I've got the filter that's given in the picture below, they calculate the pole and the zero of the transfer function but what do these two values for omega say, is it kind of a cut-off frequency? (kantelpunt is Dutch and if I translate it literally it means 'tipping point')
Notice that, for the omega:
- $$|\text{s}_{\text{z}}|=\omega_{\text{z}}=4\cdot10^4\text{rad/s}$$
- $$|\text{s}_{\text{p}}|=\omega_{\text{p}}=8\cdot10^3\text{rad/s}$$
Best Answer
First note that R2 is smaller than R1, for this reason we can first ignore R2 to get an idea what the circuit does.
For very low frequencies the impedance of the capacitor is larger than the resistance R2. R1 and C form a low pass filter, the corner frequency is given by the first frequency (the pole).
For very high frequencies the capacitor is a short, the circuit acts approximately as an ohmic voltage divider.
The point where the low pass behavior turns into voltage divider behavior is given by the second frequency (the zero).
The behavior is shown in the plot below.