Topics:

• Math of CT images – the straight and filtered back-projection

Today's homework is subdivided into three parts (and one additional bonus part) and its main topic is the inverse of the forward (Radon) projection transform, that was the topic of last lab. The inverse transform starts with a sinogram, i.e. the projection image $J(\theta, p)$ and reconstructs the input image $f(x, y)$ in Cartesian space.

Naive approach The naive approach (the unfiltered back-projection) is just the reverse process of the projection – we smear the values of each projection back to the reconstructed image and add them together. We can write it as a sum over projection images $f_{\theta}(x, y)$ which we get from the measured projection $J(\theta, p)$ via $$f_{\theta}(x, y) = J(\theta, x \cos \theta + y \sin \theta)$$ The reconstructed image is then $$f_{\text{unfilt}}(x, y) = \mathcal{B}[\mathcal{R}f] = \sum_\theta {f_{\theta}(x, y)}$$

Optimal reconstruction From theoretical point of view, the central slice theorem gives us the optimal solution. The 2D-Fourier Transform $\mathcal{F}_2$ of the input image $f(x, y)$ is the same as the set of 1D-Fourier transformed projections $\mathcal{F}_1 [J(\theta, p)]$. However, since we have only very sparse samples in $\theta, p$, direct application of the inverse transform $\mathcal{F}_2^{-1}$ is impracticable.

Nonetheless, we can use the central slice theorem to rewrite the Fourier equality $f(x, y) = \mathcal{F}_2^{-1} [\mathcal{F}_2 f]$ and with help of the Radon transform – $\mathcal{R}f$ to finally arrive at the formula:\begin{equation} f(x, y) = \mathcal{B} \{ \mathcal{F}_1^{-1} [|\omega| \cdot \mathcal{F}_1[\mathcal{R} f] \} \end{equation} i.e. the reconstructed image is obtained by back-projection $\mathcal{B}$ of the measured projections filtered by the ramp filter $|\omega|$ in Fourier domain.

Compared to the naive approach, the filtered back-projection computes $f_{\theta}$ from the filtered projection signal $J^{*}(\theta, p)$. To get $J^{*}(\theta, p)$ from the input projection data $J(\theta, p)$ we take a look at the inner part of equation (1) and write it only for a single projection:

$$\mathcal{B} \{ \mathcal{F}_1^{-1} [|\omega| \cdot \mathcal{F}_1[J(\theta, p)] \}$$

So we get either by multiplication in the Fourier domain or convolution in the spatial domain: $$J^{*}(\theta, p) = \mathcal{F}_1^{-1} [|\omega| \cdot \mathcal{F}_1[J(\theta, p)] = \mathcal{F}_1^{-1} [|\omega|] * J(\theta, p)$$

The ramp filter $|\omega|$ leads to optimal reconstruction, but when applied in Fourier-domain, it also serves as a high-pass filter (since unbounded on both ends) and is therefore magnifying noise frequencies. To minimize this effect, the filter is applied in combination with windows that reduce its frequency span. We will work with Hamming window (hamming) and the Shepp-Logan, also sinc window (shepp-logan). The last filter is ram-lak which does no windowing, but only cuts of high frequencies. Use the provided function H = designFilter(f_type, N, f_d) to get the frequency-domain filter - plot both $H$ and $\mathcal{F}_1^{-1}[H]$ to get better comprehension of the filtering functions.

Now we can summarize the (filtered) back-projection algorithm: * compute the output image size * initialize the target image $f(x, y) = 0$ * (get the filter H = designFilter(f_type, N, f_d) ) * for each projection angle $\theta$

• ( compute the filtered $J^{*}(\theta, p) = \mathcal{F}_1^{-1}[H * \mathcal{F}_1[J(\theta, p)]]$ or by convolution)
• ( compute the back-projection image $f_{\theta}(x, y)$ from $J*(\theta, p)$ )
• add the back-projection image $f_{\theta}(x, y)$ to $f(x, y)$

Get the archive with input sinogram (noisy_radon.mat), reference image (phantom.mat) and the designFilter.m script.

1. [3 pts] Implement the inverse Radon transform as a function with signature

   myIradon( projim, thetas, f_type, f_d)

   where projim is the projection image of size $[m \times n_{\theta}]$ and theta is vector of length $n_{\theta}$ containing the projection angles. The last parameter f_type is optional and if set, it specifies the filtration to be applied during the back-projection, f_d is a parameter of the cut-off frequency of the filter f_type.
2. [1.5 pts] Take the input radon image which was created with the set of projection angles ${0, 1, ..., 179^{\circ}}$ and reconstruct the original image with different back-projection setting. Visualize the reconstructed image along with the difference image to the provided phantom image for each of the 7 settings:
   1. unfiltered back-projection
   2. filtered back-projection with filters – hamming, ram-lak and shepp-logan and with filter cut-off frequencies – $d \in {0.7, 1.0}$
   3. for filtered BP: visualize the filter in the frequency and the spatial domain alongside with the reconstructed and error image
3. [0.5 pts] Report the relative reconstruction error \begin{equation} R=\sqrt{ \frac{1}{\|x\| \cdot \|y\|} \cdot \frac{ \sum_{\forall x, \forall y}(\hat{f}(x,y)-f(x,y))^2}{\sum_{\forall x, \forall y} f(x,y)^2} } \end{equation} for reconstruction from previous subtask.
4. BONUS 1 pt: Select the best filter (with respect to the minimal reconstruction error) and reconstruct the image again, but use fewer projection angles (try with each 2nd, 5th and 10th angle), report the reconstructed images.

• use the same meshgrid X, Y as last time for positioning the output image $f(x, y)$
• each time, we interpolate only the 1D signal ( $J(\theta_0, p)$ for a fixed $\theta_0$ ). With meshgrids P and[X, Y] reflecting the spatial relation between $p$ and $(x, y)$, computation of the back-projected $f_{\theta}$ can be done with interp1 function).
• the designFilter returns array of even length, if the radon image has odd lenght (in $p$), first enlarge $J(\theta_0, p)$ and then apply Fourier transform and the filter $H$.