# FFT frequency bands and filtering

I don't really understand FFTs, and filtering neither. I hope I can explain my confusion in a way that is not too confusing for you to understand.

consider the following example: I have an EEG signal, and am interested in the frequency bands from 1 to 40 Hz.

• if I apply a LP filter at 40, and a HP at 1, my signal of interest would be distorted around the beginning and end because of the filter's transition band. right? why does it have a transition band at all, and how does it work? is a filter a sum of sine waves in the specified frequency range? (sorry, seems like a basic question, still I couldn't find an answer to it)

• now to get the FFT of different frequency bands in the 1-40Hz range, I could either first bandpass filter the specific ranges and apply FFT to each (e.g. to 1-4,4-8,8-12,12-20). however, this would lead to a distorted result in each band, because of the transition bands, right?

if on the other hand, I do FFT over the whole 1-40 range, and then just 'extract' the results from 8-12 etc, the results would be more accurate? and how would the extraction mathematically work, is it just the mean of all values at the frequencies between 8 and 12?

I hope you can help me understand these things

Regarding your second question on the spectral analysis of the filtered signal through its FFT, you first obtain a long enough signal observation window and then take a dense enough FFT of the filter output, so that you can focus on any region withn 1 Hz to 40 Hz depending on your interest. Assuming that your sampling rate is $F_s$ samples per second, then an $N$-point FFT would provide you an equivalent analog frequency resolution of $$\Delta f = \frac{F_s}{N}$$ where each frequency bin, in Hz, would be located at $$f_k = \frac{F_s}{N} k ~~,~~ \text{for}~ k=0,1,...,N-1$$ Note that FFT of real valued input signal would have conjugate symmetric complex output, so only the first half (roughly) carries relevant information from $f=0$ to $f=F_s/2$