relation between DFT and continuous Fourier Transform

random800 · #1 $Old$ October 20th, 2009, 06:27 PM

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?

CaptainBlack · #2 $Old$ October 21st, 2009, 12:58 AM

Quote:

Originally Posted by random800 $View Post$

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

random800 · #3 $Old$ October 21st, 2009, 09:46 AM

thank you, CaptainBlack, for your reply.

Quote:

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?

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

Do you know a good source to read about this thing? (I am intrested in the derivation of this)

Quote:

(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

Quote:

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.

CaptainBlack · #4 $Old$ October 21st, 2009, 10:49 AM

Quote:

Originally Posted by random800 $View Post$

thank you, CaptainBlack, for your reply.

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

Quote:

Do you know a good source to read about this thing? (I am intrested in the derivation of this)

virtually any book on signal processing (or digital signal processing), but they will all use zero base array indexing (except for titles that have MATLAB in them anyway.

Quote:

Probably, you are right here too, but I have to work with the given formulas, it is not my initiative

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

It is still worth putting the zeros in, since the DFT is periodic, and the spectrum of you signal is probably not of finite bandwidth, though it may decay fast enough to get away with not padding there is no guarantee.

CB

random800 · #5 $Old$ October 21st, 2009, 11:51 AM

Thanks for the help!