<div dir="ltr">hey, starting to see what you mean much more clear, cool, really excited. Thanks a lot!!!!!</div><div class="gmail_extra"><br><br><div class="gmail_quote">2013/9/24 Alexandre Torres Porres <span dir="ltr"><<a href="mailto:porres@gmail.com" target="_blank">porres@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">so you're basically saying all i need to use is use only the real part, right?<div><br></div><div>my frankenstein was working and alive for several times until i tried some bandpass coeff, let's se if i fix this now :)</div>
</div><div class="HOEnZb"><div class="h5"><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>
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<div dir="ltr"><div class="gmail_default" style="font-family:arial,helvetica,sans-serif"><br></div><div class="gmail_extra"><br><br><div class="gmail_quote"><div>On Tue, Sep 24, 2013 at 3:08 PM, Funs Seelen <span dir="ltr"><<a href="mailto:funsseelen@gmail.com" target="_blank">funsseelen@gmail.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex"><div dir="ltr"><div class="gmail_extra"><div>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>
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<blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);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>
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<br></div><div><b style="font-family:arial,helvetica,sans-serif;font-size:12.8px">b1</b><span style="font-family:arial,helvetica,sans-serif;font-size:12.8px"> = -(P0 + P1)</span><br style="font-family:arial,helvetica,sans-serif;font-size:12.8px">
<b style="font-family:arial,helvetica,sans-serif;font-size:12.8px">b2</b><span style="font-family:arial,helvetica,sans-serif;font-size:12.8px"> = (P0*P1)</span><br></div></div></div></blockquote>
</div><div><br><div 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>
</div></div></div></div></div></blockquote></div><div><br><div class="gmail_default" style="font-family:arial,helvetica,sans-serif">I remembered again, it's called the complex conjugate.</div><div class="gmail_default" style="font-family:arial,helvetica,sans-serif">
<a href="http://en.wikipedia.org/wiki/Complex_conjugate" target="_blank">http://en.wikipedia.org/wiki/Complex_conjugate</a></div><br> </div><div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
<div dir="ltr"><div class="gmail_extra"><div class="gmail_quote"><div><div style="font-family:arial,helvetica,sans-serif">
<br></div><div 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>
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