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Excerpt from Prof. Bouman's Lecture

Accompanying Lecture Notes

Definition

Let
$ \begin{align} P_{\theta}(\rho) &= CTFT \{p_\theta(r)\} \\ F(u,v) &= CSFT\{f(x,y)\} \end{align} $

where $ \rho $ is the frequency variable corresponding to $ r $ just as $ u $ and $ v $ are the frequency variables corresponding to $ x $ and $ y $ respectively.

Then
$ P_{\theta}(\rho) = F(\rho\cos(\theta),\rho\sin(\theta)) \ $

Recall that $ p_{\theta}(r) $ is the projection of image $ f(x,y) $ at angle $ \theta $. $ P_{\theta}(\rho) $ is its 1-D Fourier transform. $ F(u,v) $ on the other hand, is the 2-D Fourier transform of image $ f(x,y) $.

So essentially, the theorem tells us that $ P_{\theta}(\rho) $ is $ F(u,v) $ in polar coordinates.

Fig 1: $ P_{\theta}(\rho) $ is $ F(u,v) $ in polar coordinates

Let us look at the simple example in figure 2 to illustrate this relationship. For the given $ f(x,y) $, the projection for $ \theta = 0 $° is a 1-D rect function. Let us assume it is of unit area.

Fig 2: $ p_{\theta}(r) $ at angle $ \theta = 0 $°

So we have that
$ \begin{align} p_{\theta}(r) |_{\theta = 0} &= rect(r) \\ \Rightarrow P_{\theta}(\rho) |_{\theta = 0} &= CTFT\{p_{\theta}(r) \}|_{\theta = 0} \\ &= sinc(\rho) \end{align} $

But $ f(x,y) $ is a 2-D rect so
$ \begin{align} f(x,y) &= rect(x)rect(y) \\ \Rightarrow F(u,v) &= CSFT\{f(x,y)\} \\ &= sinc(u)sinc(v) \end{align} $

Now lets convert $ F $ from cartesian coordinates $ u,v $ to polar coordinates $ (\rho,\theta) $ where $ \theta = 0 $°
$ \begin{align} F(u,v) &= sinc(u)sinc(v) \\ &= sinc(\rho \cos\theta)sinc(\rho \sin\theta)|_{\theta = 0} \\ &= sinc(\rho) \\ &= P_{\theta}(\rho)|_{\theta = 0} \end{align} $

So we see that the theorem holds for the above scenario. But it does not prove the theorem, only verifies it for a particular situation. The theorem can be proved in different ways. Two of the methods are presented below.

Proof

Method 1

The first method relies on plugging variable substitutions into the defintion.

By definition,
$ \begin{align} P_{\theta}(\rho) &= CTFT\{p_{\theta}(r)\} \\ &= \int_{-\infty}^{\infty}p_{\theta}(r)e^{-j2\pi\rho r}dr \\ &=\int_{-\infty}^{\infty} [\int_{-\infty}^{\infty}f(\mathbf{A_{\theta}} \begin{bmatrix} r \\ z \end{bmatrix}) dz]e^{-j2\pi\rho r}dr \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(\mathbf{A_{\theta}} \begin{bmatrix} r \\ z \end{bmatrix})e^{-j2\pi\rho r}dzdr \end{align} $

Next we make the following change of variables
$ \begin{bmatrix} r \\ z \end{bmatrix} = \mathbf{A_{-\theta}}\begin{bmatrix} r \\ z \end{bmatrix} $

Notice that the Jacobian is $ |{A_{-\theta}}| $$ =1 $ since
$ \begin{align} \frac{\partial (r,z)}{\partial (x,y)}| &= det \begin{bmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ \frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} \end{bmatrix} \\ &= det \begin{bmatrix} \frac{\partial (x\cos(\theta)+y\sin(\theta))}{\partial x} & \frac{\partial (x\cos(\theta)+y\sin(\theta))}{\partial y} \\ \frac{\partial (-x\sin(\theta)+y\cos(\theta))}{\partial x} & \frac{\partial (-x\sin(\theta)+y\cos(\theta))}{\partial y} \end{bmatrix} \\ &= det \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \\ &= \cos^2\theta + \sin^2\theta \\ &=1 \end{align} $

Then,
$ drdz = |\frac{\partial(r,z)}{\partial(x,y)}|dxdy = dxdy $

Also notice that $ r=x\cos(\theta)+y\sin(\theta) $. So finally we have that
$ \begin{align} P_{\theta}(\rho) &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)e^{-j2\pi\rho[x\cos(\theta)+y\sin(\theta)]}dxdy \\ &= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y)e^{-j2\pi[x\rho\cos(\theta)+y\rho\sin(\theta)]}dxdy \\ &= F(\rho\cos(\theta),\rho\sin(\theta))_{\blacksquare} \end{align} $

Method 2

The second method is slightly more compact than the first.

First, let $ \theta = 0 $, then
$ \begin{align} p_0(r) &= \int_{-\infty}^{\infty}f(r,y)dy \\ \Rightarrow P_0(\rho) &= \int_{-\infty}^{\infty}p_0(r)e^{-2\pi jr\rho}dr \\ &= \int_{-\infty}^{\infty}[\int_{-\infty}^{\infty}f(r,y)dy]e^{-2\pi jr\rho}dr \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(r,y)e^{(-2\pi j(r\rho+y0)}drdy \\ &= F(\rho,0) \end{align} $

By the rotation property of the CSFT, it must hold for any $ \theta $.

References

• C. A. Bouman. ECE 637. Class Lecture. Digital Image Processing I. Faculty of Electrical Engineering, Purdue University. Spring 2013.

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