On Tue, Sep 24, 2013 at 6:32 PM, Simon Wise <[email protected]> wrote:
> On 24/09/13 21:46, Funs Seelen wrote: > >> On Tue, Sep 24, 2013 at 3:35 PM, Alexandre Torres Porres >> <[email protected]>wrote: >> >> so you're basically saying all i need to use is use only the real part, >>> right? >>> >>> >> No, I meant that I have the idea that the imaginary part in the calculated >> coefficients will disappear automatically if you add complex conjugates >> for >> all poles and zeros, probably when somehow i^2 gets -1 somewhere. But I >> must say I'm not a mathematician and not sure at all. >> > > indeed it will ... a conjugate is the number with the imaginary part > negated ... so adding a number and its conjugate will certainly end up with > a real part only. > Yes, true, and the imaginary part disappears as well when multiplying if the real parts are equal, e.g.: i^2 = -1, so ... (0.5 + 0.5i) * (0.5 - 0.5i) = 0.25 + 0.25i - 0.25i - 0.25i^2 = 0.5 > > > Simon > > > ______________________________**_________________ > [email protected] mailing list > UNSUBSCRIBE and account-management -> http://lists.puredata.info/** > listinfo/pd-list <http://lists.puredata.info/listinfo/pd-list> >
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