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

Question

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.

Answer 1

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.

If this doesn't clear up your confusion, leave a comment below.