How does 2D spatial Fourier (kx-ky) transform result responds to rotation of the original?

Question

I have a 2D function $f:R\times R \rightarrow R$ that represents periodical axis-aligned spatial bumps at specific spatial periods (frequencies), like $f=\sin^2(2\pi \nu_1 x) \sin^2(2 \pi \nu_2 y)$. I expect the bump frequencies to be clearly visible as maximums on 2D Fourier transform of $F(kx,kx)= Fourier(f)$ at points $kx=2\pi\nu_1, ky=2\pi\nu_2$.

How does $F$ responds to the rotation of the $f$ around the origin? Would I see the maximums of the spatial frequencies on their correct places? Is there any other transformation that is invariant against the rotation and can show the spatially-periodic background of the function $f$?

Answer 1

Suppose $A:\mathbb R^2\to\mathbb R^2$ is a linear transformation. Then $\mathcal F(f\circ A)$ is computed as $$\mathcal F(f\circ A)(\xi)=\int f(Ax)e^{-2\pi i \xi\cdot x}\,dx = \frac{1}{\det A}\int f(y)e^{-2\pi i \xi\cdot A^{-1}y}\,dy $$ by the change of variables. Also, $\xi\cdot A^{-1}y = ((A^{-1})^* \xi )\cdot y$ by the definition of adjoint. Therefore, $$\mathcal F(f\circ A)=\frac{1}{\det A} \mathcal{F}(f)\circ (A^{-1})^* $$ In your case the matrix $A$ is orthogonal, hence $(A^{-1})^* =A$ and $\det A=1$. In this special case, the Fourier transform commutes with composition: $\mathcal F(f\circ A)= \mathcal{F}(f)\circ A$. The frequency maxima will be rotated in the same way as the function $f$ itself.