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Why the figure of the phase discontinuous?

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I ploted these function by MATLAB
p=[ ( 0.9000 - 0.0010i) (0.4243 + 0.0017i) (0.1000 + 0.3000i) ]; %this function denoted by (f_k(z)).
p1=[(0.9000 + 0.0010i) (0.2121 - 0.0008i) (0.1000 - 0.3000i) (0)] ; %this function denoted by (f_b(1/z)).
As you can see from the following figures:
In the first figure I ploted the phase of the function, but it look likes discontinuous (some experts here explained that it is not discontinouos but it related to the period of the phase), but I am still confuse from this point, and to escape from this case, I ploted (in the second figure ) the cos(phase) which seems smooth. But I am still do not under stand Why the phase of these function seem as discontinus.
(If the reason related with the period of the phase) WHY the cos(phase) is smooth ,although the cos function also periodic function?
Why the figures of the phase in the first figure look like discontinuouse?
I would appreciate if someone could further elaborate an explanation regarding these cases.

Accepted Answer

Star Strider
Star Strider on 21 Aug 2022
The phase is likely wrapped. Use the unwrap function to provide a continuous phase.
It may be necessary to experiment with it if you are using it on a matrix so use the third dimension argument as necessary. (I have only used it on a vector.)
Aisha Mohamed
Aisha Mohamed on 22 Aug 2022
Thanks Star
As I understood now (if I am correct), this discontinuouty hapened because the phase graph having jumps from +180 to -180 but the entire time when we move from right to left across the plane of the phase, we get more and mor shifts. So we have to use unwarp function to solve this problem.
My questions is
1-Is my explination correct?
2- how can I unwrap this phase?
Iwill appreciate any help
Star Strider
Star Strider on 22 Aug 2022
My pleasure!
1. Essentially, although the unwrap function only works on radian phase angles, so unwrap it as radian angles first and then convert to degrees.
2. See the documentation on unwrap that I linked to earlier.

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