It seems with bilinear transforms the back-and-forth can be omitted. Because
eventually one always transforms 1 to 1 and -1 to -1 in the z-plane, the
only thing left for bilinear transforms is replacing z by some (az+1)/(z+a).
That "a" can be found out by matching frequencies on the unit circle.
-----Original Message-----
On Wed, 16 Jan 2013 06:07:51 -0500, robert bristow-johnson wrote:
if i were to try to re-calculate the coefficients, i would first factor
out the constant gain, then factor both numerator and denominator into
discrete-time poles and zeros. then map those poles and zeros back to
analog poles and zeros using, i suppose the inverse bilinear transform
(with warping). then re-transform back with the bilinear transform with
the new sampling rate.
i dunno. that's how i might approach it.
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