Changing the atan function so that it ranges from 0 to 2*pi

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

[wenjun kou]

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);

[Stephen23]

See Daniel Svedbrand's answer for the simplest solution.

Answer 1

[Daniel Svedbrand]

Adding mod 2*pi to atan2 should work just fine

z = mod(atan2(y,x),2*pi);

[Stephen23]

+1 simple and efficient. Someone sees the wood regardless of the trees!

[Jyoti Sharma]

+1, worked for me.

[Hassan Abdelazeem]

perfect and easy , thank you indeed

[Paul Safier]

This is great! Thank you!

[Feruza Amirkulova]

Yes, mod(atan2(y,x),2*pi) worked and its gradients are the same as for (atan2(y,x)).

[John D'Errico]

Edited to remove profanity in the answer.

Answer 2

[Walter Roberson]

Use atan2() instead.

[KA]

i am not allowed to use atan2(), which is why im having difficulty

[Image Analyst]

Why the heck not? Why is atan OK but atan2 not? Is this one of those homework problems where the instructor asks you to basically replicate a built-in MATLAB function from lower level functions (in which case Paulo did your homework for you)?

[KA]

yes it is one of those "homework problems," and no Paulo did not do my homework, instead it helped me better understand how matlab works, and going through his code i noticed he was missing another if statement which is:

if y==0 && x==0

error('Invalid computation')

end

I also tried doing it a different way by using the derivation of atan2 on the wikipedia link he sent. So his answer actually helped me understand matlab, since it is my first time having to write a program in Matlab.

[Paulo Silva]

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

[KA]

ok, did not know that, thanks again

Answer 3

[Paulo Silva]

The Wikipedia got all explained, you just need to do the code, it's very simple.

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

[KA]

thanks, very helpful

[Mehmet Can Türk]

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.

Answer 4

[theodore panagos]

You can use the formula:

x = x2-x1;
y = y2-y1;
th = pi/2*(1-sign(x))*(1-sign(y^2)) + pi/4*(2-sign(x))*sign(y) - sign(x*y)*atan((abs(x)-abs(y))/(abs(x)+abs(y)));

[DGM]

To demonstrate for nonscalar inputs:

% fake xy data
x = randn(10,1);
y = randn(10,1);

% the reference
th0 = atan2(y,x);
% the given implementation
th = pi/2*(1 - sign(x)).*(1 - sign(y.^2)) ...
   + pi/4*(2 - sign(x)).*sign(y) ...
   - sign(x.*y).*atan((abs(x)-abs(y))./(abs(x)+abs(y)));

% compare
immse(th0,th)

ans = 1.5484e-32

Answer 5

[Kent Leung]

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

[Sharad Keshari]

Thank you very much. Your code was helpful.

Answer 6

[Ali Ali]

If you want to use atan2(y,x) (atan2(Y,X), returns values in the closed interval [-pi,pi]), considering that atan(b)=atan(b+pi), you can use this equation (use atan2(Y,X) instead of atan(y/x) in this equation) for your work.

[DGM]

what?

atan(b) ~= atan(b+pi)

atan(b) ~= atan(b)+pi

atan2(y,x) ~= atan2(y,x)+pi

The angle between the x-axis and a unit vector along x is 0 degrees, not 90 degrees.

atan2d(0,1) + 90 % NO

ans = 90

mod(atan2d(0,1),360) % YES

ans = 0

[Stephen23]

" atan(b)=atan(b+pi) "

Lets check that right now:

b = linspace(-5,5,100);
X = atan(b);
Y = atan(b+pi);
plot(b(:),[X(:),Y(:)])

Nope, not the same. Not even close.