# Why the figure of the phase discontinuous?

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Aisha Mohamed . 2022년 8월 21일
댓글: Star Strider . 2022년 8월 22일
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?
MYQUESTION IS
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.
THIS IS THE FIRST FIGURE THIS IS THE SECOND FIGUR 댓글을 달려면 로그인하십시오.

### 채택된 답변

Star Strider 2022년 8월 21일
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.)
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Star Strider 2022년 8월 22일
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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