Dear All, This is the my first mail to post our group.Iam working on subband acoustic echo cancellation.In that for subbanding the input signal 2 path polyphase IIR filter is used.I searched for polyphase IIR filter design in net I got some techginical documents.In those documents one documents for 2 path polyphase IIR filter design related to acoustic echgo cancellation, The filter coeffecient values also mention in this document.But here there is no time domain difference equation.I derived the difference equation from the transfer function is given in the document.But the output is cleared.I mentioned the links related to the document below. 1 http://users.cscs.wmin.ac.uk/~krukowa/pdf/Paper27.pdf 2 http://ieeexplore.ieee.org/iel5/8570/27140/01205861.pdf please give some information related to timedomain difference equation of polyphase IIr filter design. Thanks in advance. venkat.

# regarding polyphase IIR filter.

Started by ●January 15, 2007

Reply by ●January 15, 20072007-01-15

ch.venkat wrote:> Dear All, > > This is the my first mail to post our group.Iam working on subband > acoustic echo cancellation.In that for subbanding the input signal 2 path > polyphase IIR filter is used.I searched for polyphase IIR filter design in > net I got some techginical documents.In those documents one documents for 2 > path polyphase IIR filter design related to acoustic echgo cancellation, > The filter coeffecient values also mention in this document.But here there > is no time domain difference equation.I derived the difference equation > from the transfer function is given in the document.But the output is > cleared.I mentioned the links related to the document below. > 1 http://users.cscs.wmin.ac.uk/~krukowa/pdf/Paper27.pdf > 2 http://ieeexplore.ieee.org/iel5/8570/27140/01205861.pdf > > please give some information related to timedomain difference > equation of polyphase IIr filter design. > > Thanks in advance. > venkat.In the first document, the filter coefficients they give are the values of alpha[n,k] in their formula for the transfer function An(z^-2). Going from a transfer function to the difference equation is trivial; since H(z) = Y(z) / X(z) (where x[n] <-> X(z) is the input and y[n] <-> Y(z) is the output), solve for Y(z). From there, you can directly write the difference equation; each factor of z^-1 corresponds to a sample delay and the coefficients map over directly. Jason

Reply by ●January 15, 20072007-01-15

cincydsp@gmail.com wrote:> ch.venkat wrote: > > Dear All, > > > > This is the my first mail to post our group.Iam working on subb=and> > acoustic echo cancellation.In that for subbanding the input signal 2 pa=th> > polyphase IIR filter is used.I searched for polyphase IIR filter design=in> > net I got some techginical documents.In those documents one documents f=or 2> > path polyphase IIR filter design related to acoustic echgo cancellation, > > The filter coeffecient values also mention in this document.But here th=ere> > is no time domain difference equation.I derived the difference equation > > from the transfer function is given in the document.But the output is > > cleared.I mentioned the links related to the document below. > > 1 http://users.cscs.wmin.ac.uk/~krukowa/pdf/Paper27.pdf > > 2 http://ieeexplore.ieee.org/iel5/8570/27140/01205861.pdf > > > > please give some information related to timedomain differe=nce> > equation of polyphase IIr filter design. > > > > Thanks in advance. > > venkat. > > In the first document, the filter coefficients they give are the values > of alpha[n,k] in their formula for the transfer function An(z^-2). > Going from a transfer function to the difference equation is trivial; > since H(z) =3D Y(z) / X(z) (where x[n] <-> X(z) is the input and y[n] <-> > Y(z) is the output), solve for Y(z). From there, you can directly write > the difference equation; each factor of z^-1 corresponds to a sample > delay and the coefficients map over directly. > > JasonDear Jason, the transfer function is H(Z) =3D 1/N (=CE=A3 An( Z-N ) Z-n) Here N is the order of the filter and n varries from 0 to N-1. here An( Z-N ) is given by An( Z-N ) =3D =CE=A0((an k + Z -N) / (1 + an k Z -N)) here a n k is filter coeffecients from this i derived the diffrence equation but the and also wrote program. But how i can deside the output of the filter is correct or not.Give some suggestions. Thnks for ur information. venkat.

Reply by ●January 16, 20072007-01-16

venkat wrote:> Dear Jason, > the transfer function is H(Z) =3D 1/N (=CE=A3 An( Z-N ) Z-n) Here N is the > order of the filter and n varries from 0 to N-1. here An( Z-N ) is > given by > An( Z-N ) =3D =CE=A0((an k + Z -N) / (1 + an k Z -N)) here a n k is filt=er> coeffecients > > from this i derived the diffrence equation but the and also wrote > program. But how i can deside the output of the filter is correct or > not.Give some suggestions. >The goal of the polyphase IIR filter bank was to provide simultaneous low-pass and high-pass filtering while cutting the sample rate in half. To see if your polyphase implementation is correct, take the filter coefficients and generate the two equivalent low-pass and high-pass filters. Cascade the equivalent filter with a downsampling-by-2 step, and compare the outputs of those filters to those of your polyphase filter bank. They should be the same; that tells you if you implemented it right. Jason

Reply by ●January 17, 20072007-01-17

cincydsp@gmail.com wrote:> venkat wrote: > > Dear Jason, > > the transfer function is H(Z) =3D 1/N (=CE=A3 An( Z-N ) Z-n) Here N is =the> > order of the filter and n varries from 0 to N-1. here An( Z-N ) is > > given by > > An( Z-N ) =3D =CE=A0((an k + Z -N) / (1 + an k Z -N)) here a n k is fi=lter> > coeffecients > > > > from this i derived the diffrence equation but the and also wrote > > program. But how i can deside the output of the filter is correct or > > not.Give some suggestions. > > > > The goal of the polyphase IIR filter bank was to provide simultaneous > low-pass and high-pass filtering while cutting the sample rate in half. > To see if your polyphase implementation is correct, take the filter > coefficients and generate the two equivalent low-pass and high-pass > filters. Cascade the equivalent filter with a downsampling-by-2 step, > and compare the outputs of those filters to those of your polyphase > filter bank. They should be the same; that tells you if you implemented > it right. > > JasonThanks for your suggestion. downsampling-by-2 step meens? Is there any other approch for checking the filter outputs?

Reply by ●January 17, 20072007-01-17

venkat wrote:> cincydsp@gmail.com wrote: > > venkat wrote: > > > Dear Jason, > > > the transfer function is H(Z) =3D 1/N (=CE=A3 An( Z-N ) Z-n) Here N i=s the> > > order of the filter and n varries from 0 to N-1. here An( Z-N ) is > > > given by > > > An( Z-N ) =3D =CE=A0((an k + Z -N) / (1 + an k Z -N)) here a n k is =filter> > > coeffecients > > > > > > from this i derived the diffrence equation but the and also wrote > > > program. But how i can deside the output of the filter is correct or > > > not.Give some suggestions. > > > > > > > The goal of the polyphase IIR filter bank was to provide simultaneous > > low-pass and high-pass filtering while cutting the sample rate in half. > > To see if your polyphase implementation is correct, take the filter > > coefficients and generate the two equivalent low-pass and high-pass > > filters. Cascade the equivalent filter with a downsampling-by-2 step, > > and compare the outputs of those filters to those of your polyphase > > filter bank. They should be the same; that tells you if you implemented > > it right. > > > > Jason > > Thanks for your suggestion. > downsampling-by-2 step meens? > Is there any other approch for checking the filter outputs?Dear Jason I have one more dought.Please give me reply? That Is how we reconstruct the input signal from polyphase IIR filter outputs?

Reply by ●January 17, 20072007-01-17

On 16 Jan 2007 21:29:33 -0800, "venkat" <venkatakrishnachunduri@gmail.com> wrote:> >cincydsp@gmail.com wrote: >> venkat wrote: >> > Dear Jason, >> > the transfer function is H(Z) = 1/N (? An( Z-N ) Z-n) Here N is the >> > order of the filter and n varries from 0 to N-1. here An( Z-N ) is >> > given by >> > An( Z-N ) = ?((an k + Z -N) / (1 + an k Z -N)) here a n k is filter >> > coeffecients >> > >> > from this i derived the diffrence equation but the and also wrote >> > program. But how i can deside the output of the filter is correct or >> > not.Give some suggestions. >> > >> >> The goal of the polyphase IIR filter bank was to provide simultaneous >> low-pass and high-pass filtering while cutting the sample rate in half. >> To see if your polyphase implementation is correct, take the filter >> coefficients and generate the two equivalent low-pass and high-pass >> filters. Cascade the equivalent filter with a downsampling-by-2 step, >> and compare the outputs of those filters to those of your polyphase >> filter bank. They should be the same; that tells you if you implemented >> it right. >> >> Jason > >Thanks for your suggestion. >downsampling-by-2 step meens?take every other sample, i.e. cut the rate in half.>Is there any other approch for checking the filter outputs?code everything in C, and compare your dsp outputs to the C outputs