# 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 $$\boldsymbol{n}' =R\boldsymbol{n} = \begin{bmatrix} \cos\theta & -\sin\theta \\[0.3em] \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} n_1 \\n_2\end{bmatrix}$$

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}' &=R\boldsymbol{n}\\ R^{-1}\boldsymbol{n}' &=R^{-1}R\boldsymbol{n}\\ R^{-1}\boldsymbol{n}' &=\boldsymbol{n}\\ R^{T}\boldsymbol{n}' &=\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?

• Your derivation is sound: the discretisation you use is a special case of the continuous version with values concentrated on Dirac distributions on a grid. The rotation basically rotates this grid such that the equivalence is still true if you keep your representation continuous. The problem comes when you discretize your rotated grid on a discrete grid... – meduz Sep 15 '16 at 20:02