PROCESSOR-SDK-TDAX: atan2sp implementation issue in mathlib c66x Vs 3_1_0_0

Deepak Poddar

Part Number: PROCESSOR-SDK-TDAX
Other Parts Discussed in Thread: MATHLIB

I was going through the implementation of atan2sp present in the mathlib package vs. 3.1.0.0. It has definition of "atan2sp_i (float a, float b)" which is suppose to calculate arctan(a/b).

when I change the sign of 'a' and 'b' just after entering this API, I get different result. I mean arctan(-a/-b) != arctan(a/b).

Is this expected behavior, or there is some potential bug ?

Regards

Deepak Poddar

over 6 years ago

0 Yordan Kamenov over 6 years ago

TI__Mastermind 42515 points

Hi Deepak,

I have forwarded your question to an expert to comment.

Regards,
Yordan

0 Yordan Kamenov over 6 years ago in reply to Yordan Kamenov

TI__Mastermind 42515 points

Hi Deepak,

I moved your post to C66x forum as discussed.
You can get better support there.

Regards,
Yordan

0 Rahul Prabhu over 6 years ago

TI__Guru** 114410 points

Can you please elaborate or provide sample code for use to reproduce this issue that you are seeing

I tried this with a simple example and didn`t see any issues.

#include <stdio.h>
#include <ti/mathlib/src/atansp/c66/atansp_i.h>
/*
 * mathlib test.c
 */
int main(void) {
    float i, j, k;

    i= 0.45;
    j =  0.567;
    k = atansp_i((float)(i/j));

    printf("Result: %f\n", k);

    i= - 0.45;
    j =  - 0.567;
    k = atansp_i((float)(i/j));
    printf("Result: %f\n", k);
	
	return 0;
}

0 Deepak Poddar over 6 years ago in reply to Rahul Prabhu

TI__Expert 4305 points

I used the existing project of atan2sp (LE , elf one) in mathlib package.

I imported the project and changed the file atan2sp_i.h. Just after the function entry, I changed the sign of the arguments 'a' and 'b' by placing the code

a= 0.0f -a;

b =0.0f-b;

And ran the setup, and I could see that failure messages is printed.

Please try the same.

0 Victor Kazmirenko over 6 years ago

Guru 13042 points

Hello!

To my knowledge, arctan function has range of -Pi/2 to Pi/2, while atan2 has its range in whole circle. Then, supposing your a and b are both positive, and you'll get angle of the first quadrant. However, complex point with both coordinates negative belongs to the third quadrant. I am too lazy to look inside mathlib, but wish you comprehend just mentioned first.

Hope that helps.

0 Deepak Poddar over 6 years ago in reply to Victor Kazmirenko

TI__Expert 4305 points

Yes. it makes sense.

I also checked in mathlib for failing test cases, always difference is of PI.

Thanks for clarification.

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PROCESSOR-SDK-TDAX: atan2sp implementation issue in mathlib c66x Vs 3_1_0_0