Filtered back-projection

Remarks

The Fourier convention is $\hat g(\sigma) = \int g(s)\, e^{-is\sigma}\, ds$. The projection-slice theorem gives $\hat f_b(\xi) = \Phi(\abs{\xi}/b)\, \hat f(\xi)$, so $f_b \to f$ as $b \to \infty$, uniformly on $\R^2$. With $\Phi$ the indicator of $[-1, 1]$ the filter is the Ram–Lak filter, a band-limited ramp.

Example: if $\Phi$ is the indicator of $[-1, 1]$ and $\hat f$ vanishes for $\abs{\xi} > b$, then $\hat f_b = \hat f$ and the reconstruction is exact: $f_b = f$.

Non-example: back-projection without filter, $\frac{1}{4\pi} R^* R f$, is not a reconstruction of $f$. It corresponds to $\hat w(\sigma) = 1$ instead of $\abs{\sigma}$, and its Fourier transform is $\hat f(\xi)/\abs{\xi}$, a blurred image.

In practice $R f$ is known only at finitely many angles and offsets, and the integral and the convolution are replaced by sums.