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.

  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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.