Euclidean Radon transform and filtered backprojection — Examples and Counterexamples

For $u>0$, repeated differentiation gives ${\eta }^{(k)}(u)=P_k(1\mathrm{/}u)e^{-1\mathrm{/}u}$ for a polynomial $P_k$. For every integer $m\ge 0$, $v^m{e}^{-v}\to 0$ as $v\to \mathrm{\infty }$: if $v>0$, the exponential series gives $e^v\ge v^{m+1}\mathrm{/}(m+1)!$. Hence each right derivative tends to $0$ at $u=0$, and the quotient ${\eta }^{(k)}(u)\mathrm{/}u$ also tends to $0$ because it is again a polynomial in $1\mathrm{/}u$ times $e^{-1\mathrm{/}u}$. Induction on $k$ shows the zero extension is smooth on $\mathbb{R}$.

The composition $f(x)=\eta (1-\mathrm{\mid }x{\mathrm{\mid }}^2)$ is smooth and radial; it vanishes whenever $\mathrm{\mid }x\mathrm{\mid }\ge 1$ and is positive at $x=0$. Thus it is nonzero and has compact support in the closed unit disk.

For every $\theta \in S^1$, orthogonality gives $\mathrm{\mid }s\theta +t{\theta }^{\perp }{\mathrm{\mid }}^2=s^2+t^2$, so [F1] yields $Rf(\theta ,s)={\int }_{\mathbb{R}}\eta (1-s^2-t^2)dt$. If $\mathrm{\mid }s\mathrm{\mid }\ge 1$, then $1-s^2-t^2\le 0$ for every $t$, so the line integral is $0$; if $\mathrm{\mid }s\mathrm{\mid }<1$, the integrand vanishes outside $\mathrm{\mid }t\mathrm{\mid }<\sqrt{1-s^2}$ and equals the exponential displayed in the Example inside that interval, with value $0$ at its endpoints because $\eta (0)=0$. It depends on $\theta$ only through the norm identity, so the sinogram is radial-data independent of orientation. The continuous compactly supported integrands are Riemann integrable by [F2].

For $u>0$, repeated differentiation gives ${\eta }^{(k)}(u)=P_k(1\mathrm{/}u)e^{-1\mathrm{/}u}$ for a polynomial $P_k$. The exponential-series bound $e^v\ge v^{m+1}\mathrm{/}(m+1)!$ for $v>0$ shows that each such expression and its difference quotient tend to $0$ as $u\downarrow 0$; induction on $k$ proves that the zero extension is smooth. Hence $\psi$ is smooth, nonnegative, vanishes for $\mathrm{\mid }x\mathrm{\mid }\ge 1$, and is positive near $x=0$.

By [F2], $\psi$ is Riemann integrable on a bounding rectangle; $0\le \psi \le 1$ and its support lies in $[-1,1{]}^2$, so [F3] gives $J<\mathrm{\infty }$. On $[-1\mathrm{/}4,1\mathrm{/}4{]}^2$ one has $1-\mathrm{\mid }x{\mathrm{\mid }}^2\ge 7\mathrm{/}8$, hence $\psi \ge e^{-8\mathrm{/}7}$; monotonicity in [F3] gives $J\ge e^{-8\mathrm{/}7}\mathrm{/}4>0$. Thus $\varphi$ is defined, radial, smooth, compactly supported, nonnegative, and has integral one. Applying [F5] to $x=\epsilon y$ gives ${\int }_{{\mathbb{R}}^2}{\varphi }_{\epsilon }(x)dx={\epsilon }^{-2}{\epsilon }^2{\int }_{{\mathbb{R}}^2}\varphi (y)dy=1$ for every $\epsilon >0$.

The scaling $t=\epsilon u$ in the line integral [F1] gives $R{\varphi }_{\epsilon }(\theta ,s)={\epsilon }^{-1}\rho (s\mathrm{/}\epsilon )$. Radiality gives $\varphi (s\theta +t{\theta }^{\perp })=J^{-1}\eta (1-s^2-t^2)$, so $\rho$ is independent of $\theta$ and vanishes for $\mathrm{\mid }s\mathrm{\mid }>1$. It is continuous: for $s^{\mathrm{\prime }}\to s$, $\mathrm{\mid }\rho (s^{\mathrm{\prime }})-\rho (s)\mathrm{\mid }\le 2{\sup }_{\mathrm{\mid }t\mathrm{\mid }\le 1}\mathrm{\mid }\varphi (s^{\mathrm{\prime }}\theta +t{\theta }^{\perp })-\varphi (s\theta +t{\theta }^{\perp })\mathrm{\mid }⟶0,$ since $\varphi$ is uniformly continuous. Thus $\rho$ is compactly supported and Riemann integrable by [F2]. Riemann-Fubini [F4] and the orthogonal change of variables $(s,t)\mapsto s\theta +t{\theta }^{\perp }$, whose absolute Jacobian is $1$, give ${\int }_{\mathbb{R}}\rho (s)ds={\int }_{{\mathbb{R}}^2}\varphi (x)dx=1$.

By step 3.1, $\rho$ is continuous and supported on $[-1,1]$, so $R{\varphi }_{\epsilon }(\theta ,\cdot )h$ is compactly supported and continuous. If $h$ is complex-valued, apply [F2] and the compact-support change of variables [F5] separately to its real and imaginary parts; thus with $s=\epsilon u$ we get ${\int }_{\mathbb{R}}R{\varphi }_{\epsilon }(\theta ,s)h(s)ds={\int }_{-1}^1\rho (u)h(\epsilon u)du$. Uniform continuity of $h$ on $[-1,1]$ implies ${\sup }_{\mathrm{\mid }u\mathrm{\mid }\le 1}\mathrm{\mid }h(\epsilon u)-h(0)\mathrm{\mid }\to 0$. Since $\rho \ge 0$ and $\int \rho =1$ by step 3.1, the integral tends to $h(0)$.

The functional $h\mapsto h(0)$ is the point mass ${\delta }_0$ by definition of the Dirac distribution. Step 4.1 proves the asserted distributional limit for every fixed normal $\theta$.

The only nonzero detector sample is at $m=2$, so the engineering DFT is $X_k=e^{-2\pi ik(2)\mathrm{/}4}=(-1{)}^k$. By [F2] and [F3], the length-four radix-two FFT returns $X=(1,-1,1,-1)$. The sampled ramp is $\lambda =(0,\pi \mathrm{/}2,\pi ,\pi \mathrm{/}2)$, hence its filtered frequency array is $Y=(0,-\pi \mathrm{/}2,\pi ,-\pi \mathrm{/}2)$.

The inverse formula [F4] gives $Q_m=\frac{1}{4}{\sum }_{k=0}^3{Y}_k{e}^{2\pi ikm\mathrm{/}4}$. For $m=0,1,2,3$ the sums are respectively $0,-\pi ,2\pi ,-\pi$, so $Q=(0,-\pi \mathrm{/}4,\pi \mathrm{/}2,-\pi \mathrm{/}4)$.

At $x=0$, the detector coordinate is $s=0$, which is bin $m=2$ for every direction. Therefore the declared angular quadrature and backprojection scale give $\tilde{f}(0)=\frac{1}{4\pi }\frac{\pi }{4}{\sum }_{j=0}^7{Q}_j[2]=\frac{1}{4\pi }\frac{\pi }{4}(8)(\pi \mathrm{/}2)=\pi \mathrm{/}4$. This verifies the displayed finite output only; no equality with a continuous source value is used.