Why does EOG (eye EEG) first spike and then immediately fall back down when you look to one direction instead of staying spiked and only falling back down when your eyes go back to neutral?

From what I understand the corner (front of eye) is positive while the retina (back of eye) is negative. So if I have electrodes on each temple and I look left I would expect the EOG graph to spike (or dip depending on the electrode placement) and stay up until I move my positive cornea away from the left electrode.

Video explaining EOG

But instead the graph spikes and then dips back to neutral even though my positively charged cornea is still close to the left electrode since I have not returned my eye back to neutral.

To be clear this graph is correct I just don't understand why.

Ex: In the picture below I am holding my eyes left until I look right. But for some reason the graph immediately drops back to neutral even though I'm still looking left.

Ex 2: Here's a video of a live EOG recording and a person moving their eyes showing the same graph behavior.

enter image description here

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    $\begingroup$ The screenshot is mine using Backyard Brains Spike Recorder but here's a video demo from the company that makes the device and software I used. You can see really clearly the same pattern. youtube.com/watch?v=7eOkTvC7UNc&t=2s I also see a similar peak and drop pattern in the lab I volunteer at. $\endgroup$ Commented Jul 7, 2022 at 3:59
  • $\begingroup$ Welcome. Can you add the duration of the gaze shift in your question, and preferably with a time-synced horizontal line [or something] in your EOG trace? Add a time axis as well. $\endgroup$
    – AliceD
    Commented Jul 7, 2022 at 7:35
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    $\begingroup$ Hey @AliceD I instead added a link to a video recording showing a person moving their eyes and the live EOG. The graph isn't particular to one situation. It's expected with every EOG. So the details of my graph don't seem relavent. Does the video help clarify? $\endgroup$ Commented Jul 7, 2022 at 19:16

2 Answers 2


If you had an EOG signal in a perfect world, it would show constant deflections based on how far the eyes are directed left/right.

In the real world, measuring any electrical signal is quite difficult; there are lots of interfering signals.

My best guess, without further methodological details, is that the equipment here is using a high-pass filter (almost certainly a band-pass filter, but the high-pass part is the part you're asking about; the low-pass part is the reason the deflections are so smooth and slow). This removes slow drift that you expect to find when you record electrical signals and for a static signal will give you a nice, constant "0 V". However, if you hold your eyes at some deflection, that gets filtered out, too. Here's what high-pass filtering does to a square-wave signal:

enter image description here

The input signal, in red, steps up to 5V, but then stays constant; the high-pass filter returns this constant deflection towards 0V (blue trace). Then when the input steps back down to 0V, the output is a negative deflection, which again returns towards 0V as the input stays constant. In other words, a high-pass filter turns the circuit into a change detector.

This paper:

Ryu, J., Lee, M., & Kim, D. H. (2019). EOG-based eye tracking protocol using baseline drift removal algorithm for long-term eye movement detection. Expert Systems with Applications, 131, 275-287.

discusses some of the types of baseline drift you might expect (figure 2) vs. how the signal would look in an ideal world (figure 1):

EOG samples

They propose a particular algorithm to help deal with this, but in any event, it seems the device you're looking at is suitable for a "cool, I can detect the movement of the eyes!"-type home experiment; it's not suitable without substantial upgrades to processing the signal to actually identify the current position of the eyes.

It looks like the directions posted here recommend using a bandpass filter from 1-100 Hz. A constant signal is a "DC (direct current)" or "0 Hz" signal, so it's not in the passband and gets filtered out.


I captured the same signals in 1974 during Biomedic Engineering demonstrations during a festival at U of M.

I attributed that to [galvanic skin response][1] due to progressive differential temple pressures on the electrodes due to eye muscle tension. Given a constant charge Q = CV in each electrode as the differential pressure changes so does the capacitance change incrementally and the voltage decreases incrementally (the absolute value does not change polarity, just the delta V). By watching girls walking left and others right, I was able to make a linear triangle waveform. Applying slight pressure had the same effect.

With a slow AC response (0.1Hz) one might be able possibly to calibrate it in vertical and horizontal directions with 4 electrodes. If there was a need one might possibly find a useful prosthetic use for quadriplegics.

Your experimental equipment must have been a faster return to 0V such as a 1Hz cutoff. A step motion 10 to 90% fall time indicates your cutoff. , Tr = 0.35/ f-3dB is an easy way to estimate bandwidth for either LPF or HPF. [1]: https://www.ihtbio.com/galvanic-skin-response-gsr/

  • $\begingroup$ I wish who ever downvoted you would have commented why they thought this was a poor answer. We are now left wondering.... $\endgroup$ Commented Aug 7, 2022 at 3:18
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    $\begingroup$ My galvanic skin response indicates it's a troll lacking the spine to justify their negativity. There is no other explanation. $\endgroup$ Commented Aug 7, 2022 at 13:19

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