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(FPGA) high throughput inverse tangent 2 input
Hello! I would just like to ask what is the effect of having a 'negative x' on the high throughput inverse tangent (2 input)? It's because when I have 'negative y' (and 'positive x'), the answer is correct. However, whenever I have 'negative x' (regardless of the sign of y), the answer would always be wrong. Thank you for your time!
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