When doing a frequency sweep simulation of the circuit below in CircuitLab, I can see that at \$V_{out1}\$ there is a phase invertion and at \$V_{out2}\$ there is not. I know that the inverting amplifier do phase shift by 180 degrees and that two of these cascaded will give a 360 degree phase shift that equals to 0 degrees and a delay. What I don't understand is how the two inverting amplifiers cascaded "cancel eachother out" when looking at the frequency sweep. Can someone try to explain this to me?


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ -10 * -10 = +100. \$\endgroup\$ – The Photon May 5 '14 at 0:56
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    \$\begingroup\$ @ThePhoton: Fun, but not helpful even though it's true :) \$\endgroup\$ – iQt May 5 '14 at 1:26

For your frequency sweep at frequencies much lower than the gain-bandwidth of the amplifiers divided by the gain (10) the magnitude will be a constant gain of 100 and the phase will be 0 degrees. (Or 360 if you prefer.) The time delay is negligible at the lower frequencies so you won't see it.

However, at higher frequencies you will see the contribution of both amplifiers to the rolloff and phase shift, which will be different from a single amplifier with another 180 degree phase shift. So they don't really cancel each other out except at low frequencies.

So for low frequencies gain is constant and phase is pretty much the same as the input signal. Why would you expect something different?

  • \$\begingroup\$ I was clearly tired. In the simulation I thought I was looking at Vout1, but it was Vout2 and vice versa. The graphs makes a lot more sense now.. good night! \$\endgroup\$ – iQt May 5 '14 at 1:31

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