>My filter has 2 poles and 1 zero. Unlike the Cookbook filter, which
>has 2 poles and 2 zeros.
>I think that automatically assumes, the transfer function cannot be
equivalent.

No, that does not follow. A filter with two zeros can produce all of the
transfer functions that a filter with one zero can, and many more besides.
You just stick the extra zero(s) off at z=infinity.

>That also means, that my filter is also faster to process.

First order filters are typically cheaper to process than second order
filters. I wouldn't say that is a good argument that they are preferable to
second order filters, though.

E

On Wed, Feb 4, 2015 at 5:25 PM, Peter S <peter.schoffhau...@gmail.com>
wrote:

> On 04/02/2015, robert bristow-johnson <r...@audioimagination.com> wrote:
> > i'm only saying that
> > with a 2nd-order filter, there are only 5 degrees of freedom.  only 5
> > knobs.  so when someone says they came up with a different or better
> > method of computing coefficients for *whatever* 2nd-order filter, my
> > first curiosity will be what is the transfer function and from that, we
> > usually find out it is equivalent to the Cookbook transfer function
> > except Q or bandwidth is defined differently.
>
> My filter has 2 poles and 1 zero. Unlike the Cookbook filter, which
> has 2 poles and 2 zeros.
> I think that automatically assumes, the transfer function cannot be
> equivalent.
> (And strictly speaking, the transfer curve of my filter is not
> biquadratic, just BLT.)
> That also means, that my filter is also faster to process.
>
> Best regards,
> Peter
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