Physical Meaning of 2DFT

Consider the Fourier transform of continuous but non-periodic signal (the result should be easily generalized to other cases):

$$f(x,y)=\int \int_{-\infty}^{\infty} F(u,v) e^{j2\pi(xu+yv)} du\;dv$$

where $u$ and $v$ are the frequencies in the directions of $x$ and $y$, respectively. This double integration is a linear combination (with weights $F(u,v)$) of the function

$$e^{j2\pi(xu+yv)}=cos(2\pi(xu+yv))+j\sin(2\pi(xu+yv))$$

To understand the physical meaning of this 2D Fourier transform, we first consider the complex exponential $e^{j2\pi(xu+yv)}$ and its complex weight $F(u,v)$ separately:

• The complex exponential:

  The complex exponential is composed of two sinusoidal functions: $e^{j2\pi(xu+vy)}=cos(2\pi(xu+vy))+j\;sin(2\pi(xu+vy))$. First we define the following:
  

  $$\left\{ \begin{array}{l} w \stackrel{\triangle}{=}\sqrt{u^2+... ...} u=w \; cos \theta v=w \; sin \theta \end{array} \right.$$
  
  
  We can now define two vectors:
  • The vector associated with point $(x, y)$ in the 2D spatial domain:
    
    $$\vec{r} \stackrel{\triangle}{=}(x, y)$$
    
    

  • The unit vector along the direction of point $(u,v)$ in the 2D spatial frequency domain:
    
    $$\vec{n} \stackrel{\triangle}{=}(cos \theta, sin \theta) =(\frac{u}{w},\frac{v}{w})$$
    
    

  Now we have:
  
  $$e^{j2\pi(xu+yv)}=cos(2\pi(xu+yv))+j\;sin(2\pi(xu+yv))= cos(2\... ...\vec{r} \cdot \vec{n}))+j\;sin(2\pi w (\vec{r} \cdot \vec{n}))$$
  
  
  where $(\vec{r} \cdot \vec{n})$ is the inner product of the two vectors, representing the projection of point $\vec{r}=(x,y)$ onto the direction of $\vec{n}$. The function value $cos(2\pi(xu+yv))=cos(2\pi w (\vec{r} \cdot \vec{n}))$ at any spatial point $\vec{r}=(x,y)$ is the same as that of its projection $(\vec{r} \cdot \vec{n})$ on the direction $\vec{n}$). In other words, the function $cos(2\pi(ux+vy))$ represents a planar sinusoid in the x-y plane with
  • frequency $w=\sqrt{u^2+v^2}$
  • direction $\theta=tan^{-1}(v/u)$ (i.e. $\vec{n}$).

  

  

• The complex coefficient:

  The complex coefficient $F(u,v)=F_r(u,v)+jF_j(u,v)$ can also be represented in polar form in terms of its amplitude and phase:
  

  $$\left\{ \begin{array}{l} \vert F(u,v)\vert=\sqrt{F_r(u,v)^2+... ...v)=\vert F(u,v)\vert sin(\angle F(u,v)) \end{array} \right.$$
  
  

Now the 2DFT of a real signal $f(x,y)$ becomes

$$f(x,y) = \int \int_{-\infty}^{\infty} F(u,v) e^{j2\pi(xu+yv)}du\;dv$$

$$= \int \int_{-\infty}^{\infty} F_r(u,v) cos(2\pi w(\vec{n} \cdot \vec{r})) -F_j(u,v)sin(2\pi w(\vec{n} \cdot \vec{r})) du\;dv$$

$$= \int \int_{-\infty}^{\infty} \vert F(u,v)\vert [ cos\angle F(u,v)... ...n} \cdot \vec{r})) -sin\angle F(u,v) sin(2\pi w(\vec{n} \cdot \vec{r})) ]du\;dv$$

$$= \int \int_{-\infty}^{\infty} \vert F(u,v)\vert cos (2\pi w(\vec{n} \cdot \vec{r})+\angle F(u,v)) du\;dv$$

The 2D Fourier transform represents a real signal $f(x,y)$ as a linear combination (integration) of infinite 2D spatial sinusoids with

• amplitude $\vert F(u,v)\vert=\sqrt{F_r(u,v)^2+F_j(u,v)}$ and phase $\angle F(u,v)=tan^{-1}(F_j(u,v)/F_r(u,v))$ determined by the complex coefficient $F(u,v)$, and
• frequency $w=\sqrt{u^2+v^2}$ and direction $\vec{n}=tan^{-1}(v/u)$ determined by $u$ and $v$, projected on the direction determined by $x$ and $y$.

Example 0

The 2D function shown below has three frequency components (2D sinusoidal waves) of different directions:

Example 1

$$f(x,y)=\left\{ \begin{array}{ll} 1 & if\;(-\frac{a}{2}<x<\fra... ...; -\frac{b}{2}<y<\frac{b}{2}) 0 & else \end{array} \right.$$

$$F(u,v) = \int \int_{-\infty}^{\infty} f(x,y) e^{-j2\pi(ux+vy)} dx dy = \int_{-a/2}^{a/2} e^{-j2\pi ux}dx \int_{-b/2}^{b/2} e^{-j2\pi vy}dy$$

$$= \frac{sin(\pi ua)}{\pi u} \frac{sin(\pi vb)}{\pi v}$$

Example 2

$$f(x,y)=\left\{ \begin{array}{ll} 1 & x^2+y^2<R^2 \\ 0 & else \end{array} \right.$$

It is more convenient to use polar coordinate system in both spatial and frequency domains. Let

$$\left\{ \begin{array}{ll} x=r cos\theta, & y=r sin\theta \\ r=\sqrt{x^2+y^2}, & \theta=tan^{-1}(y/x) \end{array} \right.$$

$$dx dy=rdr d\theta$$

and

$$\left\{ \begin{array}{ll} u=\rho cos\phi, & v=\rho sin\phi ... ...\rho=\sqrt{u^2+v^2}, & \phi=tan^{-1}(v/u) \end{array} \right.$$

$$du dv=\rho d\rho d\phi$$

we have:

$$F(u,v) = \int \int_{-\infty}^{\infty} f(x,y) e^{-j2\pi(ux+vy)} dx dy = \i... ...\pi} e^{-j2\pi r\rho(cos \theta cos \phi + sin \theta sin \phi)} d\theta ] r dr$$

$$= \int_0^R [ \int_0^{2\pi} e^{-j2\pi r\rho cos (\theta-\phi)} d\theta ] r dr = \int_0^R [ \int_0^{2\pi} e^{-j2\pi r\rho cos \theta} d\theta ] r dr$$

To continue, we need to use 0th order Bessel function $J_0(x)$ defined as

$$J_0(x)\stackrel{\triangle}{=}\frac{1}{2\pi}\int_0^{2\pi} e^{-jx cos \theta} d\theta$$

which is related to the 1st order Bessel function $J_1(x)$ by

$$\frac{d}{dx}(x J_1(x))=x J_0(x)$$

i.e.

$$\int_0^x x J_0(x) dx=x J_1(x)$$

Substituting $2\pi r \rho$ for $x$, we have

$$F(u,v)=F(\rho,\phi)=\int_0^R 2\pi r J_0(2\pi r \rho) dr =\frac{1}{\rho} R J_1(2\pi \rho R) \$$

We see that the spectrum $F(u,v)=F(\rho,\phi)$ is independent of angle $\phi$ and therefore is central symmetric.

Example 3 2D DFT of an image: