2D Fourier Transform of Rotated Discrete Domain Signal

Question

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?

Answer 1

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.

Answer 2

I've got a bit confused by some details in your proof:

1.By definiton of 2d fourier transform we have:

$\mathcal{F}(x(\boldsymbol{n}))=X(\omega_1,\omega_2)=X(\boldsymbol{\omega})=\sum_\boldsymbol{n}x(\boldsymbol{n})e^{j\boldsymbol{\omega}^{T}\boldsymbol{n}}$

Substitute $\boldsymbol{n}$ with $\boldsymbol{n}'$,we yield:

$\mathcal{F}(x(\boldsymbol{n}'))=X(\omega_1,\omega_2)=X(\boldsymbol{\omega})=\sum_\boldsymbol{n}'x(\boldsymbol{n}')e^{j\boldsymbol{\omega}^{T}\boldsymbol{n}'}$

where $\boldsymbol{n}'=R\boldsymbol{n}$.

However , in Step 3 you wrote:

$\mathcal{F}(x(\boldsymbol{n}'))=\sum x(\boldsymbol{n}')e^{j\boldsymbol{\omega}^{T}\boldsymbol{n}}$

So the first point I got confused is that you didn't substitute the $\boldsymbol{n}$ in $e^{j\boldsymbol{\omega}^{T}\boldsymbol{n}}$ with $\boldsymbol{n}'$. Shouldn't we change all the notations of the original variable to the new one if we apply fourier transform on a new variable?

I think it may be crucial that we should figure out this issue because you continue your proof by pluging in $\boldsymbol{n} = R^{T}\boldsymbol{n}'$,However if we change all the $\boldsymbol{n}$ to $\boldsymbol{n}'$ when we apply fourier transform on $\boldsymbol{n}'$,then we cannot do this subsitute.

2.In my point of view, if we let $X(\boldsymbol{\omega})=\mathcal{F}(x(n))=\sum_\boldsymbol{n}x(\boldsymbol{n})e^{j\boldsymbol{\omega}^{T}\boldsymbol{n}}$,then if we want to apply fourier transform on a new variable $\boldsymbol{n}'$,we let $X'(\boldsymbol{\omega})=\mathcal{F}(x(n'))=\sum_\boldsymbol{n}'x(\boldsymbol{n}')e^{j\boldsymbol{\omega}^{T}\boldsymbol{n}'}$,where $X'$ and $X$ is not the same map unless $\boldsymbol{n}'$ totally equals $\boldsymbol{n}$.

So in my view, the expression $\sum x(\boldsymbol{n'})e^{j(R\boldsymbol{\omega})^{T}\boldsymbol{n}'}$ should equal to $X'(R\boldsymbol{\omega})$,not $X(R\boldsymbol{\omega})$.

Hopefully you could explain your proof in a more detailed way and help me get rid of my confusions, thank you!