Sampling Function is its own Fourier Transform

Theorem

Consider the sampling function $\operatorname {III}: \R \to \R$.

Then:

$\map \FF {\operatorname {III} } = \operatorname {III}$

where $\FF$ denotes the Fourier transform.

Proof

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Sources

• 1978: Ronald N. Bracewell: The Fourier Transform and its Applications (2nd ed.) ... (previous) ... (next): Preface to the Second Edition
• 1978: Ronald N. Bracewell: The Fourier Transform and its Applications (2nd ed.) ... (previous) ... (next): Chapter $1$: Introduction