# Given a discrete time signal, what is the sequence of possible frequencies I can get from DTFT?

I know that when I have a discrete time signal, let's say:

$$f(t_n),\;t_n=\dfrac{n}{F_s}$$

The definition of the DTFT is given by:

$$F_n(\omega)=\sum_{n=0}^{N-1}f(t_n)\cdot e^{-i\omega_nt_n}$$

Now, my question is regarding $$\omega_n$$.

I know the frequencies will be discret because we can't measure lower frequencies than the difference between $$t(n+1)$$ and $$t(n)$$.

However, I am not clear how exactly this works, and how can one determine the "frequency vector". I am not sure I got the expression for $$t(n)$$ correct.

I am sorry this question isn't 100% clear, it's because I'm trying to make heads and tails from this definition and articles online are not very clear.

Thank you.

The discrete-time Fourier transform (DTFT) is defined by

$$F(\omega)=\sum_{n=-\infty}^{\infty}f[n]e^{-jn\omega}\tag{1}$$

Note that this is the common definition, and it is different from the one in your question. First of all, the sequence $f[n]$ doesn't need to be of finite length. Second, the frequency variable $\omega$ is a continuous variable. Your formula can be derived from $(1)$ by assuming that $f[n]$ has a finite length, and by computing the DTFT on a finite set of discrete (angular) frequencies. (Note that the frequency variable $\omega$ in $(1)$ is normalized by the sampling frequency, unlike the discrete frequencies $\omega_n$ in your question.)

Another related transform is the discrete Fourier transform (DFT):

$$\tilde{F}[k]=\sum_{n=0}^{N-1}f[n]e^{-j2\pi kn/N},\qquad 0\le k<N\tag{2}$$

which, for finite length sequences of length $N$, is just a sampled version of the DTFT:

$$\tilde{F}[k]=F\left(\omega_k\right),\qquad \omega_k=\frac{2\pi k}{N}$$

Unlike the DTFT, the DFT can be computed efficiently. Efficient algorithms for computing the DFT are referred to as Fast Fourier transform (FFT) algorithms.