<div dir="ltr">well, not sure what you mean, again way over my head, but I was giving it a hard shot in the dark and it seemed to have worked out :)<div><br></div><div>I just summed both parts of Z0, for instance, and tried the given math, numbers came out!</div>
<div><br></div><div>now to make more tests and see if this is consistent, then finish the patch ;)</div><div><br></div><div>thanks!</div></div><div class="gmail_extra"><br><br><div class="gmail_quote">2013/9/24 Funs Seelen <span dir="ltr"><<a href="mailto:funsseelen@gmail.com" target="_blank">funsseelen@gmail.com</a>></span><br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr"><div class="gmail_extra"><div class="im">On Tue, Sep 24, 2013 at 2:50 PM, Alexandre Torres Porres <span dir="ltr"><<a href="mailto:porres@gmail.com" target="_blank">porres@gmail.com</a>></span> wrote:<br>
</div><div class="gmail_quote"><div class="im">
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="ltr">one doubt emerges really soon anyway. Since they are complex (there are two coordinate numbers for each pole and zero) how do I get only one number by, for example, summing or multiplying one pole to the other? as in:<div>
<div>
<br></div><div><b style="font-family:arial,helvetica,sans-serif;font-size:12.800000190734863px">b1</b><span style="font-family:arial,helvetica,sans-serif;font-size:12.800000190734863px"> = -(P0 + P1)</span><br style="font-family:arial,helvetica,sans-serif;font-size:12.800000190734863px">
<b style="font-family:arial,helvetica,sans-serif;font-size:12.800000190734863px">b2</b><span style="font-family:arial,helvetica,sans-serif;font-size:12.800000190734863px"> = (P0*P1)</span><br></div></div></div></blockquote>
</div><div><br><div class="gmail_default" style="font-family:arial,helvetica,sans-serif">You don't, the coefficients can be complex too. However, I discovered that mirroring (*) every pole and zero results in just real values without imaginary part. I don't have any mathematical proof for this, but it probably wouldn't be too hard to find such.<br>
<br></div><div class="gmail_default" style="font-family:arial,helvetica,sans-serif">*) adding another pole/zero for each complex one, like z=-j if you already have a z=j.<br></div></div></div></div></div>
</blockquote></div><br></div>