I have drawn two circuits that I believe behave similar. Do they both have the same cut-off frequency? Is there any pro/cons of the two circuits, except the one is inverting and the other is not?
Schematic 1 (R = Rf):
simulate this circuit – Schematic created using CircuitLab
Schematic 2:
Yes, they have the same cutoff frequency. In the first, since the inverting input is held essentially to ground, the current through the resistor is essentially the same as in the second. The output voltage is then the current times the feedback resistor. In the second, the voltage is the current times the input resistor.
As to pro's and con's, the first offers the ability to change the gain without changing the frequency response, although the inversion is equivalent to a phase shift of 180 degrees, while the second has a gain fixed at 1 for low frequencies. If a gain of 1 is acceptable, or the phase shift of the first is not, then the second circuit is simpler and more reliable.
Both circuits appear the same to whatever is driving the input, and have the same high pass frequency rolloff. As you say, the main difference is that the first inverts while the second doesn't.
Even in a application like audio, for example, where you don't usually care about a inversion, there are practical differences between these two circuits.
The first circuit can be easily expanded to sum a number of signals. Each only needs its own impedance feeding into the inverting input of the opamp.
The second circuit is often used to exploit the very high input impedance of the opamp. In your case you used the same capacitor and resistor on the input, but this impedance could have been much higher in the second. That can be useful when dealing with high impedance signals that you don't want to load much. This circuit can also be configured for gain over 1 by adding a resistor divider in the feedback path.
