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As defined in this post, the monopolar EMG can reduce the common mode noise by taking the difference of the two inputs, i.e. EMG_mono = k * (V1 - V2). And, for bipolar mode,EMG_bi = k3 * [k1 * (V1 - Vref) - k2 * (V2 - Vref)].

To simplify the equations, we assume k1 = k2 = k3 = 1. Then monopolar is simplified as: EMG_mono = V1 - V2, and bipolar is simplified as EMG_bi = (V1 -Vref) - (V2 - Vref) = V1 -V2. That is to say the overall effect is the same for both monopolar and bipolar modes.

The question is: why is bipolar better than monopolar at reducing (common mode) noises?

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  • \$\begingroup\$ Does bipolar use TWO wires plus Ground? \$\endgroup\$ – analogsystemsrf Nov 24 '18 at 2:32
  • \$\begingroup\$ Yes, bipolar as indicated by its name, uses two electrodes plus a ground. \$\endgroup\$ – Ray Nov 26 '18 at 10:24
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According to the link you provided, as far as my understanding goes about it, you have interpreted it wrong.


EMG_mono = k*(V1 - Vref) and EMG_bi = k*((V1 - V2) - Vref)


As explained in various answers here the bipolar electrodes further help to reduce the common noise in the two electrodes (which is basically the line noise picked up by the body), which the monopolar fails to do so since the electrodes are far apart and the (picked up signal) noise may be different which degrades the SNR. Since the bipolar electrodes are closer to each other (1 - 2 cm apart) the noise running on both the electrodes is approximately the same, and hence can be effectively removed before subtracting it with Vref.

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  • \$\begingroup\$ Thank you for the answer. Based on your answer, my understanding is that if the two electrodes are placed closer, the monopolar can also give better performance, since the noises contained in V1 and Vref will become similar. Is that so? \$\endgroup\$ – Ray Nov 26 '18 at 10:29
  • \$\begingroup\$ @Ray Sorry for the late reply, I have been busy. As far as my understanding goes, the Vref of monopolar readings is usually ground, hence the noise which is picked up by the body as some potential w.r.t. ground. Bipolar electrodes both measure the signal w.r.t. ground(Vref) but since there is a greater common voltage(noise) in them it is removed in the process. Hence bipolar is better. \$\endgroup\$ – Shreyas Nov 28 '18 at 22:29
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The idea is that you arrange your lead such that the same common mode noise is on both, then an amp with good CMRR kills the noise. It's not that the noise is reduced -- it's that it's the same on all leads

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  • \$\begingroup\$ Thank you first. If the noised can be killed by arranging leads of the monopolar EMG, why do people still use bipolar? What on earth makes bipolar better than monopolar? \$\endgroup\$ – Ray Nov 26 '18 at 10:33
  • \$\begingroup\$ It can't be killed. @Ray. It can be made the same, and then canceled by subtraction if you have a diiferential pair. No bipolar, no subtraction. \$\endgroup\$ – Scott Seidman Nov 26 '18 at 11:24
  • \$\begingroup\$ Better explained, the noise becomes common mode noise \$\endgroup\$ – Scott Seidman Nov 26 '18 at 11:39

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