why this overlap happen here?

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Aisha Mohamed
Aisha Mohamed on 22 Sep 2022
Edited: Aisha Mohamed on 23 Sep 2022
Hi all
I am working on this function
f_b(1/z) = (0.10 − 0.3i)z −1 + (0.2121 − 0.0008i)z −2 + (0.9 + 0.001i)z −3 and I wrote it in MATLAB as p1,
p1=[(0.9000 + 0.0010i) (0.2121 - 0.0008i) (0.1000 - 0.3000i) (0)] ; which has the following roots ,(Thanks Torsten)
1/( 0.0000 + 0.0000i)
1/ ( -0.4716 - 0.4706i)
1/ ( 0.2359 + 0.4717i)
when I plotted the cos(phase(f_b(1/z))) , I find the following figure
where the bright yellow colors indicate positive values of cosine function, thus phase(f_b(1/z)) =0.
Whereas the dark blue colors indicate negative values of cosine function, thus phase(f_b(1/z)) =+ or – pi.
My question is,
what happened in the center of the figure(I put it inside circle ) , why the positive (yellow) and negative (blue) values overlap by this way?
I mean why ((the interference occurs at the origin where the two peaks meet, a strip of dark blue negative values (where $phase(f_b(1/z))= \pm \pi $) permeates the top of the bright yellow positive bottom (where $ phase(f_b(1/z))= 0 $) and becomes surrounded by yellow positive values. Similarly, a bar of bright yellow positive values permeates the top of the dark blue negative to become surrounded by dark blue negative values(where $ phase(f_b(1/z))= \pm \pi $) )) this happen here?
Note that the yellow and blue figures meet at the locate of the zeros of this function.
I appreciate any help.
  2 Comments
Torsten
Torsten on 22 Sep 2022
The roots of f_b(1/z) are 1./roots(p1), not roots(p1).
Aisha Mohamed
Aisha Mohamed on 22 Sep 2022
Many Thanks Torsten. Yes you are right.

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