A Can the Unwrapped Phase Function of a Fourier Transform be Derived?

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The discussion focuses on deriving the unwrapped phase function of a Fourier Transform, starting from its polar form representation. The process involves taking the natural logarithm of the Fourier Transform and differentiating it with respect to ω. The key point of confusion arises when trying to equate the derivatives of the logarithmic expressions to find the derivative of the phase function θ(ω). The final equation presented relates the derivative of θ(ω) to the real and imaginary parts of the Fourier Transform's derivative. Ultimately, the original poster successfully solved the problem and indicated that the discussion could be deleted.
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I am stuck trying to derive the unwrapped phase function of a Fourier Transform. Here is the gist of the derivation.

We can express the Fourier Transform in polar form: X(e ) = |X(e)| e

We can take the ln of both sides, resulting in:

ln X(ejω) = ln | X(ejω) | + j θ(ω)

Taking the derivative w.r.t. ω:

d ln X(ejω) / dω = d | X(ejω)| / dω + j dθ/dω

But, if we express X(ejω) = Xre(ejω) + j Xim(ejω) then we can also find the derivative to be:

d ln X(ejω) / dω = 1/ X(ejω) [ d X(ejω)/dω] = 1/ X(ejω) [dXre(ejω)/dω + jdXim(ejω)/dω]

Here is where I cannot get to: The author then states: " Therefore, the derivative of θ(ω) with respect to ω is given by the imaginary part of the right hand side of the second equation I wrote from the top. Somehow he is finding the equation below when equating/combining the two definitions of the derivatives of the ln X(ejw),"

dθ/dω = 1/ | X(ejω)|2 [ Xre(ejω) d Xim (ejω) / dω - Xim(ejω)dXre(ejω)/dω]
 
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I was able to solve the problem, feel free to delete.
 
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