one of our homework solutions states that $$\delta(x)\equiv\frac{1}{2\pi}\int_{-\infty}^{\infty}d\omega e^{-i\omega x} $$ is the Fourier transform of the dirac delta function. But according to the definition $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}d\omega \delta(\omega)e^{-i\omega x}=\frac{1}{\sqrt{2\pi}}$$ must be the fourier transform. So what does that identity above mean and how was it derived?
0
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
5
Here's my best guess:
The statement in your homework solution is not giving you the Fourier transform of the dirac delta function. What it is saying is that we can define the dirac delta function as the inverse Fourier transform of $f(\omega) = 1$. That is, we can define $$ \delta(x) \equiv \frac 1{2 \pi} \int_{-\infty}^\infty d \omega \, 1 \cdot e^{i \omega x} $$ Note that it is much more typical to have the transforms be functions of $\omega$ than functions of $x$, hence my suspicions.
answered Jan 22, 2015 at 16:04
-
1
-
$\begingroup$ Not in the usual definition. Do you have minus there that shouldn't be there? $\endgroup$ Jan 23, 2015 at 0:40
-
$\begingroup$ In my homework solution there is one $-i\omega x$ $\endgroup$– phthJan 23, 2015 at 0:47
-
$\begingroup$ That's strange... either way, it may be a valid definition of $\delta(x)$. It's clear they're defining $\delta(x)$ and not its Fourier transform. $\endgroup$ Jan 23, 2015 at 0:48
-
$\begingroup$ The two common sign conventions come from different communities. Most physicists and mathematicians use +i for the forward transform, many engineers use -j. This is completely rationalized by letting j = -i. math.stackexchange.com/a/4604354/7882 is a decent overview over how much hassle this can cause. (I would also argue that the conjugate variable should be f rather than ω, or equivalently that a factor of 2pi needs to go in the integral. $\endgroup$– wnoiseSep 17 at 20:34
$\begingroup$
$\endgroup$
The Fourier transform is defined differently between some authors: sometimes is defined as $X(i\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty x(t) e^{-i\omega t} dt$ (symmetric way, preferred by mathematicians), sometimes as $X(i\omega) = \frac{1}{2\pi}\int_{-\infty}^\infty x(t) e^{-i\omega t} dt$ (preferred by electrical engineers, where commonly they use $j = \sqrt{-1}$ instead of "$i$"), and there are also its alternatives using the standard frequency "$f$" instead of the angular frequency "$\omega$", and also sometimes there differences on sign of the complex exponential kernel. This is why in the Wikipedia page of the Fourier transform [1], on the tables of functions vs they transforms are columns with different alternatives.
Both alternatives leads to $x(t)=\mathbb{F}^{-1}\{\mathbb{F}\{x(t)\}\}$, but if you compare $X(i\omega)$ between alternatives, you will find a difference of a factor of $\sqrt{2\pi}$.
You have to be aware of this when using software like Wolfram-Alpha [2], since its solutions could be delivered by a different definition from the one you have mind.
