So, imagine signal propagating through the 50 Ohm coax cable (TXLine) on which end there is simple transistor amplifier with really high input impedance. Unmatched, as you can imagine. What I know is that reflection is gonna happen on this end and signal will return to the TXLine. What will I see on the output of my amplifier? The amplified signal or nothing?
Please review my previous post. If you assume infinite amplifier input impedance, then the amplifier needs no input energy, it's just probing the voltage at the line end.
Thanks, this makes a lot of sense. But then again, in most of the low frequency electronics, in.e. audio cirucuits, input of the amplifier is never matched to the TXLine, but we see no such effects as reduction of the amplitude at the output. I know that people will say that on low frequency wavelength is to long compared to the cable length, but still, reflection can be calculated, right?
First remark. We can put aside the theoretical nature of the question (infinite input impedance) and assume a realistic high impedance instead, e.g. 1KOhm. This will result in a high (0.9) reflection factor, still most of the forward power will be reflected back to the generator.
Second remark. Assuming a matched generator, the voltage at the cable end will be increased by a positive reflection factor, not decreased related to the matched case.
There's no problem to apply a transmission line view, impedance matching, reflection factor etc. to the low frequency case, although there's effectively no transmission line. When you fully understood transmission lines and impedance matching, you'll see.