To fit this in with what you have been studying, maybe you could show the relevant formulas.
I think the 0.1 uF is a bit too large. Try 0.01 uF with a series 500 ohm resistor. I chose that to have 3 times as much reactance as the resistance of the series resistor at 10 KHz. This way I could retain 75% of the incoming signal as output.
The series resistor produces a
voltage divider with the capacitor, so that you can reject most of the 1 MHz component while retaining as much as possible of the 10000 KHz component.
A 0.01 uF cap has a reactance of 16 ohms at 1 MHz and 1592 ohms at 10 KHz so it should give reasonable rejection of the 1 MHz component while retaining most of the 10 KHz component.
This is a series RC circuit, so you have to treat it like in AC theory, with complex numbers, but you can see that the 1 MHz component would be greatly reduced compared with the 10 KHz one. It works out that 95 % of the 10 KHz signal is retained while only less than 1 % of the 1 MHz signal is retained.
The resistor across the
capacitor has to discharge the capacitor in the period of the 10 KHz modulation component. Otherwise the circuit would be working as a peak voltage detector.
So, if R * C = 1 / 10000 then R = ? (don't forget the R is in Mohms if C is in uF)
