Feature Selection by Filtering so there is this paper I'm reading and trying to understand fully: Towards Practical Identification of HF RFID Devices
http://dl.acm.org/citation.cfm?id=2240276.2240278

I don't want to link the PDF here, but maybe you can catch it from ACM or somewhere else.

Now what we need to know: They record a signal for $5 \mu s$ with 4 GS/s. From the normalized amplitude of the signal $l$ at time $t$ called $f(t,l)$ they obtain $F(\omega, l)$ by FT.

$$
F(\omega, l) = \frac{1}{\sqrt{M}} \sum_{m=0}^{M-1} f(t,l) \exp (-2 \pi i \frac{t\omega}{M})
$$
Where M is the length of signal considered and $0 \leq t \leq M − 1$ is time. Now quoting:


  We then remove from $F(\omega, l)$ its DC component and the redundant part of the spectrum; 
  we denote the remaining part of the spectrum by $\vec{s}_{l}$.


So now for the spectrum between 0 to 100 MHz the vector $\vec{s}_{l}$ contains amplitudes for 20000 frequencies $(D=20000)$. To reduce the dimensions the authors filter the signal with a bandpass.


  A projected vector $\vec{b}_{l}$, also called filtered features, is extracted 
  from the basic features $\vec{s}_{l}$ using a bandpass filter transformation $W_{BPF}$:
  $$
\vec{b}_{l} = W_{BPF}^{t} \vec{s}_{l}
$$
  The feature extraction from N captured responses for a given RFID device is 
  then given by $G = W_{BPF}^{t}S$ where $G$ is an array of  $\vec{b}_{l}$ and 
  $S = [\vec{s}_{0} \dots \vec{s}_{l} \dots \vec{s}_{N}]$ is a matrix that
  contains the basic features of N captured responses.
  
  In our implementation, the transformation $W_{BPF}$ consisted of a Chebyshev Type I bandpass filter with filter order of 100 [Williams and Taylors 1988]. The filter was used to keep the frequencies in a target frequency range enclosed between the upper and lower bandpass edge frequencies, denoted by Fp1 and Fp2 respectively; the latter were experimentally determined. We chose the filter order to be high for better bandpass filtering quality. Using this technique the basic feature dimensionality can be effectively reduced to only tens of frequencies (dimensions) comprised between Fp1 and Fp2.


Later on in the paper they state the following:


  Therefore, we applied feature selection by filtering with parameters Fp1 = 11 MHz and Fp2 = 15 MHz. This procedure reduced the features (fingerprint) dimensionality to D = 40.


And now my question (lately).


Why is the dimensionality now 40? If we have amplitudes of 20000 frequencies between 0-100 MHz and filter from 11 to 15 MHz shouldn't there be 800 left?
How does this bandpass matrix $W_{BPF}^{t}$ look like?


Thanks for reading and greetings,

para
The problem is that the following statement- "So now for the spectrum between 0 to 100 MHz the vector s⃗ l contains amplitudes for 20000 frequencies (D=20000)"- is faulty.

A signal that lasts $5\mu s$ is captured at 4 GS/s.  That is a total of 5e-6 * 4e9 = 20000 samples.  The Nyquist frequency is half the sample rate, or 2 GHz.  They take the Fourier transform of the samples ($F(w,l)$).  At this point, assuming the samples are real (as opposed to complex), there are 20000 frequency bins representing -2 GHz to +2 GHz.  The negative frequency information is redundant since it mirrors the positive frequencies.  I believe that when they said "We then remove from F(ω,l) its DC component and the redundant part of the spectrum" they were talking about getting rid of the 0 Hz and negative frequency information.

That would leave you with 9999 frequency bins representing 200 kHz to 2 GHz.  To make life easier, though, let's bring back the DC bin and say that we have 10000 frequency bins representing 0 Hz to 2 GHz.

Hopefully it is clear now why "So now for the spectrum between 0 to 100 MHz the vector s⃗ l contains amplitudes for 20000 frequencies (D=20000)" is faulty.  There are 10000 bins and they represent 0 to 2 GHz, not 0 to 100 MHz.  They now bandpass the signal down to 4 MHz bandwidth.  $(4e6 / 2e9) * 10000 = 20$ frequency bins.  I believe they get 40 dimensions from those 20 bins by counting both the amplitude and phase of the bins as features, $20 * 2 = 40$.