I don't get it. I have the following form of the DFT:
$$ Y_N(e^{j\omega_n})=\sum_{k=1}^{N-1}y(k)e^{j\omega_n k}\quad\omega_n=\frac{2\pi n}{N}\quad n=0,1,...,N-1 $$
But this assumes that the sampling time $T_s=1$ (1 second). I have a signal, however, that was sampled at 50 Hz so $T_s=0.02$. How does that get taken into account?
My problem is that I fit a transfer function to the DFT by first writing the continuous form and then discretizing with a sampling time of 0.02 seconds. But the result is the output being noticeably smaller amplitude than in reality:
The confusing part is that the transfer function itself, sampled at 50 Hz, fits the ETFE perfectly (the first subplot). But the output is smaller amplitude. If I sample the fitted transfer function at 1 Hz the problem is fixed... but of course in reality I should be sampling at 50 Hz. What's wrong here? Thank you!
I answered my own question. It was a MATLAB/conceptual problem.
Here's the trick. When we sample at 1 Hz, we can use:
Where $\omega_n=0,...,2\pi$. However, when we sample at frequency $1/T_s$ this is no longer the correct command, we must make sure that the frequencies go $\omega_n=0,...,2\pi/T_s$. Therefore we must write:
This solves the problem.