I understand that the F.T. of a constant signal is the Dirac. However, I cannot find anywhere showing the derivation or proof for this. I'm trying to do it myself and am getting lost. Can anyone give a worked-out derivation that the Fourier Transform of a constant signal is the Dirac? Thank You for any help!
Derivation of Fourier Transform of a constant signal
dirac deltafourier analysisfourier transformintegration
Best Answer
The derivation is very simple, provided you know what a distribution is. Very briefly: A distribution is a continuous linear functional $$ L:C_0^\infty(\mathbb{R})\rightarrow\mathbb{C}. $$ I will not specify what it means for $L$ to be continuous (it's complicated). The Dirac delta distribution is the linear functional $$ \delta(\varphi):=\varphi(0). $$ The Fourier transform is defined on a subset of the distributions called tempered distritution. The Fourier transform $\mathcal{F}(L)$ of a (tempered) distribution $L$ is again a (tempered) distribution. It is defined as the linear functional $$ \mathcal{F}(L)(\varphi):=L(\mathcal{F}(\phi)). $$ If you want to Fourier transform the constant 1, you first have to identify the constant 1 with a distribution $L_1$. This is done canonically via $$ L_1(\varphi):=\int_{\mathbb{R}} 1\cdot \varphi(x)\,dx. $$ Now you can compute the Fourier transform $\mathcal{F}(L_1)$ of $L_1$: $$ \mathcal{F}(L_1)(\phi) = L_1(\mathcal{F}(\phi)) = \int_{\mathbb{R}} 1\cdot \hat\phi(x) dx = \int_{\mathbb{R}} e^{2\pi ix\cdot 0}\cdot \hat\phi(x) dx = \mathcal{F}^{-1}(\hat\phi)(0) = \phi(0) = \delta(\phi). $$ That is it! We see that the Fourier transform for $L_1$ coincides with the Dirac delta distribution $\delta$. So in the sense of distributions, the the Fourier transform of 1 is the Dirac delta distribution.