Assume that an EEG signal is sampled at $f_s = 300$ Hz then a 10000-point segment of it is selected, called $x[n]$. The corresponding 10000-point DFT is then computed and called $X[k]$.
Assume further that famous EEG frequency bands are known as the following intervals (in Hz): $ \delta=[1,4],$ $ \theta = [4,7],$ $\alpha = [7,12],$ $ \beta = [12,30]$.
Please help me decide which of the following 4 statements are true, and which are false?
- Any 5000-point of X[k] is enough to obtain the entire frequency information.
My attempt: According to the famed Nyquist criterion, the maximum frequency we can have is $f_m = f_s / 2 = 150$Hz. The corresponding index $k$ can be found this way:
$$ \frac{k f_s}{N} = f_m = \frac{f_s}{2} \rightarrow k= \frac{N}{2} = \frac{10000}{2} = 5000.$$
This implies that the maximum frequency happens at the 5000'th frequency index. But it doesn't mean any 5000-point segment gives the frequency information so this statement is incorrect.
- $X[k]$ for 1000<k<1500 is enough to get the $\alpha$ band information.
My attempt: At $k=1000$ we get $$ f = \frac{ k f_s }{ N } = \frac{ 1000 \times 300 }{ 10000 } = 30 \textit{Hz}$$ and at $k= 1500$ we have $$ f=\frac{ 1500\times 300 }{10000 } = 45 \textit{Hz}$$ which are out of the $\alpha$ band range, so this statement is incorrect as well.
- $X[k]$ for 8000<k<10000 is enough to get the $\alpha,\beta,\delta, \theta$ bands information.
My attempt: For the very same seasons just discussed, $k=8000$ corresponds to $f= 240$ Hz and $k= 10000$ corresponds to $f= 300$ Hz which are out of all bands' ranges. As a matter of fact, 240 and 300 Hz are above the maximum frequency $f_m$ allowed by Nyquist's criterion, so I don't really know how to interpret this. Maybe I'm mistaken?
- A lowpass filter with frequency $ \frac{\pi}{2}$ removes power line frequency at $60$ Hz.
My attempt: I think this is trivially correct because $\frac{\pi}{2}$ Hz is about 1.57 Hz and the lowpass filter zeros out anything above this frequency, including 50 Hz.