I'm trying to show how the discrete Fourier transform (DFT) arises from the equation for the continuous-time Fourier Transform. I've run into an interesting caveat which I can't seem to find an explanation to anywhere. My 'derivation' goes something like this:

Lets consider the continuous time Fourier transform equation $$ F(\omega) = \int_{-\infty}^{+\infty} \, f(t) \, e^{-2\pi i \omega t} \,dt $$ If $f(t)$ was to be discretised at $N$ evenly spaced sampling points with the sampling interval of $dt$, then it could be thought of as a continuous function of time with Dirac deltas at each sampling point. The integral of each of the deltas would be equal to the value of $f(t)$ at that point. Lets denote $f(t)$ at $k$-th sampling point (0-indexed) as $f_k$. Now, the integral is non-zero only at those delta functions so the entire above equation can be written as a sum $$ F(\omega) = dt \times \sum_{k=0}^{N-1} \, f_k\, e^{-2\pi i \omega (k dt)} $$ Even though the sum is finite, this is a continuous function of $\omega$. $F(\omega)$ turns out to be periodic in $\omega$ and therefore just a single period of it is needed to be kept to retain all of the information about the sampled version of $f(t)$ it came from.

All of this is (I believe) fine. Here comes the tricky bit. I want to then say that this single period of $F(\omega)$ can be sampled down to $N$ evenly spaced points which would in fact give the equation for the DFT. I say that: "It is one of the implications of the Shannon information theory (...) that for a general case of $N$ arbitrary bits of information, there does not exist a smaller set of bits which can be used to represent all of the information of the original $N$ bits. This places a lower bound on how many samples of $F(\omega)$ one can take."

Question 1: Is this a correct statement about implications of the information theory?

Question 2: I know that N bits of $F(\omega)$ are sufficient, but how do I prove it? To put it differently: How can I one show that for any arbitrary signal I will not loose any information about $f(t)$ by sampling $F(\omega)$ at only $N$ points.

  • 1
    $\begingroup$ the most fundamental thing is that the DFT maps a periodic sequence $x_n$ having period $N$ (so that $x_{n+N}=x_n \quad \forall n \in \mathbb{Z}$) to another periodic sequence $X_k$ having the same period $N$. Relating the DFT to the DTFT or, as you are trying to do, the continuous-time Fourier transform, requires a little bit of hand-waving to put it into a form of the Riemann integral. $\endgroup$ – robert bristow-johnson Jul 8 '19 at 21:39
  • $\begingroup$ How do you take N evenly spaced samples of the function f(t), which stretches from t = -inf to +inf? All the N samples will be infinitely apart, you can never get there. $\endgroup$ – DSP Rookie Apr 4 at 6:46
  • $\begingroup$ Also N samples dt apart : dt in an integral is infinitesimal small quantity. All of those N samples will actually be confined in an small region of t $\endgroup$ – DSP Rookie Apr 4 at 6:48

Note that f() can only be sampled down to N points per period if f() is bandlimited to include only spectrum below t_period/(2*N) in frequency. And a signal exactly periodic in N, with no frequency content at or above t_period/(2*N) in frequency can be exactly represented with N complex exponential basis vectors (Fourier). And the inverse.

The number of degrees of freedom in N complex basis vectors or N complex DFT results matches the number of degrees of freedom in N complex time samples.

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