Even Impulse Pair is Fourier Transform of Cosine Function

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Theorem

Consider the (real) cosine function $\map \cos x: \R \to \R$.

$\map f x = \map \cos {\pi x}$


Then:

\(\ds \map {\hat f} s\) \(=\) \(\ds \dfrac 1 2 \map \delta {s - \dfrac 1 2} + \dfrac 1 2 \map \delta {s + \dfrac 1 2}\)
\(\ds \) \(=\) \(\ds \map {\operatorname {II} } s\)

where:

$\map {\hat f} s$ is the Fourier transform of $\map f x$.
$\operatorname {II}$ denotes the even impulse pair function.


Proof

By the definition of a Fourier transform:

\(\ds \map {\hat f} s\) \(=\) \(\ds \int_{-\infty}^\infty e^{-2 \pi i x s} \map f x \rd x\)
\(\ds \) \(=\) \(\ds \int_{-\infty}^\infty e^{-2 \pi i x s} \map \cos {\pi x} \rd x\)
\(\ds \) \(=\) \(\ds \int_{-\infty}^\infty e^{-2 \pi i x s} \dfrac 1 2 \paren {e^{i \pi x} + e^{-i \pi x} } \rd x\) Euler's Cosine Identity
\(\ds \) \(=\) \(\ds \dfrac 1 2 \int_{-\infty}^\infty e^{-2 \pi i x s} e^{i \pi x} \rd x + \dfrac 1 2 \int_{-\infty}^\infty e^{-2 \pi i x s} e^{-i \pi x} \rd x\) Linear Combination of Definite Integrals
\(\ds \) \(=\) \(\ds \dfrac 1 2 \int_{-\infty}^\infty e^{-2 \pi i x \paren {s - \frac 1 2 } } \rd x + \dfrac 1 2 \int_{-\infty}^\infty e^{-2 \pi i x \paren {s + \frac 1 2 } } \rd x\)
\(\ds \) \(=\) \(\ds \dfrac 1 2 \map \delta {s - \dfrac 1 2 } + \dfrac 1 2 \map \delta {s + \dfrac 1 2 }\) Fourier Transform of 1

$\blacksquare$


Sources

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