1
$\begingroup$

Sources like http://www.dsprelated.com/dspbooks/mdft/Relation_DFT_Fourier_Series.html explain the equivalence between FS and DFT.

However, isn't there a flaw? When I integrate over the continuous signal, there is something in between the samples (i.e., the sinc interpolation). So the document above shows only that they are equivalent after the continuous signal has been sampled (which is intuitively trivial since the sampled signal contains only Diracs)!

Is it possible to derive a relationship without the signal being sampled? Suppose for simplicity only the DC component. $\tau$ is the period, $T_s$ the (hypothetical) sampling rate, $N=\tau/T_s$ the number of samples and $x[n]=x(n T_s)$:

$$ \int_0^{\tau} x(t) dt = \int_0^{\tau} \sum_{n=0}^{N-1} x(nT_s) \mathrm{sinc}(t-nT_s) dt = \dots \sum_{n=0}^{N-1} x[n] $$

Obviously this is not true, due to sinc interpolation, so something is missing on the right side.

Improve this question
$\endgroup$
3
  • $\begingroup$ you'll need to assume that the continuous signal has a limited band width, meaning that only a finite number of terms in the Fourier series are nonzero. $\endgroup$
    – Carlo Beenakker
    Commented Feb 14, 2014 at 11:02
  • $\begingroup$ This is exactly what I mean: In the derivation in the link, the signal $x(t)$ is plugged in as $x(t) = \sum_n x(n T_s) \delta(t - n T_s)$ and therefore has not finite bandwidth! $\endgroup$
    – divB
    Commented Feb 17, 2014 at 2:09
  • $\begingroup$ And stated differently: My continuous signal has finite bandwidth by definition. $\endgroup$
    – divB
    Commented Jun 17, 2018 at 23:01

1 Answer 1

Reset to default
1
$\begingroup$

Ok, I attempt my own answer. The key is that the continuous-time signal signal is periodic. The definition of above

$$ x(t) = \sum_{n=0}^{N-1} x(n T_s) \operatorname{sinc}(t-n T_s) $$

is not a periodic signal but a signal of finite length. Suppose $N=8$ and the discrete-time signal is $x[0]=1$ and $x[n]=0$ otherwise. The continuous signal (within one period) would look like this:

enter image description here

As correctly assumed in the question, the area under this curve is not 1. However, compare this to the case when $x(t)$ is periodic:

enter image description here

Now due to the fact that $x(8 T_s)=1$ too, there is a sinc contribution from this as well. Similarly, there is a contribution from $x(-8 T_s)$, $x(16 T_s)$ etc ... and summing up all contributions gives an area of 1.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .