2
$\begingroup$

I'm reading a PDF that starts like this:

enter image description here

It sees the Discrete Fourier Transform as a "partitioning" of the continuous case, I guess. That's what I'm trying to understand.

The Fourier transform of $f(t)$ is

$$F(jw) = \int_{-\infty}^{\infty}f(t)e^{-jwt}\ dt$$

If we concentrate on a signal of finite time, I guess, we only have to integrate on this finite time, since the rest would be $0$. At first I thought he was breaking the integral

$$\int_{o}^{(n-1)T}f(t)e^{-jwt}\ dt$$

as lots of integrals with the 'size', but then I saw that it is in fact doing some other things, it's like the $e^{-jwt}$ weren't integrated at all.

Could somebody explain to me what's happening? I'm very lost.

$\endgroup$
  • $\begingroup$ Consider it is not true. The discrete Fourier transform is analogous to a Fourier series $\sum_n c_n e^{2i \pi n x}$ with finitely many terms, in that case knowing it for finitely many values of $x$ is enough. In practice, the DFT (of size $N$) is just an unitary matrix $W \in \mathbb{C}^{N \times N}, WW^* = W^* W = Id$. It is true that the Fourier series is the limiting case of the DFT. It is much less true that the Fourier transform is the limiting case of the Fourier series. You'll need the Fourier transform of distributions (with Dirac deltas) to unify everything. $\endgroup$ – reuns Jul 25 '17 at 8:02
  • $\begingroup$ It is not completely wrong neither that the FT is the limiting case of the DFT, but it is also true it is quite hard to prove. $\endgroup$ – reuns Jul 25 '17 at 8:05
0
$\begingroup$

The idea is the following: I will try to use notation similar to what was used in the picture provided. (I will use $i$ instead of $j$ for the canonical complex root of unity.)

Often in signal processing our discrete signal $\{s[0],s[1],\dots,s[N -1]\}$ was obtained by sampling a continuous signal $s : [0,(N- 1)T] \to \mathbb{R}$ with, say, $s[n] = s(nT)$.

An apparently naive but actually fairly natural way of representing the samples $\{s[0],s[1],\dots,s[N - 1]\}$ is as a measure on $[0,(N - 1)T]$ via the formula $$s_{d} = \sum_{j = 0}^{N - 1} s[j] \delta_{jT},$$ where $\delta_{x}$ denotes the Dirac mass at $x$. This is what the author means by "regarding each sample as an impulse having area ..." In fact, if one "integrates" $s[j] \delta_{jT}$ over $[0,(N - 1)T]$, one gets $s[j]$ back. (What I mean by "integrates" should be clear if you have enough real analysis background, but, if not, I can try to explain further.) In the electrical engineering literature, "impulse" means Dirac or Kronecker delta depending on the context so this is what the author means, whether or not he wants to write it (or even thinks about it) in a mathematically rigorous way.

Now take the Fourier transform of $s_{d}$. $s_{d}$ isn't a function, it's a measure. Nonetheless, we have a notion of Fourier transforms of measures and it gives us $$\hat{s}_{d}(\omega) = \sum_{j = 0}^{N - 1} s[j] \int_{-\infty}^{\infty} e^{-i \omega t} \delta_{jT}(dt) = \sum_{j = 0}^{N - 1} s[j] e^{-i \omega j T}$$ which is what the article claims.

$\endgroup$
  • $\begingroup$ I'd like to get a better understanding of this dirac thing, I don't get this integral of a function times the dirac function $\endgroup$ – Guerlando OCs Jul 26 '17 at 8:57
  • $\begingroup$ The Dirac mass $\delta_{x}$ centered at $x$ is a measure, not a function. Do you have any background in measure theory? If not, the easiest way to play with a Dirac mass is to use think of it as a linear functional that takes (sufficiently nice) continuous functions and spits out the value at $x$, i.e. $\int_{\mathbb{R}} \varphi(y) \delta_{x}(dy) = \varphi(x)$. $\endgroup$ – fourierwho Jul 26 '17 at 17:55
0
$\begingroup$

I guess that you missed the fragment "regard each sample as an impulse", which means (reading between lines) that the signal is a sum of Dirac deltas of amplitude $f[k]$.

A Dirac delta is an infinitely short signal of infinite amplitude, such that the area under its "curve" is one. When you integrate such a delta in an interval that contains it, you get one. If you integrate the product of a delta and a function, you get the value of the function at the delta. https://en.wikipedia.org/wiki/Dirac_delta_function

Hence a Dirac delta replaces an integral by an instantaneous value. This is how you trade a continous signal for a discrete one.

$\endgroup$

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.