Physical Meaning of 2-D FT

Consider the Fourier transform of a 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 complex weight $F(u,v)$, of the complex exponential composed of two sinusoidal functions:

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

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

• The complex exponential:
  • Define
    
    $$\left\{ \begin{array}{l} w \stackrel{\triangle}{=}\sqrt{u^2+... ...} u=w \; cos \theta v=w \; sin \theta \end{array} \right.$$
    
    
    The unit vector associated with point $(u,v)$ in the 2D spatial frequency domain:
    
    $${\bf n} \stackrel{\triangle}{=}[cos \theta, sin \theta]^T =\left[\frac{u}{w},\frac{v}{w}\right]^T=\frac{1}{w} [u,\;v]^T$$
    
    

  • The vector associated with point $(x, y)$ in the 2D spatial domain:
    
    $${\bf r} \stackrel{\triangle}{=}[x,\; y]^T$$
    
    

  The inner product of the two vectors defined above is
  
  $$<{\bf r},{\bf n}>={\bf r}\cdot{\bf n}={\bf r}^T{\bf n}=\frac{1}{w}(xu+yv)$$
  
  
  representing the projection of ${\bf r}=[x,\;y]^T$ onto the direction of the unit vector ${\bf n}$. Now we have:
  
  $$e^{j2\pi(xu+yv)}=e^{j2\pi w {\bf r}\cdot {\bf n} }= \cos... ...\bf r} \cdot {\bf n})+j\;\sin(2\pi w {\bf r} \cdot {\bf n})$$
  
  
  The function value $\cos(2\pi(xu+yv))=\cos(2\pi w {\bf r}\cdot {\bf n})$ at any spatial point ${\bf r}=[x,\;y]$ is the same as that of its projection ${\bf r} \cdot {\bf n}$ on the direction of ${\bf n}$. In the 2-D space, all points ${\bf r}=[x, y]^T$ on a line perpendicular to ${\bf n}$ have the same projection onto ${\bf n}$, and therefore take the same value. In other words, the function $cos(2\pi(ux+vy))$ represents a planar sinusoidal wave in the x-y plane with
  • frequency $w=\sqrt{u^2+v^2}$
  • direction ${\bf n}$ with angle $\theta=tan^{-1}(v/u)$

  

  

• 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 $\vert F(u,v)\vert$and phase $\angle F(u,v)$:
  

  $$\left\{ \begin{array}{l} \vert F(u,v)\vert=\sqrt{F_r(u,v)^... ...)=\vert F(u,v)\vert\;\sin\angle F(u,v) \end{array} \right.$$
  
  
  Now we have
  
  $$F(u,v)=F_r(u,v)+jF_j(u,v)=\vert F(u,v)\vert [\cos\angle F(u,v)+j \sin\angle F(u,v)]$$
  
  

It the 2-D signal $f(x,y)=f_r(x,y)+j f_i(x,y)=f_r(x,y)$ is real, i.e, $f_i(x,y)=0$, then its 2-D Fourier expansion can be written as

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

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

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

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

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

• amplitude $\vert F(u,v)\vert=\sqrt{F_r(u,v)^2+F_j(u,v)}$
• phase $\angle F(u,v)=tan^{-1}(F_j(u,v)/F_r(u,v))$
• frequency $w=\sqrt{u^2+v^2}$
• direction $\theta=tan^{-1}(v/u)$

The amplitude and phase are determined by the complex coefficient $F(u,v)$, while the frequency and direction are determined by the spatial frequencies $u$ and $v$.

Example 0

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

$$f(x,y)=\frac{1}{2} \sin(2\pi (x+2y)/6)+\cos(2\pi (x+y)/6) +\frac{1}{2}\sin(2\pi (2x+y)/6)$$

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:

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

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

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 More 2-D FT examples:

Example 4 2D DFT of an image: