a resistor and capacitor in series can create a filter to let either high frequency AC or low frequency AC to pass.

the high pass filter is connect to the resistor.

the low pass filter is connected to the capacitor.

the reason can be understand when consider the charging and dis-charging process of the capacitor. the charge rate is equal to 1 / RC . For a low frequency, the time vary slowly, and the capacitor can charged up and equal to to supply voltage when t >= RC or \omega <= 1/ RC . thus, the voltage across the capacitor can always equal to the supply and the voltage across the resistor is always zero. thus, low pass filter connected to capacitor.

when high frequency, time vary fast, and the capacitor cannot charge up and oppose the supply voltage. thus, current always flow in an out and make the resistor voltage follow the supply voltage. thus, for frequency \omega >= 1/RC , only high  frequency can “live” in the resistor and therefore the high pass filter is connected to the resistor.

in mathematic, one can use impedance for calculation, that i don’t show in here.

to calculate the output of an OpAmp, just need to remember 3 things:

  1. the open loop amplification is very large.  G \rightarrow \infty
  2. the impedance of the input is very large, no current flow between V_+ and V_-
  3. the output is the different of the input times amplification.  V_{out} = G (V_+ - V_-)

and OpAmp can be modified and only let high or low frequency to be amplified.

under the Inverting amplifing, such that V+ in grounded.

add a parallel capacitor with the Rf give you a low pass filter.

add a series capacitor with the Rin give you a high pass filter.