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Computed fast transforms and convolution — Examples

1 · Prerequisites

2 · Summary

These examples check three conventions used on the A page: Bluestein’s prime-length chirp identity, the difference between cyclic and zero-padded linear convolution, and DCT-II coefficient normalization at Chebyshev roots. The prime-length example checks the exact Bluestein identity at a supported padded length; the convolution example compares exact cyclic and linear operations. Neither claims that a finite-precision execution must return a successful status.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

The length-five all-ones vector through Bluestein

Example

Take $N = 5$ and $x = (1, 1, 1, 1, 1)$ . Choose $M = 16$ , the least power of two at least $2 N - 1 = 9$ . The direct unnormalised DFT is $X = (5, 0, 0, 0, 0)$ , and the exact Bluestein chirp convolution with this padding represents the same vector. This example checks the prime-length indexing and the padding threshold; it makes no floating-point accuracy claim.

Facts & Assumptions

Given: $N = 5$ , $M = 16$ , and $x_{j} = 1$ for $0 \leq j < 5$ .

[F1]

The Bluestein identity represents the DFT through the zero-padded cyclic chirp convolution with padding $M \geq 2 N - 1$ (Bluestein’s chirp identity embeds an arbitrary DFT in a padded cyclic convolution).

[F2]

$e^{z + w} = e^{z} e^{w}$ and $e^{2 π i n} = 1$ for integers $n$ ; $e^{z} = 1$ exactly when $z \in 2 π i Z$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

Verification

1.1givenF1algebra

Since $2 N - 1 = 9$ , the greatest power of two below $9$ is $8$ and the next is $16$ , so $M = 16$ is the least qualifying power. It satisfies the no-aliasing condition [F1].

1.2givenF2algebra

The direct unnormalised transform is $X_{k} = \sum_{j = 0}^{4} e^{- 2 π i j k / 5}$ . For $k = 0$ , all five terms equal $1$ , so $X_{0} = 5$ . For $1 \leq k \leq 4$ , let $r = e^{- 2 π i k / 5}$ . The addition law in [F2] gives $r^{5} = e^{- 2 π i k} = 1$ . If $r = 1$ , the kernel statement in [F2] gives $- 2 π i k / 5 \in 2 π i Z$ , so $k / 5$ is an integer, contradicting $1 \leq k \leq 4$ . Therefore $r \neq 1$ , and $(1 - r) \sum_{j = 0}^{4} r^{j} = 1 - r^{5} = 0$ gives $X_{k} = 0$ after division by $1 - r$ .

2.1step 1.1step 1.2F1∎

By [F1], the exact chirp convolution with $M = 16$ equals this same DFT; hence its output is $(5, 0, 0, 0, 0)$ .

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Two-point circular convolution aliases the linear endpoint term

Example

Let $a = b = (1, 1)$ . The exact length-two cyclic convolution is $(2, 2)$ , while the linear convolution is $(1, 2, 1)$ . Padding both inputs to length four makes the first three cyclic outputs equal the linear convolution and leaves the last output zero. The length-two circular result has wrapped the final linear coefficient into its index-zero output. This compares exact convolution operations and does not assert an output for the computed routine.

Facts & Assumptions

Given: The exact cyclic and linear convolution operations, $a_{0} = a_{1} = b_{0} = b_{1} = 1$ , and zeros outside the stated two-entry supports.

[F1]

$(a *_{2} b)_{k} = \sum_{j = 0}^{1} a_{j} b_{[k - j]_{2}}$ (The unnormalised cyclic convolution on $Z / N Z$ ).

[F2]

Zero padding to any $M \geq n + m - 1$ gives the full linear convolution in the first $n + m - 1$ cyclic outputs and zero thereafter (Zero padding embeds linear convolution in cyclic convolution without aliasing).

Verification

1.1givenF1algebra

From [F1], $(a *_{2} b)_{0} = a_{0} b_{0} + a_{1} b_{1} = 2$ and $(a *_{2} b)_{1} = a_{0} b_{1} + a_{1} b_{0} = 2$ . Direct linear convolution has coefficients $c_{0} = a_{0} b_{0} = 1$ , $c_{1} = a_{0} b_{1} + a_{1} b_{0} = 2$ , and $c_{2} = a_{1} b_{1} = 1$ .

2.1givenF2step 1.1∎

Here $n = m = 2$ , so $n + m - 1 = 3$ . Choose $M = 4 \geq 3$ . By [F2], the exact length-four cyclic operation on the padded inputs has outputs $(c_{0}, c_{1}, c_{2}, 0) = (1, 2, 1, 0)$ . Since its index-zero value is $1$ rather than the unpadded circular value $2$ , the latter contains the wrapped endpoint term $c_{2}$ .

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Three Chebyshev-root samples recover a quadratic by DCT-II

Example

Let $p (t) = 2 + T_{1} (t) - T_{2} (t)$ , where $T_{k}$ is the first-kind Chebyshev polynomial, and sample it at the three DCT-II nodes $ξ_{j} = \cos (π (j + 1 / 2) / 3)$ for $j = 0, 1, 2$ . The DCT-II output is $(6, 3 / 2, - 3 / 2)$ , and the inverse coefficient rule recovers $(2, 1, - 1)$ .

Facts & Assumptions

Given: The polynomial $p (t) = 2 + T_{1} (t) - T_{2} (t)$ and the samples $x_{j} = p (ξ_{j})$ at $ξ_{j} = \cos (π (j + 1 / 2) / 3)$ .

[F1]

At these DCT-II nodes, a polynomial $p = \sum_{k = 0}^{N - 1} a_{k} T_{k}$ has DCT values $X_{0} = N a_{0}$ and $X_{k} = (N / 2) a_{k}$ for $k \geq 1$ (DCT-II samples extract Chebyshev coefficients at first-kind roots).

[F2]

Chebyshev coefficient recovery from the DCT-II samples is $a_{0} = X_{0} / N$ and $a_{k} = 2 X_{k} / N$ for $1 \leq k < N$ (DCT-II samples extract Chebyshev coefficients at first-kind roots).

[F3]

$T_{0} (t) = 1$ , $T_{1} (t) = t$ , and $T_{2} (t) = 2 t^{2} - 1$ (Chebyshev polynomials of the first and second kinds by their three-term recurrences).

Verification

1.1givenF1F3algebra

By [F3], $p (t) = 2 + T_{1} (t) - T_{2} (t) = 2 + 1 \cdot T_{1} (t) - 1 \cdot T_{2} (t)$ has Chebyshev coefficient vector $(a_{0}, a_{1}, a_{2}) = (2, 1, - 1)$ and degree less than $3$ . Applying [F1] with $N = 3$ gives $X_{0} = 3 \cdot 2 = 6$ , $X_{1} = (3 / 2) \cdot 1 = 3 / 2$ , and $X_{2} = (3 / 2) (- 1) = - 3 / 2$ .

2.1step 1.1F2algebra∎

Applying [F2] gives $a_{0} = 6 / 3 = 2$ , $a_{1} = 2 (3 / 2) / 3 = 1$ , and $a_{2} = 2 (- 3 / 2) / 3 = - 1$ , recovering the original coefficients.

Computed fast transforms and convolution — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The length-five all-ones vector through Bluestein

Example

Facts & Assumptions

Verification

Two-point circular convolution aliases the linear endpoint term

Example

Facts & Assumptions

Verification

Three Chebyshev-root samples recover a quadratic by DCT-II

Example

Facts & Assumptions

Verification

Sources