/2) = [0  h1 h3 h1];

  you may want to perform all calculations using non-causal
  approach as equation talks about n = 0, +/- I ........

Regards
Bharat Pathak

Arithos Designs
www.Arithos.com

DSP design consultancy and training company.



>Hi, > >The question is as follows: > >10.19 Prove the following expressions for an interpolator of order I. >(a) The impulse response h(n) can be expressed as > > >h(n)=&sum;_(k=0)^(l-1)&#9618;P_(k(n-k)) > >where > > P_k (n)=P_k (n/I),for n=0,&plusmn;I,&plusmn;2I&hellip;.and 0 otherwise >
Reply by bharat pathak April 10, 20082008-04-10
dbd,

   if and only if the following two conditions are met
   we get half band filter with even coefficients being
   zero except at n = 0.

   1. the passband freq fp and stop band freq fs are symmetric
      w.r.t Fs/4, i think u agree on this. 

   2. passband deviation d1 and stopband deviation d2 are same.
      
   here we are not characterizing half band filter. We are trying
   to design half band filter with special requirement like even
   coefficients being 0, except for n = 0. 

   a half band filter which is just symmetric w.r.t Fs/4 does
   not offer any hardware advantage. specially true when designing
   with remez design algo, kaiser or chebwin.

   anyway the whole discusssion is pointless as the OP has given
   the question which is different then what we are debating here.

rgds
bharat pathak

Arithos Designs
www.Arithos.com

DSP design consultancy and training company.
>