Understanding Discrete Fourier Transform based on continuous Fourier Transform

Question

I'm reading a PDF that starts like this:

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.

Answer 1

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.

Answer 2

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.