(simulation not needed, I just need to understand logic of how RC filtering works)
Assuming transistor gain=100, then Re(1K) seen from the base = approx 100k (impedance reflection). Therefore total impedance looking into base is approx 10k || 20k || 100k = 6.24k. This can be modelled as the 1uF input capacitor (Cin) in series with 6.24K equivalent resistance. Because of arrangement of Cin and 6.24k, this forms a high-pass filter action. We can find -3dB cut-off by 1/2pi(R)(Cin)=1/2pi(6.24k)(1uF)=25Hz. This I understand and makes sense.
My question:
I was told the output is also acting as a high-pass filter. I don't understand how this is when 1uF Cout capacitor and 1k (Re) resistor are not arranged in high-pass series configuration like at input (which can be visualised easily by drawing equivalent resistance and putting in series with Cin). I was told -3dB cut-off for output is calculated as 1/2pi(Re)(Cout) = 1/2pi(1k)(1uF). I don't understand also why 1K Re resistor is used as it's not in series with Cout. Please can someone explain I'm so confused
