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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)
>Tnew<Told
>but 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
>
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