Changing the atan function so that it ranges from 0 to 2*pi
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I know that the matlab atan function returns values in the range of -pi/2 to pi/2. How do i change it so that it goes over the full range 0 to 2*pi?
My first attempt was using a while loop, but it was incorrect.
I need to write a function mfile to set the built-in matlab function atan in the range of 0 to 2*pi without using atan2. im new to matlab so im unsure of what to do.
Thank you
2 件のコメント
wenjun kou
2017 年 3 月 8 日
編集済み: wenjun kou
2017 年 3 月 8 日
Although you don't want to use atan2, I thought I might just put this out there since atan2 returns a range between -pi to pi:
a = atan2(y, x);
a = a .* (a >= 0) + (a + 2 * pi) .* (a < 0);
採用された回答
Daniel Svedbrand
2018 年 9 月 14 日
編集済み: John D'Errico
2023 年 8 月 3 日
Adding mod 2*pi to atan2 should work just fine
z = mod(atan2(y,x),2*pi);
6 件のコメント
Feruza Amirkulova
2023 年 8 月 3 日
Yes, mod(atan2(y,x),2*pi) worked and its gradients are the same as for (atan2(y,x)).
その他の回答 (4 件)
Walter Roberson
2011 年 6 月 12 日
5 件のコメント
Paulo Silva
2011 年 6 月 12 日
I didn't include that statement on purpose, when none of the others if statements are true the value of v is NaN, you could also do this:
if isnan(v)
error('Arguments must be different from zero')
end
Paulo Silva
2011 年 6 月 12 日
function v=myatan(y,x)
if nargin==1 %just in case the user only gives the value of y myatan(y)
x=1;
end
v=nan;
if x>0
v=atan(y/x);
end
if y>=0 & x<0
v=pi+atan(y/x);
end
if y<0 & x<0
v=-pi+atan(y/x);
end
if y>0 & x==0
v=pi/2;
end
if y<0 & x==0
v=-pi/2;
end
if v<0
v=v+2*pi;
end
end
2 件のコメント
Mehmet Can Türk
2022 年 4 月 9 日
編集済み: Mehmet Can Türk
2022 年 4 月 9 日
I checked the Wikipedia link and tested the code. First of all, thank you so much for the contribution.
I wanted to convert atan2 function from Matlab into another environment which supports only atan function. So I deleted the if block and everything worked perfectly.
Kent Leung
2018 年 3 月 21 日
編集済み: Kent Leung
2018 年 3 月 21 日
Better late than never. (Also posting as a future reference to myself.) The function below accepts y & x as vectors in Matlab. Rather than using 'if' statements, the below might be faster if there is some parallelization implemented in the built-in index searching.
Note: I have a slight disagreement with the above for the x>0 & y<0 case, as well as the for x=0 & y<0 case. The code below gives 0 to 2pi.
function v=myatan(y,x)
%---returns an angle in radians between 0 and 2*pi for atan
v=zeros(size(x));
v(x>0 & y>=0) = atan( y(x>0 & y>=0) ./ x(x>0 & y>=0) );
v(x>0 & y<0) = 2*pi+atan( y(x>0 & y<0) ./ x(x>0 & y<0) );
v(x<0 & y>=0) = pi+atan( y(x<0 & y>=0) ./ x(x<0 & y>=0) );
v(x<0 & y<0) = pi+atan( y(x<0 & y<0) ./ x(x<0 & y<0) );
v(x==0 & y>=0) = pi/2;
v(x==0 & y<0) = 3/2*pi;
end
theodore panagos
2018 年 10 月 27 日
You can use the formula:
atan(x,y)=pi-pi/2*(1+sgn(x))*(1-sgn(y^2))-pi/4*(2+sgn(x))*sgn(y) -sgn(x*y)*atan((abs(x)-abs(y))/(abs(x)+abs(y)))
x=x2-x1 and y=y2-y1
1 件のコメント
theodore panagos
2023 年 8 月 6 日
atan2(x,y)=pi/2*(1-sign(x))*(1-sgn(y^2))+pi()/4*(2-sgn(x))*sign(y)-sign(x*y)*atan((abs(x)-abs(y))/(abs(x)+abs(y)))
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