Re: Single pole lowpass --> highpass - z transform?

On 1 Aug, 04:53, "C Warwick" <c...@xxxxxxxxx> wrote:
Ok, say i have a single pole lowpass filter, 1 pole, no zero.

y[0] = c.x[0] + (1-c).y[-1]

This is basicly a single pole moving from (0,0) to (1,0) on the real axis.
And normalized at DC.

Now by doing this seperately after the lowpass.

highpass = x[0] - y[0]

You get a high pass. But i cant get my hed round where the poles (and zeros)
would be? I dont really see how to combine it into one difference equation
so i can work backwards to the z-transform / poles/zeros.

There is a standard way to transform a low-pass prototype filter
with cut-off at w1 to a high-pass filter with cut-off at w2.

It's been a while since I played around with those things, but
try to look up 'spectral transforms' in the book by Proakis &
Manolakis. I know they have a table with these transforms
in the chapter on IIR filters, and I also happen to know
that there were a couple of typos in that table in their 3rd
edition (and presumably earlier editions as well). The typos
were corrected in their 4th edition.

BTW, the formulas are quite elaborate. If you want to use them
to transform filters of arbitrary order, implement the transforms
for 1st and 2nd order sections, and represent the higher-order
filters (N > 2) as a sequence of 1st and 2nd order sections
(is that 'cascade form'?).

Rune
.

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