## 4 Discussion
### Reduction of Power Line Interference
For example, even if the current is as low as 0.1 \(\mu\)A, and the two electrodes have an impedance of only 5 k\(\Omega\), the differential 50 Hz signal on the bipolar input will be as large as 1 mV, which is a considerable amount of disturbance with the amplitude even larger than that of the ECG signal. In contrast, Figure 11B shows the capacitive coupling model of the Active-Shield method. In this case, the signal of the inner wire is fed back to the shield using a low impedance output of an amplifier. Note that there is still capacitive coupling between the mains power and the outer shields of both electrodes. Based on the Figure 11B, the equivalent circuit of the amplifier, as well as the capacitive coupling, is shown in Figure 12a below (also in the revised manuscript) to examine whether the capacitive coupling between the mains power and the shields would introduce significant flux to the loop for the Active-Shield method. The amplifier that drives the outer shield could be equalized to a high input impedance \(Z_{\text{in}}\), a low output impedance \(Z_{0}\) (usually less than 100 \(\Omega\)), and a voltage source with the same voltage with the input \(\text{V}_{\text{i}}\) (in reference to the isolated ground of the power supply). The output voltage \(\text{V}_{\text{o}}\) was composed of two independent components \(\text{V}_{1}\) and \(\text{V}_{2}\), where \(\text{V}_{1}\) is attributed to the voltage source \(\text{V}_{\text{i}}\) (voltage follower) and \(\text{V}_{2}\) is caused by the capacitive coupling between the mains power and the outer shield. When considering the second part \(\text{V}_{2}\), the voltage source could be regarded as a short circuit according to Thevenin's Theorem and therefore the equivalent circuit is shown in Figure 12b, which is actually a voltage divider circuit between the mains power voltage \(\text{V}_{\text{pow}}\) and the isolated ground, with a series circuit by \(Z_{1}\) and \(Z_{0}\). According to the voltage dividing rule, the \(\text{V}_{2}\) voltage could be written as:  
\[\text{V}_{2}=\frac{Z_{0}}{Z_{0}+Z_{1}}(\text{V}_{\text{pow}}-\text{V}_{ \text{isognd}})\]  
Figure 11: The capacitive coupling model for the power line interference for GND-Shield and Active-Shield methods. Given that \(Z_{1}\) is the coupling impedance between the power line and the shield and could be a few hundred M\(\Omega\) or even larger, whereas the amplifier output impedance \(Z_{0}\) is usually less than 100 \(\Omega\), the corresponding V\({}_{2}\) is usually a few dozen \(\mu\)V or even less in reference to the isolated ground. That is to say, the power line only introduces an interference V\({}_{2}\) of a few \(\mu\)V for the Active-Shield method, because of the low output impedance of the amplifier. Similar analysis could be applied to the other electrode of the Active-Shield method, and it could also be concluded that the presence of the power line only introduces a few dozen \(\mu\)V (or less) of interference V\({}_{2}^{\prime}\) to the other outer shield. Although the potential of the two shields are different for the Active-Shield method, we can see that the actual potential difference (V\({}_{2}-\)V\({}_{2}^{\prime}\)) caused by the power line interference is quite limited, since both V\({}_{2}\) and V\({}_{2}^{\prime}\) are the order of \(\mu\)V and their difference is much smaller. On the other hand, since the impedance \(Z_{\rm L}\) between the outer shield and the inner core wire is usually quite high (in the order of T\(\Omega\)) and the other ends of both shields are open (not connected to human body or electrodes), little leakage current would flow from the outer shield to the inner wire or elsewhere, leading to negligible effects of the power line induced V\({}_{2}\) on the voltage of the inner wire (Figure 12). Therefore, the differential voltage between the two inner core wires introduced by the power line interference is also negligible.