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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?

  1. 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.

  1. $X[k]$ 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.

  1. $X[k]$ 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?

  1. 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.

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  1. The first $5000$ points (well, $5001$ to be exact) contain all frequency information, because the signal is real-valued and, consequently, the spectrum is conjugate symmetric: $$X[k]=X^*[N-k]\tag{1}$$ where $N$ is the DFT length. But from $(1)$ we cannot infer all values of $X[k]$ from any arbitrary collection of $N/2$ (consecutive) points of $X[k]$. So you're right.

  2. You got that right.

  3. This is a bit tricky, and you'll need Eq. $(1)$. Think about which values of $X[k]$ you can derive from the index range $k\in[8001,9999]$. It should turn out that in fact you got everything you need.

  4. They should specify what they actually mean by $\pi/2$. I would expect this to be a normalized frequency in radians, i.e., $2\pi f/f_s$, where $f_s$ is the sampling frequency. If that is the case, then $\pi/2$ is just half the Nyquist frequency. Draw your own conclusions.

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    $\begingroup$ Thank you for the wonderful answer. I didn't know about the conjugate symmetry. Upon reflecting on it, this is what I found for statement 3: $X[k] = X[N-k] $ where the signal being real-valued played the key role. Now $X[8001] = X [1999]$ which corresponds to $60$ Hz. Similarly, $X[9999] = X [1]$ which corresponds to almost 0 Hz. Putting these together, 0-60 Hz frequency range is covered, which contains the entire spectrum for all bands!! $\endgroup$ – user54853 Dec 28 '20 at 20:18
  • $\begingroup$ @user54853: You're right, but don't forget the complex conjugation, which I denoted by $*$, so in order to get $X[1]$ from $X[9999]$ you need to take the complex conjugate: $X[1]=X^*[9999]$. $\endgroup$ – Matt L. Dec 28 '20 at 22:10
  • $\begingroup$ Yes you're right. I should have taken the conjugate. However that hardly changes anything about the answer to the problem; the entire band can still be decided from the given indices. $\endgroup$ – user54853 Dec 29 '20 at 6:33

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