# How would you reconstruct a CT image from two sinograms?

Say you perform a dual-energy x-ray CT scan using a high and low energy spectrum on three different materials. Since you're using two different energy levels, there will be two sinograms. But how would you reconstruct the main image of some $3$-material phantom from two separate sinograms?

• If the answer is really unknown then the corresponding patent will be worth a lot more than a 100 points of reputation here. Commented Jun 22, 2018 at 18:20

For a scan along a particular line we have $$\int_{x_0}^{x_1} A_i(x)dx = \ln\frac{I_{1,i}}{I_{0,i}},$$ where $A_i(x)$, $I_{0,i}$, and $I_{1,i}$ are the attenuation coefficient, initial intensity, and final intensity at energy $E_i$, respectively, and $i=1,2$. (This is Beer's law.) We will roughly have $A_2=\alpha A_1$, where $\alpha$ is some dimensionless constant relating the attenuation coefficients at energies $E_1$ and $E_2$. Thus, we should be able to pool the data consisting of $\alpha\ln(I_{1,1}/I_{0,1})$ and $\ln(I_{1,2}/I_{0,2})$. Then we perform an inverse Radon transform on the pooled data. One should be able to estimate $\alpha$ from the data.
Another grossly simple method that I would apply first would be to use an image processing algorithm to combine the inverse Radon transforms of the two sinograms. (Roughly, this assumes $\alpha = 1$.)
• @OliverG: However, it appears that the images have the same $t$s and $\theta$s, which simplifies things. Suppose $R_{1ij}$ and $R_{2ij}$ represent the $ij$th elements of sinograms 1 and 2, respectively. Then $\alpha = \sum_{ij}R_{2ij}/\sum_{ij}R_{1ij}$. Let $R_{ij} = (\alpha R_{1ij} + R_{2ij})/2$. Apply the inverse Radon transform to $R$. You may also wish to weight one sinogram relative to the other, in which case you might try $R'_{ij} = (1-p)\alpha R_{1ij} + p R_{2ij}$, where $0\le p\le 1$. Commented Jun 26, 2018 at 16:17