EMG signal is the signal generated by muscle flex. EEG electrodes is the scalp potential generated by neuron firing.

EEG is on the order of micro Volts whereas EMG is on the order of mili Volts, this implies that EMG is 3 order greater than EEG.

The problem is that when doing EEG experiments, some EMG signals are transferred onto the EEG electrodes. The question is that how can the circuit designed to handle signals at the micro Volts capable of withstanding a mili Volt signal. This cannot be done by some sort of power blocker because the EMG signal is still picked up (that's how we know it is 3 order higher).

Imagine if I had a 250 kV power line. Now the voltage on the line swings upward by 3 order in magnitude. Nothing can measure this voltage and survive...but why is the EEG circuit capable of withstanding a surge of 3 order higher?


2 Answers 2


In your hypothetical example of trying to detect a 3VDC signal against a background of 250kVDC noise (say between the phase 1 and phase 2 power lines), imagine the signal is being measured by a well-insulated lineman who can put his DMM's positive lead on the "phase 1" 250.003kVDC wire, and the negative lead on the "phase 2" 250.000kVDC wire. Now this floating DMM can measure the 3VDC difference between the two 250kV wires. Even though the lineman and his DMM are now at 250kV potential w/respect to ground, his DMM is measuring only the difference between its positive and negative leads. The DMM doesn't blow up, because the entire thing is within an acceptable "common mode" range due to voltage isolation.

(I don't want to get sidetracked on the actual details of AC vs DC HV power distribution systems, just giving an idea how differential measurement works.)

ECG/EKG (Electrocardiogram) has a similar problem to EEG (Electroencephalography): common-mode noise exceeds the differential signal by many orders of magnitude. And because the human body is a high-impedance signal source, incidental power-line noise is a big problem. The signal of interest is a very small voltage (mV), riding on top of a much larger noise voltage (several Volts).

Differential amplifiers are the key to extracting the signal from the noise. Multiple electrodes are used, and the signal is measured in the voltage difference between electrodes. As long as the common-mode input range provides enough dynamic range to accommodate the power-line noise and unwanted EMG (muscle) artifacts (which must be presented equally at both the positive and negative inputs), the small differential signal is amplified to a useful level without saturation.

I don't know about EEG (Electroencephalography) specifically, but ECG (Electrocardiogram) defines its bipolar limb leads (lead I, lead II, lead III) in terms of voltage differences between the various electrodes (LL left arm, RA right arm, LL left leg). Even the so-called unipolar leads are actually differential, as their negative reference (Wilson's Central Terminal) is a virtual ground formed from the average of LL, RA, and LL.

For this differential measurement to be possible, there has to be balance. It's critical that both the positive and negative leads have to pick up exactly the same common-mode noise. Any mismatch between the input leads, will cause mismatched common-mode noise, which is indistinguishable from the differential signal. So physical symmetry is important. And both the positive and negative signals have to be within an acceptable common-mode input range (determined by the op-amp's input common-mode range and the input network impedance).

Although your example of a 250kV power line is hypothetical, in a real ECG system they do have to worry about not only measuring differential signals in the mV range, but they also have to be able to withstand hundreds of volts and many joules from an externally applied defibrillator. If the patient has a heart attack, the doctor won't bother to disconnect the poor ECG machine before applying the defibrillator shock to revive the patient. Anyone and anything connected to the patient, will get shocked by the defibrillator. So at least in the medical electronics world, there does have to be some circuitry to protect against high voltage, yet not mess up the measurement of high-impedance, low-voltage signals of interest. Using symmetrical defibrillator protection circuitry on each electrode input, makes the protection circuit's error voltage contribute to the common-mode noise, which is rejected by the differential amplifier.

  • \$\begingroup\$ EMG will not be common mode on EEG \$\endgroup\$ Mar 10, 2015 at 3:06
  • \$\begingroup\$ @ScottSeidman yes, that's apparently one of the things that makes EEG hard to interpret (per Wikipedia). Eye muscles twitch as the eyes saccade. I was thinking more along the lines of the "virtual ground" reference point, WCT of the heart, I imagine there would be a similar common reference node for the brain. I haven't worked on any EEG products yet (though hopefully soon), so it's a bit outside my area at the moment. \$\endgroup\$
    – MarkU
    Mar 10, 2015 at 3:18
  • 1
    \$\begingroup\$ @MarkU Not only do the eye muscles twitch with saccades, but there's the electroretinal potential - leveraged for electro oculography (EOG). Eye movements are horrible EEG contaminants - they generate both slow (EOG) and fast (EMG) signals that are easily picked up by the electrodes on the front of the head. \$\endgroup\$ Mar 10, 2015 at 14:45

EEG is low freq compared to EMG. You can isolate EEG with a high CMRR amp with a small enough gain to not saturate given EMG and other noise, then low pass filter and then amplify more.

This is not perfect, and I suspect a good deal of EEG studies may show EMG contamination, particularly early brain-machine interface studies. In fact, many years ago I appeared on a membership flyer for the Society for Neuroscience in a photo arguing this point in front of a poster at their Annual Meeeting.

If situation permits, you can pull very small signals out of very big noise using ensemble averaging techniques. This is often done for EEG in evoked potential studies.


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