Evaluating regular Fourier Transform from DFT

Question

Suppose we know the DFT of a discrete limited sequence, some $X[k],\ k = 0, 1,\dots ,N-1$. How can we calculate the Fourier Transform of the same signal for a random frequency $\Omega$?

EDIT:

How about this, we can get the starting signal with inverse DFT, like: $$x[n] = \frac{1}{N}\sum\limits_{k=0}^{N-1} X[k]e^{j\frac{2\pi}{N}kn}$$ and then get the Fourier Transform normally with: $$\begin{align} X(e^{j\Omega}) &= \sum\limits_{n = -\infty}^{+\infty} x[n]e^{-j\Omega n}\\ &= \frac{1}{N}\sum\limits_{n = -\infty}^{+\infty} e^{-j\Omega n}\sum\limits_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}kn} \end{align}$$

Is this correct?

EDIT2: As @Matt L. correctly states, I want the Discrete-time Fourier transform rather than the normal Fourier Transform, of course. In my native-language textbook Fourier Transform means both FT for continuous signals and DTFT for discrete signals. I hoped that that was clear from the edit. :)

Answer 1

Your question makes perfect sense, yet your solution will not work. Note that the expression for $x[n]$

$$\begin{align} x[n]=\frac{1}{N}\sum\limits_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}kn} \end{align}$$

is periodic with period $N$. This means that the sum

$$ \frac{1}{N}\sum\limits_{n = -\infty}^{+\infty} e^{-j\Omega n}\sum\limits_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}kn}$$

does not converge. Let's assume that $x[n]$ is defined for $0\le n< N$ and is zero otherwise. Then we get for the discrete-time Fourier transform of $x[n]$

$$X(e^{j\Omega})=\sum_{n=0}^{N-1}x[n]e^{-jn\Omega}= \frac{1}{N}\sum_{n=0}^{N-1}\sum\limits_{k=0}^{N-1}X[k]e^{j\frac{2\pi}{N}kn}e^{-jn\Omega}\tag{1}$$

This formula actually works and can be used if only the DFT $X[k]$ is given. Of course this only works for finite length signals $x[n]$. Equation (1) simply means that the DFT contains the complete information about the (finite-length) signal $x[n]$ and any frequency point $\Omega$ of the DTFT can be evaluated using the discrete values $X[k]$.

Answer 2

There are an infinite number of continuous waveforms that have the same DFT after sampling, which then have different FTs. So you need to specify more. If some continuous signal you want is strictly band-limited, then Sinc interpolation will give you one smaller family of possible solutions from which you might be able to derive an FT.

Answer 3

If the frequency, f, is an integer multiple of SampleRate, fs, divided by N than it's simply the X[f/fs]. If not you have to do an sin(x)/x interpolation (see http://en.wikipedia.org/wiki/Whittaker%E2%80%93Shannon_interpolation_formula)