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I was told that the bottom half of the circuit wouldn't matter due to having a corner frequency being significantly higher compared to the top half, or due to time constant (somewhere along this line..). Can you please help me understand why?

How would I go about reducing this to a first order low pass filter? (just one resister, one capacitor).

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Thanks in advance.

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    \$\begingroup\$ Hi! I couldn't figure this out. What is your end goal? If you only want to understand why the oscilloscope doesn't load the function generator, some simulation tool might be helpful. If you need an equivalent circuit, you might want to check en.wikipedia.org/wiki/Miller_theorem and en.wikipedia.org/wiki/Y-%CE%94_transform. You can also plot the transfer function without the simples L network. \$\endgroup\$ – user24368 Sep 15 '19 at 6:30
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The simple answer is to recognise that C1/R1 and 1Meg/13pf is a special case, given the application : it is almost certainly a 10:1 oscilloscope probe.

1 Meg and some small C is a standard scope channel input : the precise value of C is undefined but happens to be 13 pF fro this scope.

Then R1 is 9 Megohms to give the correct DC attenuation, and C1 is adjusted for a flat frequency response (actually, an accurate square wave response) each time you plug that particular 10:1 probe into that particular scope. (Calibrating the probes is just a routine part of using a scope. The scope will provide a "CAL" output for that purpose, and the correct plastic trimming tool lives in that pouch on top of the scope.)

The probe output is then a perfectly scaled version of the probe input, with the same frequency response. Which means you can model the entire circled block as Cp in parallel with 1.3 pF and 10 megohms (followed by a perfect 20dB attenuator)

Given the source impedance of 50R, you can for all practical purposes ignore the 10 megohm probe resistance, giving an attenuator composed of 50R and (Cp + 1.3 pF) to which you now have to add the impedance of the cable.

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