# A basic question regarding frequency analysis of an EEG signal

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 Answer

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.

• 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 = X $ which corresponds to $60$ Hz. Similarly, $X = X $ which corresponds to almost 0 Hz. Putting these together, 0-60 Hz frequency range is covered, which contains the entire spectrum for all bands!! – user54853 Dec 28 '20 at 20:18
• @user54853: You're right, but don't forget the complex conjugation, which I denoted by $*$, so in order to get $X$ from $X$ you need to take the complex conjugate: $X=X^*$. – Matt L. Dec 28 '20 at 22:10
• 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. – user54853 Dec 29 '20 at 6:33