From: mmoctar on 4 Jun 2010 18:31 I think that as for signals you can use a sinc funtion to approximate your new coefficients. Moc>I have a difference equation >y(n)=3DA1*y(n-1)+A2*y(n-2)...+B1*x(n-1)+B2*Y(n-2)=85+C >The coefficients A1,A2,...B1,B2,.... were computed with a known sample >time. C is a null offset. >Is there an easy way to recalculate A1,A2,..,B1,B2,... for another >sample time so the response is the same? >Doing this for something simple like >y(n)=3DA1*y(n-1)+B1*x(n-1)+C >A1new=3DA1old^(Tnew/Told) >Tnewbut what if there are more coefficients? >My fall back solution is to use z transform tables to convert to the s >domain and then back to the z domain with the new time period. I can >find the equations using Mathcad but I can't believe this hasn't been >done before and there isn't a more general way of doing this other >than going through each case I may encounter. I don't wont to re- >invent or rediscover the solution if it already exists. A more general >case would involve calculating all the z domain poles and zeros, >converting the s domain and then back. That is messy and then there >is that pesky offset C. > >Thanks, > >Peter Nachtwey >