Hello friends. I have an apparently unsolvable problem: I have a given IIR filter (they give me the poles and the zeros). The filter was originally sampled at a given (very high) frequency. I also have a signal, which is sampled at a much lower rate. When I filter my signal, I have to resample it at the same (high and unuseful) frequency of the filter, in order to have a correct filtering. I'd like instead to filter my signal at its original (low) rate. Is it possible to lower the frequency of an IIR (just repositioning the zeros and the poles)? If that's possible, I could filter my signal at the original rate, saving much computational time! Please, let me know if there's any solution ;-) Bye Stefano
resampling of an IIR filter
Started by ●November 28, 2002
Reply by ●November 30, 20022002-11-30
Interesting problem stefan!!!!!!! when ever a signal or system is down sampled ie its sampling rate is lowrered from its high value, then in its Z-Transform function the Z is replaced by Z raised to the power of inverse of M. That shows that poles and zeroes are changed by (increased or decreased I haven't worked out) M times and the positioning I feel remains the same. Take a 10th order IIR filter and replace Z with Z raised to the power of inverse of M and see how many poles and zeroes u are now getting and also u can see that the position of Poles and Zeroes remain the same. This is just a response which came spontaneously after looking at ur query. I haven't seriously worket it out. Hope my answer could help atleast partially to u. with Regards Madhusudhan stefanosorrentino <shepan@shep...> wrote:Hello friends. I have an apparently unsolvable problem: I have a given IIR filter (they give me the poles and the zeros). The filter was originally sampled at a given (very high) frequency. I also have a signal, which is sampled at a much lower rate. When I filter my signal, I have to resample it at the same (high and unuseful) frequency of the filter, in order to have a correct filtering. I'd like instead to filter my signal at its original (low) rate. Is it possible to lower the frequency of an IIR (just repositioning the zeros and the poles)? If that's possible, I could filter my signal at the original rate, saving much computational time! Please, let me know if there's any solution ;-) Bye Stefano _____________________________________ Note: If you do a simple "reply" with your email client, only the author of this message will receive your answer. You need to do a "reply all" if you want your answer to be distributed to the entire group. _____________________________________ About this discussion group: To Join: audiodsp-subscribe@audi... To Post: audiodsp@audi... To Leave: audiodsp-unsubscribe@audi... Archives: http://groups.yahoo.com/group/audiodsp Other DSP-Related Groups: http://www.dsprelated.com --------------------------------- Do you Yahoo!? Yahoo! Mail Plus - Powerful. Affordable. Sign up now
Reply by ●November 30, 20022002-11-30
Since Z is replaced by Z raised to the power of inverse of M ( where M is the decimation factor ), so all the poles and zeroes not only radially move but also their position with respect to the real axis also changes (angular displacement). But it ensures stable filter but since poles move towards unit circle there is a possibility of the system to have tendency to go into oscillations. in my earlier mail I wrote that the no of Poles and Zeroes may get altered. with best regards Madhusudhan stefanosorrentino <shepan@shep...> wrote:Hello friends. I have an apparently unsolvable problem: I have a given IIR filter (they give me the poles and the zeros). The filter was originally sampled at a given (very high) frequency. I also have a signal, which is sampled at a much lower rate. When I filter my signal, I have to resample it at the same (high and unuseful) frequency of the filter, in order to have a correct filtering. I'd like instead to filter my signal at its original (low) rate. Is it possible to lower the frequency of an IIR (just repositioning the zeros and the poles)? If that's possible, I could filter my signal at the original rate, saving much computational time! Please, let me know if there's any solution ;-) Bye Stefano _____________________________________ Note: If you do a simple "reply" with your email client, only the author of this message will receive your answer. You need to do a "reply all" if you want your answer to be distributed to the entire group. _____________________________________ About this discussion group: To Join: audiodsp-subscribe@audi... To Post: audiodsp@audi... To Leave: audiodsp-unsubscribe@audi... Archives: http://groups.yahoo.com/group/audiodsp Other DSP-Related Groups: http://www.dsprelated.com --------------------------------- Do you Yahoo!? Yahoo! Mail Plus - Powerful. Affordable. Sign up now
