2
$\begingroup$

I'm currently trying to understand time continuous Fourier transformations for my Signal Analysis class (Electrical Engineering major). In my Textbook we have a set of "useful FT-pairs", one of which being $x(t)<=FT=>X(\omega)$

$\sin(\omega_0*t)<=>i\pi[\delta(\omega+\omega_0)-\delta(\omega-\omega_0)]$

I'm trying to derive this on my own, but I keep running into the same dead end. I replace the sine with $\frac 1{2i}(e^\left(i\omega_0t\right)-e^\left(-i\omega_0t\right))$ and drop that into the $\int f(x) e^\left(-i\omega t\right)$. Multiply the functions together by adding together the exponents and then finding the anti-derivative isn't hard, but don't understand how you get from that to the $\delta(x)$. I played around on wolfram alpha and found out that apparently $\int_{-\infty}^{\infty} e^\left(it(\omega + \omega_0\right)dt$ doesn't converge except for $\omega = -\omega_0$, and then it converges to infinity. This seems close to the behavior of the delta function, but I don't think it's enough for me to replace them?

Could really use some help understanding this.

$\endgroup$
1
$\begingroup$

You should understand that the Dirac-delta $\delta(\omega)$ is not a conventional function. It is a generalized function or a distribution. The mathematical theory behind it is really involved and the guy who explained everything(Laurent Schwartz) got a Field's Medal for it.

The best way to go around it is to find out the inverse Fourier Transform of $i\pi[\delta(\omega+\omega_0)-\delta(\omega-\omega_0)]$. It comes out to be $\sin(\omega_0t)$ without any problems.

This gives you the mathematical closure over an intuitive understanding.

The wiki page on Dirac Delta mentions the same integral that you are asking, but the IFT approach looks more natural to me.

$\endgroup$
0
$\begingroup$

Your observations are right. The Fourier transform of $x\mapsto\sin(ax)$ is not defined by the integral but rather in a weak sense or in the sense of distributions (http://en.wikipedia.org/wiki/Distribution_(mathematics)#Tempered_distributions_and_Fourier_transform). This is also explained on the Wikipedia page on the Fourier transform (http://en.wikipedia.org/wiki/Fourier_transform#Tempered_distributions).

$\endgroup$
0
$\begingroup$

Here are two ways to look at it.

  • Assume that whatever mathematical trickery is used to define it (distribution theory) the result should be consistent with the inverse Fourier transform formula, and $$\int_{-\infty}^{\infty} dk \delta(k-k_0) \exp(ikx)\frac{1}{2\pi} = \exp(ik_0x)\frac{1}{2\pi}$$
  • Think about this happening in a finite range. $$\int_{-L}^{L}dx \exp(-ikx)\exp(ik_0x) = \frac{2\sin(k_0-k)L}{k_0-k}$$ is sharply peaked at $k=k_0$ with a peak of size $2L$, the length of the interval. For large $L$ this acts just like a $\delta$ function. (Integrate it against another continuous function and take $L\to\infty$ to prove this and check out the normalization.)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.