Electronic – Second order all-pass filter input impedance

active-filteranaloginput-impedanceltspice

I have a question about the second order all-pass filter input impedance when ω = ∞ and ω = 0 and I would appreciate if you could help me to understand the logic.

  1. When ω = 0, all capacitors can be seen as open circuit, because we are at DC level. In this case I expect the Zin to be R3+R4, what would be 3 kOhms, because the voltage at opamp inputs would be 0V. From the simulation I see that the expected input impedance is 1.5 kOhms. Why?
  2. When ω = ∞, capacitors can be seen as short circuits, and therefore the current will travel along C with no electrical impedance and R3, R4 are the only impedances that separates V1 from ground. In this case I do not understand why the input impedance is 3 kOhms.

Thank you in advance!
Input impedance simulation

Circuit setup

Best Answer

At DC \$\omega = 0\$ (all capacitors can be seen as open circuit) we have this situation:

schematic

simulate this circuit – Schematic created using CircuitLab

And $$R_{IN_{DC}} = R_3 + R_4 = 3k\Omega $$

But at high frequency (\$\omega = ∞\$) when all capacitors can be seen as short circuits we have this situation:

schematic

simulate this circuit

Therefore the input resistance is now equal to:

$$R_{IN_{HF}} = \left[R_1 \times\left(1 + \frac{R_4}{R_3}\right)\right]||(R_3+R_4) = 3k\Omega||3k\Omega = 1.5k\Omega$$

Why? because now the voltage across \$R_1\$ resistor is no longer equal to \$V_{IN}\$ but to the difference between \$V_{IN}\$ and op-amp output voltage. And the op-amp is working as a voltage follower, meaning that the op-amp output voltage is the same as the input voltage (at non-inverting input). And the input voltage is the output voltage produced by the voltage divider build around \$R_3, R_4\$.