24 Sep
2013
24 Sep
'13
6:32 p.m.
On 24/09/13 21:46, Funs Seelen wrote:
On Tue, Sep 24, 2013 at 3:35 PM, Alexandre Torres Porres porres@gmail.comwrote:
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.
Simon