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Question on Digital Signal Processing (3rd Editions)- Proakis & Manolakis

Started by marathonBoon April 9, 2008
Does anyone know how to do and explain question 10.19 from Digital Signal
Processing (3rd Editions)- Proakis & Manolakis under multirate digital
signal processing?
 



Hi,

  I have 4th edition. so if you could tell me what is the
  question, i could answer it.

  If i do blind mapping, then is the question?

  prove that the half band filter that satisfies eq .....
  is always odd and the even coefficient are zero.

  if this is the question, then first of all the question
  itself needs correction,
  
  "even coefficients except at n = 0 are zero" is right way.

  the way to solve this is using ideal filter equation and
  rectangular window based half band design.

  hn = sin(2*pi*fc*n/Fs)/(pi*n)

  if you substitute fc = Fs/4 you will start to get even
  coefficients = 0 except at n = 0.

  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.

  secondly if you have coefficient set like

  h0    0    h1     h2     h1    0   h0

  this is possible only with odd length..

  another equation that holds good for half band filter is

  (nc + 1) modulus 4 = 0

  where nc is number of coefficients (example case nc = 7)

regards
bharat pathak

Arithos Designs
www.Arithos.com

DSP design consultancy and training company.

>
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. Dale B. Dalrymple http://dbdimages.com
>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. >
dbd, i was talking about following windows. rect, hann, hamm, balckman, harris, nutall, etc. i was NOT talking about kaiser, chebywin and likes where extra control is possible. anyway, thx for pointing. rgds bharat pathak Arithos Designs www.Arithos.com DSP design consultancy and training company.
On Apr 9, 9:04 pm, "bharat pathak" <bha...@arithos.com> wrote:
> >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. > > dbd, > > i was talking about following windows. > > rect, hann, hamm, balckman, harris, nutall, etc. > > i was NOT talking about kaiser, chebywin and likes > where extra control is possible. > > anyway, thx for pointing. > > rgds > bharat pathak > > Arithos Designswww.Arithos.com > > DSP design consultancy and training company.
I was talking about triangle^N:N=2,3,... and other window families that have no sidelobes. Kaiser-Bessel and Dolph-Chebyshev will have ripples. You can also avoid ripples with optimization algorithms like Parks- McCellan-Steiglitz's METEOR. The point is that half-band filters are characterized by their symmetries, not the presence of ripples. Dale B. Dalrymple
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
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.
>
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 >
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
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. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;