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Not sure by what you mean by impedance wise.Well it does confuse the issue (for me)
I admit the "virtual ground" at the op-amp input is a mystery to me.
I know it is related to the negative feedback etc. and (for me) a lot of math but what I have trouble with is even if we are feeding a bjt common emitter stage, if the cap is large enough,
whatever is on the input side, impedance wize, it's the output side of the cap that will determine if the circuits will behave differently.
Do you follow me? Am I wrong?
But to try to clarify, a virtual ground point means that the apparent impedance at that point to ground is essentially zero ohms. Thus you really need to look at the input and output currents to determine the circuit behavior. For example, if you have a 1kΩ input resistor going to a virtual ground input on the op amp, then a +1V input will give +1ma through the resistor to the virtual ground. The op amp absorbs this current by the negative feedback from the output to the virtual ground. Thus, in this example, if you had a 10k ohm resistor from the output to the virtual ground, then the op amp output will be at -10V (inverting gain of 10) to sink the 1ma of current from the virtual ground and keep the virtual ground at zero volts.
Of course, if you add capacitors or other impedances into the circuit, then you have to account for their effect at the frequency of interest.
Does that make any more sense?