Even Impulse Pair is Fourier Transform of Cosine Function
Jump to navigation
Jump to search
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
- 1978: Ronald N. Bracewell: The Fourier Transform and its Applications (2nd ed.) ... (previous) ... (next): Chapter $1$: Introduction
- Weisstein, Eric W. "Fourier Transform." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/FourierTransform.html