I do not get exactly what does atan2d(0,0) means?
2 ビュー (過去 30 日間)
古いコメントを表示
採用された回答
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.
2 件のコメント
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.
その他の回答 (1 件)
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.
0 件のコメント
参考
カテゴリ
Help Center および File Exchange で Resizing and Reshaping Matrices についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!