2D Fourier Transform of Rotated Discrete Domain Signal Assume we know that the Fourier transform of a signal $x(n_1,n_2)$ is $\mathcal{F}(x(n_1,n_2))=X(\omega_1,\omega_2)$. What is the Fourier transform of the signal after being transformed by a rotation matrix?

I have found a property of the DFT that states that rotating the spatial domain contents rotates the frequency domain contents which is called the Rotation Property.  I am assuming this extends to DDFT's (Discrete Domain Fourier Transforms).  However, I would like to derive it.

We can represent our signal using vector notation as $x(\boldsymbol{n})$, where $\boldsymbol{n} = \begin{bmatrix} n_1 \\n_2\end{bmatrix}$.

The rotation of $\boldsymbol{n}$, denoted $\boldsymbol{n}' =\begin{bmatrix} n_1' \\n_2'\end{bmatrix}$, can be expressed as 
$$\boldsymbol{n}' =R\boldsymbol{n} = \begin{bmatrix}
       \cos\theta &amp; -\sin\theta \\[0.3em]
       \sin\theta &amp; \cos\theta 
     \end{bmatrix} \begin{bmatrix} n_1 \\n_2\end{bmatrix} 
$$

So I am trying to find the Fourier transform of $x(R\boldsymbol{n}$).

My attempt is as follows:


Step 1: Write the Fourier transform in vector notation.

Define $\boldsymbol{\omega} = \begin{bmatrix} \omega_1 \\ \omega_2\end{bmatrix}$.  We can now write the 2D Discrete Domain Fourier transform as: $$\mathcal{F}(x(\boldsymbol{n}))=X(\omega_1,\omega_2)=X(\boldsymbol{\omega})=\sum_\boldsymbol{n}x(\boldsymbol{n})e^{j\boldsymbol{\omega}^{T}\boldsymbol{n}}$$ because $\boldsymbol{\omega}^{T}\boldsymbol{n}=\omega_1 n_1 + \omega_2 n_2$.
Step 2: Solve for $\boldsymbol{n}$:
\begin{align}
\boldsymbol{n}' &amp;=R\boldsymbol{n}\\
R^{-1}\boldsymbol{n}' &amp;=R^{-1}R\boldsymbol{n}\\
R^{-1}\boldsymbol{n}' &amp;=\boldsymbol{n}\\
R^{T}\boldsymbol{n}' &amp;=\boldsymbol{n}
\end{align}
Noting that $R^{-1}=R^{T}$ for rotation matrices.
Step 3: Apply Fourier transform to rotated signal:
$$\mathcal{F}(x(\boldsymbol{n}'))=\sum x(\boldsymbol{n'})e^{j\boldsymbol{\omega}^{T}\boldsymbol{n}}$$
Plug in $\boldsymbol{n} = R^{T}\boldsymbol{n}'$
$$\mathcal{F}(x(\boldsymbol{n}'))=\sum x(\boldsymbol{n'})e^{j\boldsymbol{\omega}^{T}R^{T}\boldsymbol{n}'}=\sum x(\boldsymbol{n'})e^{j(R\boldsymbol{\omega})^{T}\boldsymbol{n}'}$$
And therefore
$$\mathcal{F}(x(R\boldsymbol{n}))=X(R\boldsymbol{\omega})$$

Which I am interpreting as meaning the original statement of rotating the spatial domain contents rotates the frequency domain contents.


Can anyone confirm this is correct?
The Rotation Property means that the choice of coordinate direction would not affect the spectrum of the signal itself.

To perform the rotation, I think it is better to do in the spatial domain since frequency domain is a complex one and it will just increase the complexity.