Hi,Your suspicion the source impedance affects the signal is correct. Essentially it forms a voltage divider with the first stage input impeadance.
It is that reduced signal that is amplified.
Another way to look at this is the source impeadance is in series with the amp input resistance, so the effective R1 is greater, hence the gain is lower.
That is two ways to analyze the circuit that should give the same resultant gain of the first stage.
For part II I have no idea what loop is being closed.
Part iii leaves me perplexed as I would first determine the gain of the system to determine the equivalent.
As I read the problem, only the signal source has 50k output impedance; the opamps have zero output impedance.Hi,
The only thing that would make sense is if the output was connected to the input to form the loop. The output resistance would be connected to the input forming a closed loop. That would be a typical connection scheme too.
Hello,As I read the problem, only the signal source has 50k output impedance; the opamps have zero output impedance.
by Duane Benson
by Jake Hertz
by Jake Hertz
by Duane Benson