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According to Wolfram, Fourier transform of $1$ is $\sqrt{2\pi} \, \delta(\omega)$. Can someone explain what this means?

The only frequency the function $1$ should just be $0$ since $\cos(0) = 1$.

So, shouldn't the result have just been $0$ instead of that expression?

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  • $\begingroup$ I learned this a while ago so I'm trying to regain my intuitive understanding $\endgroup$ – Goldname Apr 26 '18 at 17:12
  • $\begingroup$ $\delta(w)$ is zero for all $w$ except $w=0$. This is what you are expecting. $\endgroup$ – Math Lover Apr 26 '18 at 17:13
  • $\begingroup$ @MathLover Not understand. When w is 0, $\delta(w)$ is infinite?? $\endgroup$ – Goldname Apr 26 '18 at 17:15
  • $\begingroup$ Yes, it is right. Note that the energy of the signal is infinite too. $\endgroup$ – Math Lover Apr 26 '18 at 17:17
  • $\begingroup$ @MathLover Ignore the previous comment. I'm not understanding why the energy of the signal is infinite. Also, what does the $\sqrt{2\pi}$ do? $\endgroup$ – Goldname Apr 26 '18 at 17:25
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Short intuitive answer: The Fourier transform breaks functions down into their constituent frequencies, and frequency is the inverse of wavelength. A constant function $1$ is so spread out that it effectively has infinite wavelength, or zero frequency. Hence its Fourier transform is concentrated at $0$, giving a spike (e.g. a delta function). Conversely, the Fourier transform of $\delta$ is constant.


Long technical answer: The Fourier transform is most easily defined on $\mathcal{S}$, the space of Schwarz functions, where we have a formula that

$$\hat{f}(\xi) = \int_{\mathbb{R}} f(x) e^{-2\pi i x \xi} \, dx.$$

This definition leads to a lot of useful properties, such as

$$\int_{\mathbb{R}} \hat{f} g \, dx = \int_{\mathbb{R}} f \hat{g} \, dx.$$

We can use this to define the distributional Fourier transform by taking this as a definition: Given a (tempered) distribution $T$ that acts on Schwarz functions $g$ via the pairing $\langle T, g\rangle$, we define the Fourier transform of $T$ to be the distrubtion that acts via

$$\langle \hat{T}, g\rangle = \langle T, \hat{g}\rangle.$$

Basically, all this means is that we can move the hat back and forth just like with "nice" functions.


Given all this, the delta function is best viewed as the distribution which acts via $$\langle \delta, g\rangle = g(0).$$

Thinking of $\delta$ as a spike at zero with area one and integrating this against $g$ also makes sense. Then the Fourier transform of $\delta$ is the distribution that acts via

$$\langle \hat{\delta}, g\rangle = \langle \delta, \hat{g}\rangle = \hat{g}(0) = \int_{\mathbb{R}} g \, dx = \int_{\mathbb{R}} 1 \cdot g \, dx.$$

Therefore, we can think of $\hat{\delta}$ as simply integrating against a constant function $1$, and identify $\hat{\delta} = 1$. (Depending, of course, on the normalization of the Fourier transform).

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  • $\begingroup$ The function itself recovered (the fourier coefficients) by the fourier transform is the energy of the signal right? $\endgroup$ – Goldname Apr 26 '18 at 20:49
  • $\begingroup$ @Goldname Energy is most meaningful when you're talking about square integrable functions, because then it is just the $L^2$ norm $(\int |f|^2 \, dx)^{1/2} = (\int |\hat{f}|^2 \, dx)^{1/2}$. In the case of the delta function or another distribution, it's a lot less meaningful. $\endgroup$ – user296602 Apr 26 '18 at 21:00
  • $\begingroup$ Then how should one interpret the delta meaning? $\endgroup$ – Goldname Apr 26 '18 at 21:03
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    $\begingroup$ Is it that there is a total area of $\sqrt(2*\pi)$ according to wolfram, and since there are no other frequencies, we place all the area on the 0 frequency? $\endgroup$ – Goldname Apr 26 '18 at 21:03
  • $\begingroup$ Trying to interpret the delta function as having an area under its graph is risky at best (in a mathematical context; in engineering, ymmv). It really is much better to think of it as an operator that takes functions and outputs numbers. $\endgroup$ – user296602 Apr 26 '18 at 21:10

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