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Calculation of tan_inverse(gy/gx) in VHDL.

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Nandita16

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Code for Edge Direction "tan_inverse(Gy/Gx)" in VHDL for implementing NMS in Canny.

I am implementing canny edge detection in VHDL.
I have successfully calculated gradient ie. G=|Gx+Gy| but now to proceed further to Non-Maximum Suppression ,I need to calculate edge direction ie.

Thetha=Tan_inverse(Gy/Gx).

Please help me how i can do this as in need floating Point division for Gy/Gx.
Thanking you in anticipation.
 

Re: Code for Edge Direction "tan_inverse(Gy/Gx)" in VHDL for implementing NMS in Cann

You surely want fixed point rather than floating point arithmetic. Thinking about reasonable number formats and required accuracy would be a first step to approach a VHDL implementation of your problem. Also intended processing speed plays a role, do you need a pipelined parallel divider or can a serial divider be used?

I implemented a similar atan2(a/b) function with an octant (0..45°) atan ROM table, a/b <> b/a permutation, linear interpolation and the below mapping
Code: 
--sign u_a sign_ub |u_b|>|u_a|  gamma
--   0        0        0        0   + phi_8
--   0        0        1        90  - phi_8 
--   1        0        1        90  + phi_8
--   1        0        0        180 - phi_8
--   1        1        0        180 + phi_8
--   1        1        1        270 - phi_8
--   0        1        1        270 + phi_8     
--   0        1        0        0   - phi_8

I'm not sure if atan2 is actually required for edge detection, I guess there are simpler methods.
 

Re: Code for Edge Direction "tan_inverse(Gy/Gx)" in VHDL for implementing NMS in Cann

Sir right now i just need a code to proceed further. Either its serial or its parallel. I want to implement Tan_inverse(Gy/Gx). And I think for this we need floating point division only as Gy/Gx can be any no. either integer or float. So sir kindly suggest some code snippets to help me.
Thanks for your reply. Hope to have sooner response.
 

I want to calculate Tan_inverse(gy/gx) in VHDL. Please help me with the code of the same.
Thank you.
 

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