I want to know that why large laplacian spatial filtering is done on EEG signals? I tried a lot but could not find material on LLSF. And what is meant by spatial filter ( in context of Digital Signal Processing)? Why is it called "Large" and "laplacian"?



It is a 2nd derivative of the recorded voltage for each electrode. Suppose the recordings at position $(i,j)$ is $V(i,j)$, then Laplacian operation at $(i,j)$ is

$(V(i,j)-2V(i-1,j)+V(i-2,j))^2 + (V(i,j)-2V(i,j-1)+V(i,j-2))^2$

And it serves as a high-pass filter that enhances localized activity while suppresses the diffusion activity. The implementation of the Laplacian in EEG filtering on the voltage at each electrode is to subtract the weighted voltages from the surrounding electrodes from the voltage recording at current electrode, where the weight is electrode distance dependent.


The difference between large and small is determined by how you select the electrodes. For the small Laplacian, you select the nearest-neighbor electrodes, while for the large Laplacian, they are the next-nearest-neighbor electrodes. See the figure below. If the signal you are interested in is less highly localized and/or varies in exact location over time, the large Laplacian is expected.

enter image description here

Read the paper Spatial filter selection for EEG-based communication by Dennis J. McFarland et all for more details.

  • $\begingroup$ Thanks. Please tell me what is meant by "spatial" filter in context of dsp? And what is meant by second derivative of voltage, in a layman language? $\endgroup$ Feb 5 '14 at 17:20
  • $\begingroup$ @MuhammadAkmal Thank you Muhammad. Spatial means the positions (coordinates) that you obtain the signal, here refers to the electrodes distribution. Please see my update in answer for the laplacian operator definition. Thanks $\endgroup$
    – lennon310
    Feb 5 '14 at 17:40
  • $\begingroup$ you are very welcome! $\endgroup$
    – lennon310
    Feb 5 '14 at 17:52
  • $\begingroup$ The equation which you wrote for laplacian operator, Does that defines second derivative? What does the second derivative do? $\endgroup$ Feb 5 '14 at 18:16
  • $\begingroup$ That is the definition, and also the digital implement of second derivative: v''(t)=d(v'(t))/dt=d(v(t)-v(t-1))/dt^2 $\endgroup$
    – lennon310
    Feb 5 '14 at 18:19

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