Q1.1

The values R1: 1 kΩ and R2: 3.3 kΩ should result in a ratio pretty close to 0.75: $\frac{R2}{R1 + R2} = \frac{3.3}{4.3} = 0.77$.

Alternatively, 3.3 kΩ and 10 kΩ also result in a ratio close to 0.75.

Q1.2 & Q1.3

The measured signal amplitude is 0.5 V.

The input impedance of the oscilloscope must be ~1 MΩ because it forms a resistive divider with the ~1 MΩ resistor on the breadboard that results in a ratio of 1/2.

Q1.4 a & b

From the lecture we know that the the signal will be attenuated by a factor of $R_{in}/(R_{in} + R_{out})$. In the case of a 100 kΩ electrode and the oscilloscope (with an $R_{in}$ of ~1 MΩ), this attenuation factor will be ~0.9. This is not acceptable as our measurement would be off by 10%.

In contrast, the attenuation factor for the Arduino output would be 0.99999 (using the same formula). So the measurement would be essentially perfect.

Q1.5

Circuit 1 attenuates high frequencies and circuit 2 attenuates low frequencies. This is because the capacitor has low impedance for high frequencies. In circuit 1, this makes a very low divider ratio for high frequencies.

Q1.6

The phase of the filtered waveforms also shifts for some frequencies.

Q1.7

The capacitively coupled noise shows up as spikes at a regular interval of 5 ms (200 Hz). The spike amplitude (peak-to-peak) is ~350 mV for my test conditions. (Note that the noise is sensitive to experimental conditions and may show up differently for others.)

Q1.8

The noise amplitude decreases slightly to ~250 mV. By increasing the distance between the two wires, we have reduced the capacitance between them, and thus reduced the capacitive coupling.