- #1
Cyrus
- 3,238
- 17
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 jω) = |X(ejω)| ejω
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]
We can express the Fourier Transform in polar form: X(e jω) = |X(ejω)| ejω
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]