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Exact Gaussian grid alias identity and explicit alias bound

Statement

Assume Countable Choice. Let $d \geq 1$ , $N \geq 0$ , $c_{j} \in C$ , $x_{j} \in [0, 1)^{d}$ , integers $K_{r} \geq 0$ , positive integers $M_{r} > 2 K_{r}$ , and $τ_{r} > 0$ . Write $P = \prod_{r = 1}^{d} M_{r}$ , define $Φ_{τ} (z) = \prod_{r = 1}^{d} ϕ_{τ_{r}} (z_{r})$ , and let $g_{ℓ} = \sum_{j = 0}^{N - 1} c_{j} Φ_{τ} (ℓ_{1} / M_{1} - x_{j, 1}, \dots, ℓ_{d} / M_{d} - x_{j, d}), 0 \leq ℓ_{r} < M_{r} .$ For $k \in K = \prod_{r} {- K_{r}, \dots, K_{r}}$ define the normalized grid DFT ${\tilde{g}}_{k} = \frac{1}{P} \sum_{ℓ_{1} = 0}^{M_{1} - 1} \dots \sum_{ℓ_{d} = 0}^{M_{d} - 1} g_{ℓ} \exp ⁣ (- 2 π i \sum_{r = 1}^{d} k_{r} ℓ_{r} / M_{r}) .$ Put $\hat{Φ} (m) = \prod_{r} \sqrt{τ_{r}} \exp (- π τ_{r} m_{r}^{2})$ . Then ${\tilde{g}}_{k} = \sum_{q \in Z^{d}} \hat{Φ} (k + q ⊙ M) \sum_{j = 0}^{N - 1} c_{j} \exp (- 2 π i (k + q ⊙ M) \cdot x_{j}),$ where $(q ⊙ M)_{r} = q_{r} M_{r}$ . In particular, $\hat{Φ} (k) > 0$ and $\frac{{\tilde{g}}_{k}}{\hat{Φ} (k)} = F_{k} + \sum_{q \neq 0} [\prod_{r = 1}^{d} \exp ⁣ (- π τ_{r} ((k_{r} + q_{r} M_{r})^{2} - k_{r}^{2}))] \sum_{j = 0}^{N - 1} c_{j} \exp (- 2 π i (k + q ⊙ M) \cdot x_{j}) .$ For $s_{r} = \exp (- π τ_{r} M_{r} (M_{r} - 2 K_{r})) \in (0, 1)$ and $A = \prod_{r = 1}^{d} (1 + \frac{2 s_{r}}{1 - s_{r}}) - 1,$ the alias error satisfies $∣ \frac{{\tilde{g}}_{k}}{\hat{Φ} (k)} - F_{k} ∣ \leq A \sum_{j = 0}^{N - 1} ∣ c_{j} ∣ .$

Facts & Assumptions

Given: The finite data above and Countable Choice.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice, inherited through the periodized-Gaussian Fourier-series supplier.

[F1]

Fourier series of the periodized Gaussian: $ϕ_{τ} (x) = \sum_{m \in Z} \sqrt{τ} e^{- π τ m^{2}} e^{2 π i m x}$ , absolutely and uniformly.

[F2]

Type-1, type-2, and type-3 nonuniform Fourier transforms: $F_{k} = \sum_{j} c_{j} e^{- 2 π i k \cdot x_{j}}$ for $k \in K$ .

[F3]

$\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential and $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ : the exponential is multiplicative under addition, $e^{z + w} = e^{z} e^{w}$ , and $e^{z} = 1$ exactly when $z \in 2 π i Z$ .

[F4]

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ : $∣ e^{i t} ∣ = 1$ for real $t$ .

[F5]

For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges: for $0 < s < 1$ , $\sum_{h = 1}^{\infty} s^{h} = s / (1 - s)$ .

[F6]

Real elementary functions and their domains, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ , The exponential function is strictly increasing, and Nonnegative real square roots exist uniquely: $π \geq 2 > 0$ , $\exp (0) = 1$ , the exponential is positive and strictly increasing, and $\sqrt{τ_{r}} > 0$ for $τ_{r} > 0$ .

[F7]

Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive: $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ for complex numbers.

Proof

1.1A1F1F3givenconstructalgebra

Substitute [F1] into each factor of $Φ_{τ}$ and then into ${\tilde{g}}_{k}$ . Absolute uniform convergence in [F1] permits interchanging the finite grid/source sums with the absolutely convergent Fourier series. For an integer $m_{r}$ , the coordinate root sum is $\frac{1}{M_{r}} \sum_{ℓ_{r} = 0}^{M_{r} - 1} e^{2 π i (m_{r} - k_{r}) ℓ_{r} / M_{r}} = {\begin{cases} 1, & m_{r} - k_{r} \in M_{r} Z, \\ 0, & m_{r} - k_{r} \notin M_{r} Z . \end{cases}$ Indeed, in the first case every term is $1$ ; in the second, $z = e^{2 π i (m_{r} - k_{r}) / M_{r}} \neq 1$ by [F3] while $z^{M_{r}} = 1$ , so the finite geometric sum is $(1 - z^{M_{r}}) / (1 - z) = 0$ . Thus only $m_{r} = k_{r} + q_{r} M_{r}$ remain in each coordinate, giving the displayed exact alias identity.

2.1F1F2F6step 1.1algebra

The $q = 0$ term of that identity is $\hat{Φ} (k) F_{k}$ by [F1]–[F2]. Each factor $\sqrt{τ_{r}} \exp (- π τ_{r} k_{r}^{2})$ is positive by [F6], so division by $\hat{Φ} (k)$ is valid. For each $q$ , its coefficient ratio is $\frac{\hat{Φ} (k + q ⊙ M)}{\hat{Φ} (k)} = \prod_{r} \exp ⁣ (- π τ_{r} ((k_{r} + q_{r} M_{r})^{2} - k_{r}^{2})) .$ Separating $q = 0$ proves the deconvolution identity.

3.1givenF4F5F6F7step 1.1step 2.1algebra∎

Since $M_{r} > 2 K_{r}$ , $τ_{r} > 0$ and $π > 0$ by [F6], the exponent parameter $λ_{r} = π τ_{r} M_{r} (M_{r} - 2 K_{r})$ is positive. Positivity and strict monotonicity of $\exp$ in [F6] imply $0 < s_{r} = e^{- λ_{r}} < 1$ . Fix one coordinate $r$ . If $q_{r} = \pm h$ with integer $h \geq 1$ and $∣ k_{r} ∣ \leq K_{r}$ , then $(k_{r} + q_{r} M_{r})^{2} - k_{r}^{2} = q_{r}^{2} M_{r}^{2} + 2 q_{r} M_{r} k_{r} \geq h^{2} M_{r}^{2} - 2 h M_{r} K_{r} \geq h M_{r} (M_{r} - 2 K_{r}) > 0.$ Hence the absolute alias multiplier for that coordinate is at most $s_{r}^{h}$ . By [F5] its sum over nonzero $q_{r}$ is at most $2 s_{r} / (1 - s_{r})$ , so the sum over all coordinate multi-indices other than zero is at most the product of the coordinate sums including their unit terms, minus the all-zero term, namely $A$ . The unit-modulus fact [F4], multiplicativity, and the triangle inequality in [F7] bound each source phase sum by $\sum_{j} ∣ c_{j} ∣$ . This gives the asserted alias bound.

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Sources

Jens Keiner, Stefan Kunis, and Antje Vollrath, The Nonequispaced FFT and its Applications, Part I, slides 44–58 (standard reference, not scraped)
Leslie Greengard and June-Yub Lee, Accelerating the Nonuniform Fast Fourier Transform (standard reference, not scraped)