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Dirac delta distribution is defined as

$f(t_{0})=\int_{-\infty }^{\infty } \! f(t)\delta(t-t_{0}) \, dt $ where $f(t)$ is smooth function. Then my question is:

:Calculate Fourier transform $\hat \delta(\omega)$ from $\delta (t-t_{0})$

Solution:

$$\hat \delta(\omega)=\frac{1}{\sqrt{2 \pi} }\int_{-\infty }^{\infty } \! \delta (t-t_{0}) e^{-j \omega t}\, dt $$

$$\hat \delta(\omega)=\frac {1}{\sqrt{2 \pi}}e^{-j \omega t_{0}}$$

Can someone explain me how they got this solution and write what are the steps between? On internet I always find some general formulas and I don't know how to use them.

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    $\begingroup$ Just follow the given definition... $\endgroup$ – Ian Oct 24 '16 at 21:12
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Let $f$ be a smooth integrable function as you commented. Then we can define the Fourier Transform as

$$\mathcal{F}[f](y) := \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{-iyx}\mathrm{d}x$$

We can also derive the inverse of Fourier Transform as:

$$\mathcal{F}^c[f](y):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(x)e^{+iyx}\mathrm{d}x$$

Now lets see what happen wen we apply both to a function

$$f(y) = \mathcal{F}^c[\mathcal{F}[f]](y) = \frac{1}{2\pi}\int_\mathbb{R}\left(\int_{\mathbb{R}} f(\omega)e^{-i\omega x}\mathrm{d}\omega \right)e^{+iyx}\mathrm{d}x = \int_{\mathbb{R}}f(\omega)\left(\int_{\mathbb{R}}\frac{1}{2\pi}e^{-i(\omega-y)x}\mathrm{d}x\right)\mathrm{d}\omega$$

We can see in this last equality that the function in brakets acts as a Dirac Delta. So we can relate this as de Dirac as a notation (because the Dirac Delta just have formal sense in distribution theory)

$$\delta(\omega-y) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-i(\omega-y)x}\mathrm{d}x$$

This is an intuitive constructive way of seen that the Dirac is related to Fourier Transforms. Now for your question we can use the definition on how an Dirac Delta acts

$\hat{\delta}(\omega) = \frac{1}{(2\pi)^{1/2}}\int \delta(t-t_0)e^{-i\omega t}\mathrm{d}t = \frac{1}{(2\pi)^{1/2}}e^{-\omega t_0}$

Where we just used the above

$$f(y) = \int f (\omega)\delta(\omega-y)\mathrm{d}\omega$$

For $f(t):=e^{-i\omega t}/\sqrt{2\pi}$

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We have \begin{align}f(t_{0})=:\int_{-\infty }^{\infty } \! f(t)\delta(t-t_{0}) \, \text{d}t &&(1) \\ \hat g(\omega):=\widehat{g(\cdot)}(\omega):=\int_{-\infty }^{\infty }e^{-j \omega t}g(t) \ \text{d}t &&(2) \end{align} and therefore we get: $$\widehat{\delta(\cdot-t_0)}(\omega) \stackrel{(2)}{=} \frac{1}{\sqrt{2 \pi} }\int_{-\infty }^{\infty } \underbrace{e^{-j \omega t}}_{=:f(t)} \delta (t-t_{0})\ \text{d}t\stackrel{(1)}{=}\frac{1}{\sqrt{2 \pi}}f(t_0)=\frac{1}{\sqrt{2 \pi}}e^{-j \omega t_{0}}$$

To be precise this is the Fourier transform of $\delta(\cdot-t_0)$ and not $\delta$. For $\delta=\delta(\cdot)$ we'd have to set $t_0=0$ in above formula and we get $\hat\delta(\omega)=(2\pi)^{-1/2}.$

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You indeed have to start with the definition. The action of the Fourier transform of the distribution $f$ on a test function $\phi$ is defined as $$(\mathcal F[f], \phi) = (f, \mathcal F[\phi]),$$ which is valid because if $\phi$ is a test function of rapid decay, then so is $\mathcal F[\phi]$, and the functional defined in this way is linear and continuous. From this we obtain $$(\mathcal F[\delta], \phi) = (\delta, \mathcal F[\phi]) = \mathcal F[\phi](0) = \frac 1 {\sqrt {2 \pi}} \int_{-\infty}^\infty \phi(t) dt,$$ which means that $\mathcal F[\delta]$ is the same functional as $1/\sqrt {2 \pi}$.

The transform of $\delta(t - t_0)$ can be found either in the same way directly or by applying the translation property, which is valid for singular distributions because a linear transformation is defined so as to be consistent with a linear transformation of a regular distribution.

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