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10.4: Basic Properties of the Fourier Transform

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    34575
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    The Fourier transform has several important properties. These can all be derived from the definition of the Fourier transform; the proofs are left as exercises.

    1. The Fourier transform is linear: if we have two functions \(f(x)\) and \(g(x)\), whose Fourier transforms are \(F(k)\) and \(G(k)\) respectively, then for any constants \(a, b \in \mathbb{C}\), \[a f(x) + b g(x) \;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; a F(k) + b G(k).\]

    2. Performing a coordinate translation on a function causes its Fourier transform to be multiplied by a phase factor: \[f(x+b) \;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; e^{ikb} \, F(k).\] As a consequence, translations leave the Fourier spectrum \(|F(k)|^2\) unchanged.

    3. If the Fourier transform of \(f(x)\) is \(F(k)\), then \[f^*(x) \quad \overset{\mathrm{FT}}{\longrightarrow} \;\; F^*(-k).\] As a consequence, the Fourier transform of a real function must satisfy the symmetry relation \(F(k) = F^*(-k)\), meaning that the Fourier spectrum is symmetric about the origin in k-space: \(\big|\,F(k)\,\big|^2 = \big|\,F(-k)\,\big|^2.\)

    4. When you take the derivative of a function, that is equivalent to multiplying its Fourier transform by a factor of \(ik\): \[\frac{d}{dx} f(x) \,\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; ik F(k).\]

      For functions of time, because of the difference in sign convention discussed in Section 10.3, there is an extra minus sign: \[\frac{d}{dt} f(t) \;\;\;\; \overset{\mathrm{FT}}{\longrightarrow} \;\;\; -i\omega F(\omega).\]


    This page titled 10.4: Basic Properties of the Fourier Transform is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Y. D. Chong via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.