You know the current through R4:

\$i_4 = \dfrac{A}{R_4}\$

Thus, you know the current through R3:

\$i_3 = i_4 \$

Thus, you know the output voltage of the 2nd op-amp:

\$v_{O2} = i_4(R_4 + R_3) = A(1 + \dfrac{R_3}{R_4})\$

Thus, you know the voltage across R2:

\$v_{R2} = A - v_{O2} = -A\dfrac{R_3}{R_4} \$

Thus, you know the current through R2 which is identical to the current through the capacitor:

\$i_C = i_{R2} = -\dfrac{A}{R_2}\dfrac{R_3}{R_4}\$

Now recall:

\$i_C = C \dfrac{dv_C}{dt}\$

Can you take it from here?

I can't find an error in my calcs, so I think I'm just
misinterpreting what Vs and Is actually are.

Switching to the phasor domain, we have:

\$I_c = -\dfrac{A}{R_2}\dfrac{R_3}{R_4} = j \omega C V_c\$

or

\$V_c = -\dfrac{A}{j \omega}\dfrac{R_3}{R_2R_4C}\$

Thus, the output voltage of the first op-amp is:

\$V_{o1} = A + V_c = A - \dfrac{A}{j \omega}\dfrac{R_3}{R_2R_4C}\$

And the voltage across R1 is:

\$V_{r1} = A - V_{o1} = \dfrac{A}{j \omega}\dfrac{R_3}{R_2R_4C}\$

Finally, the current through R1 is:

\$I_{r1} = \dfrac{V_{r1}}{R_1} = \dfrac{A}{j \omega}\dfrac{R_3}{R_1R_2R_4C} \$

But the source current is identical to \$I_{r1}\$, thus:

\$\dfrac{V_s}{I_s} = \dfrac{A}{I_{r1}} = j\omega \dfrac{R_1R_2R_4C}{R_3} = j \omega L_{eq} \$

## Best Answer

The gyrator simulates only some of the properties of the inductor, e.g. its frequency characteristics and Q factor, but not all of them, e.g. not energy storage.