# Thread: relation between DFT and continuous Fourier Transform

1. ## relation between DFT and continuous Fourier Transform

Greetings,

I have a function with a bounded support. For simplicity, let it be $\mathrm{supp} f(x) \subset (0,1)$. I would like to find $\hat f(z)$:
$\hat f(z) = \int_0^1 e^{-2 \pi i x z} f(x) dx$.

First, I thought that DFT defined by
$\hat F_k = \frac{1}{\sqrt{N}} \sum_{j=1}^N e^{-2 \pi i \frac{(j-1)(k-1)}{N}} F_j$
gave the following relation
if
$F_j = f\left(\frac{j-1}{N}\right)$
then
$\hat f(k-1) \approx \frac{\hat F_k}{\sqrt{N}}$
but, apparently, I was wrong.

So, what is the relation between Fourier transform and DFT?

2. Originally Posted by random800
Greetings,

I have a function with a bounded support. For simplicity, let it be $\mathrm{supp} f(x) \subset (0,1)$. I would like to find $\hat f(z)$:
$\hat f(z) = \int_0^1 e^{-2 \pi i x z} f(x) dx$.

First, I thought that DFT defined by
$\hat F_k = \frac{1}{\sqrt{N}} \sum_{j=1}^N e^{-2 \pi i \frac{(j-1)(k-1)}{N}} F_j$
gave the following relation
if
$F_j = f\left(\frac{j-1}{N}\right)$
then
$\hat f(k-1) \approx \frac{\hat F_k}{\sqrt{N}}$
but, apparently, I was wrong.

So, what is the relation between Fourier transform and DFT?
In what way do you think this is wrong?

There are two possibilities: the normalisation, that only the first $\lfloor N/2 \rfloor$ points are valid.

Or is the problem something else?

(you will find the equations are simpler if you use indices running from 0 to N-1)

You may also find it advantageous to use an interval 0-2 for x in the DFT to capture the spectral influence of the discontinuities better - if that makes any sense (consider the case where f(x)=1 for x in (0,1) and 0 otherwise).

CB

There are two possibilities: the normalisation, that only the first points are valid.
Yes, your are right, only the first half of points were valid. Is the following relation correct?

$\hat F_k \approx \left\{\begin{array}{ll}
\sqrt N \hat f(k-1) & \mbox{ if } 1 \leq k \leq \lfloor N/2 \rfloor \\
\sqrt N \hat f(k-1-N) & \mbox{ if } \lfloor N/2 \rfloor < k \leq N
\end{array}\right.
$

(you will find the equations are simpler if you use indices running from 0 to N-1)
Probably, you are right here too, but I have to work with the given formulas, it is not my initiative

You may also find it advantageous to use an interval 0-2 for x in the DFT to capture the spectral influence of the discontinuities better - if that makes any sense (consider the case where f(x)=1 for x in (0,1) and 0 otherwise).
Well, in my case all functions are smooth on $(-\infty, +\infty)$ and they have at least two derivatives.

4. Originally Posted by random800

Yes, your are right, only the first half of points were valid. Is the following relation correct?

$\hat F_k \approx \left\{\begin{array}{ll}
\sqrt N \hat f(k-1) & \mbox{ if } 1 \leq k \leq \lfloor N/2 \rfloor \\
\sqrt N \hat f(k-1-N) & \mbox{ if } \lfloor N/2 \rfloor < k \leq N
\end{array}\right.
$
The top half of the DFT is the image of the negative frequency half of the spectrum

Well, in my case all functions are smooth on $(-\infty, +\infty)$ and they have at least two derivatives.