Reposting from https://physics.stackexchange.com/questions/736554/how-to-calculate-voltage-change-across-electrodes-in-tank-of-water because apparently this is an engineering question and not an experimental physics question.
Apologies for the basic question. I am a neuroscience student and not that familiar with physics.
I am delivering a 1 V bipolar sinusoidal stimulus at 842Hz to a tank of deionized water at 26C with two electrodes for one minute. I am also recording from two electrodes with a sampling rate of 20kHz inside that tank approximately 0.3 m away. I would like to calculate the voltage recorded by the recording electrodes that is due to the sinusoidal stimulus as a function of time. However, I am not sure how to go about doing this. I would greatly appreciate any help in calculating this.
My purpose for this is to separate my electrical stimulus from the electric organ discharge produced by a black ghost knifefish in the tank. My recording electrodes pick up both the discharge produced by the fish and the output of my stimulating electrodes. I have chosen the frequency of my electrical stimulus to be within 1-2Hz of the frequency produced by the fish, in hopes of observing a jamming avoidance response by the fish (i.e. the fish shifting its discharge frequency away from the stimulus frequency). This makes it hard to distinguish my stimulus from the fish's discharge in the spectrum view of my data (i.e. after Fourier transforming). Therefore it would be very helpful if I could calculate the voltage recorded by the recording electrodes in response to the stimulus voltage, given the stimulus voltage as a function of time. Then I would be able to subtract out this voltage from my data to get the discharge from the fish. I have recordings from when there is no stimulus, and by merely eyeballing my data I can see that the voltage change in the recording electrodes in response to the stimulus is somewhere around 3-4% of the stimulus voltage. But it would be great if there were a more methodical way to do this.
If more information is needed, let me know and I can try to provide it.