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Consider the following convention for defining the Fourier transform

$\hat{f}(\omega) = \int f(x) e^{-2 \pi i x \omega } d\omega $.

Why is the Fourier transform of 1 equal to $\delta(\omega)$. Loosely, I see that when $\omega \neq 0$ the integral will be zero and when $\omega = 0$ the integral diverges but does anyone have a more rigorous way of showing this? Also, how do we know there is not some scaling factor in front or something?

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  • $\begingroup$ Consider the Fourier transform of $f(x) = \exp(-\epsilon x^2)$. It is proportional to $\epsilon^{-1/2} \exp( -\pi^2 \omega^2 / \epsilon )$. A family of smooth functions $f_\epsilon(\omega ) = \epsilon^{-1} f(x/\epsilon)$ is a "nascent delta function". That is, when $\epsilon \to 0$, $f_\epsilon \to A\delta$ where $A$ is some constant. This constant will depend on your convention for the Fourier transform. $\endgroup$ – Robin Ekman Jan 27 '15 at 23:30
  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Jan 27 '15 at 23:41
  • $\begingroup$ The level of rigor in Robin's answer was all I was looking for but probably wouldn't be enough to satisfy a mathematician. $\endgroup$ – Laplacian Jan 28 '15 at 0:00
  • $\begingroup$ This is similar to how a wavefunction's position eigenstate (which is a delta function) corresponds to the momentum being equally spread over the entire domain of possible momenta . $\endgroup$ – M.M Jan 28 '15 at 0:40
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Why is the Fourier transform of 1 equal to δ(ω)

$$f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}F(\omega)e^{i\omega t}\mathrm d\omega = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\delta(\omega)e^{i\omega t}\mathrm d\omega = \frac{1}{\sqrt{2\pi}}e^{i0 t} = \frac{1}{\sqrt{2\pi}}$$

This follows from the sifting property of $\delta(x)$:

$$f(x) = \int f(\tau)\delta(\tau - x)d\tau$$

So, it's clear that the (inverse) Fourier transform of a $\delta$ distribution is a constant. The value of the constant $(1, 2\pi, \frac{1}{2\pi}, \frac{1}{\sqrt{2\pi}})$ etc., depends on the convention.

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Consider the Fourier transform of $f(x) = \exp(-\epsilon x^2)$. It is proportional to $\epsilon^{-1/2} \exp( -\pi^2 \omega^2 / \epsilon )$ (the factor for $\omega^2/\epsilon$ varies with convention - with the convention in your question it is this) . A family of smooth functions $f_\epsilon(\omega ) = \epsilon^{-1} f(x/\epsilon)$ is a "nascent delta function". That is, $$f_\epsilon \to A\delta \text{ when } \epsilon \to 0$$ where $A$ is some constant. This constant will depend on your convention for the Fourier transform.

This is the rigorous way to see that the Fourier transform of a constant is a delta function. The formula $$\hat f(\omega) = \int e^{2\pi i \omega x} f(x)\, dx$$ only makes sense for integrable $f$. For functions that are not integrable, the Fourier transform has to be defined by continuous extension from integrable functions to some larger function space. See also this answer.

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Try taking the Fourier transform of $f(x)$. Then take the inverse Fourier transform of that. Obviously you get $f(x)$ on one side. On the other you get the integral of the delta function.

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  • $\begingroup$ Huh? I don't know what you mean. $\endgroup$ – Laplacian Jan 27 '15 at 23:43
  • $\begingroup$ @Laplacian: What I just outlined gives you the integral representation of the delta function, which is just the transform of $1$. $\endgroup$ – Ryan Unger Jan 28 '15 at 0:14
  • $\begingroup$ @Laplacian: See, for instance, dlmf.nist.gov/1.17. $\endgroup$ – Ryan Unger Jan 28 '15 at 0:16
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For a less technical answer: generally speaking the more peaked your function is, the more its Fourier transform tends to spread out, and the more spread out your function is the more peaked its Fourier transform is. Since the function $f(x) = 1$ is just a horizontal line, maximally spread, its Fourier transform must but infinitely narrow, a delta spike.

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