why the give code doesn't meet my requirements?

I want to Wrap angles to the range [-180, 180] degrees. For this I have the following code:

function wrapped_angles = wrapTo180(angles)
    % Wrap angles to the range [-180, 180] degrees
    wrapped_angles = mod(angles + 180, 360) - 180;
end

Likewise, I want Wrap angles to the range [-90, 90] degrees. For this I have the following code:

function wrapped_angles = wrapTo90(angles)
    % Wrap angles to the range [-90, 90] degrees
    wrapped_angles = mod(angles + 90, 180) - 90;
end

The 2nd function works correctly but the 1st one doesn't work correctly. Why it is so?

[Stephen23]

"The 2nd function works correctly but the 1st one doesn't work correctly"

Please given an example input value and the corresponding incorrect output value.

[Sadiq Akbar]

@Stephen23

Thank you very much for your kind response. see below:

clear;clc

angles = [-182 -185 189 184];
% angles = [-150 -32 32 189];
wrapped_angles = wrapTo180(angles);
% wrapped_angles = wrapTo90(angles);
disp('Final Wrapped Angles:');
disp(wrapped_angles);

This is a test code. When I give the top angles to warp180(angles), it should give: -2 -5 9 4 but it gives: 178 175 -171 -176

[Torsten]

The interval you transform the angle to should have length 360 degrees, shouldn't it ? So I don't understand how you could transform angles uniquely to the interval [-90 90].

For the interval [-180 180], it's ok.

angles = [-182 -185 189 184];
wrapped_angle180 = mod(angles,360) - 180

wrapped_angle180 = 1×4

-2 -5 9 4

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wrapped_angle90 = mod(angles,180) - 90

wrapped_angle90 = 1×4

88 85 -81 -86

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Answer 1

[dpb]

angles = [-182 -185 189 184];
rem(angles,180)

ans = 1×4

-2 -5 9 4

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From the doc for rem one learns...

"The concept of remainder after division is not uniquely defined, and the two functions mod and rem each compute a different variation. The mod function produces a result that is either zero or has the same sign as the divisor. The rem function produces a result that is either zero or has the same sign as the dividend."

To use MATLAB mod as you wish, you would have to write your own version of rem --

sign(angles).*mod(abs(angles),180)

ans = 1×4

-2 -5 9 4

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[dpb]

The same information as above is also in the doc for mod

[Sadiq Akbar]

@Stephen23

@Torsten

@dpb

Thanks to all of your for your kind responses. As I was busy in some other activities, so couldn't give attention here. Let me explain my issue in simple words: I have two vectors u and b. I want to restric their entries either within 0-360 or within -180 180. i.e.,

clear;clc
    u = [-35 -45 -55 35 45 55];  % 6 variables within [-180,180]
    b = [30 45 55 135 245 355];  % 6 variables within [0,360]
    P = numel(b) / 2;  % Number of sources
    M = numel(b);      % Total number of elements
    % Precompute angles for efficiency
    var1_u = u(1:P);
    var2_u = u(P+1:end);
    var1_b = b(1:P);
    var2_b = b(P+1:end);

    % Wrap angles between 0 and 180 degrees
    var1_u = rem(var1_u, -180);
    var2_u = mod(var2_u, 180);
    var1_b = mod(var1_b, 0);
    var2_b = mod(var2_b, 360);

I tried differently but in vain.

[Torsten]

mod(x,360) 

restricts x to be within 0 and 360,

mod(x,360) - 180 

restricts x to be within -180 and 180.

Of course, there are different ways to do with different results ( I say this just in case that you want to respond that you don't like what you get ).

u = [-35 -45 -55 35 45 55];  % 6 variables within [-180,180]
b = [30 45 55 135 245 355];  % 6 variables within [0,360]
mod(u,360) - 180

ans = 1×6

145 135 125 -145 -135 -125

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mod(b,360)

ans = 1×6

30 45 55 135 245 355

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[dpb]

    u = [-35 -45 -55 35 45 55];  % 6 variables within [-180,180]
    b = [30 45 55 135 245 355];  % 6 variables within [0,360]
    
    rem(u,180)

ans = 1×6

-35 -45 -55 35 45 55

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    rem(b,360)

ans = 1×6

30 45 55 135 245 355

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What would you expect the answers to be if not the above (and, more importantly, why/how do you determine that?). You can't write an algorithm until you can define the problem uniquely.

[Sadiq Akbar]

@Torsten

@dpb

Thanks a lot to both of you for your kind responses.

[dpb]

So what was/is the correct answer?

[Torsten]

I think the correct answer is that there is no "correct" answer.

One has to decide whether to use "rem" or "mod" or some other normalization to the respective interval.

[dpb]

That's the point of the Q? -- trying to determine what was/is OP's basis for determining what result satisifed -- and maybe, thereby, clarifying his/her thinking.