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

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Navid 2013-12-10

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编辑： Matt J 2013-12-10

采纳的回答： Matt J

I know what does "atan2d" do, but why atan2d(0,0) give us zero? shouldn't it be NaN?

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Answer 1

Matt J 2013-12-10

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编辑：Matt J 2013-12-10

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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 2013-12-10

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 2013-12-10

编辑：Matt J 2013-12-10

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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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Answer 2

Matt J 2013-12-10

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编辑：Matt J 2013-12-10

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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.