And in this case let's say we know everything but the value of C2, and we want to set C2 to a value to obtain the right low-pass filter effect. But to calculate the cut-off frequency one needs to know the equivalent resistor value of all the network before C2 (only at the left side of C2 the equivalent resistance matters since the signal is coming from left?).
Only after knowing this approximate equivalent resistance one can set C2 to obtain a desired low-pass filter.
First of all for simplicity the unknown output impedance of the input signal is neglected(even-though it might have a big impact if it is big). Here are the steps I followed to thevenize this circuit.
And now I want to compare the frequency response of the original and thevenized circuits:
My questions are:
The cut-off frequency definition is where the ouput is 3dB less than the output at DC. Following this definition yileds very similar cut off frequencies. The cut-off freq. in both are very close. One is around 7.8Hz the other is around 9.3Hz. I think that's the effect of neglecting C3 at the beginning.
1-) The resistors in the original network drops the input voltage at each stage until the output. But in the Thevenin circuit we have no similar loss. The final Thevenin voltage is 5/32 of the original one. So I would expect a power ratio (5/32)^2 which is around 32dB. But the difference at DC is around 18dB. What could be the reason?
2-) Is my way of using Thevenin here correct for practical rough estimate? And similarly can I use the Thevenin with R7 R9 C3 (ignoring the circuit after C3) to see the high-pass filter effects roughly?