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I do not get exactly what does atan2d(0,0) means?

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Navid
Navid 2013 年 12 月 10 日
編集済み: Matt J 2013 年 12 月 10 日
I know what does "atan2d" do, but why atan2d(0,0) give us zero? shouldn't it be NaN?

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Matt J
Matt J 2013 年 12 月 10 日
編集済み: Matt J 2013 年 12 月 10 日
I think the reasoning is this,
t=atan2d(y,x)
should satisfy
norm([x,y])*cosd(t)=x
norm([x,y])*sind(t)=y
In the case where x=y=0, any t would satisfy this, but the convention t=0 ensures continuity of atan2d(y,x) over x>=0.
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Navid
Navid 2013 年 12 月 10 日
Matt,
Thank you for your answer and it was a good discussion, but I am not satisfied yet.
If we choose t=pi/2, it will give us continuity over y>=0!!! in fact as you said, any t would satisfy it and we can say that it has singularity at x=y=0. So I believe that it should be NaN.
Matt J
Matt J 2013 年 12 月 10 日
編集済み: Matt J 2013 年 12 月 10 日
Hi Navid,
If we choose t=pi/2, it will give us continuity over y greater than or equal 0
That is true. It is something of an arbitrary choice, I imagine, but less arbitrary than, say t=pi/sqrt(3).
in fact as you said, any t would satisfy it and we can say that it has singularity at x=y=0. So I believe that it should be NaN.
We can only guess the minds of the software developers. I don't think it could precisely be called a singularity, though, because there exist real values that satisfy the defining equations, if you're willing to accept
norm([x,y])*cosd(t)=x
norm([x,y])*sind(t)=y
as that definition.

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その他の回答 (1 件)

Matt J
Matt J 2013 年 12 月 10 日
編集済み: Matt J 2013 年 12 月 10 日
As further motivation, consider also the similar convention used by the angle() function
>> angle(0+0i)
ans =
0
In Fourier Analysis, this makes sense because, even though Fourier spectra are in general complex-valued, there are prominent cases where a spectrum S(f) is real-valued and you want the phase response to be semi-continuous at zero crossings S(f)=0.

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