ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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A finite complete-angle filtered-backprojection calculation

Example

Assume Countable Choice for the continuous backprojection convention. As a finite-grid illustration, take eight equally spaced full-circle directions $θ_{j} = (\cos (2 π j / 8), \sin (2 π j / 8))$ , $j = 0, \dots, 7$ , four detector offsets $s_{m} = m - 2$ for $m = 0, 1, 2, 3$ , detector spacing $Δ s = 1$ , and the synthetic row $G_{j} = (0, 0, 1, 0)$ at every direction. Apply the length-four radix-two FFT to each row, multiply frequency $k$ by the declared sampled ramp $λ_{k} = 2 π \min (k, 4 - k) / 4$ , and invert the finite transform. Each filtered row is $Q = (0, - π / 4, π / 2, - π / 4)$ . With the eight-point trapezoidal angular rule of weight $Δ θ = π / 4$ , the one-pixel backprojection at $x = 0$ is $\tilde{f} (0) = \frac{1}{4 π} Δ θ \sum_{j = 0}^{7} Q_{j} [2] = \frac{π}{4} .$ This is the exact output of the specified finite computation. The array is a synthetic discrete sinogram; no claim is made that the finite detector or angle grid exactly represents a continuous phantom or that this output converges to one. The continuous theorem assumes full continuous data and makes no finite-sampling, truncation, or noise guarantee.

Example references

Frank Natterer, “X-ray Tomography,” §3.1

Facts & Assumptions

Given: The finite array and quadrature convention in the Example, Countable Choice, and the recursive radix-two FFT on length four.

[A1]

Countable Choice, written ${A C}_{ω}$ , means every sequence of nonempty sets has a choice function. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

Complete-data filtered backprojection: take the negative-sign $2 π$ Fourier transform in detector offset, multiply by $2 π ∣ σ ∣$ , inverse transform, backproject over the full circle, and scale by $1 / (4 π)$ .

[F2]

The recursive radix-two fast Fourier transform: ${FFT}_{m}$ is the recursive even/odd transform on arrays of length $2^{m}$ .

[F3]

Correctness of the recursive radix-two FFT: for every length $2^{m}$ input, ${FFT}_{m}$ returns the unnormalised engineering DFT.

[F4]

The unnormalised engineering DFT and its conversion to the unitary transform: $X_{k} (G) = \sum_{m = 0}^{N - 1} G_{m} e^{- 2 π i k m / N}$ and $G_{m} = N^{- 1} \sum_{k = 0}^{N - 1} X_{k} (G) e^{2 π i k m / N}$ .

[F5]

Finite Fourier inversion for the unitary transform on $Z / N Z$ : the finite inverse transform has positive sign and factor $N^{- 1 / 2}$ in the unitary convention; after the engineering rescaling the inverse formula has factor $1 / N$ .

Verification

1.1givenF2F3F4algebra

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

2.1F4F5step 1.1algebra

The inverse formula [F4] gives $Q_{m} = \frac{1}{4} \sum_{k = 0}^{3} Y_{k} e^{2 π i k m / 4}$ . For $m = 0, 1, 2, 3$ the sums are respectively $0, - π, 2 π, - π$ , so $Q = (0, - π / 4, π / 2, - π / 4)$ .

3.1givenA1F1step 2.1algebra∎

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 $\tilde{f} (0) = \frac{1}{4 π} \frac{π}{4} \sum_{j = 0}^{7} Q_{j} [2] = \frac{1}{4 π} \frac{π}{4} (8) (π / 2) = π / 4$ . This verifies the displayed finite output only; no equality with a continuous source value is used.

Depends on

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Sources

Frank Natterer, X-ray Tomography, in Inverse Problems and Imaging (standard reference, not scraped)