Polarity of inputs to do measure (P-N or N-P?)

GEngGEng
in Cyton
Hello guys!
One more doubt: in the datasheet page 22 we can find the following information:
"The differential input voltage (VINxP – VINxN)". So the input differential amplifier measure the P - N. 
And so, if I'm using SRB2, all the P inputs are connected to SRB2 and inputs are now the N pins.

Knowing this I have the following quetion:

If the differential amplifier do P-N, and the P is connected with SRB2, so the resulting signal will be
Reference(P) - Signal(N), instead the habitual Signal - Reference, so the signal will be inverted, right?
Why It is not a problem?

Comments

  • wjcroftwjcroft Mount Shasta, CA
    You are correct, using the SRB2 reference (a positive bus consisting of IN*P pins) and the IN1N, IN2N, IN*N pins as channels, does invert the EEG signal. Most EEG signal processing is based on FFT, or filters, or coherence -- where only frequency or phase alignment matters -- not absolute polarity. In some applications such as P300, polarity is important. In which case the inversion is obvious and easily reversed.

    The default of SRB2 as reference was chosen over SRB1, for a number of technical reasons, having to do with flexibility of settings.

    William

  • GEngGEng
    Hi wjcroft!
    Thanks a lot for your reply!
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