2
$\begingroup$

I know that there is a complex and a polar formulation of the Fourier series. The polar form of can be written as a reformulation of the real valued Fourier series for the 1D case: $$f(x)=a_{0}+\sum_{k}^{\infty} a_{k}\cos(k\omega_{0}x)+b_{k} \sin(k\omega_{0}x)= d_{0}+\sum_{k=1}^{\infty}d_{k}cos(k\omega_{0}x-\phi_{k})$$ with $\omega_{0} := 2\pi /T$ and $$d_{0}=a_{0}, d_k:=\sqrt{a^2_k +b^2_k}\quad\text{as magnitude and the phase}\quad\phi_{k}:=\arctan{\frac{b_{k}}{a_{k}}}$$ , which can be calculated as follows for a dft: $$d_k \cos\phi_k = \sqrt{2/N}\sum^{N}_{j=1}z_j \cos(2\pi k(j-1)/N)$$ $$d_k \sin\phi_k = \sqrt{2/N}\sum^{N}_{j=1}z_j \sin(2\pi k(j-1)/N)$$ It's essentially taking the real and the imaginary parts separately and using trigonometric relationships to calculate the coefficients. Now for the 2D case I have seen this approach in a paper which delivers consistent results: $$d_k \cos\phi_k = 1/N\sum^{N}_{j=1}\left(x_j\cos\left(\frac{2\pi (k+1)(j-1)}{N}\right)-y_j\sin\left(\frac{2\pi (k+1)(j-1)}{N}\right)\right)$$ $$d_k \sin\phi_k = 1/N\sum^{N}_{j=1}\left(x_j\sin\left(\frac{2\pi (k+1)(j-1)}{N}\right)+y_j\cos\left(\frac{2\pi (k+1)(j-1)}{N}\right)\right)$$ And I just do not understand, where this is coming from. I also didn't encounter any book on DFT, which explains this. There were rare ones which mentioned at least the polar formulation. I know how the complex case works. Does someone know how the above formulas for the 2d and 1d case are derived or any literature about the polar form? I would really appreciate any information.

New contributor
Maxim J is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$

0

Your Answer

Maxim J is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.