Attenuated Radon and Abel transforms

Anne V. Clough; Harrison H. Barrett

doi:10.1364/JOSA.73.001590

Journal of the Optical Society of America
Vol. 73,
Issue 11,
pp. 1590-1595
(1983)
•https://doi.org/10.1364/JOSA.73.001590

Attenuated Radon and Abel transforms

Anne V. Clough and Harrison H. Barrett

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Anne V. Clough and Harrison H. Barrett, "Attenuated Radon and Abel transforms," J. Opt. Soc. Am. 73, 1590-1595 (1983)

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Abstract

The attenuated Radon transform is the mathematical basis of single-photon emission-computed tomography. The case of constant attenuation is reviewed, and a new proof of the Tretiak–Metz algorithm is presented. A space-domain version of the inverse attenuated Radon transform is derived. A special case of this transform that is applicable when the object is rotationally symmetric, the attenuated Abel transform, is derived, and its inverse is found.

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Type of Transform	Unattenuated	Attenuated
Randon	$λ_{ϕ} (p) = \int_{\infty} d^{2} r f (r) δ (p - r \cdot \hat{n})$	$λ_{ϕ} (p) = \int_{\infty} d^{2} r f (r) exp [- μ (r \cdot {\hat{n}}^{'} + L_{2})] δ (p - r \cdot \hat{n})$
Inverse Randon
Space Domain	$f (r) = \frac{1}{2 π^{2}} \int_{0}^{π} d ϕ {[λ_{ϕ} (p) * \frac{d}{d p} P (\frac{1}{p})]}_{p = r \cdot n}$	$f (r) = \frac{1}{4 π^{2}} \int_{0}^{2 π} d ϕ exp (μ r \cdot {\hat{n}}^{'}) {[{λ^{†}}_{ϕ} (p) * \frac{d}{d p} \frac{cos μ p}{p}]}_{p = r \cdot \hat{n}}$
Frequency Domain	$f (r) = \int_{0}^{π} d ϕ \int_{- \infty}^{\infty} d ν Λ_{ϕ} (ν) ∣ ν ∣ exp (2 π i ν r \cdot \hat{n})$	$f (r) = \frac{1}{2} \int_{0}^{2 π} d ϕ exp (μ r \cdot n^{'}) \int_{- \infty}^{\infty} {Λ^{†}}_{ϕ} (ν) ∣ ν ∣ [1 - rect \frac{π ν}{μ}] exp (2 π i ν r \cdot \hat{n}) d ν$
Abel	$λ (x) = 2 \int_{∣ x ∣}^{r_{0}} r f (r) \frac{1}{{(r^{2} - x^{2})}^{1 / 2}} d r$	$λ (x) = 2 exp [- μ {({r_{0}}^{2} - x^{2})}^{1 / 2}] \int_{∣ x ∣}^{r_{0}} r f (r) \frac{cosh [μ {(r^{2} - x^{2})}^{1 / 2}]}{{(r^{2} - x^{2})}^{1 / 2}} d r$
Inverse Abel	$f (r) = \frac{- 1}{π} \int_{r}^{\infty} \frac{d λ (x)}{d x} \frac{1}{{(x^{2} - r^{2})}^{1 / 2}} d x$	$f (r) = \frac{- 1}{π} \int_{r}^{\infty} \frac{d}{d x} {λ (x) exp [μ {({r_{0}}^{2} - x^{2})}^{1 / 2}]} \frac{cos [μ {(x^{2} - r^{2})}^{1 / 2}]}{{(x^{2} - r^{2})}^{1 / 2}} d x$

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Journal of the Optical Society of America